3.193 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^4 \, dx\)

Optimal. Leaf size=327 \[ -\frac {a^2 \left (4 c^2-48 c d-55 d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^3}{120 d f}-\frac {a^2 \left (4 c^3-48 c^2 d-123 c d^2-64 d^3\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{120 d f}+\frac {a^2 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \tanh ^{-1}(\sin (e+f x))}{16 f}-\frac {a^2 \left (8 c^4-96 c^3 d-438 c^2 d^2-464 c d^3-165 d^4\right ) \tan (e+f x) \sec (e+f x)}{240 f}-\frac {a^2 \left (4 c^5-48 c^4 d-311 c^3 d^2-448 c^2 d^3-288 c d^4-64 d^5\right ) \tan (e+f x)}{60 d f}+\frac {a^2 \tan (e+f x) (c+d \sec (e+f x))^5}{6 d f}-\frac {a^2 (c-12 d) \tan (e+f x) (c+d \sec (e+f x))^4}{30 d f} \]

[Out]

1/16*a^2*(24*c^4+64*c^3*d+84*c^2*d^2+48*c*d^3+11*d^4)*arctanh(sin(f*x+e))/f-1/60*a^2*(4*c^5-48*c^4*d-311*c^3*d
^2-448*c^2*d^3-288*c*d^4-64*d^5)*tan(f*x+e)/d/f-1/240*a^2*(8*c^4-96*c^3*d-438*c^2*d^2-464*c*d^3-165*d^4)*sec(f
*x+e)*tan(f*x+e)/f-1/120*a^2*(4*c^3-48*c^2*d-123*c*d^2-64*d^3)*(c+d*sec(f*x+e))^2*tan(f*x+e)/d/f-1/120*a^2*(4*
c^2-48*c*d-55*d^2)*(c+d*sec(f*x+e))^3*tan(f*x+e)/d/f-1/30*a^2*(c-12*d)*(c+d*sec(f*x+e))^4*tan(f*x+e)/d/f+1/6*a
^2*(c+d*sec(f*x+e))^5*tan(f*x+e)/d/f

________________________________________________________________________________________

Rubi [A]  time = 0.42, antiderivative size = 371, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3987, 100, 153, 147, 50, 63, 217, 203} \[ \frac {a^2 \left (84 c^2 d^2+64 c^3 d+24 c^4+48 c d^3+11 d^4\right ) \tan (e+f x)}{16 f}+\frac {a^3 \left (84 c^2 d^2+64 c^3 d+24 c^4+48 c d^3+11 d^4\right ) \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a (\sec (e+f x)+1)}}\right )}{8 f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {\left (84 c^2 d^2+64 c^3 d+24 c^4+48 c d^3+11 d^4\right ) \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{48 f}+\frac {d \tan (e+f x) (a \sec (e+f x)+a)^2 \left (d \left (48 c^2+32 c d+19 d^2\right ) \sec (e+f x)+2 \left (56 c^2 d+52 c^3+48 c d^2+9 d^3\right )\right )}{120 f}+\frac {d \tan (e+f x) (a \sec (e+f x)+a)^2 (c+d \sec (e+f x))^3}{6 f}+\frac {d (9 c+2 d) \tan (e+f x) (a \sec (e+f x)+a)^2 (c+d \sec (e+f x))^2}{30 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c + d*Sec[e + f*x])^4,x]

[Out]

(a^2*(24*c^4 + 64*c^3*d + 84*c^2*d^2 + 48*c*d^3 + 11*d^4)*Tan[e + f*x])/(16*f) + (a^3*(24*c^4 + 64*c^3*d + 84*
c^2*d^2 + 48*c*d^3 + 11*d^4)*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f*x])/(8*f*Sq
rt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + ((24*c^4 + 64*c^3*d + 84*c^2*d^2 + 48*c*d^3 + 11*d^4)*(a^2
+ a^2*Sec[e + f*x])*Tan[e + f*x])/(48*f) + (d*(9*c + 2*d)*(a + a*Sec[e + f*x])^2*(c + d*Sec[e + f*x])^2*Tan[e
+ f*x])/(30*f) + (d*(a + a*Sec[e + f*x])^2*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(6*f) + (d*(a + a*Sec[e + f*x]
)^2*(2*(52*c^3 + 56*c^2*d + 48*c*d^2 + 9*d^3) + d*(48*c^2 + 32*c*d + 19*d^2)*Sec[e + f*x])*Tan[e + f*x])/(120*
f)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3987

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[(a^2*g*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x
]]), Subst[Int[((g*x)^(p - 1)*(a + b*x)^(m - 1/2)*(c + d*x)^n)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])

Rubi steps

\begin {align*} \int \sec (e+f x) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^4 \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{3/2} (c+d x)^4}{\sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{6 f}+\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {(a+a x)^{3/2} (c+d x)^2 \left (-a^2 \left (6 c^2+2 c d+3 d^2\right )-a^2 d (9 c+2 d) x\right )}{\sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{6 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {d (9 c+2 d) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{30 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{6 f}-\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {(a+a x)^{3/2} (c+d x) \left (a^4 \left (30 c^3+28 c^2 d+37 c d^2+4 d^3\right )+a^4 d \left (48 c^2+32 c d+19 d^2\right ) x\right )}{\sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{30 a^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {d (9 c+2 d) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{30 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{6 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (52 c^3+56 c^2 d+48 c d^2+9 d^3\right )+d \left (48 c^2+32 c d+19 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{120 f}-\frac {\left (a^2 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{3/2}}{\sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{24 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {\left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{48 f}+\frac {d (9 c+2 d) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{30 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{6 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (52 c^3+56 c^2 d+48 c d^2+9 d^3\right )+d \left (48 c^2+32 c d+19 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{120 f}-\frac {\left (a^3 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+a x}}{\sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{16 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {a^2 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \tan (e+f x)}{16 f}+\frac {\left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{48 f}+\frac {d (9 c+2 d) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{30 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{6 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (52 c^3+56 c^2 d+48 c d^2+9 d^3\right )+d \left (48 c^2+32 c d+19 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{120 f}-\frac {\left (a^4 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-a x} \sqrt {a+a x}} \, dx,x,\sec (e+f x)\right )}{16 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {a^2 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \tan (e+f x)}{16 f}+\frac {\left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{48 f}+\frac {d (9 c+2 d) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{30 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{6 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (52 c^3+56 c^2 d+48 c d^2+9 d^3\right )+d \left (48 c^2+32 c d+19 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{120 f}+\frac {\left (a^3 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 a-x^2}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{8 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {a^2 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \tan (e+f x)}{16 f}+\frac {\left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{48 f}+\frac {d (9 c+2 d) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{30 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{6 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (52 c^3+56 c^2 d+48 c d^2+9 d^3\right )+d \left (48 c^2+32 c d+19 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{120 f}+\frac {\left (a^3 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}}\right )}{8 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {a^2 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \tan (e+f x)}{16 f}+\frac {a^3 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}}\right ) \tan (e+f x)}{8 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {\left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{48 f}+\frac {d (9 c+2 d) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{30 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{6 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (52 c^3+56 c^2 d+48 c d^2+9 d^3\right )+d \left (48 c^2+32 c d+19 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{120 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 2.03, size = 460, normalized size = 1.41 \[ -\frac {a^2 (\cos (e+f x)+1)^2 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sec ^6(e+f x) \left (240 \left (24 c^4+64 c^3 d+84 c^2 d^2+48 c d^3+11 d^4\right ) \cos ^6(e+f x) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-2 \sin (e+f x) \left (1200 c^4 \cos (3 (e+f x))+120 c^4 \cos (4 (e+f x))+240 c^4 \cos (5 (e+f x))+360 c^4+4640 c^3 d \cos (3 (e+f x))+960 c^3 d \cos (4 (e+f x))+800 c^3 d \cos (5 (e+f x))+2880 c^3 d+6720 c^2 d^2 \cos (3 (e+f x))+1260 c^2 d^2 \cos (4 (e+f x))+960 c^2 d^2 \cos (5 (e+f x))+5220 c^2 d^2+32 \left (75 c^4+310 c^3 d+480 c^2 d^2+336 c d^3+88 d^4\right ) \cos (e+f x)+20 \left (24 c^4+192 c^3 d+324 c^2 d^2+240 c d^3+55 d^4\right ) \cos (2 (e+f x))+4032 c d^3 \cos (3 (e+f x))+720 c d^3 \cos (4 (e+f x))+576 c d^3 \cos (5 (e+f x))+4080 c d^3+896 d^4 \cos (3 (e+f x))+165 d^4 \cos (4 (e+f x))+128 d^4 \cos (5 (e+f x))+1255 d^4\right )\right )}{15360 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c + d*Sec[e + f*x])^4,x]

[Out]

-1/15360*(a^2*(1 + Cos[e + f*x])^2*Sec[(e + f*x)/2]^4*Sec[e + f*x]^6*(240*(24*c^4 + 64*c^3*d + 84*c^2*d^2 + 48
*c*d^3 + 11*d^4)*Cos[e + f*x]^6*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*
x)/2]]) - 2*(360*c^4 + 2880*c^3*d + 5220*c^2*d^2 + 4080*c*d^3 + 1255*d^4 + 32*(75*c^4 + 310*c^3*d + 480*c^2*d^
2 + 336*c*d^3 + 88*d^4)*Cos[e + f*x] + 20*(24*c^4 + 192*c^3*d + 324*c^2*d^2 + 240*c*d^3 + 55*d^4)*Cos[2*(e + f
*x)] + 1200*c^4*Cos[3*(e + f*x)] + 4640*c^3*d*Cos[3*(e + f*x)] + 6720*c^2*d^2*Cos[3*(e + f*x)] + 4032*c*d^3*Co
s[3*(e + f*x)] + 896*d^4*Cos[3*(e + f*x)] + 120*c^4*Cos[4*(e + f*x)] + 960*c^3*d*Cos[4*(e + f*x)] + 1260*c^2*d
^2*Cos[4*(e + f*x)] + 720*c*d^3*Cos[4*(e + f*x)] + 165*d^4*Cos[4*(e + f*x)] + 240*c^4*Cos[5*(e + f*x)] + 800*c
^3*d*Cos[5*(e + f*x)] + 960*c^2*d^2*Cos[5*(e + f*x)] + 576*c*d^3*Cos[5*(e + f*x)] + 128*d^4*Cos[5*(e + f*x)])*
Sin[e + f*x]))/f

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 387, normalized size = 1.18 \[ \frac {15 \, {\left (24 \, a^{2} c^{4} + 64 \, a^{2} c^{3} d + 84 \, a^{2} c^{2} d^{2} + 48 \, a^{2} c d^{3} + 11 \, a^{2} d^{4}\right )} \cos \left (f x + e\right )^{6} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, {\left (24 \, a^{2} c^{4} + 64 \, a^{2} c^{3} d + 84 \, a^{2} c^{2} d^{2} + 48 \, a^{2} c d^{3} + 11 \, a^{2} d^{4}\right )} \cos \left (f x + e\right )^{6} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (40 \, a^{2} d^{4} + 32 \, {\left (15 \, a^{2} c^{4} + 50 \, a^{2} c^{3} d + 60 \, a^{2} c^{2} d^{2} + 36 \, a^{2} c d^{3} + 8 \, a^{2} d^{4}\right )} \cos \left (f x + e\right )^{5} + 15 \, {\left (8 \, a^{2} c^{4} + 64 \, a^{2} c^{3} d + 84 \, a^{2} c^{2} d^{2} + 48 \, a^{2} c d^{3} + 11 \, a^{2} d^{4}\right )} \cos \left (f x + e\right )^{4} + 64 \, {\left (5 \, a^{2} c^{3} d + 15 \, a^{2} c^{2} d^{2} + 9 \, a^{2} c d^{3} + 2 \, a^{2} d^{4}\right )} \cos \left (f x + e\right )^{3} + 10 \, {\left (36 \, a^{2} c^{2} d^{2} + 48 \, a^{2} c d^{3} + 11 \, a^{2} d^{4}\right )} \cos \left (f x + e\right )^{2} + 96 \, {\left (2 \, a^{2} c d^{3} + a^{2} d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{480 \, f \cos \left (f x + e\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c+d*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

1/480*(15*(24*a^2*c^4 + 64*a^2*c^3*d + 84*a^2*c^2*d^2 + 48*a^2*c*d^3 + 11*a^2*d^4)*cos(f*x + e)^6*log(sin(f*x
+ e) + 1) - 15*(24*a^2*c^4 + 64*a^2*c^3*d + 84*a^2*c^2*d^2 + 48*a^2*c*d^3 + 11*a^2*d^4)*cos(f*x + e)^6*log(-si
n(f*x + e) + 1) + 2*(40*a^2*d^4 + 32*(15*a^2*c^4 + 50*a^2*c^3*d + 60*a^2*c^2*d^2 + 36*a^2*c*d^3 + 8*a^2*d^4)*c
os(f*x + e)^5 + 15*(8*a^2*c^4 + 64*a^2*c^3*d + 84*a^2*c^2*d^2 + 48*a^2*c*d^3 + 11*a^2*d^4)*cos(f*x + e)^4 + 64
*(5*a^2*c^3*d + 15*a^2*c^2*d^2 + 9*a^2*c*d^3 + 2*a^2*d^4)*cos(f*x + e)^3 + 10*(36*a^2*c^2*d^2 + 48*a^2*c*d^3 +
11*a^2*d^4)*cos(f*x + e)^2 + 96*(2*a^2*c*d^3 + a^2*d^4)*cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^6)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c+d*sec(f*x+e))^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*((-11*a^2*d^4-48*a^2*d^3*c-84*a^2*d^2*c^2-64*a^2*d*c^3-24*
a^2*c^4)/32*ln(abs(tan((f*x+exp(1))/2)-1))-(-11*a^2*d^4-48*a^2*d^3*c-84*a^2*d^2*c^2-64*a^2*d*c^3-24*a^2*c^4)/3
2*ln(abs(tan((f*x+exp(1))/2)+1))-(165*tan((f*x+exp(1))/2)^11*a^2*d^4+720*tan((f*x+exp(1))/2)^11*a^2*d^3*c+1260
*tan((f*x+exp(1))/2)^11*a^2*d^2*c^2+960*tan((f*x+exp(1))/2)^11*a^2*d*c^3+360*tan((f*x+exp(1))/2)^11*a^2*c^4-93
5*tan((f*x+exp(1))/2)^9*a^2*d^4-4080*tan((f*x+exp(1))/2)^9*a^2*d^3*c-7140*tan((f*x+exp(1))/2)^9*a^2*d^2*c^2-54
40*tan((f*x+exp(1))/2)^9*a^2*d*c^3-2040*tan((f*x+exp(1))/2)^9*a^2*c^4+1986*tan((f*x+exp(1))/2)^7*a^2*d^4+10272
*tan((f*x+exp(1))/2)^7*a^2*d^3*c+15480*tan((f*x+exp(1))/2)^7*a^2*d^2*c^2+13440*tan((f*x+exp(1))/2)^7*a^2*d*c^3
+4560*tan((f*x+exp(1))/2)^7*a^2*c^4-3006*tan((f*x+exp(1))/2)^5*a^2*d^4-11232*tan((f*x+exp(1))/2)^5*a^2*d^3*c-1
9080*tan((f*x+exp(1))/2)^5*a^2*d^2*c^2-17280*tan((f*x+exp(1))/2)^5*a^2*d*c^3-5040*tan((f*x+exp(1))/2)^5*a^2*c^
4+1305*tan((f*x+exp(1))/2)^3*a^2*d^4+7440*tan((f*x+exp(1))/2)^3*a^2*d^3*c+13980*tan((f*x+exp(1))/2)^3*a^2*d^2*
c^2+11200*tan((f*x+exp(1))/2)^3*a^2*d*c^3+2760*tan((f*x+exp(1))/2)^3*a^2*c^4-795*tan((f*x+exp(1))/2)*a^2*d^4-3
120*tan((f*x+exp(1))/2)*a^2*d^3*c-4500*tan((f*x+exp(1))/2)*a^2*d^2*c^2-2880*tan((f*x+exp(1))/2)*a^2*d*c^3-600*
tan((f*x+exp(1))/2)*a^2*c^4)*1/240/(tan((f*x+exp(1))/2)^2-1)^6)

________________________________________________________________________________________

maple [A]  time = 1.96, size = 602, normalized size = 1.84 \[ \frac {20 a^{2} c^{3} d \tan \left (f x +e \right )}{3 f}+\frac {2 a^{2} c \,d^{3} \tan \left (f x +e \right ) \left (\sec ^{3}\left (f x +e \right )\right )}{f}+\frac {12 a^{2} c \,d^{3} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{5 f}+\frac {3 a^{2} c \,d^{3} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{f}+\frac {a^{2} c^{4} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{2 f}+\frac {4 a^{2} c^{3} d \sec \left (f x +e \right ) \tan \left (f x +e \right )}{f}+\frac {21 a^{2} c^{2} d^{2} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{4 f}+\frac {4 a^{2} c^{2} d^{2} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{f}+\frac {4 a^{2} c \,d^{3} \tan \left (f x +e \right ) \left (\sec ^{4}\left (f x +e \right )\right )}{5 f}+\frac {3 a^{2} c^{2} d^{2} \tan \left (f x +e \right ) \left (\sec ^{3}\left (f x +e \right )\right )}{2 f}+\frac {4 a^{2} c^{3} d \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{3 f}+\frac {2 a^{2} c^{4} \tan \left (f x +e \right )}{f}+\frac {3 a^{2} c^{4} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2 f}+\frac {11 a^{2} d^{4} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16 f}+\frac {16 a^{2} d^{4} \tan \left (f x +e \right )}{15 f}+\frac {11 a^{2} d^{4} \tan \left (f x +e \right ) \left (\sec ^{3}\left (f x +e \right )\right )}{24 f}+\frac {2 a^{2} d^{4} \tan \left (f x +e \right ) \left (\sec ^{4}\left (f x +e \right )\right )}{5 f}+\frac {8 a^{2} d^{4} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{15 f}+\frac {21 a^{2} c^{2} d^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{4 f}+\frac {11 a^{2} d^{4} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{16 f}+\frac {4 a^{2} c^{3} d \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {3 a^{2} c \,d^{3} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {24 a^{2} c \,d^{3} \tan \left (f x +e \right )}{5 f}+\frac {8 a^{2} c^{2} d^{2} \tan \left (f x +e \right )}{f}+\frac {a^{2} d^{4} \tan \left (f x +e \right ) \left (\sec ^{5}\left (f x +e \right )\right )}{6 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c+d*sec(f*x+e))^4,x)

[Out]

20/3/f*a^2*c^3*d*tan(f*x+e)+4/f*a^2*c^2*d^2*tan(f*x+e)*sec(f*x+e)^2+2/f*a^2*c*d^3*tan(f*x+e)*sec(f*x+e)^3+3/f*
a^2*c*d^3*sec(f*x+e)*tan(f*x+e)+12/5/f*a^2*c*d^3*tan(f*x+e)*sec(f*x+e)^2+1/2*a^2*c^4*sec(f*x+e)*tan(f*x+e)/f+4
/f*a^2*c^3*d*sec(f*x+e)*tan(f*x+e)+21/4/f*a^2*c^2*d^2*sec(f*x+e)*tan(f*x+e)+4/5/f*a^2*c*d^3*tan(f*x+e)*sec(f*x
+e)^4+3/2/f*a^2*c^2*d^2*tan(f*x+e)*sec(f*x+e)^3+4/3/f*a^2*c^3*d*tan(f*x+e)*sec(f*x+e)^2+2*a^2*c^4*tan(f*x+e)/f
+3/2/f*a^2*c^4*ln(sec(f*x+e)+tan(f*x+e))+11/16/f*a^2*d^4*ln(sec(f*x+e)+tan(f*x+e))+16/15/f*a^2*d^4*tan(f*x+e)+
21/4/f*a^2*c^2*d^2*ln(sec(f*x+e)+tan(f*x+e))+11/24/f*a^2*d^4*tan(f*x+e)*sec(f*x+e)^3+11/16/f*a^2*d^4*sec(f*x+e
)*tan(f*x+e)+4/f*a^2*c^3*d*ln(sec(f*x+e)+tan(f*x+e))+3/f*a^2*c*d^3*ln(sec(f*x+e)+tan(f*x+e))+2/5/f*a^2*d^4*tan
(f*x+e)*sec(f*x+e)^4+8/15/f*a^2*d^4*tan(f*x+e)*sec(f*x+e)^2+24/5/f*a^2*c*d^3*tan(f*x+e)+8/f*a^2*c^2*d^2*tan(f*
x+e)+1/6/f*a^2*d^4*tan(f*x+e)*sec(f*x+e)^5

________________________________________________________________________________________

maxima [B]  time = 0.48, size = 683, normalized size = 2.09 \[ \frac {640 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{3} d + 1920 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{2} d^{2} + 128 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{2} c d^{3} + 640 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c d^{3} + 64 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{2} d^{4} - 5 \, a^{2} d^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (f x + e\right )^{5} - 40 \, \sin \left (f x + e\right )^{3} + 33 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1} - 15 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 15 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 180 \, a^{2} c^{2} d^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 240 \, a^{2} c d^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 30 \, a^{2} d^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 120 \, a^{2} c^{4} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 960 \, a^{2} c^{3} d {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 720 \, a^{2} c^{2} d^{2} {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 480 \, a^{2} c^{4} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 960 \, a^{2} c^{4} \tan \left (f x + e\right ) + 1920 \, a^{2} c^{3} d \tan \left (f x + e\right )}{480 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c+d*sec(f*x+e))^4,x, algorithm="maxima")

[Out]

1/480*(640*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c^3*d + 1920*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c^2*d^2 +
128*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^2*c*d^3 + 640*(tan(f*x + e)^3 + 3*tan(f*x + e))
*a^2*c*d^3 + 64*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^2*d^4 - 5*a^2*d^4*(2*(15*sin(f*x +
e)^5 - 40*sin(f*x + e)^3 + 33*sin(f*x + e))/(sin(f*x + e)^6 - 3*sin(f*x + e)^4 + 3*sin(f*x + e)^2 - 1) - 15*lo
g(sin(f*x + e) + 1) + 15*log(sin(f*x + e) - 1)) - 180*a^2*c^2*d^2*(2*(3*sin(f*x + e)^3 - 5*sin(f*x + e))/(sin(
f*x + e)^4 - 2*sin(f*x + e)^2 + 1) - 3*log(sin(f*x + e) + 1) + 3*log(sin(f*x + e) - 1)) - 240*a^2*c*d^3*(2*(3*
sin(f*x + e)^3 - 5*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1) - 3*log(sin(f*x + e) + 1) + 3*log(sin
(f*x + e) - 1)) - 30*a^2*d^4*(2*(3*sin(f*x + e)^3 - 5*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1) -
3*log(sin(f*x + e) + 1) + 3*log(sin(f*x + e) - 1)) - 120*a^2*c^4*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(si
n(f*x + e) + 1) + log(sin(f*x + e) - 1)) - 960*a^2*c^3*d*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x +
e) + 1) + log(sin(f*x + e) - 1)) - 720*a^2*c^2*d^2*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1
) + log(sin(f*x + e) - 1)) + 480*a^2*c^4*log(sec(f*x + e) + tan(f*x + e)) + 960*a^2*c^4*tan(f*x + e) + 1920*a^
2*c^3*d*tan(f*x + e))/f

________________________________________________________________________________________

mupad [B]  time = 5.35, size = 484, normalized size = 1.48 \[ \frac {\left (-3\,a^2\,c^4-8\,a^2\,c^3\,d-\frac {21\,a^2\,c^2\,d^2}{2}-6\,a^2\,c\,d^3-\frac {11\,a^2\,d^4}{8}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}+\left (17\,a^2\,c^4+\frac {136\,a^2\,c^3\,d}{3}+\frac {119\,a^2\,c^2\,d^2}{2}+34\,a^2\,c\,d^3+\frac {187\,a^2\,d^4}{24}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9+\left (-38\,a^2\,c^4-112\,a^2\,c^3\,d-129\,a^2\,c^2\,d^2-\frac {428\,a^2\,c\,d^3}{5}-\frac {331\,a^2\,d^4}{20}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+\left (42\,a^2\,c^4+144\,a^2\,c^3\,d+159\,a^2\,c^2\,d^2+\frac {468\,a^2\,c\,d^3}{5}+\frac {501\,a^2\,d^4}{20}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+\left (-23\,a^2\,c^4-\frac {280\,a^2\,c^3\,d}{3}-\frac {233\,a^2\,c^2\,d^2}{2}-62\,a^2\,c\,d^3-\frac {87\,a^2\,d^4}{8}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (5\,a^2\,c^4+24\,a^2\,c^3\,d+\frac {75\,a^2\,c^2\,d^2}{2}+26\,a^2\,c\,d^3+\frac {53\,a^2\,d^4}{8}\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (24\,c^4+64\,c^3\,d+84\,c^2\,d^2+48\,c\,d^3+11\,d^4\right )}{4\,\left (6\,c^4+16\,c^3\,d+21\,c^2\,d^2+12\,c\,d^3+\frac {11\,d^4}{4}\right )}\right )\,\left (24\,c^4+64\,c^3\,d+84\,c^2\,d^2+48\,c\,d^3+11\,d^4\right )}{8\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(e + f*x))^2*(c + d/cos(e + f*x))^4)/cos(e + f*x),x)

[Out]

(tan(e/2 + (f*x)/2)*(5*a^2*c^4 + (53*a^2*d^4)/8 + 26*a^2*c*d^3 + 24*a^2*c^3*d + (75*a^2*c^2*d^2)/2) - tan(e/2
+ (f*x)/2)^11*(3*a^2*c^4 + (11*a^2*d^4)/8 + 6*a^2*c*d^3 + 8*a^2*c^3*d + (21*a^2*c^2*d^2)/2) + tan(e/2 + (f*x)/
2)^9*(17*a^2*c^4 + (187*a^2*d^4)/24 + 34*a^2*c*d^3 + (136*a^2*c^3*d)/3 + (119*a^2*c^2*d^2)/2) - tan(e/2 + (f*x
)/2)^3*(23*a^2*c^4 + (87*a^2*d^4)/8 + 62*a^2*c*d^3 + (280*a^2*c^3*d)/3 + (233*a^2*c^2*d^2)/2) - tan(e/2 + (f*x
)/2)^7*(38*a^2*c^4 + (331*a^2*d^4)/20 + (428*a^2*c*d^3)/5 + 112*a^2*c^3*d + 129*a^2*c^2*d^2) + tan(e/2 + (f*x)
/2)^5*(42*a^2*c^4 + (501*a^2*d^4)/20 + (468*a^2*c*d^3)/5 + 144*a^2*c^3*d + 159*a^2*c^2*d^2))/(f*(15*tan(e/2 +
(f*x)/2)^4 - 6*tan(e/2 + (f*x)/2)^2 - 20*tan(e/2 + (f*x)/2)^6 + 15*tan(e/2 + (f*x)/2)^8 - 6*tan(e/2 + (f*x)/2)
^10 + tan(e/2 + (f*x)/2)^12 + 1)) + (a^2*atanh((tan(e/2 + (f*x)/2)*(48*c*d^3 + 64*c^3*d + 24*c^4 + 11*d^4 + 84
*c^2*d^2))/(4*(12*c*d^3 + 16*c^3*d + 6*c^4 + (11*d^4)/4 + 21*c^2*d^2)))*(48*c*d^3 + 64*c^3*d + 24*c^4 + 11*d^4
+ 84*c^2*d^2))/(8*f)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int c^{4} \sec {\left (e + f x \right )}\, dx + \int 2 c^{4} \sec ^{2}{\left (e + f x \right )}\, dx + \int c^{4} \sec ^{3}{\left (e + f x \right )}\, dx + \int d^{4} \sec ^{5}{\left (e + f x \right )}\, dx + \int 2 d^{4} \sec ^{6}{\left (e + f x \right )}\, dx + \int d^{4} \sec ^{7}{\left (e + f x \right )}\, dx + \int 4 c d^{3} \sec ^{4}{\left (e + f x \right )}\, dx + \int 8 c d^{3} \sec ^{5}{\left (e + f x \right )}\, dx + \int 4 c d^{3} \sec ^{6}{\left (e + f x \right )}\, dx + \int 6 c^{2} d^{2} \sec ^{3}{\left (e + f x \right )}\, dx + \int 12 c^{2} d^{2} \sec ^{4}{\left (e + f x \right )}\, dx + \int 6 c^{2} d^{2} \sec ^{5}{\left (e + f x \right )}\, dx + \int 4 c^{3} d \sec ^{2}{\left (e + f x \right )}\, dx + \int 8 c^{3} d \sec ^{3}{\left (e + f x \right )}\, dx + \int 4 c^{3} d \sec ^{4}{\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**2*(c+d*sec(f*x+e))**4,x)

[Out]

a**2*(Integral(c**4*sec(e + f*x), x) + Integral(2*c**4*sec(e + f*x)**2, x) + Integral(c**4*sec(e + f*x)**3, x)
+ Integral(d**4*sec(e + f*x)**5, x) + Integral(2*d**4*sec(e + f*x)**6, x) + Integral(d**4*sec(e + f*x)**7, x)
+ Integral(4*c*d**3*sec(e + f*x)**4, x) + Integral(8*c*d**3*sec(e + f*x)**5, x) + Integral(4*c*d**3*sec(e + f
*x)**6, x) + Integral(6*c**2*d**2*sec(e + f*x)**3, x) + Integral(12*c**2*d**2*sec(e + f*x)**4, x) + Integral(6
*c**2*d**2*sec(e + f*x)**5, x) + Integral(4*c**3*d*sec(e + f*x)**2, x) + Integral(8*c**3*d*sec(e + f*x)**3, x)
+ Integral(4*c**3*d*sec(e + f*x)**4, x))

________________________________________________________________________________________