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

[Out]

(a^2*(24*c^4 + 64*c^3*d + 84*c^2*d^2 + 48*c*d^3 + 11*d^4)*ArcTanh[Sin[e + f*x]])/(16*f) - (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[e + f*x])/(60*d*f) - (a^2*(8*c^4 - 96*c^3*d - 438*c^2
*d^2 - 464*c*d^3 - 165*d^4)*Sec[e + f*x]*Tan[e + f*x])/(240*f) - (a^2*(4*c^3 - 48*c^2*d - 123*c*d^2 - 64*d^3)*
(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(120*d*f) - (a^2*(4*c^2 - 48*c*d - 55*d^2)*(c + d*Sec[e + f*x])^3*Tan[e +
f*x])/(120*d*f) - (a^2*(c - 12*d)*(c + d*Sec[e + f*x])^4*Tan[e + f*x])/(30*d*f) + (a^2*(c + d*Sec[e + f*x])^5
*Tan[e + f*x])/(6*d*f)

________________________________________________________________________________________

Rubi [A]  time = 0.42421, 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 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])

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 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 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 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 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 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])

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 = 2.0183, size = 460, normalized size = 1.41 \[ -\frac{a^2 (\cos (e+f x)+1)^2 \sec ^4\left (\frac{1}{2} (e+f x)\right ) \sec ^6(e+f x) \left (240 \left (84 c^2 d^2+64 c^3 d+24 c^4+48 c d^3+11 d^4\right ) \cos ^6(e+f x) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-2 \sin (e+f x) \left (6720 c^2 d^2 \cos (3 (e+f x))+1260 c^2 d^2 \cos (4 (e+f x))+960 c^2 d^2 \cos (5 (e+f x))+32 \left (480 c^2 d^2+310 c^3 d+75 c^4+336 c d^3+88 d^4\right ) \cos (e+f x)+20 \left (324 c^2 d^2+192 c^3 d+24 c^4+240 c d^3+55 d^4\right ) \cos (2 (e+f x))+5220 c^2 d^2+4640 c^3 d \cos (3 (e+f x))+960 c^3 d \cos (4 (e+f x))+800 c^3 d \cos (5 (e+f x))+2880 c^3 d+1200 c^4 \cos (3 (e+f x))+120 c^4 \cos (4 (e+f x))+240 c^4 \cos (5 (e+f x))+360 c^4+4032 c d^3 \cos (3 (e+f x))+720 c d^3 \cos (4 (e+f x))+576 c d^3 \cos (5 (e+f x))+4080 c d^3+896 d^4 \cos (3 (e+f x))+165 d^4 \cos (4 (e+f x))+128 d^4 \cos (5 (e+f x))+1255 d^4\right )\right )}{15360 f} \]

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]

-(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)] + 1
200*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*Cos[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]))/(15360*f)

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Maple [A]  time = 0.075, size = 602, normalized size = 1.8 \begin{align*} \text{result too large to display} \end{align*}

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]

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))+2/f*a^2*c^4*tan(f*x+e)+16/15
/f*a^2*d^4*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))+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+20/3/f*a^2*c^3*d*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+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)+4/3/f*a^2*c^3*d*t
an(f*x+e)*sec(f*x+e)^2+3/2/f*a^2*c^2*d^2*tan(f*x+e)*sec(f*x+e)^3+12/5/f*a^2*c*d^3*tan(f*x+e)*sec(f*x+e)^2+4/f*
a^2*c^2*d^2*tan(f*x+e)*sec(f*x+e)^2+4/5/f*a^2*c*d^3*tan(f*x+e)*sec(f*x+e)^4+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)+21/4/f*a^2*c^2*d^2*sec(f*x+e)*tan(f*x+e)+4/f*a^2*c^3*d*sec(f*x+e)*tan(
f*x+e)+1/2*a^2*c^4*sec(f*x+e)*tan(f*x+e)/f

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Maxima [B]  time = 1.02794, size = 922, normalized size = 2.82 \begin{align*} \text{result too large to display} \end{align*}

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

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Fricas [A]  time = 0.56677, size = 884, normalized size = 2.7 \begin{align*} \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}} \end{align*}

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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 ) \end{align*}

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))

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Giac [B]  time = 1.4544, size = 1038, normalized size = 3.17 \begin{align*} \text{result too large to display} \end{align*}

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]

1/240*(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)*log(abs(tan(1/2*f*x + 1/2*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)*log(abs(tan(1/2*f*x + 1/
2*e) - 1)) - 2*(360*a^2*c^4*tan(1/2*f*x + 1/2*e)^11 + 960*a^2*c^3*d*tan(1/2*f*x + 1/2*e)^11 + 1260*a^2*c^2*d^2
*tan(1/2*f*x + 1/2*e)^11 + 720*a^2*c*d^3*tan(1/2*f*x + 1/2*e)^11 + 165*a^2*d^4*tan(1/2*f*x + 1/2*e)^11 - 2040*
a^2*c^4*tan(1/2*f*x + 1/2*e)^9 - 5440*a^2*c^3*d*tan(1/2*f*x + 1/2*e)^9 - 7140*a^2*c^2*d^2*tan(1/2*f*x + 1/2*e)
^9 - 4080*a^2*c*d^3*tan(1/2*f*x + 1/2*e)^9 - 935*a^2*d^4*tan(1/2*f*x + 1/2*e)^9 + 4560*a^2*c^4*tan(1/2*f*x + 1
/2*e)^7 + 13440*a^2*c^3*d*tan(1/2*f*x + 1/2*e)^7 + 15480*a^2*c^2*d^2*tan(1/2*f*x + 1/2*e)^7 + 10272*a^2*c*d^3*
tan(1/2*f*x + 1/2*e)^7 + 1986*a^2*d^4*tan(1/2*f*x + 1/2*e)^7 - 5040*a^2*c^4*tan(1/2*f*x + 1/2*e)^5 - 17280*a^2
*c^3*d*tan(1/2*f*x + 1/2*e)^5 - 19080*a^2*c^2*d^2*tan(1/2*f*x + 1/2*e)^5 - 11232*a^2*c*d^3*tan(1/2*f*x + 1/2*e
)^5 - 3006*a^2*d^4*tan(1/2*f*x + 1/2*e)^5 + 2760*a^2*c^4*tan(1/2*f*x + 1/2*e)^3 + 11200*a^2*c^3*d*tan(1/2*f*x
+ 1/2*e)^3 + 13980*a^2*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 + 7440*a^2*c*d^3*tan(1/2*f*x + 1/2*e)^3 + 1305*a^2*d^4*t
an(1/2*f*x + 1/2*e)^3 - 600*a^2*c^4*tan(1/2*f*x + 1/2*e) - 2880*a^2*c^3*d*tan(1/2*f*x + 1/2*e) - 4500*a^2*c^2*
d^2*tan(1/2*f*x + 1/2*e) - 3120*a^2*c*d^3*tan(1/2*f*x + 1/2*e) - 795*a^2*d^4*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*
x + 1/2*e)^2 - 1)^6)/f