3.194 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=242 \[ \frac {3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {a^2 \left (c^2-10 c d-12 d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{20 d f}-\frac {a^2 \left (2 c^3-20 c^2 d-57 c d^2-30 d^3\right ) \tan (e+f x) \sec (e+f x)}{40 f}-\frac {a^2 \left (c^4-10 c^3 d-44 c^2 d^2-40 c d^3-12 d^4\right ) \tan (e+f x)}{10 d f}+\frac {a^2 \tan (e+f x) (c+d \sec (e+f x))^4}{5 d f}-\frac {a^2 (c-10 d) \tan (e+f x) (c+d \sec (e+f x))^3}{20 d f} \]

[Out]

3/8*a^2*(2*c+d)*(2*c^2+3*c*d+2*d^2)*arctanh(sin(f*x+e))/f-1/10*a^2*(c^4-10*c^3*d-44*c^2*d^2-40*c*d^3-12*d^4)*t
an(f*x+e)/d/f-1/40*a^2*(2*c^3-20*c^2*d-57*c*d^2-30*d^3)*sec(f*x+e)*tan(f*x+e)/f-1/20*a^2*(c^2-10*c*d-12*d^2)*(
c+d*sec(f*x+e))^2*tan(f*x+e)/d/f-1/20*a^2*(c-10*d)*(c+d*sec(f*x+e))^3*tan(f*x+e)/d/f+1/5*a^2*(c+d*sec(f*x+e))^
4*tan(f*x+e)/d/f

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Rubi [A]  time = 0.34, antiderivative size = 277, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3987, 100, 147, 50, 63, 217, 203} \[ \frac {3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)}{8 f}+\frac {3 a^3 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a (\sec (e+f x)+1)}}\right )}{4 f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{8 f}+\frac {d \tan (e+f x) (a \sec (e+f x)+a)^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right )}{20 f}+\frac {d \tan (e+f x) (a \sec (e+f x)+a)^2 (c+d \sec (e+f x))^2}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c + d*Sec[e + f*x])^3,x]

[Out]

(3*a^2*(2*c + d)*(2*c^2 + 3*c*d + 2*d^2)*Tan[e + f*x])/(8*f) + (3*a^3*(2*c + d)*(2*c^2 + 3*c*d + 2*d^2)*ArcTan
[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f*x])/(4*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*S
ec[e + f*x]]) + ((2*c + d)*(2*c^2 + 3*c*d + 2*d^2)*(a^2 + a^2*Sec[e + f*x])*Tan[e + f*x])/(8*f) + (d*(a + a*Se
c[e + f*x])^2*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(5*f) + (d*(a + a*Sec[e + f*x])^2*(2*(8*c^2 + 5*c*d + 2*d^2
) + d*(7*c + 2*d)*Sec[e + f*x])*Tan[e + f*x])/(20*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 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))^3 \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{3/2} (c+d x)^3}{\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))^2 \tan (e+f x)}{5 f}+\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {(a+a x)^{3/2} (c+d x) \left (-a^2 \left (5 c^2+2 c d+2 d^2\right )-a^2 d (7 c+2 d) x\right )}{\sqrt {a-a x}} \, dx,x,\sec (e+f x)\right )}{5 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))^2 \tan (e+f x)}{5 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}-\frac {\left (a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\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 )}{4 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}-\frac {\left (3 a^3 (2 c+d) \left (2 c^2+3 c d+2 d^2\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 )}{8 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)}{8 f}+\frac {(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}-\frac {\left (3 a^4 (2 c+d) \left (2 c^2+3 c d+2 d^2\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 )}{8 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)}{8 f}+\frac {(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}+\frac {\left (3 a^3 (2 c+d) \left (2 c^2+3 c d+2 d^2\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 )}{4 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)}{8 f}+\frac {(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}+\frac {\left (3 a^3 (2 c+d) \left (2 c^2+3 c d+2 d^2\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 )}{4 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)}{8 f}+\frac {3 a^3 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}}\right ) \tan (e+f x)}{4 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac {d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac {d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}\\ \end {align*}

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Mathematica [A]  time = 1.39, size = 326, normalized size = 1.35 \[ -\frac {a^2 (\cos (e+f x)+1)^2 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sec ^5(e+f x) \left (120 \left (4 c^3+8 c^2 d+7 c d^2+2 d^3\right ) \cos ^5(e+f x) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-2 \sin (e+f x) \left (20 c^3 \cos (3 (e+f x))+40 c^3 \cos (4 (e+f x))+120 c^3+120 c^2 d \cos (3 (e+f x))+100 c^2 d \cos (4 (e+f x))+380 c^2 d+5 \left (12 c^3+72 c^2 d+87 c d^2+34 d^3\right ) \cos (e+f x)+16 \left (10 c^3+30 c^2 d+30 c d^2+9 d^3\right ) \cos (2 (e+f x))+105 c d^2 \cos (3 (e+f x))+80 c d^2 \cos (4 (e+f x))+400 c d^2+30 d^3 \cos (3 (e+f x))+24 d^3 \cos (4 (e+f x))+152 d^3\right )\right )}{1280 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c + d*Sec[e + f*x])^3,x]

[Out]

-1/1280*(a^2*(1 + Cos[e + f*x])^2*Sec[(e + f*x)/2]^4*Sec[e + f*x]^5*(120*(4*c^3 + 8*c^2*d + 7*c*d^2 + 2*d^3)*C
os[e + f*x]^5*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) - 2*(120*c
^3 + 380*c^2*d + 400*c*d^2 + 152*d^3 + 5*(12*c^3 + 72*c^2*d + 87*c*d^2 + 34*d^3)*Cos[e + f*x] + 16*(10*c^3 + 3
0*c^2*d + 30*c*d^2 + 9*d^3)*Cos[2*(e + f*x)] + 20*c^3*Cos[3*(e + f*x)] + 120*c^2*d*Cos[3*(e + f*x)] + 105*c*d^
2*Cos[3*(e + f*x)] + 30*d^3*Cos[3*(e + f*x)] + 40*c^3*Cos[4*(e + f*x)] + 100*c^2*d*Cos[4*(e + f*x)] + 80*c*d^2
*Cos[4*(e + f*x)] + 24*d^3*Cos[4*(e + f*x)])*Sin[e + f*x]))/f

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fricas [A]  time = 0.45, size = 294, normalized size = 1.21 \[ \frac {15 \, {\left (4 \, a^{2} c^{3} + 8 \, a^{2} c^{2} d + 7 \, a^{2} c d^{2} + 2 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )^{5} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, {\left (4 \, a^{2} c^{3} + 8 \, a^{2} c^{2} d + 7 \, a^{2} c d^{2} + 2 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )^{5} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (8 \, a^{2} d^{3} + 8 \, {\left (10 \, a^{2} c^{3} + 25 \, a^{2} c^{2} d + 20 \, a^{2} c d^{2} + 6 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )^{4} + 5 \, {\left (4 \, a^{2} c^{3} + 24 \, a^{2} c^{2} d + 21 \, a^{2} c d^{2} + 6 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )^{3} + 8 \, {\left (5 \, a^{2} c^{2} d + 10 \, a^{2} c d^{2} + 3 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )^{2} + 10 \, {\left (3 \, a^{2} c d^{2} + 2 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{80 \, f \cos \left (f x + e\right )^{5}} \]

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))^3,x, algorithm="fricas")

[Out]

1/80*(15*(4*a^2*c^3 + 8*a^2*c^2*d + 7*a^2*c*d^2 + 2*a^2*d^3)*cos(f*x + e)^5*log(sin(f*x + e) + 1) - 15*(4*a^2*
c^3 + 8*a^2*c^2*d + 7*a^2*c*d^2 + 2*a^2*d^3)*cos(f*x + e)^5*log(-sin(f*x + e) + 1) + 2*(8*a^2*d^3 + 8*(10*a^2*
c^3 + 25*a^2*c^2*d + 20*a^2*c*d^2 + 6*a^2*d^3)*cos(f*x + e)^4 + 5*(4*a^2*c^3 + 24*a^2*c^2*d + 21*a^2*c*d^2 + 6
*a^2*d^3)*cos(f*x + e)^3 + 8*(5*a^2*c^2*d + 10*a^2*c*d^2 + 3*a^2*d^3)*cos(f*x + e)^2 + 10*(3*a^2*c*d^2 + 2*a^2
*d^3)*cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^5)

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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))^3,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*(-(6*a^2*d^3+21*a^2*d^2*c+24*a^2*d*c^2+12*a^2*c^3)/16*ln(a
bs(tan((f*x+exp(1))/2)-1))+(6*a^2*d^3+21*a^2*d^2*c+24*a^2*d*c^2+12*a^2*c^3)/16*ln(abs(tan((f*x+exp(1))/2)+1))-
(30*tan((f*x+exp(1))/2)^9*a^2*d^3+105*tan((f*x+exp(1))/2)^9*a^2*d^2*c+120*tan((f*x+exp(1))/2)^9*a^2*d*c^2+60*t
an((f*x+exp(1))/2)^9*a^2*c^3-140*tan((f*x+exp(1))/2)^7*a^2*d^3-490*tan((f*x+exp(1))/2)^7*a^2*d^2*c-560*tan((f*
x+exp(1))/2)^7*a^2*d*c^2-280*tan((f*x+exp(1))/2)^7*a^2*c^3+288*tan((f*x+exp(1))/2)^5*a^2*d^3+800*tan((f*x+exp(
1))/2)^5*a^2*d^2*c+1120*tan((f*x+exp(1))/2)^5*a^2*d*c^2+480*tan((f*x+exp(1))/2)^5*a^2*c^3-180*tan((f*x+exp(1))
/2)^3*a^2*d^3-790*tan((f*x+exp(1))/2)^3*a^2*d^2*c-1040*tan((f*x+exp(1))/2)^3*a^2*d*c^2-360*tan((f*x+exp(1))/2)
^3*a^2*c^3+130*tan((f*x+exp(1))/2)*a^2*d^3+375*tan((f*x+exp(1))/2)*a^2*d^2*c+360*tan((f*x+exp(1))/2)*a^2*d*c^2
+100*tan((f*x+exp(1))/2)*a^2*c^3)*1/40/(tan((f*x+exp(1))/2)^2-1)^5)

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maple [A]  time = 1.62, size = 420, normalized size = 1.74 \[ \frac {3 a^{2} c^{3} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2 f}+\frac {5 a^{2} c^{2} d \tan \left (f x +e \right )}{f}+\frac {21 a^{2} c \,d^{2} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{8 f}+\frac {21 a^{2} c \,d^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8 f}+\frac {6 a^{2} d^{3} \tan \left (f x +e \right )}{5 f}+\frac {3 a^{2} d^{3} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{5 f}+\frac {2 a^{2} c^{3} \tan \left (f x +e \right )}{f}+\frac {3 a^{2} c^{2} d \sec \left (f x +e \right ) \tan \left (f x +e \right )}{f}+\frac {3 a^{2} c^{2} d \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {4 a^{2} c \,d^{2} \tan \left (f x +e \right )}{f}+\frac {2 a^{2} c \,d^{2} \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{f}+\frac {a^{2} d^{3} \tan \left (f x +e \right ) \left (\sec ^{3}\left (f x +e \right )\right )}{2 f}+\frac {3 a^{2} d^{3} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{4 f}+\frac {3 a^{2} d^{3} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{4 f}+\frac {c^{3} a^{2} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{2 f}+\frac {a^{2} c^{2} d \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{f}+\frac {3 a^{2} c \,d^{2} \tan \left (f x +e \right ) \left (\sec ^{3}\left (f x +e \right )\right )}{4 f}+\frac {a^{2} d^{3} \tan \left (f x +e \right ) \left (\sec ^{4}\left (f x +e \right )\right )}{5 f} \]

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))^3,x)

[Out]

3/2/f*a^2*c^3*ln(sec(f*x+e)+tan(f*x+e))+5/f*a^2*c^2*d*tan(f*x+e)+21/8/f*a^2*c*d^2*sec(f*x+e)*tan(f*x+e)+21/8/f
*a^2*c*d^2*ln(sec(f*x+e)+tan(f*x+e))+6/5/f*a^2*d^3*tan(f*x+e)+3/5/f*a^2*d^3*tan(f*x+e)*sec(f*x+e)^2+2*a^2*c^3*
tan(f*x+e)/f+3/f*a^2*c^2*d*sec(f*x+e)*tan(f*x+e)+3/f*a^2*c^2*d*ln(sec(f*x+e)+tan(f*x+e))+4/f*a^2*c*d^2*tan(f*x
+e)+2/f*a^2*c*d^2*tan(f*x+e)*sec(f*x+e)^2+1/2/f*a^2*d^3*tan(f*x+e)*sec(f*x+e)^3+3/4/f*a^2*d^3*sec(f*x+e)*tan(f
*x+e)+3/4/f*a^2*d^3*ln(sec(f*x+e)+tan(f*x+e))+1/2/f*c^3*a^2*sec(f*x+e)*tan(f*x+e)+1/f*a^2*c^2*d*tan(f*x+e)*sec
(f*x+e)^2+3/4/f*a^2*c*d^2*tan(f*x+e)*sec(f*x+e)^3+1/5/f*a^2*d^3*tan(f*x+e)*sec(f*x+e)^4

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maxima [B]  time = 0.41, size = 469, normalized size = 1.94 \[ \frac {240 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{2} d + 480 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c d^{2} + 16 \, {\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^{3} + 80 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} d^{3} - 45 \, a^{2} c 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 )} - 30 \, a^{2} 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 )} - 60 \, a^{2} c^{3} {\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 )} - 360 \, a^{2} c^{2} 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 )} - 180 \, a^{2} c 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 )} + 240 \, a^{2} c^{3} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 480 \, a^{2} c^{3} \tan \left (f x + e\right ) + 720 \, a^{2} c^{2} d \tan \left (f x + e\right )}{240 \, 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))^3,x, algorithm="maxima")

[Out]

1/240*(240*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c^2*d + 480*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c*d^2 + 16*
(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^2*d^3 + 80*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*d^
3 - 45*a^2*c*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)) - 30*a^2*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)) - 60*a^2*c^3*(2*sin(f*x + e)/(sin(f*
x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) - 360*a^2*c^2*d*(2*sin(f*x + e)/(sin(f*x + e)^2
- 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) - 180*a^2*c*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)) + 240*a^2*c^3*log(sec(f*x + e) + tan(f*x + e)) + 480*a^2*c^3*ta
n(f*x + e) + 720*a^2*c^2*d*tan(f*x + e))/f

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mupad [B]  time = 5.49, size = 394, normalized size = 1.63 \[ \frac {3\,a^2\,\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c+d\right )\,\left (2\,c^2+3\,c\,d+2\,d^2\right )}{2\,\left (6\,c^3+12\,c^2\,d+\frac {21\,c\,d^2}{2}+3\,d^3\right )}\right )\,\left (2\,c+d\right )\,\left (2\,c^2+3\,c\,d+2\,d^2\right )}{4\,f}-\frac {\left (3\,a^2\,c^3+6\,a^2\,c^2\,d+\frac {21\,a^2\,c\,d^2}{4}+\frac {3\,a^2\,d^3}{2}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9+\left (-14\,a^2\,c^3-28\,a^2\,c^2\,d-\frac {49\,a^2\,c\,d^2}{2}-7\,a^2\,d^3\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+\left (24\,a^2\,c^3+56\,a^2\,c^2\,d+40\,a^2\,c\,d^2+\frac {72\,a^2\,d^3}{5}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+\left (-18\,a^2\,c^3-52\,a^2\,c^2\,d-\frac {79\,a^2\,c\,d^2}{2}-9\,a^2\,d^3\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (5\,a^2\,c^3+18\,a^2\,c^2\,d+\frac {75\,a^2\,c\,d^2}{4}+\frac {13\,a^2\,d^3}{2}\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )} \]

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))^3)/cos(e + f*x),x)

[Out]

(3*a^2*atanh((3*tan(e/2 + (f*x)/2)*(2*c + d)*(3*c*d + 2*c^2 + 2*d^2))/(2*((21*c*d^2)/2 + 12*c^2*d + 6*c^3 + 3*
d^3)))*(2*c + d)*(3*c*d + 2*c^2 + 2*d^2))/(4*f) - (tan(e/2 + (f*x)/2)^9*(3*a^2*c^3 + (3*a^2*d^3)/2 + (21*a^2*c
*d^2)/4 + 6*a^2*c^2*d) - tan(e/2 + (f*x)/2)^7*(14*a^2*c^3 + 7*a^2*d^3 + (49*a^2*c*d^2)/2 + 28*a^2*c^2*d) - tan
(e/2 + (f*x)/2)^3*(18*a^2*c^3 + 9*a^2*d^3 + (79*a^2*c*d^2)/2 + 52*a^2*c^2*d) + tan(e/2 + (f*x)/2)^5*(24*a^2*c^
3 + (72*a^2*d^3)/5 + 40*a^2*c*d^2 + 56*a^2*c^2*d) + tan(e/2 + (f*x)/2)*(5*a^2*c^3 + (13*a^2*d^3)/2 + (75*a^2*c
*d^2)/4 + 18*a^2*c^2*d))/(f*(5*tan(e/2 + (f*x)/2)^2 - 10*tan(e/2 + (f*x)/2)^4 + 10*tan(e/2 + (f*x)/2)^6 - 5*ta
n(e/2 + (f*x)/2)^8 + tan(e/2 + (f*x)/2)^10 - 1))

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

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))**3,x)

[Out]

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

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