3.457 \(\int \frac{\tan ^5(e+f x)}{a+b \sec ^3(e+f x)} \, dx\)

Optimal. Leaf size=219 \[ \frac{\left (a^{2/3}-2 b^{2/3}\right ) \log \left (a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}\right )}{6 \sqrt [3]{a} b^{4/3} f}-\frac{\left (a^{2/3}-2 b^{2/3}\right ) \log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{4/3} f}-\frac{\left (a^{2/3}+2 b^{2/3}\right ) \tan ^{-1}\left (\frac{\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{\sqrt{3} \sqrt [3]{b}}\right )}{\sqrt{3} \sqrt [3]{a} b^{4/3} f}-\frac{\log \left (a \cos ^3(e+f x)+b\right )}{3 a f}+\frac{\sec (e+f x)}{b f} \]

[Out]

-(((a^(2/3) + 2*b^(2/3))*ArcTan[(b^(1/3) - 2*a^(1/3)*Cos[e + f*x])/(Sqrt[3]*b^(1/3))])/(Sqrt[3]*a^(1/3)*b^(4/3
)*f)) - ((a^(2/3) - 2*b^(2/3))*Log[b^(1/3) + a^(1/3)*Cos[e + f*x]])/(3*a^(1/3)*b^(4/3)*f) + ((a^(2/3) - 2*b^(2
/3))*Log[b^(2/3) - a^(1/3)*b^(1/3)*Cos[e + f*x] + a^(2/3)*Cos[e + f*x]^2])/(6*a^(1/3)*b^(4/3)*f) - Log[b + a*C
os[e + f*x]^3]/(3*a*f) + Sec[e + f*x]/(b*f)

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Rubi [A]  time = 0.320877, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {4138, 1834, 1871, 1860, 31, 634, 617, 204, 628, 260} \[ \frac{\left (a^{2/3}-2 b^{2/3}\right ) \log \left (a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}\right )}{6 \sqrt [3]{a} b^{4/3} f}-\frac{\left (a^{2/3}-2 b^{2/3}\right ) \log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{4/3} f}-\frac{\left (a^{2/3}+2 b^{2/3}\right ) \tan ^{-1}\left (\frac{\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{\sqrt{3} \sqrt [3]{b}}\right )}{\sqrt{3} \sqrt [3]{a} b^{4/3} f}-\frac{\log \left (a \cos ^3(e+f x)+b\right )}{3 a f}+\frac{\sec (e+f x)}{b f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^3),x]

[Out]

-(((a^(2/3) + 2*b^(2/3))*ArcTan[(b^(1/3) - 2*a^(1/3)*Cos[e + f*x])/(Sqrt[3]*b^(1/3))])/(Sqrt[3]*a^(1/3)*b^(4/3
)*f)) - ((a^(2/3) - 2*b^(2/3))*Log[b^(1/3) + a^(1/3)*Cos[e + f*x]])/(3*a^(1/3)*b^(4/3)*f) + ((a^(2/3) - 2*b^(2
/3))*Log[b^(2/3) - a^(1/3)*b^(1/3)*Cos[e + f*x] + a^(2/3)*Cos[e + f*x]^2])/(6*a^(1/3)*b^(4/3)*f) - Log[b + a*C
os[e + f*x]^3]/(3*a*f) + Sec[e + f*x]/(b*f)

Rule 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

Rule 1834

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[((c*x)^m*Pq)/(a + b*
x^n), x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IntegerQ[n] &&  !IGtQ[m, 0]

Rule 1871

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rule 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s
*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/
b]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{\tan ^5(e+f x)}{a+b \sec ^3(e+f x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^2 \left (b+a x^3\right )} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{1}{b x^2}+\frac{-2 b-a x+b x^2}{b \left (b+a x^3\right )}\right ) \, dx,x,\cos (e+f x)\right )}{f}\\ &=\frac{\sec (e+f x)}{b f}-\frac{\operatorname{Subst}\left (\int \frac{-2 b-a x+b x^2}{b+a x^3} \, dx,x,\cos (e+f x)\right )}{b f}\\ &=\frac{\sec (e+f x)}{b f}-\frac{\operatorname{Subst}\left (\int \frac{x^2}{b+a x^3} \, dx,x,\cos (e+f x)\right )}{f}-\frac{\operatorname{Subst}\left (\int \frac{-2 b-a x}{b+a x^3} \, dx,x,\cos (e+f x)\right )}{b f}\\ &=-\frac{\log \left (b+a \cos ^3(e+f x)\right )}{3 a f}+\frac{\sec (e+f x)}{b f}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt [3]{b} \left (-a \sqrt [3]{b}-4 \sqrt [3]{a} b\right )+\sqrt [3]{a} \left (-a \sqrt [3]{b}+2 \sqrt [3]{a} b\right ) x}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx,x,\cos (e+f x)\right )}{3 \sqrt [3]{a} b^{5/3} f}-\frac{\left (a^{2/3}-2 b^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{b}+\sqrt [3]{a} x} \, dx,x,\cos (e+f x)\right )}{3 b^{4/3} f}\\ &=-\frac{\left (a^{2/3}-2 b^{2/3}\right ) \log \left (\sqrt [3]{b}+\sqrt [3]{a} \cos (e+f x)\right )}{3 \sqrt [3]{a} b^{4/3} f}-\frac{\log \left (b+a \cos ^3(e+f x)\right )}{3 a f}+\frac{\sec (e+f x)}{b f}+\frac{\left (a^{2/3}-2 b^{2/3}\right ) \operatorname{Subst}\left (\int \frac{-\sqrt [3]{a} \sqrt [3]{b}+2 a^{2/3} x}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx,x,\cos (e+f x)\right )}{6 \sqrt [3]{a} b^{4/3} f}+\frac{\left (a^{2/3}+2 b^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2} \, dx,x,\cos (e+f x)\right )}{2 b f}\\ &=-\frac{\left (a^{2/3}-2 b^{2/3}\right ) \log \left (\sqrt [3]{b}+\sqrt [3]{a} \cos (e+f x)\right )}{3 \sqrt [3]{a} b^{4/3} f}+\frac{\left (a^{2/3}-2 b^{2/3}\right ) \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+a^{2/3} \cos ^2(e+f x)\right )}{6 \sqrt [3]{a} b^{4/3} f}-\frac{\log \left (b+a \cos ^3(e+f x)\right )}{3 a f}+\frac{\sec (e+f x)}{b f}+\frac{\left (a^{2/3}+2 b^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{a} \cos (e+f x)}{\sqrt [3]{b}}\right )}{\sqrt [3]{a} b^{4/3} f}\\ &=-\frac{\left (a^{2/3}+2 b^{2/3}\right ) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{a} \cos (e+f x)}{\sqrt [3]{b}}}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{a} b^{4/3} f}-\frac{\left (a^{2/3}-2 b^{2/3}\right ) \log \left (\sqrt [3]{b}+\sqrt [3]{a} \cos (e+f x)\right )}{3 \sqrt [3]{a} b^{4/3} f}+\frac{\left (a^{2/3}-2 b^{2/3}\right ) \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+a^{2/3} \cos ^2(e+f x)\right )}{6 \sqrt [3]{a} b^{4/3} f}-\frac{\log \left (b+a \cos ^3(e+f x)\right )}{3 a f}+\frac{\sec (e+f x)}{b f}\\ \end{align*}

Mathematica [C]  time = 0.365992, size = 251, normalized size = 1.15 \[ \frac{-\text{RootSum}\left [\text{$\#$1}^3 a-6 \text{$\#$1}^2 a-\text{$\#$1}^3 b+12 \text{$\#$1} a-8 a\& ,\frac{\text{$\#$1}^2 a b \log \left (-\text{$\#$1}+\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )-\text{$\#$1}^2 b^2 \log \left (-\text{$\#$1}+\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )-4 a^2 \log \left (-\text{$\#$1}+\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )+2 \text{$\#$1} a^2 \log \left (-\text{$\#$1}+\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )+4 a b \log \left (-\text{$\#$1}+\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )-8 \text{$\#$1} a b \log \left (-\text{$\#$1}+\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )}{\text{$\#$1}^2 a-\text{$\#$1}^2 b-4 \text{$\#$1} a+4 a}\& \right ]+3 a \sec (e+f x)+3 b \log \left (\sec ^2\left (\frac{1}{2} (e+f x)\right )\right )}{3 a b f} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^3),x]

[Out]

(3*b*Log[Sec[(e + f*x)/2]^2] - RootSum[-8*a + 12*a*#1 - 6*a*#1^2 + a*#1^3 - b*#1^3 & , (-4*a^2*Log[1 - #1 + Ta
n[(e + f*x)/2]^2] + 4*a*b*Log[1 - #1 + Tan[(e + f*x)/2]^2] + 2*a^2*Log[1 - #1 + Tan[(e + f*x)/2]^2]*#1 - 8*a*b
*Log[1 - #1 + Tan[(e + f*x)/2]^2]*#1 + a*b*Log[1 - #1 + Tan[(e + f*x)/2]^2]*#1^2 - b^2*Log[1 - #1 + Tan[(e + f
*x)/2]^2]*#1^2)/(4*a - 4*a*#1 + a*#1^2 - b*#1^2) & ] + 3*a*Sec[e + f*x])/(3*a*b*f)

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Maple [A]  time = 0.063, size = 274, normalized size = 1.3 \begin{align*}{\frac{2}{3\,fa}\ln \left ( \cos \left ( fx+e \right ) +\sqrt [3]{{\frac{b}{a}}} \right ) \left ({\frac{b}{a}} \right ) ^{-{\frac{2}{3}}}}-{\frac{1}{3\,fa}\ln \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}-\sqrt [3]{{\frac{b}{a}}}\cos \left ( fx+e \right ) + \left ({\frac{b}{a}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{b}{a}} \right ) ^{-{\frac{2}{3}}}}+{\frac{2\,\sqrt{3}}{3\,fa}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\cos \left ( fx+e \right ){\frac{1}{\sqrt [3]{{\frac{b}{a}}}}}}-1 \right ) } \right ) \left ({\frac{b}{a}} \right ) ^{-{\frac{2}{3}}}}-{\frac{1}{3\,fb}\ln \left ( \cos \left ( fx+e \right ) +\sqrt [3]{{\frac{b}{a}}} \right ){\frac{1}{\sqrt [3]{{\frac{b}{a}}}}}}+{\frac{1}{6\,fb}\ln \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}-\sqrt [3]{{\frac{b}{a}}}\cos \left ( fx+e \right ) + \left ({\frac{b}{a}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{b}{a}}}}}}+{\frac{\sqrt{3}}{3\,fb}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\cos \left ( fx+e \right ){\frac{1}{\sqrt [3]{{\frac{b}{a}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{b}{a}}}}}}-{\frac{\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{3} \right ) }{3\,fa}}+{\frac{1}{fb\cos \left ( fx+e \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^5/(a+b*sec(f*x+e)^3),x)

[Out]

2/3/f/a/(b/a)^(2/3)*ln(cos(f*x+e)+(b/a)^(1/3))-1/3/f/a/(b/a)^(2/3)*ln(cos(f*x+e)^2-(b/a)^(1/3)*cos(f*x+e)+(b/a
)^(2/3))+2/3/f/a/(b/a)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(b/a)^(1/3)*cos(f*x+e)-1))-1/3/f/b/(b/a)^(1/3)*ln(c
os(f*x+e)+(b/a)^(1/3))+1/6/f/b/(b/a)^(1/3)*ln(cos(f*x+e)^2-(b/a)^(1/3)*cos(f*x+e)+(b/a)^(2/3))+1/3/f/b*3^(1/2)
/(b/a)^(1/3)*arctan(1/3*3^(1/2)*(2/(b/a)^(1/3)*cos(f*x+e)-1))-1/3*ln(b+a*cos(f*x+e)^3)/a/f+1/f/b/cos(f*x+e)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 3.62116, size = 10531, normalized size = 48.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^3),x, algorithm="fricas")

[Out]

-1/36*(2*((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(
a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(
3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a
^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))*a*b*f*cos(f*x + e)*log(1/36*((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*
a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3)
- 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)
/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))
^2*a^2*b^3*f^2 - ((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 +
8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*
(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a
^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))*a*b^3*f + 4*a^2*b + 5*b^3 + (a^3 + 8*a*b^2)*cos(f*x + e)
) - (((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^
4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) +
1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b
^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))*a*b*f*cos(f*x + e) + 3*sqrt(1/3)*a*b*f*sqrt(-(((-I*sqrt(3) + 1)*(1/(
a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)
/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54
*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1
/3) + 6/(a*f))^2*a^2*b^2*f^2 - 12*((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^
3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4
*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b
^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))*a*b^2*f + 288*a^2 + 36*b^2)/(a^2*b^2*f^
2))*cos(f*x + e) - 18*b*cos(f*x + e))*log(1/36*((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(
-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 +
b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2
+ b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))^2*a^2*b^3*f^2 - ((-I*sqrt(
3) + 1)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(
2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3
*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*
b^4*f^3))^(1/3) + 6/(a*f))*a*b^3*f + 4*a^2*b + 5*b^3 - 1/12*sqrt(1/3)*(((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2
+ b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) -
1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a
*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))*a^
2*b^3*f^2 + 18*a*b^3*f)*sqrt(-(((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3)
+ 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^
3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*
f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))^2*a^2*b^2*f^2 - 12*((-I*sqrt(3) + 1)*(1/(a
^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/
(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*
(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/
3) + 6/(a*f))*a*b^2*f + 288*a^2 + 36*b^2)/(a^2*b^2*f^2)) - 2*(a^3 + 8*a*b^2)*cos(f*x + e)) - (((-I*sqrt(3) + 1
)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2
+ b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3)
+ 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^
3))^(1/3) + 6/(a*f))*a*b*f*cos(f*x + e) - 3*sqrt(1/3)*a*b*f*sqrt(-(((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b
^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54
*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4
*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))^2*a^2*
b^2*f^2 - 12*((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^
2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*s
qrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 -
2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))*a*b^2*f + 288*a^2 + 36*b^2)/(a^2*b^2*f^2))*cos(f*x + e) - 18
*b*cos(f*x + e))*log(-1/36*((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/
54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^
(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3)
- 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))^2*a^2*b^3*f^2 + ((-I*sqrt(3) + 1)*(1/(a^2*f^2)
- (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^
2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 +
8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/
(a*f))*a*b^3*f - 4*a^2*b - 5*b^3 - 1/12*sqrt(1/3)*(((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2
))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^
2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*
a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))*a^2*b^3*f^2 + 18*a*b^3
*f)*sqrt(-(((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)
/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqr
t(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2
*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))^2*a^2*b^2*f^2 - 12*((-I*sqrt(3) + 1)*(1/(a^2*f^2) - (2*a^2 + b
^2)/(a^2*b^2*f^2))/(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54
*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 9*(I*sqrt(3) + 1)*(-1/27/(a^3*f^3) + 1/54*(a^2 + 8*b^2)/(a*b^4
*f^3) + 1/18*(2*a^2 + b^2)/(a^3*b^2*f^3) - 1/54*(a^4 - 2*a^2*b^2 + b^4)/(a^3*b^4*f^3))^(1/3) + 6/(a*f))*a*b^2*
f + 288*a^2 + 36*b^2)/(a^2*b^2*f^2)) + 2*(a^3 + 8*a*b^2)*cos(f*x + e)) - 36*a)/(a*b*f*cos(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**5/(a+b*sec(f*x+e)**3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{5}}{b \sec \left (f x + e\right )^{3} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^3),x, algorithm="giac")

[Out]

integrate(tan(f*x + e)^5/(b*sec(f*x + e)^3 + a), x)