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

Optimal. Leaf size=166 \[ \frac{\log \left (a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}\right )}{6 \sqrt [3]{a} b^{2/3} f}-\frac{\log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{2/3} f}+\frac{\tan ^{-1}\left (\frac{\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{\sqrt{3} \sqrt [3]{b}}\right )}{\sqrt{3} \sqrt [3]{a} b^{2/3} f}+\frac{\log \left (a \cos ^3(e+f x)+b\right )}{3 a f} \]

[Out]

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

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Rubi [A]  time = 0.146733, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {4138, 1871, 200, 31, 634, 617, 204, 628, 260} \[ \frac{\log \left (a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}\right )}{6 \sqrt [3]{a} b^{2/3} f}-\frac{\log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{2/3} f}+\frac{\tan ^{-1}\left (\frac{\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{\sqrt{3} \sqrt [3]{b}}\right )}{\sqrt{3} \sqrt [3]{a} b^{2/3} f}+\frac{\log \left (a \cos ^3(e+f x)+b\right )}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(b^(1/3) - 2*a^(1/3)*Cos[e + f*x])/(Sqrt[3]*b^(1/3))]/(Sqrt[3]*a^(1/3)*b^(2/3)*f) - Log[b^(1/3) + a^(1/
3)*Cos[e + f*x]]/(3*a^(1/3)*b^(2/3)*f) + 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^(2/3)*f) + Log[b + a*Cos[e + f*x]^3]/(3*a*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 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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

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

Mathematica [C]  time = 0.244182, size = 242, normalized size = 1.46 \[ \frac{\text{RootSum}\left [\text{$\#$1}^3 a-3 \text{$\#$1}^2 a-\text{$\#$1}^3 b-3 \text{$\#$1}^2 b+3 \text{$\#$1} a-3 \text{$\#$1} b-a-b\& ,\frac{\text{$\#$1}^2 a \log \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-\text{$\#$1}\right )-\text{$\#$1}^2 b \log \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-\text{$\#$1}\right )-4 \text{$\#$1} a \log \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-\text{$\#$1}\right )-a \log \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-\text{$\#$1}\right )-2 \text{$\#$1} b \log \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-\text{$\#$1}\right )-b \log \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-\text{$\#$1}\right )}{\text{$\#$1}^2 a-\text{$\#$1}^2 b-2 \text{$\#$1} a-2 \text{$\#$1} b+a-b}\& \right ]-3 \log \left (\sec ^2\left (\frac{1}{2} (e+f x)\right )\right )}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.066, size = 141, normalized size = 0.9 \begin{align*} -{\frac{1}{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}{6\,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{\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{\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{3} \right ) }{3\,fa}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 72.2434, size = 5415, normalized size = 32.62 \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)^3/(a+b*sec(f*x+e)^3),x, algorithm="fricas")

[Out]

-1/12*(6*sqrt(1/3)*a*f*sqrt(((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^
2*f^3))^(1/3) - 2/(a*f))^2*a^2*f^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^
2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*f + 4)/(a^2*f^2))*arctan(-1/8*(2*sqrt(1/3)*sqrt((3*(I*sqrt(3) + 1)*(-1/54
/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))^2*a^2*b^2*f^2 + 4*a^2*cos(f*x
+ e)^2 - 4*a*b*cos(f*x + e) - 2*(a^2*b*f*cos(f*x + e) - 2*a*b^2*f)*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54
/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f)) + 4*b^2)*((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3)
+ 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*b*f^2 + 2*b*f)*sqrt(((3*(I*sqrt(3) + 1
)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))^2*a^2*f^2 + 4*(3*(I*s
qrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*f + 4)/(a
^2*f^2)) + sqrt(1/3)*((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))
^(1/3) - 2/(a*f))^2*a^2*b^2*f^3 - 8*a*b*f*cos(f*x + e) + 4*b^2*f - 4*(a^2*b*f^2*cos(f*x + e) - a*b^2*f^2)*(3*(
I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f)))*sqrt(((
3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))^2*a^2
*f^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a
*f))*a*f + 4)/(a^2*f^2)))/a) - 6*sqrt(1/3)*a*f*sqrt(((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) -
1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))^2*a^2*f^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b
^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*f + 4)/(a^2*f^2))*arctan(-1/8*(2*sqrt(1/3)*sqrt((
3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))^2*a^2
*b^2*f^2 + 4*a^2*cos(f*x + e)^2 - 4*a*b*cos(f*x + e) - 2*(a^2*b*f*cos(f*x + e) - 2*a*b^2*f)*(3*(I*sqrt(3) + 1)
*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f)) + 4*b^2)*((3*(I*sqrt(3
) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*b*f^2 + 2*b*f)
*sqrt(((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f
))^2*a^2*f^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3
) - 2/(a*f))*a*f + 4)/(a^2*f^2)) - sqrt(1/3)*((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a
^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))^2*a^2*b^2*f^3 - 8*a*b*f*cos(f*x + e) + 4*b^2*f - 4*(a^2*b*f^2*cos(f*
x + e) - a*b^2*f^2)*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(
1/3) - 2/(a*f)))*sqrt(((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3)
)^(1/3) - 2/(a*f))^2*a^2*f^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2)/(a^
3*b^2*f^3))^(1/3) - 2/(a*f))*a*f + 4)/(a^2*f^2)))/a) + (3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3)
- 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*f*log(1/4*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b
^2*f^3) - 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))^2*a^2*b^2*f^2 + a^2*cos(f*x + e)^2 + 2*a*b*cos(f*x
+ e) + (a^2*b*f*cos(f*x + e) + a*b^2*f)*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b
^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f)) + b^2) - ((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a
^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))*a*f + 6)*log((3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3)
- 1/54*(a^2 - b^2)/(a^3*b^2*f^3))^(1/3) - 2/(a*f))^2*a^2*b^2*f^2 + 4*a^2*cos(f*x + e)^2 - 4*a*b*cos(f*x + e) -
2*(a^2*b*f*cos(f*x + e) - 2*a*b^2*f)*(3*(I*sqrt(3) + 1)*(-1/54/(a^3*f^3) + 1/54/(a*b^2*f^3) - 1/54*(a^2 - b^2
)/(a^3*b^2*f^3))^(1/3) - 2/(a*f)) + 4*b^2))/(a*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.88372, size = 1077, normalized size = 6.49 \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)^3/(a+b*sec(f*x+e)^3),x, algorithm="giac")

[Out]

1/6*(2*sqrt(3)*(-1/(a*b^2))^(1/3)*arctan((((a^2*b)^(2/3)*(sqrt(3)*a^2 - sqrt(3)*a*b) - (sqrt(3)*a^3 - sqrt(3)*
a^2*b)*(a^2*b)^(1/3))*cos(f*x + e) + (a^2*b)^(2/3)*(sqrt(3)*a^2 - sqrt(3)*a*b) - (sqrt(3)*a^2*b - sqrt(3)*a*b^
2)*(a^2*b)^(1/3))/(2*a^3*b - 2*a^2*b^2 + (2*a^3*b - 2*a^2*b^2 - (a^2*b)^(2/3)*(a^2 - a*b) - (a^3 - a^2*b)*(a^2
*b)^(1/3))*cos(f*x + e) - (a^2*b)^(2/3)*(a^2 - a*b) - (a^2*b - a*b^2)*(a^2*b)^(1/3))) - (-1/(a*b^2))^(1/3)*log
(144*(2*a^3*b - 2*a^2*b^2 + (2*a^3*b - 2*a^2*b^2 - (a^2*b)^(2/3)*(a^2 - a*b) - (a^3 - a^2*b)*(a^2*b)^(1/3))*co
s(f*x + e) - (a^2*b)^(2/3)*(a^2 - a*b) - (a^2*b - a*b^2)*(a^2*b)^(1/3))^2 + 144*(((a^2*b)^(2/3)*(sqrt(3)*a^2 -
sqrt(3)*a*b) - (sqrt(3)*a^3 - sqrt(3)*a^2*b)*(a^2*b)^(1/3))*cos(f*x + e) + (a^2*b)^(2/3)*(sqrt(3)*a^2 - sqrt(
3)*a*b) - (sqrt(3)*a^2*b - sqrt(3)*a*b^2)*(a^2*b)^(1/3))^2) + 2*(-1/(a*b^2))^(1/3)*log(abs(24*a^3*b - 24*a^2*b
^2 + 24*(a^3*b - a^2*b^2 + (a^2*b)^(2/3)*(a^2 - a*b) + (a^3 - a^2*b)*(a^2*b)^(1/3))*cos(f*x + e) + 24*(a^2*b)^
(2/3)*(a^2 - a*b) + 24*(a^2*b - a*b^2)*(a^2*b)^(1/3))) - 6*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)/a +
2*log(abs(a + b + 3*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 3*a
*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + a*(cos(f*x + e) -
1)^3/(cos(f*x + e) + 1)^3 - b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3))/a)/f