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]

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

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Rubi [A]  time = 0.15, antiderivative size = 166, normalized size of antiderivative = 1.00, 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 31

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

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 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 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]

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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.25, size = 242, normalized size = 1.46 \[ \frac {\text {RootSum}\left [\text {$\#$1}^3 a-\text {$\#$1}^3 b-3 \text {$\#$1}^2 a-3 \text {$\#$1}^2 b+3 \text {$\#$1} a-3 \text {$\#$1} b-a-b\& ,\frac {\text {$\#$1}^2 a \log \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\text {$\#$1}\right )-\text {$\#$1}^2 b \log \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\text {$\#$1}\right )-4 \text {$\#$1} a \log \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\text {$\#$1}\right )-a \log \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\text {$\#$1}\right )-2 \text {$\#$1} b \log \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\text {$\#$1}\right )-b \log \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\text {$\#$1}\right )}{\text {$\#$1}^2 a-\text {$\#$1}^2 b-2 \text {$\#$1} a-2 \text {$\#$1} b+a-b}\& \right ]-3 \log \left (\sec ^2\left (\frac {1}{2} (e+f x)\right )\right )}{3 a f} \]

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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fricas [C]  time = 36.12, size = 2278, normalized size = 13.72 \[ \text {result too large to display} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (f x + e\right )^{3}}{b \sec \left (f x + e\right )^{3} + a}\,{d x} \]

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]

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

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maple [A]  time = 0.77, size = 141, normalized size = 0.85 \[ -\frac {\ln \left (\cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 f a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\ln \left (\cos ^{2}\left (f x +e \right )-\left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 f a \left (\frac {b}{a}\right )^{\frac {2}{3}}}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \left (f x +e \right )}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 f a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\ln \left (b +a \left (\cos ^{3}\left (f x +e \right )\right )\right )}{3 a f} \]

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

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maxima [A]  time = 0.44, size = 159, normalized size = 0.96 \[ -\frac {\frac {2 \, \sqrt {3} {\left (a {\left (3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}} - \frac {2 \, b}{a}\right )} + 2 \, b\right )} \arctan \left (-\frac {\sqrt {3} {\left (\left (\frac {b}{a}\right )^{\frac {1}{3}} - 2 \, \cos \left (f x + e\right )\right )}}{3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{a b} - \frac {3 \, {\left (2 \, \left (\frac {b}{a}\right )^{\frac {2}{3}} + 1\right )} \log \left (\cos \left (f x + e\right )^{2} - \left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x + e\right ) + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{a \left (\frac {b}{a}\right )^{\frac {2}{3}}} - \frac {6 \, {\left (\left (\frac {b}{a}\right )^{\frac {2}{3}} - 1\right )} \log \left (\left (\frac {b}{a}\right )^{\frac {1}{3}} + \cos \left (f x + e\right )\right )}{a \left (\frac {b}{a}\right )^{\frac {2}{3}}}}{18 \, f} \]

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]

-1/18*(2*sqrt(3)*(a*(3*(b/a)^(1/3) - 2*b/a) + 2*b)*arctan(-1/3*sqrt(3)*((b/a)^(1/3) - 2*cos(f*x + e))/(b/a)^(1
/3))/(a*b) - 3*(2*(b/a)^(2/3) + 1)*log(cos(f*x + e)^2 - (b/a)^(1/3)*cos(f*x + e) + (b/a)^(2/3))/(a*(b/a)^(2/3)
) - 6*((b/a)^(2/3) - 1)*log((b/a)^(1/3) + cos(f*x + e))/(a*(b/a)^(2/3)))/f

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mupad [B]  time = 8.54, size = 1620, normalized size = 9.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log(262144*(a - b)^2*(8*a - 8*b + 4*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)
*a^2 + 4*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*b^2 - 3*root(27*a^3*b^2*z^3 - 27*
a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a^3 - 24*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 -
b^2, z, k)^2*a*b^2 - 36*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^3*a^3*b + 28*root
(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*a*b + 36*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2
+ 9*a*b^2*z + a^2 - b^2, z, k)^3*a^2*b^2)*(16*a^2*tan(e/2 + (f*x)/2)^2 + 32*b^2*tan(e/2 + (f*x)/2)^2 - 4*root(
27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*a^3 - 4*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 +
9*a*b^2*z + a^2 - b^2, z, k)*b^3 - 8*a^2 + 8*b^2 + 3*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 -
b^2, z, k)^2*a^4 - 3*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a^4*tan(e/2 + (f*x)
/2)^2 + 24*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a*b^3 + 3*root(27*a^3*b^2*z^3
- 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a^3*b + 36*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z
+ a^2 - b^2, z, k)^3*a^4*b - 48*a*b*tan(e/2 + (f*x)/2)^2 + 24*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*
z + a^2 - b^2, z, k)^2*a^2*b^2 - 36*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^3*a^2*
b^3 + 14*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*a^3*tan(e/2 + (f*x)/2)^2 - 4*root
(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*b^3*tan(e/2 + (f*x)/2)^2 - 32*root(27*a^3*b^2*
z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*a*b^2 - 32*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*
z + a^2 - b^2, z, k)*a^2*b - 146*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*a*b^2*tan
(e/2 + (f*x)/2)^2 + 64*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)*a^2*b*tan(e/2 + (f*
x)/2)^2 + 24*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a*b^3*tan(e/2 + (f*x)/2)^2
+ 57*root(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a^3*b*tan(e/2 + (f*x)/2)^2 - 54*roo
t(27*a^3*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^3*a^4*b*tan(e/2 + (f*x)/2)^2 + 84*root(27*a^3
*b^2*z^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^2*a^2*b^2*tan(e/2 + (f*x)/2)^2 - 36*root(27*a^3*b^2*z
^3 - 27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^3*a^2*b^3*tan(e/2 + (f*x)/2)^2 + 198*root(27*a^3*b^2*z^3 -
27*a^2*b^2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k)^3*a^3*b^2*tan(e/2 + (f*x)/2)^2))*root(27*a^3*b^2*z^3 - 27*a^2*b^
2*z^2 + 9*a*b^2*z + a^2 - b^2, z, k), k, 1, 3)/f - log(tan(e/2 + (f*x)/2)^2 + 1)/(a*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{3}{\left (e + f x \right )}}{a + b \sec ^{3}{\left (e + f x \right )}}\, dx \]

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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