3.92 \(\int \frac {\tan ^3(x)}{a+b \cos ^3(x)} \, dx\)

Optimal. Leaf size=153 \[ -\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cos (x)+b^{2/3} \cos ^2(x)\right )}{6 a^{5/3}}+\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \cos (x)\right )}{3 a^{5/3}}-\frac {b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \cos (x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}-\frac {\log \left (a+b \cos ^3(x)\right )}{3 a}+\frac {\sec ^2(x)}{2 a}+\frac {\log (\cos (x))}{a} \]

[Out]

ln(cos(x))/a+1/3*b^(2/3)*ln(a^(1/3)+b^(1/3)*cos(x))/a^(5/3)-1/6*b^(2/3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*cos(x)+b^(2
/3)*cos(x)^2)/a^(5/3)-1/3*ln(a+b*cos(x)^3)/a+1/2*sec(x)^2/a-1/3*b^(2/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*cos(x))/
a^(1/3)*3^(1/2))/a^(5/3)*3^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3230, 1834, 1871, 200, 31, 634, 617, 204, 628, 260} \[ -\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cos (x)+b^{2/3} \cos ^2(x)\right )}{6 a^{5/3}}+\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \cos (x)\right )}{3 a^{5/3}}-\frac {b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \cos (x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}-\frac {\log \left (a+b \cos ^3(x)\right )}{3 a}+\frac {\sec ^2(x)}{2 a}+\frac {\log (\cos (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^3/(a + b*Cos[x]^3),x]

[Out]

-((b^(2/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Cos[x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(5/3))) + Log[Cos[x]]/a + (b^(2/
3)*Log[a^(1/3) + b^(1/3)*Cos[x]])/(3*a^(5/3)) - (b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Cos[x] + b^(2/3)*Cos[x]
^2])/(6*a^(5/3)) - Log[a + b*Cos[x]^3]/(3*a) + Sec[x]^2/(2*a)

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 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 3230

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + b*(c*ff*x)^n)^p)/(1 - ff^2*x^2)^(
(m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && ILtQ[(m - 1)/2, 0]

Rubi steps

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

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Mathematica [C]  time = 0.30, size = 217, normalized size = 1.42 \[ \frac {-2 \text {RootSum}\left [\text {$\#$1}^3 a-\text {$\#$1}^3 b+3 \text {$\#$1}^2 a+3 \text {$\#$1}^2 b+3 \text {$\#$1} a-3 \text {$\#$1} b+a+b\& ,\frac {\text {$\#$1}^2 a \log \left (\tan ^2\left (\frac {x}{2}\right )-\text {$\#$1}\right )-\text {$\#$1}^2 b \log \left (\tan ^2\left (\frac {x}{2}\right )-\text {$\#$1}\right )+2 \text {$\#$1} a \log \left (\tan ^2\left (\frac {x}{2}\right )-\text {$\#$1}\right )+a \log \left (\tan ^2\left (\frac {x}{2}\right )-\text {$\#$1}\right )+4 \text {$\#$1} b \log \left (\tan ^2\left (\frac {x}{2}\right )-\text {$\#$1}\right )+b \log \left (\tan ^2\left (\frac {x}{2}\right )-\text {$\#$1}\right )}{\text {$\#$1}^2 a-\text {$\#$1}^2 b+2 \text {$\#$1} a+2 \text {$\#$1} b+a-b}\& \right ]+3 \sec ^2(x)+6 \left (\log \left (\sec ^2\left (\frac {x}{2}\right )\right )+\log (\cos (x))\right )}{6 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^3/(a + b*Cos[x]^3),x]

[Out]

(6*(Log[Cos[x]] + Log[Sec[x/2]^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[x/2]^2] + b*Log[-#1 + Tan[x/2]^2] + 2*a*Log[-#1 + Tan[x/2]^2]*#1 + 4*b*Log[-#1 + Tan[x/2
]^2]*#1 + a*Log[-#1 + Tan[x/2]^2]*#1^2 - b*Log[-#1 + Tan[x/2]^2]*#1^2)/(a - b + 2*a*#1 + 2*b*#1 + a*#1^2 - b*#
1^2) & ] + 3*Sec[x]^2)/(6*a)

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fricas [C]  time = 46.28, size = 1690, normalized size = 11.05 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+b*cos(x)^3),x, algorithm="fricas")

[Out]

1/12*(6*sqrt(1/3)*a*sqrt((((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^2
- 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a + 4)/a^2)*arctan(-1/8*(2*s
qrt(1/3)*sqrt(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^4 + 4*b^2*cos(
x)^2 - 4*a*b*cos(x) + 2*(a^2*b*cos(x) - 2*a^3)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5
)^(1/3) + 2/a) + 4*a^2)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a^3 - 2
*a^2)*sqrt((((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^2 - 4*((1/2)^(1/
3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a + 4)/a^2) + sqrt(1/3)*(((1/2)^(1/3)*(I*s
qrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^5 - 8*a^2*b*cos(x) + 4*a^3 + 4*(a^3*b*cos(x)
- a^4)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a))*sqrt((((1/2)^(1/3)*(I*sq
rt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^
2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a + 4)/a^2))/b^2)*cos(x)^2 - 6*sqrt(1/3)*a*sqrt((((1/2)^(1/3)*(I*sqrt(3)
+ 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5
- (a^2 - b^2)/a^5)^(1/3) + 2/a)*a + 4)/a^2)*arctan(-1/8*(2*sqrt(1/3)*sqrt(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3
+ b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^4 + 4*b^2*cos(x)^2 - 4*a*b*cos(x) + 2*(a^2*b*cos(x) - 2*a^3)*((
1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a) + 4*a^2)*(((1/2)^(1/3)*(I*sqrt(3)
+ 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a^3 - 2*a^2)*sqrt((((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 +
b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)
/a^5)^(1/3) + 2/a)*a + 4)/a^2) - sqrt(1/3)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(
1/3) + 2/a)^2*a^5 - 8*a^2*b*cos(x) + 4*a^3 + 4*(a^3*b*cos(x) - a^4)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/
a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a))*sqrt((((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1
/3) + 2/a)^2*a^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a + 4)/a^2)
)/b^2)*cos(x)^2 - ((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a*cos(x)^2*log
(1/4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)^2*a^4 + b^2*cos(x)^2 + 2*a*
b*cos(x) - (a^2*b*cos(x) + a^3)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)
+ a^2) + 12*cos(x)^2*log(-cos(x)) + (((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) +
2/a)*a*cos(x)^2 - 6*cos(x)^2)*log(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a
)^2*a^4 + 4*b^2*cos(x)^2 - 4*a*b*cos(x) + 2*(a^2*b*cos(x) - 2*a^3)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a
^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a) + 4*a^2) + 6)/(a*cos(x)^2)

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giac [A]  time = 2.13, size = 143, normalized size = 0.93 \[ -\frac {b \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {a}{b}\right )^{\frac {1}{3}} + \cos \relax (x) \right |}\right )}{3 \, a^{2}} - \frac {\log \left ({\left | b \cos \relax (x)^{3} + a \right |}\right )}{3 \, a} + \frac {\log \left ({\left | \cos \relax (x) \right |}\right )}{a} + \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, \cos \relax (x)\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{2}} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (\cos \relax (x)^{2} + \left (-\frac {a}{b}\right )^{\frac {1}{3}} \cos \relax (x) + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{2}} + \frac {1}{2 \, a \cos \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+b*cos(x)^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.11, size = 125, normalized size = 0.82 \[ \frac {\ln \left (\cos \relax (x )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\cos ^{2}\relax (x )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \cos \relax (x )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \relax (x )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (a +b \left (\cos ^{3}\relax (x )\right )\right )}{3 a}+\frac {\ln \left (\cos \relax (x )\right )}{a}+\frac {1}{2 a \cos \relax (x )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.59, size = 151, normalized size = 0.99 \[ \frac {\sqrt {3} {\left (b {\left (3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}} - \frac {2 \, a}{b}\right )} + 2 \, a\right )} \arctan \left (-\frac {\sqrt {3} {\left (\left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, \cos \relax (x)\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2}} - \frac {{\left (2 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} + 1\right )} \log \left (\cos \relax (x)^{2} - \left (\frac {a}{b}\right )^{\frac {1}{3}} \cos \relax (x) + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (\left (\frac {a}{b}\right )^{\frac {2}{3}} - 1\right )} \log \left (\left (\frac {a}{b}\right )^{\frac {1}{3}} + \cos \relax (x)\right )}{3 \, a \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {\log \left (\cos \relax (x)\right )}{a} + \frac {1}{2 \, a \cos \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3/(a+b*cos(x)^3),x, algorithm="maxima")

[Out]

1/9*sqrt(3)*(b*(3*(a/b)^(1/3) - 2*a/b) + 2*a)*arctan(-1/3*sqrt(3)*((a/b)^(1/3) - 2*cos(x))/(a/b)^(1/3))/a^2 -
1/6*(2*(a/b)^(2/3) + 1)*log(cos(x)^2 - (a/b)^(1/3)*cos(x) + (a/b)^(2/3))/(a*(a/b)^(2/3)) - 1/3*((a/b)^(2/3) -
1)*log((a/b)^(1/3) + cos(x))/(a*(a/b)^(2/3)) + log(cos(x))/a + 1/2/(a*cos(x)^2)

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mupad [B]  time = 5.21, size = 1281, normalized size = 8.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^3/(a + b*cos(x)^3),x)

[Out]

(2*tan(x/2)^2)/(a - 2*a*tan(x/2)^2 + a*tan(x/2)^4) + log(tan(x/2)^2 - 1)/a + symsum(log((262144*(9*a*b^10 - b^
11 - 37*a^2*b^9 + 85*a^3*b^8 - 107*a^4*b^7 + 43*a^5*b^6 + 73*a^6*b^5 - 121*a^7*b^4 + 72*a^8*b^3 - 16*a^9*b^2))
/a^6 + root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*(root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2
- b^2, z, k)*(root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*((262144*(72*a^5*b^9 - 96*a^6*b^8 + 1
428*a^7*b^7 - 3684*a^8*b^6 + 612*a^9*b^5 + 3972*a^10*b^4 - 2112*a^11*b^3 - 192*a^12*b^2))/a^6 + root(27*a^5*z^
3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*(root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*((26214
4*(5184*a^10*b^6 - 3024*a^9*b^7 + 1728*a^11*b^5 - 6048*a^12*b^4 + 1296*a^13*b^3 + 864*a^14*b^2))/a^6 - root(27
*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*((262144*(1296*a^10*b^7 - 3888*a^11*b^6 + 2592*a^12*b^5 + 2
592*a^13*b^4 - 3888*a^14*b^3 + 1296*a^15*b^2))/a^6 - (262144*tan(x/2)^2*(1296*a^10*b^7 - 11016*a^11*b^6 + 2721
6*a^12*b^5 - 28512*a^13*b^4 + 12960*a^14*b^3 - 1944*a^15*b^2))/a^6) + (262144*tan(x/2)^2*(4104*a^9*b^7 - 16740
*a^10*b^6 + 18468*a^11*b^5 - 1836*a^12*b^4 - 5292*a^13*b^3 + 1296*a^14*b^2))/a^6) + (262144*(288*a^7*b^8 - 183
6*a^8*b^7 - 1692*a^9*b^6 + 6084*a^10*b^5 + 108*a^11*b^4 - 4248*a^12*b^3 + 1296*a^13*b^2))/a^6 + (262144*tan(x/
2)^2*(4392*a^8*b^7 - 360*a^7*b^8 + 3366*a^9*b^6 - 29934*a^10*b^5 + 35946*a^11*b^4 - 15354*a^12*b^3 + 1944*a^13
*b^2))/a^6) - (262144*tan(x/2)^2*(72*a^5*b^9 - 162*a^6*b^8 + 3780*a^7*b^7 - 20160*a^8*b^6 + 30276*a^9*b^5 - 14
526*a^10*b^4 + 432*a^11*b^3 + 288*a^12*b^2))/a^6) + (262144*(68*a^4*b^9 - 436*a^5*b^8 + 903*a^6*b^7 - 55*a^7*b
^6 - 1579*a^8*b^5 + 987*a^9*b^4 + 608*a^10*b^3 - 496*a^11*b^2))/a^6 - (262144*tan(x/2)^2*(90*a^4*b^9 - 666*a^5
*b^8 + 3753*a^6*b^7 - 5925*a^7*b^6 - 1311*a^8*b^5 + 8919*a^9*b^4 - 5604*a^10*b^3 + 744*a^11*b^2))/a^6) - (2621
44*(8*a^2*b^10 - 26*a^3*b^9 - 30*a^4*b^8 + 292*a^5*b^7 - 540*a^6*b^6 + 230*a^7*b^5 + 402*a^8*b^4 - 496*a^9*b^3
+ 160*a^10*b^2))/a^6 + (262144*tan(x/2)^2*(10*a^2*b^10 - 54*a^3*b^9 + 52*a^4*b^8 + 920*a^5*b^7 - 4214*a^6*b^6
+ 7442*a^7*b^5 - 6168*a^8*b^4 + 2252*a^9*b^3 - 240*a^10*b^2))/a^6) - (262144*tan(x/2)^2*(11*a*b^10 - b^11 - 8
7*a^2*b^9 + 391*a^3*b^8 - 1045*a^4*b^7 + 1705*a^5*b^6 - 1677*a^6*b^5 + 941*a^7*b^4 - 262*a^8*b^3 + 24*a^9*b^2)
)/a^6)*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k), k, 1, 3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{3}{\relax (x )}}{a + b \cos ^{3}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)**3/(a+b*cos(x)**3),x)

[Out]

Integral(tan(x)**3/(a + b*cos(x)**3), x)

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