3.378 \(\int \frac{1}{a+b \tan ^3(c+d x)} \, dx\)

Optimal. Leaf size=256 \[ \frac{\sqrt [3]{b} \left (a^{4/3}-b^{4/3}\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \tan (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{2/3} d \left (a^2+b^2\right )}-\frac{\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tan (c+d x)+b^{2/3} \tan ^2(c+d x)\right )}{6 a^{2/3} d \left (a^2+b^2\right )}+\frac{\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tan (c+d x)\right )}{3 a^{2/3} d \left (a^2+b^2\right )}-\frac{b \log \left (a \cos ^3(c+d x)+b \sin ^3(c+d x)\right )}{3 d \left (a^2+b^2\right )}+\frac{a x}{a^2+b^2} \]

[Out]

(a*x)/(a^2 + b^2) + (b^(1/3)*(a^(4/3) - b^(4/3))*ArcTan[(a^(1/3) - 2*b^(1/3)*Tan[c + d*x])/(Sqrt[3]*a^(1/3))])
/(Sqrt[3]*a^(2/3)*(a^2 + b^2)*d) - (b*Log[a*Cos[c + d*x]^3 + b*Sin[c + d*x]^3])/(3*(a^2 + b^2)*d) + (b^(1/3)*(
a^(4/3) + b^(4/3))*Log[a^(1/3) + b^(1/3)*Tan[c + d*x]])/(3*a^(2/3)*(a^2 + b^2)*d) - (b^(1/3)*(a^(4/3) + b^(4/3
))*Log[a^(2/3) - a^(1/3)*b^(1/3)*Tan[c + d*x] + b^(2/3)*Tan[c + d*x]^2])/(6*a^(2/3)*(a^2 + b^2)*d)

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Rubi [A]  time = 0.380429, antiderivative size = 256, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.857, Rules used = {3661, 6725, 635, 203, 260, 1871, 1860, 31, 634, 617, 204, 628} \[ \frac{\sqrt [3]{b} \left (a^{4/3}-b^{4/3}\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \tan (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{2/3} d \left (a^2+b^2\right )}-\frac{\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tan (c+d x)+b^{2/3} \tan ^2(c+d x)\right )}{6 a^{2/3} d \left (a^2+b^2\right )}+\frac{\sqrt [3]{b} \left (a^{4/3}+b^{4/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tan (c+d x)\right )}{3 a^{2/3} d \left (a^2+b^2\right )}-\frac{b \log \left (a \cos ^3(c+d x)+b \sin ^3(c+d x)\right )}{3 d \left (a^2+b^2\right )}+\frac{a x}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x]^3)^(-1),x]

[Out]

(a*x)/(a^2 + b^2) + (b^(1/3)*(a^(4/3) - b^(4/3))*ArcTan[(a^(1/3) - 2*b^(1/3)*Tan[c + d*x])/(Sqrt[3]*a^(1/3))])
/(Sqrt[3]*a^(2/3)*(a^2 + b^2)*d) - (b*Log[a*Cos[c + d*x]^3 + b*Sin[c + d*x]^3])/(3*(a^2 + b^2)*d) + (b^(1/3)*(
a^(4/3) + b^(4/3))*Log[a^(1/3) + b^(1/3)*Tan[c + d*x]])/(3*a^(2/3)*(a^2 + b^2)*d) - (b^(1/3)*(a^(4/3) + b^(4/3
))*Log[a^(2/3) - a^(1/3)*b^(1/3)*Tan[c + d*x] + b^(2/3)*Tan[c + d*x]^2])/(6*a^(2/3)*(a^2 + b^2)*d)

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 635

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

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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.575201, size = 278, normalized size = 1.09 \[ \frac{-3 a^{2/3} b \tan ^2(c+d x) \text{Hypergeometric2F1}\left (\frac{2}{3},1,\frac{5}{3},-\frac{b \tan ^3(c+d x)}{a}\right )-b^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tan (c+d x)+b^{2/3} \tan ^2(c+d x)\right )-2 a^{2/3} b \log \left (a+b \tan ^3(c+d x)\right )+3 a^{2/3} b \log (-\tan (c+d x)+i)+3 a^{2/3} b \log (\tan (c+d x)+i)-3 i a^{5/3} \log (-\tan (c+d x)+i)+3 i a^{5/3} \log (\tan (c+d x)+i)-2 \sqrt{3} b^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \tan (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )+2 b^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tan (c+d x)\right )}{6 a^{2/3} d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x]^3)^(-1),x]

[Out]

(-2*Sqrt[3]*b^(5/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Tan[c + d*x])/(Sqrt[3]*a^(1/3))] - (3*I)*a^(5/3)*Log[I - Tan[c
+ d*x]] + 3*a^(2/3)*b*Log[I - Tan[c + d*x]] + (3*I)*a^(5/3)*Log[I + Tan[c + d*x]] + 3*a^(2/3)*b*Log[I + Tan[c
+ d*x]] + 2*b^(5/3)*Log[a^(1/3) + b^(1/3)*Tan[c + d*x]] - b^(5/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Tan[c + d*x]
+ b^(2/3)*Tan[c + d*x]^2] - 2*a^(2/3)*b*Log[a + b*Tan[c + d*x]^3] - 3*a^(2/3)*b*Hypergeometric2F1[2/3, 1, 5/3,
-((b*Tan[c + d*x]^3)/a)]*Tan[c + d*x]^2)/(6*a^(2/3)*(a^2 + b^2)*d)

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Maple [A]  time = 0.027, size = 355, normalized size = 1.4 \begin{align*}{\frac{b}{3\,d \left ({a}^{2}+{b}^{2} \right ) }\ln \left ( \tan \left ( dx+c \right ) +\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{b}{6\,d \left ({a}^{2}+{b}^{2} \right ) }\ln \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}-\sqrt [3]{{\frac{a}{b}}}\tan \left ( dx+c \right ) + \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{b\sqrt{3}}{3\,d \left ({a}^{2}+{b}^{2} \right ) }\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\tan \left ( dx+c \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{a}{3\,d \left ({a}^{2}+{b}^{2} \right ) }\ln \left ( \tan \left ( dx+c \right ) +\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{a}{6\,d \left ({a}^{2}+{b}^{2} \right ) }\ln \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}-\sqrt [3]{{\frac{a}{b}}}\tan \left ( dx+c \right ) + \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{a\sqrt{3}}{3\,d \left ({a}^{2}+{b}^{2} \right ) }\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\tan \left ( dx+c \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{b\ln \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{3} \right ) }{3\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{b\ln \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{a\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(d*x+c)^3),x)

[Out]

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

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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(1/(a+b*tan(d*x+c)^3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 10.1303, size = 10283, normalized size = 40.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)^3),x, algorithm="fricas")

[Out]

-1/24*(2*(a^2 + b^2)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2
- b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 +
a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d))*
d*log(-1/4*(4*b^2*tan(d*x + c)^2 - ((a^4 + a^2*b^2)*d^2*tan(d*x + c)^2 - (a^4 + a^2*b^2)*d^2)*((1/2)^(1/3)*(I*
sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/
3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^
3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d))^2 + 2*(a^2*b*d*tan(d*x + c)^2 - a^2*b
*d + 2*(a^3 - a*b^2)*d*tan(d*x + c))*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d +
b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^
2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(
a^2*d + b^2*d)) - 4*a^2)/(tan(d*x + c)^2 + 1)) - 24*a*d*x - ((a^2 + b^2)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*
d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^
2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a
^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d))*d - 3*sqrt(1/3)*(a^2 + b^2)*d*sqrt(-((a^4 + 2*a^2*b^2 + b^
4)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b
^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b
^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d))^2*d^2 - 4*(a^2*b +
b^3)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2
+ b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) -
2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d))*d - 12*b^2)/((a
^4 + 2*a^2*b^2 + b^4)*d^2)) - 6*b)*log(1/4*(8*a^4 - 16*a^2*b^2 - ((a^6 + 2*a^4*b^2 + a^2*b^4)*d^2*tan(d*x + c)
^2 - (a^6 + 2*a^4*b^2 + a^2*b^4)*d^2)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d +
b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)
^2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/
(a^2*d + b^2*d))^2 + 8*(2*a^2*b^2 - b^4)*tan(d*x + c)^2 + 2*((a^4*b + a^2*b^3)*d*tan(d*x + c)^2 + 2*(a^5 - a*b
^4)*d*tan(d*x + c) - (a^4*b + a^2*b^3)*d)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2
*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^
2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) +
2*b/(a^2*d + b^2*d)) + 3*sqrt(1/3)*(4*(a^4*b + a^2*b^3)*d*tan(d*x + c)^2 - 4*(a^5 - a*b^4)*d*tan(d*x + c) - ((
a^6 + 2*a^4*b^2 + a^2*b^4)*d^2*tan(d*x + c)^2 - (a^6 + 2*a^4*b^2 + a^2*b^4)*d^2)*((1/2)^(1/3)*(I*sqrt(3) + 1)*
(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^
(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^
2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d)) - 4*(a^4*b + a^2*b^3)*d)*sqrt(-((a^4 + 2*a^2*b^2 +
b^4)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2
+ b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) -
2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d))^2*d^2 - 4*(a^2*
b + b^3)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a
^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3)
- 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d))*d - 12*b^2)/
((a^4 + 2*a^2*b^2 + b^4)*d^2)) - 24*(a^3*b - a*b^3)*tan(d*x + c))/(tan(d*x + c)^2 + 1)) - ((a^2 + b^2)*((1/2)^
(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*
d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d
+ b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d))*d + 3*sqrt(1/3)*(a^2 + b^2)*
d*sqrt(-((a^4 + 2*a^2*b^2 + b^4)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*
d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b
/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*
d + b^2*d))^2*d^2 - 4*(a^2*b + b^3)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b
^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2
*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a
^2*d + b^2*d))*d - 12*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - 6*b)*log(-1/4*(8*a^4 - 16*a^2*b^2 - ((a^6 + 2*a^4*
b^2 + a^2*b^4)*d^2*tan(d*x + c)^2 - (a^6 + 2*a^4*b^2 + a^2*b^4)*d^2)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3
+ a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-
I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 +
b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d))^2 + 8*(2*a^2*b^2 - b^4)*tan(d*x + c)^2 + 2*((a^4*b + a^2*b^3)*
d*tan(d*x + c)^2 + 2*(a^5 - a*b^4)*d*tan(d*x + c) - (a^4*b + a^2*b^3)*d)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*
d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^
2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a
^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d)) - 3*sqrt(1/3)*(4*(a^4*b + a^2*b^3)*d*tan(d*x + c)^2 - 4*(a
^5 - a*b^4)*d*tan(d*x + c) - ((a^6 + 2*a^4*b^2 + a^2*b^4)*d^2*tan(d*x + c)^2 - (a^6 + 2*a^4*b^2 + a^2*b^4)*d^2
)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^
2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^
3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(a^2*d + b^2*d)) - 4*(a^4*b + a^2*b^
3)*d)*sqrt(-((a^4 + 2*a^2*b^2 + b^4)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d +
b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*d)^
2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*b/(
a^2*d + b^2*d))^2*d^2 - 4*(a^2*b + b^3)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d
+ b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3) + 2*(1/2)^(2/3)*b^2*(-I*sqrt(3) + 1)/((a^2*d + b^2*
d)^2*(b/(a^4*d^3 + a^2*b^2*d^3) - 2*b^3/(a^2*d + b^2*d)^3 - (a^2 - b^2)*b/((a^2 + b^2)^2*a^2*d^3))^(1/3)) + 2*
b/(a^2*d + b^2*d))*d - 12*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - 24*(a^3*b - a*b^3)*tan(d*x + c))/(tan(d*x + c)
^2 + 1)))/((a^2 + b^2)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{a + b \tan ^{3}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)**3),x)

[Out]

Integral(1/(a + b*tan(c + d*x)**3), x)

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Giac [A]  time = 1.58741, size = 451, normalized size = 1.76 \begin{align*} \frac{\frac{2 \,{\left (a^{3} b^{2} \left (-\frac{a}{b}\right )^{\frac{1}{3}} + a b^{4} \left (-\frac{a}{b}\right )^{\frac{1}{3}} - a^{2} b^{3} - b^{5}\right )} \left (-\frac{a}{b}\right )^{\frac{1}{3}} \log \left ({\left | -\left (-\frac{a}{b}\right )^{\frac{1}{3}} + \tan \left (d x + c\right ) \right |}\right )}{a^{5} b + 2 \, a^{3} b^{3} + a b^{5}} + \frac{6 \,{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (\left (-\frac{a}{b}\right )^{\frac{1}{3}}\right ) + \arctan \left (\frac{\sqrt{3}{\left (\left (-\frac{a}{b}\right )^{\frac{1}{3}} + 2 \, \tan \left (d x + c\right )\right )}}{3 \, \left (-\frac{a}{b}\right )^{\frac{1}{3}}}\right )\right )}{\left (\left (-a b^{2}\right )^{\frac{1}{3}} b^{2} + \left (-a b^{2}\right )^{\frac{2}{3}} a\right )}}{\sqrt{3} a^{3} b + \sqrt{3} a b^{3}} + \frac{6 \,{\left (d x + c\right )} a}{a^{2} + b^{2}} + \frac{{\left (\left (-a b^{2}\right )^{\frac{1}{3}} b^{2} - \left (-a b^{2}\right )^{\frac{2}{3}} a\right )} \log \left (\tan \left (d x + c\right )^{2} + \left (-\frac{a}{b}\right )^{\frac{1}{3}} \tan \left (d x + c\right ) + \left (-\frac{a}{b}\right )^{\frac{2}{3}}\right )}{a^{3} b + a b^{3}} + \frac{3 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, b \log \left ({\left | b \tan \left (d x + c\right )^{3} + a \right |}\right )}{a^{2} + b^{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(d*x+c)^3),x, algorithm="giac")

[Out]

1/6*(2*(a^3*b^2*(-a/b)^(1/3) + a*b^4*(-a/b)^(1/3) - a^2*b^3 - b^5)*(-a/b)^(1/3)*log(abs(-(-a/b)^(1/3) + tan(d*
x + c)))/(a^5*b + 2*a^3*b^3 + a*b^5) + 6*(pi*floor((d*x + c)/pi + 1/2)*sgn((-a/b)^(1/3)) + arctan(1/3*sqrt(3)*
((-a/b)^(1/3) + 2*tan(d*x + c))/(-a/b)^(1/3)))*((-a*b^2)^(1/3)*b^2 + (-a*b^2)^(2/3)*a)/(sqrt(3)*a^3*b + sqrt(3
)*a*b^3) + 6*(d*x + c)*a/(a^2 + b^2) + ((-a*b^2)^(1/3)*b^2 - (-a*b^2)^(2/3)*a)*log(tan(d*x + c)^2 + (-a/b)^(1/
3)*tan(d*x + c) + (-a/b)^(2/3))/(a^3*b + a*b^3) + 3*b*log(tan(d*x + c)^2 + 1)/(a^2 + b^2) - 2*b*log(abs(b*tan(
d*x + c)^3 + a))/(a^2 + b^2))/d