3.7.92 \(\int \frac {1}{\sqrt [3]{a+b \tan (c+d x)}} \, dx\) [692]

Optimal. Leaf size=415 \[ -\frac {x}{4 \sqrt [3]{a-\sqrt {-b^2}}}-\frac {x}{4 \sqrt [3]{a+\sqrt {-b^2}}}-\frac {\sqrt {3} b \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-\sqrt {-b^2}}}}{\sqrt {3}}\right )}{2 \sqrt {-b^2} \sqrt [3]{a-\sqrt {-b^2}} d}+\frac {\sqrt {3} b \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+\sqrt {-b^2}}}}{\sqrt {3}}\right )}{2 \sqrt {-b^2} \sqrt [3]{a+\sqrt {-b^2}} d}-\frac {b \log (\cos (c+d x))}{4 \sqrt {-b^2} \sqrt [3]{a-\sqrt {-b^2}} d}+\frac {b \log (\cos (c+d x))}{4 \sqrt {-b^2} \sqrt [3]{a+\sqrt {-b^2}} d}-\frac {3 b \log \left (\sqrt [3]{a-\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} \sqrt [3]{a-\sqrt {-b^2}} d}+\frac {3 b \log \left (\sqrt [3]{a+\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} \sqrt [3]{a+\sqrt {-b^2}} d} \]

[Out]

-1/4*x/(a-(-b^2)^(1/2))^(1/3)-1/4*b*ln(cos(d*x+c))/d/(a-(-b^2)^(1/2))^(1/3)/(-b^2)^(1/2)-3/4*b*ln((a-(-b^2)^(1
/2))^(1/3)-(a+b*tan(d*x+c))^(1/3))/d/(a-(-b^2)^(1/2))^(1/3)/(-b^2)^(1/2)-1/2*b*arctan(1/3*(1+2*(a+b*tan(d*x+c)
)^(1/3)/(a-(-b^2)^(1/2))^(1/3))*3^(1/2))*3^(1/2)/d/(a-(-b^2)^(1/2))^(1/3)/(-b^2)^(1/2)-1/4*x/(a+(-b^2)^(1/2))^
(1/3)+1/4*b*ln(cos(d*x+c))/d/(-b^2)^(1/2)/(a+(-b^2)^(1/2))^(1/3)+3/4*b*ln((a+(-b^2)^(1/2))^(1/3)-(a+b*tan(d*x+
c))^(1/3))/d/(-b^2)^(1/2)/(a+(-b^2)^(1/2))^(1/3)+1/2*b*arctan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a+(-b^2)^(1/2))
^(1/3))*3^(1/2))*3^(1/2)/d/(-b^2)^(1/2)/(a+(-b^2)^(1/2))^(1/3)

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Rubi [A]
time = 0.22, antiderivative size = 415, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3566, 726, 57, 631, 210, 31} \begin {gather*} -\frac {\sqrt {3} b \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-\sqrt {-b^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-b^2} d \sqrt [3]{a-\sqrt {-b^2}}}+\frac {\sqrt {3} b \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+\sqrt {-b^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-b^2} d \sqrt [3]{a+\sqrt {-b^2}}}-\frac {3 b \log \left (\sqrt [3]{a-\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} d \sqrt [3]{a-\sqrt {-b^2}}}+\frac {3 b \log \left (\sqrt [3]{a+\sqrt {-b^2}}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt {-b^2} d \sqrt [3]{a+\sqrt {-b^2}}}-\frac {b \log (\cos (c+d x))}{4 \sqrt {-b^2} d \sqrt [3]{a-\sqrt {-b^2}}}+\frac {b \log (\cos (c+d x))}{4 \sqrt {-b^2} d \sqrt [3]{a+\sqrt {-b^2}}}-\frac {x}{4 \sqrt [3]{a-\sqrt {-b^2}}}-\frac {x}{4 \sqrt [3]{a+\sqrt {-b^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(-1/3),x]

[Out]

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

Rule 31

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

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 210

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

Rule 631

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 726

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 3566

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[(a + x)^n/(b^2 + x^2), x], x
, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.26, size = 251, normalized size = 0.60 \begin {gather*} \frac {i \left (\frac {2 \sqrt {3} \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{\sqrt [3]{a-i b}}-\frac {2 \sqrt {3} \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{\sqrt [3]{a+i b}}+\frac {\log (i-\tan (c+d x))}{\sqrt [3]{a+i b}}-\frac {\log (i+\tan (c+d x))}{\sqrt [3]{a-i b}}+\frac {3 \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{\sqrt [3]{a-i b}}-\frac {3 \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{\sqrt [3]{a+i b}}\right )}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.31, size = 58, normalized size = 0.14

method	result	size
derivativedivides	\(\frac {b \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\textit {\_R} \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d}\)	\(58\)
default	\(\frac {b \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\textit {\_R} \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2 d}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*b*sum(_R/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R),_R=RootOf(_Z^6-2*_Z^3*a+a^2+b^2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [3]{a + b \tan {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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Mupad [B]
time = 7.24, size = 817, normalized size = 1.97 \begin {gather*} \frac {\ln \left (\frac {243\,b^5\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^5}+\frac {\frac {1944\,b^4\,\left (a^2-b^2\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^2}+1944\,a\,b^4\,{\left (\frac {1}{d^3\,\left (b+a\,1{}\mathrm {i}\right )}\right )}^{2/3}\,\left (a^2+b^2\right )}{8\,d^3\,\left (b+a\,1{}\mathrm {i}\right )}\right )\,{\left (\frac {1}{b\,d^3+a\,d^3\,1{}\mathrm {i}}\right )}^{1/3}}{2}+\ln \left (\frac {243\,b^5\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^5}+\frac {\left (\frac {1944\,b^4\,\left (a^2-b^2\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^2}+1944\,a\,b^4\,\left (a^2+b^2\right )\,{\left (\frac {b+a\,1{}\mathrm {i}}{d^3\,\left (a^2+b^2\right )}\right )}^{2/3}\right )\,\left (b+a\,1{}\mathrm {i}\right )}{8\,d^3\,\left (a^2+b^2\right )}\right )\,{\left (\frac {b+a\,1{}\mathrm {i}}{8\,a^2\,d^3+8\,b^2\,d^3}\right )}^{1/3}+\frac {\ln \left (\frac {243\,b^5\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^5}+\frac {\left (\frac {1944\,b^4\,\left (a^2-b^2\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^2}+486\,a\,b^4\,{\left (\frac {1}{d^3\,\left (b+a\,1{}\mathrm {i}\right )}\right )}^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^2+b^2\right )\right )\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^3}{64\,d^3\,\left (b+a\,1{}\mathrm {i}\right )}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{b\,d^3+a\,d^3\,1{}\mathrm {i}}\right )}^{1/3}}{4}-\frac {\ln \left (\frac {243\,b^5\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^5}-\frac {\left (\frac {1944\,b^4\,\left (a^2-b^2\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^2}+486\,a\,b^4\,{\left (\frac {1}{d^3\,\left (b+a\,1{}\mathrm {i}\right )}\right )}^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^2+b^2\right )\right )\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^3}{64\,d^3\,\left (b+a\,1{}\mathrm {i}\right )}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{b\,d^3+a\,d^3\,1{}\mathrm {i}}\right )}^{1/3}}{4}+\frac {\ln \left (\frac {243\,b^5\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^5}+\frac {\left (\frac {1944\,b^4\,\left (a^2-b^2\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^2}+1944\,a\,b^4\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^2+b^2\right )\,{\left (\frac {1{}\mathrm {i}}{8\,d^3\,\left (a+b\,1{}\mathrm {i}\right )}\right )}^{2/3}\right )\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{64\,d^3\,\left (a+b\,1{}\mathrm {i}\right )}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{8\,\left (a\,d^3+b\,d^3\,1{}\mathrm {i}\right )}\right )}^{1/3}}{2}-\frac {\ln \left (\frac {243\,b^5\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^5}-\frac {\left (\frac {1944\,b^4\,\left (a^2-b^2\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^2}+1944\,a\,b^4\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^2+b^2\right )\,{\left (\frac {1{}\mathrm {i}}{8\,d^3\,\left (a+b\,1{}\mathrm {i}\right )}\right )}^{2/3}\right )\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{64\,d^3\,\left (a+b\,1{}\mathrm {i}\right )}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{8\,\left (a\,d^3+b\,d^3\,1{}\mathrm {i}\right )}\right )}^{1/3}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tan(c + d*x))^(1/3),x)

[Out]

(log((243*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 + ((1944*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2 + 1944*
a*b^4*(1/(d^3*(a*1i + b)))^(2/3)*(a^2 + b^2))/(8*d^3*(a*1i + b)))*(1/(a*d^3*1i + b*d^3))^(1/3))/2 + log((243*b
^5*(a + b*tan(c + d*x))^(1/3))/d^5 + (((1944*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2 + 1944*a*b^4*(a^2
+ b^2)*((a*1i + b)/(d^3*(a^2 + b^2)))^(2/3))*(a*1i + b))/(8*d^3*(a^2 + b^2)))*((a*1i + b)/(8*a^2*d^3 + 8*b^2*
d^3))^(1/3) + (log((243*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 + (((1944*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/
3))/d^2 + 486*a*b^4*(1/(d^3*(a*1i + b)))^(2/3)*(3^(1/2)*1i - 1)^2*(a^2 + b^2))*(3^(1/2)*1i - 1)^3)/(64*d^3*(a*
1i + b)))*(3^(1/2)*1i - 1)*(1/(a*d^3*1i + b*d^3))^(1/3))/4 - (log((243*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 - (
((1944*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2 + 486*a*b^4*(1/(d^3*(a*1i + b)))^(2/3)*(3^(1/2)*1i + 1)
^2*(a^2 + b^2))*(3^(1/2)*1i + 1)^3)/(64*d^3*(a*1i + b)))*(3^(1/2)*1i + 1)*(1/(a*d^3*1i + b*d^3))^(1/3))/4 + (l
og((243*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 + (((1944*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2 + 1944*a
*b^4*(3^(1/2)*1i - 1)^2*(a^2 + b^2)*(1i/(8*d^3*(a + b*1i)))^(2/3))*(3^(1/2)*1i - 1)^3*1i)/(64*d^3*(a + b*1i)))
*(3^(1/2)*1i - 1)*(1i/(8*(a*d^3 + b*d^3*1i)))^(1/3))/2 - (log((243*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 - (((19
44*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2 + 1944*a*b^4*(3^(1/2)*1i + 1)^2*(a^2 + b^2)*(1i/(8*d^3*(a +
b*1i)))^(2/3))*(3^(1/2)*1i + 1)^3*1i)/(64*d^3*(a + b*1i)))*(3^(1/2)*1i + 1)*(1i/(8*(a*d^3 + b*d^3*1i)))^(1/3)
)/2

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