3.273 \(\int \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=185 \[ \frac{i \sqrt{3} \sqrt [3]{a} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{2^{2/3} d}-\frac{3 i (a+i a \tan (c+d x))^{4/3}}{4 a d}-\frac{3 i \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}-\frac{i \sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}+\frac{\sqrt [3]{a} x}{2\ 2^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.126797, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3543, 3481, 57, 617, 204, 31} \[ \frac{i \sqrt{3} \sqrt [3]{a} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{2^{2/3} d}-\frac{3 i (a+i a \tan (c+d x))^{4/3}}{4 a d}-\frac{3 i \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}-\frac{i \sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}+\frac{\sqrt [3]{a} x}{2\ 2^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(1/3),x]

[Out]

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

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rule 3481

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Dist[b/d, Subst[Int[(a + x)^(n - 1)/(a - x), x]
, x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), 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 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 31

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

Rubi steps

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

Mathematica [F]  time = 180.005, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^(1/3),x]

[Out]

$Aborted

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Maple [A]  time = 0.017, size = 161, normalized size = 0.9 \begin{align*}{\frac{-{\frac{3\,i}{4}}}{ad} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}}}-{\frac{{\frac{i}{2}}\sqrt [3]{2}}{d}\sqrt [3]{a}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ) }+{\frac{{\frac{i}{4}}\sqrt [3]{2}}{d}\sqrt [3]{a}\ln \left ( \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{2}\sqrt [3]{a}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }+{2}^{{\frac{2}{3}}}{a}^{{\frac{2}{3}}} \right ) }+{\frac{{\frac{i}{2}}\sqrt [3]{2}\sqrt{3}}{d}\sqrt [3]{a}\arctan \left ({\frac{\sqrt{3}}{3} \left ({{2}^{{\frac{2}{3}}}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [3]{a}}}}+1 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/3),x)

[Out]

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

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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(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.25733, size = 873, normalized size = 4.72 \begin{align*} \frac{{\left ({\left (i \, \sqrt{3} d - d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt{3} d - d\right )} \left (\frac{i \, a}{4 \, d^{3}}\right )^{\frac{1}{3}} \log \left (2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )} +{\left (\sqrt{3} d + i \, d\right )} \left (\frac{i \, a}{4 \, d^{3}}\right )^{\frac{1}{3}}\right ) +{\left ({\left (-i \, \sqrt{3} d - d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, \sqrt{3} d - d\right )} \left (\frac{i \, a}{4 \, d^{3}}\right )^{\frac{1}{3}} \log \left (2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )} -{\left (\sqrt{3} d - i \, d\right )} \left (\frac{i \, a}{4 \, d^{3}}\right )^{\frac{1}{3}}\right ) + 2 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \left (\frac{i \, a}{4 \, d^{3}}\right )^{\frac{1}{3}} \log \left (2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )} - 2 i \, d \left (\frac{i \, a}{4 \, d^{3}}\right )^{\frac{1}{3}}\right ) - 3 i \cdot 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{8}{3} i \, d x + \frac{8}{3} i \, c\right )}}{2 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

1/2*(((I*sqrt(3)*d - d)*e^(2*I*d*x + 2*I*c) + I*sqrt(3)*d - d)*(1/4*I*a/d^3)^(1/3)*log(2^(1/3)*(a/(e^(2*I*d*x
+ 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (sqrt(3)*d + I*d)*(1/4*I*a/d^3)^(1/3)) + ((-I*sqrt(3)*d - d)*e^
(2*I*d*x + 2*I*c) - I*sqrt(3)*d - d)*(1/4*I*a/d^3)^(1/3)*log(2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/
3*I*d*x + 2/3*I*c) - (sqrt(3)*d - I*d)*(1/4*I*a/d^3)^(1/3)) + 2*(d*e^(2*I*d*x + 2*I*c) + d)*(1/4*I*a/d^3)^(1/3
)*log(2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - 2*I*d*(1/4*I*a/d^3)^(1/3)) - 3*I*2
^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(8/3*I*d*x + 8/3*I*c))/(d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{a \left (i \tan{\left (c + d x \right )} + 1\right )} \tan ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2*(a+I*a*tan(d*x+c))**(1/3),x)

[Out]

Integral((a*(I*tan(c + d*x) + 1))**(1/3)*tan(c + d*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} \tan \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(1/3)*tan(d*x + c)^2, x)