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

Optimal. Leaf size=174 \[ -\frac{\sqrt{3} \sqrt [3]{a} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{2^{2/3} d}+\frac{3 \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac{3 \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}+\frac{\sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}+\frac{i \sqrt [3]{a} x}{2\ 2^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.118028, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3527, 3481, 57, 617, 204, 31} \[ -\frac{\sqrt{3} \sqrt [3]{a} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{2^{2/3} d}+\frac{3 \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac{3 \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}+\frac{\sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}+\frac{i \sqrt [3]{a} x}{2\ 2^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3527

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*
(a + b*Tan[e + f*x])^m)/(f*m), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 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 (c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx &=\frac{3 \sqrt [3]{a+i a \tan (c+d x)}}{d}-i \int \sqrt [3]{a+i a \tan (c+d x)} \, dx\\ &=\frac{3 \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac{a \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{2/3}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=\frac{i \sqrt [3]{a} x}{2\ 2^{2/3}}+\frac{\sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}+\frac{3 \sqrt [3]{a+i a \tan (c+d x)}}{d}-\frac{\left (3 \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 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{i \sqrt [3]{a} x}{2\ 2^{2/3}}+\frac{\sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}+\frac{3 \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 \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac{\left (3 \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{i \sqrt [3]{a} x}{2\ 2^{2/3}}-\frac{\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{\sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}+\frac{3 \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 \sqrt [3]{a+i a \tan (c+d x)}}{d}\\ \end{align*}

Mathematica [F]  time = 180.006, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [A]  time = 0.014, size = 154, normalized size = 0.9 \begin{align*} 3\,{\frac{\sqrt [3]{a+ia\tan \left ( dx+c \right ) }}{d}}+{\frac{\sqrt [3]{2}}{2\,d}\sqrt [3]{a}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ) }-{\frac{\sqrt [3]{2}}{4\,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{\sqrt [3]{2}\sqrt{3}}{2\,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)*(a+I*a*tan(d*x+c))^(1/3),x)

[Out]

3*(a+I*a*tan(d*x+c))^(1/3)/d+1/2/d*a^(1/3)*2^(1/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1/4/d*a^(1/3)*
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/d*a^(1/3)*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)*(a+I*a*tan(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.01128, size = 741, normalized size = 4.26 \begin{align*} \frac{\left (\frac{1}{4}\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} d - d\right )} \left (\frac{a}{d^{3}}\right )^{\frac{1}{3}} \log \left (\left (\frac{1}{4}\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} d + d\right )} \left (\frac{a}{d^{3}}\right )^{\frac{1}{3}} + 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 )}\right ) + \left (\frac{1}{4}\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} d - d\right )} \left (\frac{a}{d^{3}}\right )^{\frac{1}{3}} \log \left (\left (\frac{1}{4}\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} d + d\right )} \left (\frac{a}{d^{3}}\right )^{\frac{1}{3}} + 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 )}\right ) + 2 \, \left (\frac{1}{4}\right )^{\frac{1}{3}} d \left (\frac{a}{d^{3}}\right )^{\frac{1}{3}} \log \left (-2 \, \left (\frac{1}{4}\right )^{\frac{1}{3}} d \left (\frac{a}{d^{3}}\right )^{\frac{1}{3}} + 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 )}\right ) + 6 \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{2}{3} i \, d x + \frac{2}{3} i \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((1/4)^(1/3)*(-I*sqrt(3)*d - d)*(a/d^3)^(1/3)*log((1/4)^(1/3)*(I*sqrt(3)*d + d)*(a/d^3)^(1/3) + 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)) + (1/4)^(1/3)*(I*sqrt(3)*d - d)*(a/d^3)^(1/3)*log((
1/4)^(1/3)*(-I*sqrt(3)*d + d)*(a/d^3)^(1/3) + 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*(1/4)^(1/3)*d*(a/d^3)^(1/3)*log(-2*(1/4)^(1/3)*d*(a/d^3)^(1/3) + 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)) + 6*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))/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{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(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), 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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(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), x)