3.287 \(\int (a+i a \tan (c+d x))^{5/3} \, dx\)

Optimal. Leaf size=177 \[ \frac{i 2^{2/3} \sqrt{3} a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{d}+\frac{3 i a^{5/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{\sqrt [3]{2} d}+\frac{i a^{5/3} \log (\cos (c+d x))}{\sqrt [3]{2} d}-\frac{a^{5/3} x}{\sqrt [3]{2}}+\frac{3 i a (a+i a \tan (c+d x))^{2/3}}{2 d} \]

[Out]

-((a^(5/3)*x)/2^(1/3)) + (I*2^(2/3)*Sqrt[3]*a^(5/3)*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(S
qrt[3]*a^(1/3))])/d + (I*a^(5/3)*Log[Cos[c + d*x]])/(2^(1/3)*d) + ((3*I)*a^(5/3)*Log[2^(1/3)*a^(1/3) - (a + I*
a*Tan[c + d*x])^(1/3)])/(2^(1/3)*d) + (((3*I)/2)*a*(a + I*a*Tan[c + d*x])^(2/3))/d

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Rubi [A]  time = 0.100217, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {3478, 3481, 55, 617, 204, 31} \[ \frac{i 2^{2/3} \sqrt{3} a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{d}+\frac{3 i a^{5/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{\sqrt [3]{2} d}+\frac{i a^{5/3} \log (\cos (c+d x))}{\sqrt [3]{2} d}-\frac{a^{5/3} x}{\sqrt [3]{2}}+\frac{3 i a (a+i a \tan (c+d x))^{2/3}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(5/3),x]

[Out]

-((a^(5/3)*x)/2^(1/3)) + (I*2^(2/3)*Sqrt[3]*a^(5/3)*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(S
qrt[3]*a^(1/3))])/d + (I*a^(5/3)*Log[Cos[c + d*x]])/(2^(1/3)*d) + ((3*I)*a^(5/3)*Log[2^(1/3)*a^(1/3) - (a + I*
a*Tan[c + d*x])^(1/3)])/(2^(1/3)*d) + (((3*I)/2)*a*(a + I*a*Tan[c + d*x])^(2/3))/d

Rule 3478

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

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 55

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

Mathematica [C]  time = 0.623186, size = 82, normalized size = 0.46 \[ -\frac{3 i a \left (\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left (-1+\, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )\right )}{\sqrt [3]{2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(5/3),x]

[Out]

((-3*I)*a*((a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^(2/3)*(-1 + Hypergeometric2F1[2/3, 1, 5/3, E^((2
*I)*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]))/(2^(1/3)*d)

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Maple [A]  time = 0.012, size = 159, normalized size = 0.9 \begin{align*}{\frac{{\frac{3\,i}{2}}a}{d} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}}+{\frac{i{2}^{{\frac{2}{3}}}}{d}{a}^{{\frac{5}{3}}}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ) }-{\frac{{\frac{i}{2}}{2}^{{\frac{2}{3}}}}{d}{a}^{{\frac{5}{3}}}\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{i\sqrt{3}{2}^{{\frac{2}{3}}}}{d}{a}^{{\frac{5}{3}}}\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((a+I*a*tan(d*x+c))^(5/3),x)

[Out]

3/2*I*a*(a+I*a*tan(d*x+c))^(2/3)/d+I/d*a^(5/3)*2^(2/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1/2*I/d*a^
(5/3)*2^(2/3)*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))+I/d*a^(5/3
)*3^(1/2)*2^(2/3)*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((a+I*a*tan(d*x+c))^(5/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.6324, size = 780, normalized size = 4.41 \begin{align*} \frac{3 i \cdot 2^{\frac{2}{3}} a \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{2}{3}} e^{\left (\frac{4}{3} i \, d x + \frac{4}{3} i \, c\right )} + \left (-\frac{4 i \, a^{5}}{d^{3}}\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} d - d\right )} \log \left (\frac{4 \cdot 2^{\frac{1}{3}} a^{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 (i \, \sqrt{3} d^{2} - d^{2}\right )} \left (-\frac{4 i \, a^{5}}{d^{3}}\right )^{\frac{2}{3}}}{4 \, a^{3}}\right ) + \left (-\frac{4 i \, a^{5}}{d^{3}}\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} d - d\right )} \log \left (\frac{4 \cdot 2^{\frac{1}{3}} a^{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 (-i \, \sqrt{3} d^{2} - d^{2}\right )} \left (-\frac{4 i \, a^{5}}{d^{3}}\right )^{\frac{2}{3}}}{4 \, a^{3}}\right ) + 2 \, \left (-\frac{4 i \, a^{5}}{d^{3}}\right )^{\frac{1}{3}} d \log \left (\frac{2 \cdot 2^{\frac{1}{3}} a^{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 (-\frac{4 i \, a^{5}}{d^{3}}\right )^{\frac{2}{3}} d^{2}}{2 \, a^{3}}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/3),x, algorithm="fricas")

[Out]

1/2*(3*I*2^(2/3)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*e^(4/3*I*d*x + 4/3*I*c) + (-4*I*a^5/d^3)^(1/3)*(-I*sqrt
(3)*d - d)*log(1/4*(4*2^(1/3)*a^3*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (I*sqrt(3)*d^2
- d^2)*(-4*I*a^5/d^3)^(2/3))/a^3) + (-4*I*a^5/d^3)^(1/3)*(I*sqrt(3)*d - d)*log(1/4*(4*2^(1/3)*a^3*(a/(e^(2*I*
d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (-I*sqrt(3)*d^2 - d^2)*(-4*I*a^5/d^3)^(2/3))/a^3) + 2*(-4*I
*a^5/d^3)^(1/3)*d*log(1/2*(2*2^(1/3)*a^3*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (-4*I*a
^5/d^3)^(2/3)*d^2)/a^3))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(5/3),x)

[Out]

Timed out

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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{5}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/3),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/3), x)