3.4.7 \(\int \frac {1}{(a+i a \tan (c+d x))^{5/3}} \, dx\) [307]

Optimal. Leaf size=213 \[ -\frac {x}{8\ 2^{2/3} a^{5/3}}-\frac {i \sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4\ 2^{2/3} a^{5/3} d}+\frac {i \log (\cos (c+d x))}{8\ 2^{2/3} a^{5/3} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8\ 2^{2/3} a^{5/3} d}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}+\frac {3 i}{8 a d (a+i a \tan (c+d x))^{2/3}} \]

[Out]

-1/16*x*2^(1/3)/a^(5/3)+1/16*I*ln(cos(d*x+c))*2^(1/3)/a^(5/3)/d+3/16*I*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(
1/3))*2^(1/3)/a^(5/3)/d-1/8*I*arctan(1/3*(a^(1/3)+2^(2/3)*(a+I*a*tan(d*x+c))^(1/3))/a^(1/3)*3^(1/2))*3^(1/2)*2
^(1/3)/a^(5/3)/d+3/10*I/d/(a+I*a*tan(d*x+c))^(5/3)+3/8*I/a/d/(a+I*a*tan(d*x+c))^(2/3)

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Rubi [A]
time = 0.11, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3560, 3562, 59, 631, 210, 31} \begin {gather*} -\frac {i \sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4\ 2^{2/3} a^{5/3} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8\ 2^{2/3} a^{5/3} d}+\frac {i \log (\cos (c+d x))}{8\ 2^{2/3} a^{5/3} d}-\frac {x}{8\ 2^{2/3} a^{5/3}}+\frac {3 i}{8 a d (a+i a \tan (c+d x))^{2/3}}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(-5/3),x]

[Out]

-1/8*x/(2^(2/3)*a^(5/3)) - ((I/4)*Sqrt[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)*a^(5/3)*d) + ((I/8)*Log[Cos[c + d*x]])/(2^(2/3)*a^(5/3)*d) + (((3*I)/8)*Log[2^(1/3)*a^(1/3) -
(a + I*a*Tan[c + d*x])^(1/3)])/(2^(2/3)*a^(5/3)*d) + ((3*I)/10)/(d*(a + I*a*Tan[c + d*x])^(5/3)) + ((3*I)/8)/
(a*d*(a + I*a*Tan[c + d*x])^(2/3))

Rule 31

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

Rule 59

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 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 3560

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] +
Dist[1/(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] && LtQ[n
, 0]

Rule 3562

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]

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (c+d x))^{5/3}} \, dx &=\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}+\frac {\int \frac {1}{(a+i a \tan (c+d x))^{2/3}} \, dx}{2 a}\\ &=\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}+\frac {3 i}{8 a d (a+i a \tan (c+d x))^{2/3}}+\frac {\int \sqrt [3]{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}+\frac {3 i}{8 a d (a+i a \tan (c+d x))^{2/3}}-\frac {i \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{2/3}} \, dx,x,i a \tan (c+d x)\right )}{4 a d}\\ &=-\frac {x}{8\ 2^{2/3} a^{5/3}}+\frac {i \log (\cos (c+d x))}{8\ 2^{2/3} a^{5/3} d}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}+\frac {3 i}{8 a d (a+i a \tan (c+d x))^{2/3}}-\frac {(3 i) \text {Subst}\left (\int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{8\ 2^{2/3} a^{5/3} d}-\frac {(3 i) \text {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 )}{8 \sqrt [3]{2} a^{4/3} d}\\ &=-\frac {x}{8\ 2^{2/3} a^{5/3}}+\frac {i \log (\cos (c+d x))}{8\ 2^{2/3} a^{5/3} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8\ 2^{2/3} a^{5/3} d}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}+\frac {3 i}{8 a d (a+i a \tan (c+d x))^{2/3}}+\frac {(3 i) \text {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 )}{4\ 2^{2/3} a^{5/3} d}\\ &=-\frac {x}{8\ 2^{2/3} a^{5/3}}-\frac {i \sqrt {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 )}{4\ 2^{2/3} a^{5/3} d}+\frac {i \log (\cos (c+d x))}{8\ 2^{2/3} a^{5/3} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8\ 2^{2/3} a^{5/3} d}+\frac {3 i}{10 d (a+i a \tan (c+d x))^{5/3}}+\frac {3 i}{8 a d (a+i a \tan (c+d x))^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 1.54, size = 331, normalized size = 1.55 \begin {gather*} \frac {i e^{-2 i (c+d x)} \left (6+27 e^{2 i (c+d x)}+21 e^{4 i (c+d x)}-10 \sqrt {3} e^{\frac {10}{3} i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \text {ArcTan}\left (\frac {1+\frac {2 e^{\frac {2}{3} i (c+d x)}}{\sqrt [3]{1+e^{2 i (c+d x)}}}}{\sqrt {3}}\right )+10 e^{\frac {10}{3} i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \log \left (1-\frac {e^{\frac {2}{3} i (c+d x)}}{\sqrt [3]{1+e^{2 i (c+d x)}}}\right )-5 e^{\frac {10}{3} i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \log \left (\frac {e^{\frac {4}{3} i (c+d x)}+e^{\frac {2}{3} i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}}+\left (1+e^{2 i (c+d x)}\right )^{2/3}}{\left (1+e^{2 i (c+d x)}\right )^{2/3}}\right )\right ) \sec ^2(c+d x)}{80 d (a+i a \tan (c+d x))^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/80)*(6 + 27*E^((2*I)*(c + d*x)) + 21*E^((4*I)*(c + d*x)) - 10*Sqrt[3]*E^(((10*I)/3)*(c + d*x))*(1 + E^((2*
I)*(c + d*x)))^(1/3)*ArcTan[(1 + (2*E^(((2*I)/3)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))^(1/3))/Sqrt[3]] + 10*E^
(((10*I)/3)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(1/3)*Log[1 - E^(((2*I)/3)*(c + d*x))/(1 + E^((2*I)*(c + d*x)
))^(1/3)] - 5*E^(((10*I)/3)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(1/3)*Log[(E^(((4*I)/3)*(c + d*x)) + E^(((2*I
)/3)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(1/3) + (1 + E^((2*I)*(c + d*x)))^(2/3))/(1 + E^((2*I)*(c + d*x)))^(
2/3)])*Sec[c + d*x]^2)/(d*E^((2*I)*(c + d*x))*(a + I*a*Tan[c + d*x])^(5/3))

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Maple [A]
time = 0.10, size = 177, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {3 i a \left (\frac {\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}}{4 a^{2}}+\frac {1}{8 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}+\frac {1}{10 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}\right )}{d}\)	\(177\)
default	\(\frac {3 i a \left (\frac {\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}}{4 a^{2}}+\frac {1}{8 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}+\frac {1}{10 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}\right )}{d}\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(d*x+c))^(5/3),x,method=_RETURNVERBOSE)

[Out]

3*I/d*a*(1/4*(1/6*2^(1/3)/a^(2/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1/12*2^(1/3)/a^(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))-1/6*2^(1/3)/a^(2/3)*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)))/a^2+1/8/a^2/(a+I*a*tan(d*x+c))^(2/3)+1/10/a/(a+I*a
*tan(d*x+c))^(5/3))

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Maxima [A]
time = 0.51, size = 164, normalized size = 0.77 \begin {gather*} -\frac {i \, {\left (\frac {10 \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{a^{\frac {2}{3}}} + \frac {5 \cdot 2^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right )}{a^{\frac {2}{3}}} - \frac {10 \cdot 2^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}{a^{\frac {2}{3}}} - \frac {6 \, {\left (5 i \, a \tan \left (d x + c\right ) + 9 \, a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}}}\right )}}{80 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/80*I*(10*sqrt(3)*2^(1/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 2*(I*a*tan(d*x + c) + a)^(1/3))/a^(1
/3))/a^(2/3) + 5*2^(1/3)*log(2^(2/3)*a^(2/3) + 2^(1/3)*(I*a*tan(d*x + c) + a)^(1/3)*a^(1/3) + (I*a*tan(d*x + c
) + a)^(2/3))/a^(2/3) - 10*2^(1/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3))/a^(2/3) - 6*(5*I*a*tan
(d*x + c) + 9*a)/(I*a*tan(d*x + c) + a)^(5/3))/(a*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (146) = 292\).
time = 0.97, size = 336, normalized size = 1.58 \begin {gather*} \frac {{\left (80 \, a^{2} d \left (-\frac {i}{256 \, a^{5} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (8 i \, a^{2} d \left (-\frac {i}{256 \, a^{5} 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 ) - 40 \, {\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (-\frac {i}{256 \, a^{5} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \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 )} - 4 \, {\left (\sqrt {3} a^{2} d + i \, a^{2} d\right )} \left (-\frac {i}{256 \, a^{5} d^{3}}\right )^{\frac {1}{3}}\right ) - 40 \, {\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (-\frac {i}{256 \, a^{5} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \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 )} + 4 \, {\left (\sqrt {3} a^{2} d - i \, a^{2} d\right )} \left (-\frac {i}{256 \, a^{5} d^{3}}\right )^{\frac {1}{3}}\right ) - 3 \cdot 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} {\left (-7 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{80 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^(5/3),x, algorithm="fricas")

[Out]

1/80*(80*a^2*d*(-1/256*I/(a^5*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*log(8*I*a^2*d*(-1/256*I/(a^5*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)) - 40*(-I*sqrt(3)*a^2*d + a^2*d)*(-1/256*I/(a^5
*d^3))^(1/3)*e^(4*I*d*x + 4*I*c)*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) - 4*(
sqrt(3)*a^2*d + I*a^2*d)*(-1/256*I/(a^5*d^3))^(1/3)) - 40*(I*sqrt(3)*a^2*d + a^2*d)*(-1/256*I/(a^5*d^3))^(1/3)
*e^(4*I*d*x + 4*I*c)*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) + 4*(sqrt(3)*a^2*
d - I*a^2*d)*(-1/256*I/(a^5*d^3))^(1/3)) - 3*2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*(-7*I*e^(4*I*d*x + 4*
I*c) - 9*I*e^(2*I*d*x + 2*I*c) - 2*I)*e^(2/3*I*d*x + 2/3*I*c))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {5}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))**(5/3),x)

[Out]

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

[Out]

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

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Mupad [B]
time = 3.99, size = 233, normalized size = 1.09 \begin {gather*} \frac {\frac {3{}\mathrm {i}}{10\,d}+\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{8\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}}+\frac {{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,144{}\mathrm {i}-1152\,{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{7/3}\,d^2\right )}{{\left (-a\right )}^{5/3}\,d}+\frac {{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,144{}\mathrm {i}-1152\,{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{7/3}\,d^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{5/3}\,d}-\frac {{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,\ln \left (a^2\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,144{}\mathrm {i}+1152\,{\left (\frac {1}{256}{}\mathrm {i}\right )}^{1/3}\,{\left (-a\right )}^{7/3}\,d^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{5/3}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*tan(c + d*x)*1i)^(5/3),x)

[Out]

(3i/(10*d) + ((a + a*tan(c + d*x)*1i)*3i)/(8*a*d))/(a + a*tan(c + d*x)*1i)^(5/3) + ((1i/256)^(1/3)*log(a^2*d^2
*(a + a*tan(c + d*x)*1i)^(1/3)*144i - 1152*(1i/256)^(1/3)*(-a)^(7/3)*d^2))/((-a)^(5/3)*d) + ((1i/256)^(1/3)*lo
g(a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*144i - 1152*(1i/256)^(1/3)*(-a)^(7/3)*d^2*((3^(1/2)*1i)/2 - 1/2))*((3^
(1/2)*1i)/2 - 1/2))/((-a)^(5/3)*d) - ((1i/256)^(1/3)*log(a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/3)*144i + 1152*(1i
/256)^(1/3)*(-a)^(7/3)*d^2*((3^(1/2)*1i)/2 + 1/2))*((3^(1/2)*1i)/2 + 1/2))/((-a)^(5/3)*d)

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