3.102 \(\int (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=101 \[ \frac{4 i a^2 \sqrt{a+i a \tan (c+d x)}}{d}-\frac{4 i \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

((-4*I)*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((4*I)*a^2*Sqrt[a + I*a*Tan
[c + d*x]])/d + (((2*I)/3)*a*(a + I*a*Tan[c + d*x])^(3/2))/d

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Rubi [A]  time = 0.0613302, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {3478, 3480, 206} \[ \frac{4 i a^2 \sqrt{a+i a \tan (c+d x)}}{d}-\frac{4 i \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}+\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-4*I)*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((4*I)*a^2*Sqrt[a + I*a*Tan
[c + d*x]])/d + (((2*I)/3)*a*(a + I*a*Tan[c + d*x])^(3/2))/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 3480

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

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.852981, size = 140, normalized size = 1.39 \[ -\frac{\sqrt{2} a^2 e^{-i (c+2 d x)} \sqrt{1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x)) \left (12 i \sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt{1+e^{2 i (c+d x)}} (\tan (c+d x)-7 i) \sec (c+d x)\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

-(Sqrt[2]*a^2*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(Cos[d*x]
+ I*Sin[d*x])*((12*I)*ArcSinh[E^(I*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]*(-7*I + Tan[c + d*
x])))/(3*d*E^(I*(c + 2*d*x)))

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Maple [A]  time = 0.014, size = 73, normalized size = 0.7 \begin{align*}{\frac{2\,ia}{d} \left ({\frac{1}{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+2\,a\sqrt{a+ia\tan \left ( dx+c \right ) }-2\,{a}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+ia\tan \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(5/2),x)

[Out]

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

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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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.25169, size = 818, normalized size = 8.1 \begin{align*} \frac{\sqrt{2}{\left (32 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 24 i \, a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + 12 \, \sqrt{2} \sqrt{-\frac{a^{5}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 i \, \sqrt{2} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\right ) - 12 \, \sqrt{2} \sqrt{-\frac{a^{5}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (-4 i \, \sqrt{2} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\right )}{6 \,{\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((a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/6*(sqrt(2)*(32*I*a^2*e^(2*I*d*x + 2*I*c) + 24*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 12*
sqrt(2)*sqrt(-a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(4*I*sqrt(2)*sqrt(-a^5/d^2)*d*e^(2*I*d*x + 2*I*c) +
4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*
I*c)/a^2) - 12*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(-4*I*sqrt(2)*sqrt(-a^5/d^2)*d*e^(2*
I*d*x + 2*I*c) + 4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*
e^(-2*I*d*x - 2*I*c)/a^2))/(d*e^(2*I*d*x + 2*I*c) + 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))**(5/2),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}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2), x)