3.122 \(\int \frac{1}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=104 \[ -\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{i}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{i}{3 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0608089, antiderivative size = 104, 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 = {3479, 3480, 206} \[ -\frac{i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{i}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{i}{3 d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(-3/2),x]

[Out]

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

Rule 3479

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

Mathematica [A]  time = 0.512256, size = 124, normalized size = 1.19 \[ \frac{5 e^{2 i (c+d x)}+4 e^{4 i (c+d x)}-3 e^{3 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+1}{3 a d \left (1+e^{2 i (c+d x)}\right )^2 (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + 5*E^((2*I)*(c + d*x)) + 4*E^((4*I)*(c + d*x)) - 3*E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSi
nh[E^(I*(c + d*x))])/(3*a*d*(1 + E^((2*I)*(c + d*x)))^2*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]  time = 0.017, size = 78, normalized size = 0.8 \begin{align*}{\frac{2\,ia}{d} \left ( -{\frac{\sqrt{2}}{8}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{5}{2}}}}+{\frac{1}{4\,{a}^{2}}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}+{\frac{1}{6\,a} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.2299, size = 817, normalized size = 7.86 \begin{align*} \frac{{\left (-3 i \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3 i \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{12 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(-3*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(4*I*d*x + 4*I*c)*log((2*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(2
*I*d*x + 2*I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d
*x - I*c)) + 3*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(4*I*d*x + 4*I*c)*log(-(2*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2
))*e^(2*I*d*x + 2*I*c) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*
e^(-I*d*x - I*c)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(4*I*e^(4*I*d*x + 4*I*c) + 5*I*e^(2*I*d*x + 2*I*
c) + I)*e^(I*d*x + I*c))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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