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

Optimal. Leaf size=87 \[ \frac{i \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{2} \sqrt{a+i a \sinh (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}} \]

[Out]

((I/2)*ArcTanh[(Sqrt[a]*Cosh[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) + ((I/2)*Co
sh[c + d*x])/(d*(a + I*a*Sinh[c + d*x])^(3/2))

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Rubi [A]  time = 0.044912, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2650, 2649, 206} \[ \frac{i \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{2} \sqrt{a+i a \sinh (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Sinh[c + d*x])^(-3/2),x]

[Out]

((I/2)*ArcTanh[(Sqrt[a]*Cosh[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) + ((I/2)*Co
sh[c + d*x])/(d*(a + I*a*Sinh[c + d*x])^(3/2))

Rule 2650

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

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[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 \sinh (c+d x))^{3/2}} \, dx &=\frac{i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}}+\frac{\int \frac{1}{\sqrt{a+i a \sinh (c+d x)}} \, dx}{4 a}\\ &=\frac{i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}}+\frac{i \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cosh (c+d x)}{\sqrt{a+i a \sinh (c+d x)}}\right )}{2 a d}\\ &=\frac{i \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{2} \sqrt{a+i a \sinh (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.225912, size = 156, normalized size = 1.79 \[ \frac{\left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )-i \left (\sinh \left (\frac{1}{2} (c+d x)\right )+(1-i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (1-i \tanh \left (\frac{1}{4} (c+d x)\right )\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2\right )\right )}{2 a d (\sinh (c+d x)-i) \sqrt{a+i a \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*(Cosh[(c + d*x)/2] - I*((1 - I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^
(1/4)*(1 - I*Tanh[(c + d*x)/4])]*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2 + Sinh[(c + d*x)/2])))/(2*a*d*(-I
+ Sinh[c + d*x])*Sqrt[a + I*a*Sinh[c + d*x]])

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Maple [F]  time = 0.118, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\sinh \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a \sinh \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*sinh(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.19344, size = 1044, normalized size = 12. \begin{align*} \frac{\sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - i \, a}{\left (-2 i \, e^{\left (2 \, d x + 2 \, c\right )} + 2 \, e^{\left (d x + c\right )}\right )} e^{\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c\right )} + \sqrt{\frac{1}{2}}{\left (i \, a^{2} d e^{\left (3 \, d x + 3 \, c\right )} + 3 \, a^{2} d e^{\left (2 \, d x + 2 \, c\right )} - 3 i \, a^{2} d e^{\left (d x + c\right )} - a^{2} d\right )} \sqrt{\frac{1}{a^{3} d^{2}}} \log \left (\frac{\sqrt{\frac{1}{2}}{\left (a^{2} d e^{\left (d x + c\right )} - i \, a^{2} d\right )} \sqrt{\frac{1}{a^{3} d^{2}}} + \sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - i \, a} e^{\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c\right )}}{e^{\left (d x + c\right )} - i}\right ) + \sqrt{\frac{1}{2}}{\left (-i \, a^{2} d e^{\left (3 \, d x + 3 \, c\right )} - 3 \, a^{2} d e^{\left (2 \, d x + 2 \, c\right )} + 3 i \, a^{2} d e^{\left (d x + c\right )} + a^{2} d\right )} \sqrt{\frac{1}{a^{3} d^{2}}} \log \left (-\frac{\sqrt{\frac{1}{2}}{\left (a^{2} d e^{\left (d x + c\right )} - i \, a^{2} d\right )} \sqrt{\frac{1}{a^{3} d^{2}}} - \sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - i \, a} e^{\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c\right )}}{e^{\left (d x + c\right )} - i}\right )}{2 \, a^{2} d e^{\left (3 \, d x + 3 \, c\right )} - 6 i \, a^{2} d e^{\left (2 \, d x + 2 \, c\right )} - 6 \, a^{2} d e^{\left (d x + c\right )} + 2 i \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(1/2)*sqrt(I*a*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) - I*a)*(-2*I*e^(2*d*x + 2*c) + 2*e^(d*x + c))*e^(-1/2*d*
x - 1/2*c) + sqrt(1/2)*(I*a^2*d*e^(3*d*x + 3*c) + 3*a^2*d*e^(2*d*x + 2*c) - 3*I*a^2*d*e^(d*x + c) - a^2*d)*sqr
t(1/(a^3*d^2))*log((sqrt(1/2)*(a^2*d*e^(d*x + c) - I*a^2*d)*sqrt(1/(a^3*d^2)) + sqrt(1/2)*sqrt(I*a*e^(2*d*x +
2*c) + 2*a*e^(d*x + c) - I*a)*e^(-1/2*d*x - 1/2*c))/(e^(d*x + c) - I)) + sqrt(1/2)*(-I*a^2*d*e^(3*d*x + 3*c) -
3*a^2*d*e^(2*d*x + 2*c) + 3*I*a^2*d*e^(d*x + c) + a^2*d)*sqrt(1/(a^3*d^2))*log(-(sqrt(1/2)*(a^2*d*e^(d*x + c)
- I*a^2*d)*sqrt(1/(a^3*d^2)) - sqrt(1/2)*sqrt(I*a*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) - I*a)*e^(-1/2*d*x - 1/2*
c))/(e^(d*x + c) - I)))/(2*a^2*d*e^(3*d*x + 3*c) - 6*I*a^2*d*e^(2*d*x + 2*c) - 6*a^2*d*e^(d*x + c) + 2*I*a^2*d
)

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

[Out]

Integral((I*a*sinh(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 \sinh \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*sinh(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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