3.71 \(\int \frac{1}{(a+i a \sinh (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=122 \[ \frac{3 i \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{2} \sqrt{a+i a \sinh (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{3 i \cosh (c+d x)}{16 a d (a+i a \sinh (c+d x))^{3/2}}+\frac{i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}} \]

[Out]

(((3*I)/16)*ArcTanh[(Sqrt[a]*Cosh[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[c + d*x]])])/(Sqrt[2]*a^(5/2)*d) + ((I/
4)*Cosh[c + d*x])/(d*(a + I*a*Sinh[c + d*x])^(5/2)) + (((3*I)/16)*Cosh[c + d*x])/(a*d*(a + I*a*Sinh[c + d*x])^
(3/2))

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Rubi [A]  time = 0.0681726, antiderivative size = 122, 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 = {2650, 2649, 206} \[ \frac{3 i \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{2} \sqrt{a+i a \sinh (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{3 i \cosh (c+d x)}{16 a d (a+i a \sinh (c+d x))^{3/2}}+\frac{i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Sinh[c + d*x])^(-5/2),x]

[Out]

(((3*I)/16)*ArcTanh[(Sqrt[a]*Cosh[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[c + d*x]])])/(Sqrt[2]*a^(5/2)*d) + ((I/
4)*Cosh[c + d*x])/(d*(a + I*a*Sinh[c + d*x])^(5/2)) + (((3*I)/16)*Cosh[c + d*x])/(a*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))^{5/2}} \, dx &=\frac{i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}+\frac{3 \int \frac{1}{(a+i a \sinh (c+d x))^{3/2}} \, dx}{8 a}\\ &=\frac{i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}+\frac{3 i \cosh (c+d x)}{16 a d (a+i a \sinh (c+d x))^{3/2}}+\frac{3 \int \frac{1}{\sqrt{a+i a \sinh (c+d x)}} \, dx}{32 a^2}\\ &=\frac{i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}+\frac{3 i \cosh (c+d x)}{16 a d (a+i a \sinh (c+d x))^{3/2}}+\frac{(3 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 )}{16 a^2 d}\\ &=\frac{3 i \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{2} \sqrt{a+i a \sinh (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}+\frac{3 i \cosh (c+d x)}{16 a d (a+i a \sinh (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.195971, size = 210, normalized size = 1.72 \[ \frac{\left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \left (4 \sinh \left (\frac{1}{2} (c+d x)\right )+4 i \cosh \left (\frac{1}{2} (c+d x)\right )+6 \sinh \left (\frac{1}{2} (c+d x)\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2+3 \left (\sinh \left (\frac{1}{2} (c+d x)\right )-i \cosh \left (\frac{1}{2} (c+d x)\right )\right )^3+(3-3 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (1-i \tanh \left (\frac{1}{4} (c+d x)\right )\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^4\right )}{16 d (a+i a \sinh (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*((4*I)*Cosh[(c + d*x)/2] + (3 - 3*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])^4 + 4*Sinh[(c + d*x)/2] + 6*(C
osh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2*Sinh[(c + d*x)/2] + 3*((-I)*Cosh[(c + d*x)/2] + Sinh[(c + d*x)/2])^3
))/(16*d*(a + I*a*Sinh[c + d*x])^(5/2))

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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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*sinh(d*x+c))^(5/2),x)

[Out]

int(1/(a+I*a*sinh(d*x+c))^(5/2),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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.21884, size = 1395, normalized size = 11.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*sinh(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/8*(8*sqrt(1/2)*sqrt(I*a*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) - I*a)*(-3*I*e^(4*d*x + 4*c) - 11*e^(3*d*x + 3*c)
- 11*I*e^(2*d*x + 2*c) - 3*e^(d*x + c))*e^(-1/2*d*x - 1/2*c) + sqrt(1/2)*(12*I*a^3*d*e^(5*d*x + 5*c) + 60*a^3*
d*e^(4*d*x + 4*c) - 120*I*a^3*d*e^(3*d*x + 3*c) - 120*a^3*d*e^(2*d*x + 2*c) + 60*I*a^3*d*e^(d*x + c) + 12*a^3*
d)*sqrt(1/(a^5*d^2))*log(1/8*(sqrt(1/2)*(8*a^3*d*e^(d*x + c) - 8*I*a^3*d)*sqrt(1/(a^5*d^2)) + 8*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)*(-12*I*a^3*
d*e^(5*d*x + 5*c) - 60*a^3*d*e^(4*d*x + 4*c) + 120*I*a^3*d*e^(3*d*x + 3*c) + 120*a^3*d*e^(2*d*x + 2*c) - 60*I*
a^3*d*e^(d*x + c) - 12*a^3*d)*sqrt(1/(a^5*d^2))*log(-1/8*(sqrt(1/2)*(8*a^3*d*e^(d*x + c) - 8*I*a^3*d)*sqrt(1/(
a^5*d^2)) - 8*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)))/(8*a^3*d*e^(5*d*x + 5*c) - 40*I*a^3*d*e^(4*d*x + 4*c) - 80*a^3*d*e^(3*d*x + 3*c) + 80*I*a^3*d*e^(2*d*x +
2*c) + 40*a^3*d*e^(d*x + c) - 8*I*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(1/(a+I*a*sinh(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 \frac{1}{{\left (i \, a \sinh \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(1/(a+I*a*sinh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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