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

Optimal. Leaf size=123 \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{a} \coth (c+d x)}{\sqrt{a+i a \text{csch}(c+d x)}}\right )}{a^{3/2} d}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{a} \coth (c+d x)}{\sqrt{2} \sqrt{a+i a \text{csch}(c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\coth (c+d x)}{2 d (a+i a \text{csch}(c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.145928, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {3777, 3920, 3774, 203, 3795} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{a} \coth (c+d x)}{\sqrt{a+i a \text{csch}(c+d x)}}\right )}{a^{3/2} d}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{a} \coth (c+d x)}{\sqrt{2} \sqrt{a+i a \text{csch}(c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{\coth (c+d x)}{2 d (a+i a \text{csch}(c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Csch[c + d*x])^(-3/2),x]

[Out]

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

Rule 3777

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

Rule 3920

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 3774

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

Rule 203

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

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rubi steps

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

Mathematica [B]  time = 2.36687, size = 327, normalized size = 2.66 \[ \frac{\left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \left (-8 \sqrt{i a (\text{csch}(c+d x)+i)} \tan ^{-1}\left (\frac{\sqrt{i a (\text{csch}(c+d x)+i)}}{\sqrt{a}}\right )+i \text{csch}(c+d x) \left (-8 \sqrt{i a (\text{csch}(c+d x)+i)} \tan ^{-1}\left (\frac{\sqrt{i a (\text{csch}(c+d x)+i)}}{\sqrt{a}}\right )+5 \sqrt{2} \sqrt{i a (\text{csch}(c+d x)+i)} \tan ^{-1}\left (\frac{\sqrt{i a (\text{csch}(c+d x)+i)}}{\sqrt{2} \sqrt{a}}\right )+2 \sqrt{a}\right )+5 \sqrt{2} \sqrt{i a (\text{csch}(c+d x)+i)} \tan ^{-1}\left (\frac{\sqrt{i a (\text{csch}(c+d x)+i)}}{\sqrt{2} \sqrt{a}}\right )-2 \sqrt{a}\right )}{4 a^{3/2} d (\text{csch}(c+d x)+i) \sqrt{a+i a \text{csch}(c+d x)} \left (\cosh \left (\frac{1}{2} (c+d x)\right )-i \sinh \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[a] - 8*ArcTan[Sqrt[I*a*(I + Csch[c + d*x])]/Sqrt[a]]*Sqrt[I*a*(I + Csch[c + d*x])] + 5*Sqrt[2]*ArcTa
n[Sqrt[I*a*(I + Csch[c + d*x])]/(Sqrt[2]*Sqrt[a])]*Sqrt[I*a*(I + Csch[c + d*x])] + I*Csch[c + d*x]*(2*Sqrt[a]
- 8*ArcTan[Sqrt[I*a*(I + Csch[c + d*x])]/Sqrt[a]]*Sqrt[I*a*(I + Csch[c + d*x])] + 5*Sqrt[2]*ArcTan[Sqrt[I*a*(I
+ Csch[c + d*x])]/(Sqrt[2]*Sqrt[a])]*Sqrt[I*a*(I + Csch[c + d*x])]))*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]
))/(4*a^(3/2)*d*(I + Csch[c + d*x])*Sqrt[a + I*a*Csch[c + d*x]]*(Cosh[(c + d*x)/2] - I*Sinh[(c + d*x)/2]))

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Maple [F]  time = 0.176, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia{\rm csch} \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*csch(d*x+c))^(3/2),x)

[Out]

int(1/(a+I*a*csch(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 \operatorname{csch}\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*csch(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 3.06548, size = 2372, normalized size = 19.28 \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*csch(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

(sqrt(1/2)*(5*a^2*d*e^(3*d*x + 3*c) + 15*I*a^2*d*e^(2*d*x + 2*c) - 15*a^2*d*e^(d*x + c) - 5*I*a^2*d)*sqrt(1/(a
^3*d^2))*log((sqrt(1/2)*(4*I*a^2*d*e^(2*d*x + 2*c) + 4*I*a^2*d)*sqrt(1/(a^3*d^2)) + sqrt((a*e^(2*d*x + 2*c) +
2*I*a*e^(d*x + c) - a)/(e^(2*d*x + 2*c) - 1))*(-4*I*e^(2*d*x + 2*c) + 4*I))/((10*I + 5)*e^(2*d*x + 2*c) + (10*
I - 20)*e^(d*x + c) - 10*I - 5)) - sqrt(1/2)*(5*a^2*d*e^(3*d*x + 3*c) + 15*I*a^2*d*e^(2*d*x + 2*c) - 15*a^2*d*
e^(d*x + c) - 5*I*a^2*d)*sqrt(1/(a^3*d^2))*log((sqrt(1/2)*(-4*I*a^2*d*e^(2*d*x + 2*c) - 4*I*a^2*d)*sqrt(1/(a^3
*d^2)) + sqrt((a*e^(2*d*x + 2*c) + 2*I*a*e^(d*x + c) - a)/(e^(2*d*x + 2*c) - 1))*(-4*I*e^(2*d*x + 2*c) + 4*I))
/((10*I + 5)*e^(2*d*x + 2*c) + (10*I - 20)*e^(d*x + c) - 10*I - 5)) - (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)*sqrt(1/(a^3*d^2))*log(((-(I - 4)*a^2*d*e^(3*d*x + 3*c) - (4*I
+ 1)*a^2*d)*sqrt(1/(a^3*d^2)) + sqrt((a*e^(2*d*x + 2*c) + 2*I*a*e^(d*x + c) - a)/(e^(2*d*x + 2*c) - 1))*((I -
4)*e^(3*d*x + 3*c) + (4*I + 1)*e^(2*d*x + 2*c) - (I - 4)*e^(d*x + c) - 4*I - 1))/((10*I + 24)*e^(2*d*x + 2*c)
+ (24*I - 10)*e^(d*x + c))) + (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)*sqrt(1/(a^3*d^2))*log((((I - 4)*a^2*d*e^(3*d*x + 3*c) + (4*I + 1)*a^2*d)*sqrt(1/(a^3*d^2)) + sqrt((a*e
^(2*d*x + 2*c) + 2*I*a*e^(d*x + c) - a)/(e^(2*d*x + 2*c) - 1))*((I - 4)*e^(3*d*x + 3*c) + (4*I + 1)*e^(2*d*x +
2*c) - (I - 4)*e^(d*x + c) - 4*I - 1))/((10*I + 24)*e^(2*d*x + 2*c) + (24*I - 10)*e^(d*x + c))) - sqrt((a*e^(
2*d*x + 2*c) + 2*I*a*e^(d*x + c) - a)/(e^(2*d*x + 2*c) - 1))*(2*e^(3*d*x + 3*c) - 2*I*e^(2*d*x + 2*c) - 2*e^(d
*x + c) + 2*I))/(4*a^2*d*e^(3*d*x + 3*c) + 12*I*a^2*d*e^(2*d*x + 2*c) - 12*a^2*d*e^(d*x + c) - 4*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 \operatorname{csch}{\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*csch(d*x+c))**(3/2),x)

[Out]

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

[Out]

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