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(coth(d*x+c)*a^(1/2)/(a+I*a*csch(d*x+c))^(1/2))/a^(3/2)/d-1/2*coth(d*x+c)/d/(a+I*a*csch(d*x+c))^(3/2)
-5/4*arctanh(1/2*coth(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*csch(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 123, normalized size of antiderivative = 1.00, 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 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 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 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 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
]

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]

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*}

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Mathematica [B]  time = 2.45, 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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fricas [B]  time = 2.45, size = 886, normalized size = 7.20 \[ \text {result too large to display} \]

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^(2*d*x + 2*c) + 10*I*a^2*d*e^(d*x + c) - 5*a^2*d)*sqrt(1/(a^3*d^2))*log((4*sqrt(1/2)*(a
^2*d*e^(2*d*x + 2*c) - a^2*d)*sqrt(a/(e^(2*d*x + 2*c) - 1))*sqrt(1/(a^3*d^2)) + 2*a*e^(d*x + c) - 2*I*a)*e^(-d
*x - c)) - sqrt(1/2)*(5*a^2*d*e^(2*d*x + 2*c) + 10*I*a^2*d*e^(d*x + c) - 5*a^2*d)*sqrt(1/(a^3*d^2))*log(-(4*sq
rt(1/2)*(a^2*d*e^(2*d*x + 2*c) - a^2*d)*sqrt(a/(e^(2*d*x + 2*c) - 1))*sqrt(1/(a^3*d^2)) - 2*a*e^(d*x + c) + 2*
I*a)*e^(-d*x - c)) - (2*a^2*d*e^(2*d*x + 2*c) + 4*I*a^2*d*e^(d*x + c) - 2*a^2*d)*sqrt(1/(a^3*d^2))*log((2*(a*d
*e^(2*d*x + 2*c) - a*d)*sqrt(a/(e^(2*d*x + 2*c) - 1))*sqrt(1/(a^3*d^2)) + 2*e^(d*x + c) + 2*I)*e^(-d*x - c)/(a
*d)) + (2*a^2*d*e^(2*d*x + 2*c) + 4*I*a^2*d*e^(d*x + c) - 2*a^2*d)*sqrt(1/(a^3*d^2))*log(-(2*(a*d*e^(2*d*x + 2
*c) - a*d)*sqrt(a/(e^(2*d*x + 2*c) - 1))*sqrt(1/(a^3*d^2)) - 2*e^(d*x + c) - 2*I)*e^(-d*x - c)/(a*d)) - (2*a^2
*d*e^(2*d*x + 2*c) + 4*I*a^2*d*e^(d*x + c) - 2*a^2*d)*sqrt(1/(a^3*d^2))*log(((2*a^2*d*e^(2*d*x + 2*c) - 2*I*a^
2*d*e^(d*x + c) - 4*a^2*d)*sqrt(1/(a^3*d^2)) + sqrt(a/(e^(2*d*x + 2*c) - 1))*(2*e^(3*d*x + 3*c) - 4*I*e^(2*d*x
+ 2*c) - 2*e^(d*x + c) + 4*I))*e^(-2*d*x - 2*c)/(a*d)) + (2*a^2*d*e^(2*d*x + 2*c) + 4*I*a^2*d*e^(d*x + c) - 2
*a^2*d)*sqrt(1/(a^3*d^2))*log(-((2*a^2*d*e^(2*d*x + 2*c) - 2*I*a^2*d*e^(d*x + c) - 4*a^2*d)*sqrt(1/(a^3*d^2))
- sqrt(a/(e^(2*d*x + 2*c) - 1))*(2*e^(3*d*x + 3*c) - 4*I*e^(2*d*x + 2*c) - 2*e^(d*x + c) + 4*I))*e^(-2*d*x - 2
*c)/(a*d)) + sqrt(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^(2*d*x + 2*c) + 8*I*a^2*d*e^(d*x + c) - 4*a^2*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a \operatorname {csch}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [F]  time = 0.63, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +i a \,\mathrm {csch}\left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a \operatorname {csch}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+\frac {a\,1{}\mathrm {i}}{\mathrm {sinh}\left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (i a \operatorname {csch}{\left (c + d x \right )} + a\right )^{\frac {3}{2}}}\, dx \]

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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