3.51 \(\int (a+i a \text {csch}(c+d x))^{5/2} \, dx\)

Optimal. Leaf size=107 \[ \frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a+i a \text {csch}(c+d x)}}\right )}{d}+\frac {14 a^3 \coth (c+d x)}{3 d \sqrt {a+i a \text {csch}(c+d x)}}+\frac {2 a^2 \coth (c+d x) \sqrt {a+i a \text {csch}(c+d x)}}{3 d} \]

[Out]

2*a^(5/2)*arctanh(coth(d*x+c)*a^(1/2)/(a+I*a*csch(d*x+c))^(1/2))/d+14/3*a^3*coth(d*x+c)/d/(a+I*a*csch(d*x+c))^
(1/2)+2/3*a^2*coth(d*x+c)*(a+I*a*csch(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.13, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3775, 3915, 3774, 203, 3792} \[ \frac {14 a^3 \coth (c+d x)}{3 d \sqrt {a+i a \text {csch}(c+d x)}}+\frac {2 a^2 \coth (c+d x) \sqrt {a+i a \text {csch}(c+d x)}}{3 d}+\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a+i a \text {csch}(c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Csch[c + d*x])^(5/2),x]

[Out]

(2*a^(5/2)*ArcTanh[(Sqrt[a]*Coth[c + d*x])/Sqrt[a + I*a*Csch[c + d*x]]])/d + (14*a^3*Coth[c + d*x])/(3*d*Sqrt[
a + I*a*Csch[c + d*x]]) + (2*a^2*Coth[c + d*x]*Sqrt[a + I*a*Csch[c + d*x]])/(3*d)

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 3775

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[a/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 2)*(a*(n - 1) + b*(3*n - 4)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 3792

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

Rule 3915

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Dist[c, In
t[Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Sqrt[a + b*Csc[e + f*x]]*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 (a+i a \text {csch}(c+d x))^{5/2} \, dx &=\frac {2 a^2 \coth (c+d x) \sqrt {a+i a \text {csch}(c+d x)}}{3 d}+\frac {1}{3} (2 a) \int \sqrt {a+i a \text {csch}(c+d x)} \left (\frac {3 a}{2}+\frac {7}{2} i a \text {csch}(c+d x)\right ) \, dx\\ &=\frac {2 a^2 \coth (c+d x) \sqrt {a+i a \text {csch}(c+d x)}}{3 d}+\frac {1}{3} \left (7 i a^2\right ) \int \text {csch}(c+d x) \sqrt {a+i a \text {csch}(c+d x)} \, dx+a^2 \int \sqrt {a+i a \text {csch}(c+d x)} \, dx\\ &=\frac {14 a^3 \coth (c+d x)}{3 d \sqrt {a+i a \text {csch}(c+d x)}}+\frac {2 a^2 \coth (c+d x) \sqrt {a+i a \text {csch}(c+d x)}}{3 d}-\frac {\left (2 i a^3\right ) \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 )}{d}\\ &=\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \coth (c+d x)}{\sqrt {a+i a \text {csch}(c+d x)}}\right )}{d}+\frac {14 a^3 \coth (c+d x)}{3 d \sqrt {a+i a \text {csch}(c+d x)}}+\frac {2 a^2 \coth (c+d x) \sqrt {a+i a \text {csch}(c+d x)}}{3 d}\\ \end {align*}

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Mathematica [A]  time = 1.48, size = 136, normalized size = 1.27 \[ \frac {2 a^2 \sqrt {a+i a \text {csch}(c+d x)} \left (\coth (c+d x)+\frac {14 \sinh \left (\frac {1}{2} (c+d x)\right )}{\cosh \left (\frac {1}{2} (c+d x)\right )-i \sinh \left (\frac {1}{2} (c+d x)\right )}+\frac {3 (-1)^{3/4} \coth (c+d x) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\text {csch}(c+d x)+i}\right )}{(\text {csch}(c+d x)-i) \sqrt {\text {csch}(c+d x)+i}}-7 i\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*Sqrt[a + I*a*Csch[c + d*x]]*(-7*I + Coth[c + d*x] + (3*(-1)^(3/4)*ArcTanh[(-1)^(3/4)*Sqrt[I + Csch[c +
d*x]]]*Coth[c + d*x])/((-I + Csch[c + d*x])*Sqrt[I + Csch[c + d*x]]) + (14*Sinh[(c + d*x)/2])/(Cosh[(c + d*x)/
2] - I*Sinh[(c + d*x)/2])))/(3*d)

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fricas [B]  time = 0.62, size = 569, normalized size = 5.32 \[ \frac {6 \, \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 \, d x + 2 \, c\right )} - d\right )} \log \left (\frac {{\left (2 \, a^{3} e^{\left (d x + c\right )} + 2 i \, a^{3} + 2 \, \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 \, d x + 2 \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 \, d x + 2 \, c\right )} - 1}}\right )} e^{\left (-d x - c\right )}}{d}\right ) - 6 \, \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 \, d x + 2 \, c\right )} - d\right )} \log \left (\frac {{\left (2 \, a^{3} e^{\left (d x + c\right )} + 2 i \, a^{3} - 2 \, \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 \, d x + 2 \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 \, d x + 2 \, c\right )} - 1}}\right )} e^{\left (-d x - c\right )}}{d}\right ) + 6 \, \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 \, d x + 2 \, c\right )} - d\right )} \log \left (\frac {{\left (2 \, \sqrt {\frac {a^{5}}{d^{2}}} {\left (a d e^{\left (2 \, d x + 2 \, c\right )} - i \, a d e^{\left (d x + c\right )} - 2 \, a d\right )} + {\left (2 \, a^{3} e^{\left (3 \, d x + 3 \, c\right )} - 4 i \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a^{3} e^{\left (d x + c\right )} + 4 i \, a^{3}\right )} \sqrt {\frac {a}{e^{\left (2 \, d x + 2 \, c\right )} - 1}}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right ) - 6 \, \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 \, d x + 2 \, c\right )} - d\right )} \log \left (-\frac {{\left (2 \, \sqrt {\frac {a^{5}}{d^{2}}} {\left (a d e^{\left (2 \, d x + 2 \, c\right )} - i \, a d e^{\left (d x + c\right )} - 2 \, a d\right )} - {\left (2 \, a^{3} e^{\left (3 \, d x + 3 \, c\right )} - 4 i \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a^{3} e^{\left (d x + c\right )} + 4 i \, a^{3}\right )} \sqrt {\frac {a}{e^{\left (2 \, d x + 2 \, c\right )} - 1}}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right ) + {\left (64 \, a^{2} e^{\left (3 \, d x + 3 \, c\right )} - 48 i \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 48 \, a^{2} e^{\left (d x + c\right )} + 64 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 \, d x + 2 \, c\right )} - 1}}}{12 \, {\left (d e^{\left (2 \, d x + 2 \, c\right )} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*sqrt(a^5/d^2)*(d*e^(2*d*x + 2*c) - d)*log((2*a^3*e^(d*x + c) + 2*I*a^3 + 2*sqrt(a^5/d^2)*(d*e^(2*d*x +
2*c) - d)*sqrt(a/(e^(2*d*x + 2*c) - 1)))*e^(-d*x - c)/d) - 6*sqrt(a^5/d^2)*(d*e^(2*d*x + 2*c) - d)*log((2*a^3
*e^(d*x + c) + 2*I*a^3 - 2*sqrt(a^5/d^2)*(d*e^(2*d*x + 2*c) - d)*sqrt(a/(e^(2*d*x + 2*c) - 1)))*e^(-d*x - c)/d
) + 6*sqrt(a^5/d^2)*(d*e^(2*d*x + 2*c) - d)*log((2*sqrt(a^5/d^2)*(a*d*e^(2*d*x + 2*c) - I*a*d*e^(d*x + c) - 2*
a*d) + (2*a^3*e^(3*d*x + 3*c) - 4*I*a^3*e^(2*d*x + 2*c) - 2*a^3*e^(d*x + c) + 4*I*a^3)*sqrt(a/(e^(2*d*x + 2*c)
- 1)))*e^(-2*d*x - 2*c)/d) - 6*sqrt(a^5/d^2)*(d*e^(2*d*x + 2*c) - d)*log(-(2*sqrt(a^5/d^2)*(a*d*e^(2*d*x + 2*
c) - I*a*d*e^(d*x + c) - 2*a*d) - (2*a^3*e^(3*d*x + 3*c) - 4*I*a^3*e^(2*d*x + 2*c) - 2*a^3*e^(d*x + c) + 4*I*a
^3)*sqrt(a/(e^(2*d*x + 2*c) - 1)))*e^(-2*d*x - 2*c)/d) + (64*a^2*e^(3*d*x + 3*c) - 48*I*a^2*e^(2*d*x + 2*c) -
48*a^2*e^(d*x + c) + 64*I*a^2)*sqrt(a/(e^(2*d*x + 2*c) - 1)))/(d*e^(2*d*x + 2*c) - d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \operatorname {csch}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*csch(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.95, size = 0, normalized size = 0.00 \[ \int \left (a +i a \,\mathrm {csch}\left (d x +c \right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \operatorname {csch}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*csch(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {a\,1{}\mathrm {i}}{\mathrm {sinh}\left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i a \operatorname {csch}{\left (c + d x \right )} + a\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*csch(d*x+c))**(5/2),x)

[Out]

Integral((I*a*csch(c + d*x) + a)**(5/2), x)

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