3.1.71 \(\int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx\) [71]

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

[Out]

1/4*I*cosh(d*x+c)/d/(a+I*a*sinh(d*x+c))^(5/2)+3/16*I*cosh(d*x+c)/a/d/(a+I*a*sinh(d*x+c))^(3/2)+3/32*I*arctanh(
1/2*cosh(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*sinh(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 122, normalized size of antiderivative = 1.00, 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 = {2729, 2728, 212} \begin {gather*} \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}} \end {gather*}

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 212

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

Rule 2728

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 2729

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]

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) \text {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*}

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Mathematica [A]
time = 0.14, size = 210, normalized size = 1.72 \begin {gather*} \frac {\left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (4 i \cosh \left (\frac {1}{2} (c+d x)\right )+(3-3 i) \sqrt [4]{-1} \text {ArcTan}\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+4 \sinh \left (\frac {1}{2} (c+d x)\right )+6 \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2 \sinh \left (\frac {1}{2} (c+d x)\right )+3 \left (-i \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right )^3\right )}{16 d (a+i a \sinh (c+d x))^{5/2}} \end {gather*}

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 = 1.70, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (91) = 182\).
time = 0.42, size = 348, normalized size = 2.85 \begin {gather*} -\frac {3 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (4 \, d x + 4 \, c\right )} - 4 \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} + 6 i \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{3} d e^{\left (d x + c\right )} - i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (\sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}}\right ) + 3 \, \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} - 6 i \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a^{3} d e^{\left (d x + c\right )} + i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (-\sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}}\right ) - 2 \, \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}} {\left (-3 i \, e^{\left (4 \, d x + 4 \, c\right )} - 11 \, e^{\left (3 \, d x + 3 \, c\right )} - 11 i \, e^{\left (2 \, d x + 2 \, c\right )} - 3 \, e^{\left (d x + c\right )}\right )}}{16 \, {\left (a^{3} d e^{\left (4 \, d x + 4 \, c\right )} - 4 i \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} - 6 \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} + 4 i \, a^{3} d e^{\left (d x + c\right )} + a^{3} d\right )}} \end {gather*}

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]

-1/16*(3*sqrt(1/2)*(-I*a^3*d*e^(4*d*x + 4*c) - 4*a^3*d*e^(3*d*x + 3*c) + 6*I*a^3*d*e^(2*d*x + 2*c) + 4*a^3*d*e
^(d*x + c) - I*a^3*d)*sqrt(1/(a^5*d^2))*log(sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2)) + sqrt(1/2*I*a*e^(-d*x - c))) +
3*sqrt(1/2)*(I*a^3*d*e^(4*d*x + 4*c) + 4*a^3*d*e^(3*d*x + 3*c) - 6*I*a^3*d*e^(2*d*x + 2*c) - 4*a^3*d*e^(d*x +
c) + I*a^3*d)*sqrt(1/(a^5*d^2))*log(-sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2)) + sqrt(1/2*I*a*e^(-d*x - c))) - 2*sqrt(
1/2*I*a*e^(-d*x - c))*(-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)))/(a^3
*d*e^(4*d*x + 4*c) - 4*I*a^3*d*e^(3*d*x + 3*c) - 6*a^3*d*e^(2*d*x + 2*c) + 4*I*a^3*d*e^(d*x + c) + a^3*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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