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]

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)

________________________________________________________________________________________

Rubi [A]  time = 0.07, 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 = {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 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])

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 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]

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 = 0.20, 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))

________________________________________________________________________________________

fricas [B]  time = 0.53, size = 348, normalized size = 2.85 \[ \frac {\sqrt {\frac {1}{2}} {\left (12 i \, a^{3} d e^{\left (4 \, d x + 4 \, c\right )} + 48 \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} - 72 i \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} - 48 \, a^{3} d e^{\left (d x + c\right )} + 12 i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (\sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}}\right ) + \sqrt {\frac {1}{2}} {\left (-12 i \, a^{3} d e^{\left (4 \, d x + 4 \, c\right )} - 48 \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} + 72 i \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} + 48 \, a^{3} d e^{\left (d x + c\right )} - 12 i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (-\sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}}\right ) + 8 \, \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}} {\left (-3 i \, e^{\left (4 \, d x + 4 \, c\right )} - 11 \, e^{\left (3 \, d x + 3 \, c\right )} - 11 i \, e^{\left (2 \, d x + 2 \, c\right )} - 3 \, e^{\left (d x + c\right )}\right )}}{8 \, {\left (8 \, a^{3} d e^{\left (4 \, d x + 4 \, c\right )} - 32 i \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} - 48 \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} + 32 i \, a^{3} d e^{\left (d x + c\right )} + 8 \, a^{3} d\right )}} \]

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*(sqrt(1/2)*(12*I*a^3*d*e^(4*d*x + 4*c) + 48*a^3*d*e^(3*d*x + 3*c) - 72*I*a^3*d*e^(2*d*x + 2*c) - 48*a^3*d*
e^(d*x + c) + 12*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))
) + sqrt(1/2)*(-12*I*a^3*d*e^(4*d*x + 4*c) - 48*a^3*d*e^(3*d*x + 3*c) + 72*I*a^3*d*e^(2*d*x + 2*c) + 48*a^3*d*
e^(d*x + c) - 12*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)
)) + 8*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)))/(8*a^3*d*e^(4*d*x + 4*c) - 32*I*a^3*d*e^(3*d*x + 3*c) - 48*a^3*d*e^(2*d*x + 2*c) + 32*I*a^3*d*e^(d*x +
c) + 8*a^3*d)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

maple [F]  time = 0.12, 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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________