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

Optimal result

Integrand size = 17, antiderivative size = 122 \[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {3 i \text {arctanh}\left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {2} \sqrt {a+i a \sinh (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}+\frac {3 i \cosh (c+d x)}{16 a d (a+i a \sinh (c+d x))^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.72 \[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {\left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (4 i \cosh \left (\frac {1}{2} (c+d x)\right )+(3-3 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1-i \tanh \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^4+4 \sinh \left (\frac {1}{2} (c+d x)\right )+6 \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2 \sinh \left (\frac {1}{2} (c+d x)\right )+3 \left (-i \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right )^3\right )}{16 d (a+i a \sinh (c+d x))^{5/2}} \] Input:

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

Output:

((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*(Cos 
h[(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))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 3129, 3042, 3129, 3042, 3128, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+a \sin (i c+i d x))^{5/2}}dx\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {3 \int \frac {1}{(i \sinh (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int \frac {1}{(\sin (i c+i d x) a+a)^{3/2}}dx}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {i \sinh (c+d x) a+a}}dx}{4 a}+\frac {i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}}\right )}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {\sin (i c+i d x) a+a}}dx}{4 a}+\frac {i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}}\right )}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {3 \left (\frac {i \int \frac {1}{2 a-\frac {a^2 \cosh ^2(c+d x)}{i \sinh (c+d x) a+a}}d\frac {a \cosh (c+d x)}{\sqrt {i \sinh (c+d x) a+a}}}{2 a d}+\frac {i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}}\right )}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {3 \left (\frac {i \text {arctanh}\left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {2} \sqrt {a+i a \sinh (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}}\right )}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\)

Input:

Int[(a + I*a*Sinh[c + d*x])^(-5/2),x]
 

Output:

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

Defintions of rubi rules used

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128

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

rule 3129

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] + Simp[(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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (97 ) = 194\).

Time = 0.27 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.79

method	result	size
default	\(-\frac {\sqrt {\sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{2} a}\, \left (3 \ln \left (\frac {2 \sqrt {-a}\, \sqrt {\sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{2} a}-2 a}{\cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}\right ) a \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{4}-3 \sqrt {\sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{2} a}\, \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{2} \sqrt {-a}-2 \sqrt {-a}\, \sqrt {\sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{2} a}\right ) \sqrt {2}}{32 a^{3} \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{3} \sqrt {-a}\, \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{2}}\, d}\)	\(218\)

Input:

int(1/(a+I*a*sinh(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/32*(sinh(1/2*c+1/4*I*Pi+1/2*d*x)^2*a)^(1/2)*(3*ln(2*((-a)^(1/2)*(sinh(1 
/2*c+1/4*I*Pi+1/2*d*x)^2*a)^(1/2)-a)/cosh(1/2*c+1/4*I*Pi+1/2*d*x))*a*cosh( 
1/2*c+1/4*I*Pi+1/2*d*x)^4-3*(sinh(1/2*c+1/4*I*Pi+1/2*d*x)^2*a)^(1/2)*cosh( 
1/2*c+1/4*I*Pi+1/2*d*x)^2*(-a)^(1/2)-2*(-a)^(1/2)*(sinh(1/2*c+1/4*I*Pi+1/2 
*d*x)^2*a)^(1/2))/a^3/cosh(1/2*c+1/4*I*Pi+1/2*d*x)^3/(-a)^(1/2)/sinh(1/2*c 
+1/4*I*Pi+1/2*d*x)*2^(1/2)/(a*cosh(1/2*c+1/4*I*Pi+1/2*d*x)^2)^(1/2)/d
 

Fricas [B] (verification not implemented)

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 = 0.10 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.85 \[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=-\frac {3 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (4 \, d x + 4 \, c\right )} - 4 \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} + 6 i \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{3} d e^{\left (d x + c\right )} - i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (\sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}}\right ) + 3 \, \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} - 6 i \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a^{3} d e^{\left (d x + c\right )} + i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (-\sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}}\right ) - 2 \, \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}} {\left (-3 i \, e^{\left (4 \, d x + 4 \, c\right )} - 11 \, e^{\left (3 \, d x + 3 \, c\right )} - 11 i \, e^{\left (2 \, d x + 2 \, c\right )} - 3 \, e^{\left (d x + c\right )}\right )}}{16 \, {\left (a^{3} d e^{\left (4 \, d x + 4 \, c\right )} - 4 i \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} - 6 \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} + 4 i \, a^{3} d e^{\left (d x + c\right )} + a^{3} d\right )}} \] Input:

integrate(1/(a+I*a*sinh(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

-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(-sqr 
t(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)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+I*a*sinh(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+I*a*sinh(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int \frac {1}{{\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=-\frac {\int \frac {1}{\sqrt {\sinh \left (d x +c \right ) i +1}\, \sinh \left (d x +c \right )^{2}-2 \sqrt {\sinh \left (d x +c \right ) i +1}\, \sinh \left (d x +c \right ) i -\sqrt {\sinh \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:

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

Output:

( - int(1/(sqrt(sinh(c + d*x)*i + 1)*sinh(c + d*x)**2 - 2*sqrt(sinh(c + d* 
x)*i + 1)*sinh(c + d*x)*i - sqrt(sinh(c + d*x)*i + 1)),x))/(sqrt(a)*a**2)