\(\int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx\) [125]

Optimal result

Integrand size = 20, antiderivative size = 110 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\frac {(3 i A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac {(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}} \] Output:

1/32*(3*I*A+5*B)*arctanh(1/2*a^(1/2)*cosh(x)*2^(1/2)/(a+I*a*sinh(x))^(1/2) 
)*2^(1/2)/a^(5/2)+1/4*(I*A-B)*cosh(x)/(a+I*a*sinh(x))^(5/2)+1/16*(3*I*A+5* 
B)*cosh(x)/a/(a+I*a*sinh(x))^(3/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.31 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.67 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\frac {\left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right ) \left (4 i (A+i B) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )+(3 i A+5 B) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^3+(1-i) \sqrt [4]{-1} (3 A-5 i B) \arctan \left (\frac {i+\tanh \left (\frac {x}{4}\right )}{\sqrt {2}}\right ) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^4+8 (A+i B) \sinh \left (\frac {x}{2}\right )+2 (3 i A+5 B) \sinh \left (\frac {x}{2}\right ) (-i+\sinh (x))\right )}{16 (a+i a \sinh (x))^{5/2}} \] Input:

Integrate[(A + B*Sinh[x])/(a + I*a*Sinh[x])^(5/2),x]
 

Output:

((Cosh[x/2] + I*Sinh[x/2])*((4*I)*(A + I*B)*(Cosh[x/2] + I*Sinh[x/2]) + (( 
3*I)*A + 5*B)*(Cosh[x/2] + I*Sinh[x/2])^3 + (1 - I)*(-1)^(1/4)*(3*A - (5*I 
)*B)*ArcTan[(I + Tanh[x/4])/Sqrt[2]]*(Cosh[x/2] + I*Sinh[x/2])^4 + 8*(A + 
I*B)*Sinh[x/2] + 2*((3*I)*A + 5*B)*Sinh[x/2]*(-I + Sinh[x])))/(16*(a + I*a 
*Sinh[x])^(5/2))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 3229, 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.

\(\displaystyle \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A-i B \sin (i x)}{(a+a \sin (i x))^{5/2}}dx\)

\(\Big \downarrow \) 3229

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3129

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

Input:

Int[(A + B*Sinh[x])/(a + I*a*Sinh[x])^(5/2),x]
 

Output:

((I*A - B)*Cosh[x])/(4*(a + I*a*Sinh[x])^(5/2)) + ((3*A - (5*I)*B)*(((I/2) 
*ArcTanh[(Sqrt[a]*Cosh[x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[x]])])/(Sqrt[2]*a^(3 
/2)) + ((I/2)*Cosh[x])/(a + I*a*Sinh[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]
 

rule 3229

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* 
x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1))   I 
nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && 
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

Maple [F]

\[\int \frac {A +B \sinh \left (x \right )}{\left (a +i a \sinh \left (x \right )\right )^{\frac {5}{2}}}d x\]

Input:

int((A+B*sinh(x))/(a+I*a*sinh(x))^(5/2),x)
 

Output:

int((A+B*sinh(x))/(a+I*a*sinh(x))^(5/2),x)
 

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. 347 vs. \(2 (77) = 154\).

Time = 0.10 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.15 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\frac {\sqrt {\frac {1}{2}} {\left (a^{3} e^{\left (4 \, x\right )} - 4 i \, a^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 4 i \, a^{3} e^{x} + a^{3}\right )} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} \log \left (\frac {\sqrt {\frac {1}{2}} a^{3} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (3 i \, A + 5 \, B\right )}}{3 i \, A + 5 \, B}\right ) - \sqrt {\frac {1}{2}} {\left (a^{3} e^{\left (4 \, x\right )} - 4 i \, a^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 4 i \, a^{3} e^{x} + a^{3}\right )} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} \log \left (-\frac {\sqrt {\frac {1}{2}} a^{3} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} - \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (3 i \, A + 5 \, B\right )}}{3 i \, A + 5 \, B}\right ) + 2 \, {\left ({\left (-3 i \, A - 5 \, B\right )} e^{\left (4 \, x\right )} - {\left (11 \, A + 3 i \, B\right )} e^{\left (3 \, x\right )} + {\left (-11 i \, A + 3 \, B\right )} e^{\left (2 \, x\right )} - {\left (3 \, A - 5 i \, B\right )} e^{x}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}}}{16 \, {\left (a^{3} e^{\left (4 \, x\right )} - 4 i \, a^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 4 i \, a^{3} e^{x} + a^{3}\right )}} \] Input:

integrate((A+B*sinh(x))/(a+I*a*sinh(x))^(5/2),x, algorithm="fricas")
 

Output:

1/16*(sqrt(1/2)*(a^3*e^(4*x) - 4*I*a^3*e^(3*x) - 6*a^3*e^(2*x) + 4*I*a^3*e 
^x + a^3)*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)/a^5)*log((sqrt(1/2)*a^3*sqrt(- 
(9*A^2 - 30*I*A*B - 25*B^2)/a^5) + sqrt(1/2*I*a*e^(-x))*(3*I*A + 5*B))/(3* 
I*A + 5*B)) - sqrt(1/2)*(a^3*e^(4*x) - 4*I*a^3*e^(3*x) - 6*a^3*e^(2*x) + 4 
*I*a^3*e^x + a^3)*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)/a^5)*log(-(sqrt(1/2)*a 
^3*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)/a^5) - sqrt(1/2*I*a*e^(-x))*(3*I*A + 
5*B))/(3*I*A + 5*B)) + 2*((-3*I*A - 5*B)*e^(4*x) - (11*A + 3*I*B)*e^(3*x) 
+ (-11*I*A + 3*B)*e^(2*x) - (3*A - 5*I*B)*e^x)*sqrt(1/2*I*a*e^(-x)))/(a^3* 
e^(4*x) - 4*I*a^3*e^(3*x) - 6*a^3*e^(2*x) + 4*I*a^3*e^x + a^3)
 

Sympy [F(-1)]

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

integrate((A+B*sinh(x))/(a+I*a*sinh(x))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((A+B*sinh(x))/(a+I*a*sinh(x))^(5/2),x, algorithm="maxima")
 

Output:

integrate((B*sinh(x) + A)/(I*a*sinh(x) + a)^(5/2), x)
 

Giac [F]

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

integrate((A+B*sinh(x))/(a+I*a*sinh(x))^(5/2),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((B*sinh(x) + A)/(I*a*sinh(x) + a)^(5/2), x)
 

Mupad [F(-1)]

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

int((A + B*sinh(x))/(a + a*sinh(x)*1i)^(5/2),x)
 

Output:

int((A + B*sinh(x))/(a + a*sinh(x)*1i)^(5/2), x)
 

Reduce [F]

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

int((A+B*sinh(x))/(a+I*a*sinh(x))^(5/2),x)
 

Output:

( - (int(sinh(x)/(sqrt(sinh(x)*i + 1)*sinh(x)**2 - 2*sqrt(sinh(x)*i + 1)*s 
inh(x)*i - sqrt(sinh(x)*i + 1)),x)*b + int(1/(sqrt(sinh(x)*i + 1)*sinh(x)* 
*2 - 2*sqrt(sinh(x)*i + 1)*sinh(x)*i - sqrt(sinh(x)*i + 1)),x)*a))/(sqrt(a 
)*a**2)