Integrand size = 20, antiderivative size = 79 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{3/2}} \, dx=\frac {(i A+3 B) \text {arctanh}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{2 \sqrt {2} a^{3/2}}+\frac {(i A-B) \cosh (x)}{2 (a+i a \sinh (x))^{3/2}} \] Output:
1/4*(I*A+3*B)*arctanh(1/2*a^(1/2)*cosh(x)*2^(1/2)/(a+I*a*sinh(x))^(1/2))*2 ^(1/2)/a^(3/2)+1/2*(I*A-B)*cosh(x)/(a+I*a*sinh(x))^(3/2)
Time = 0.34 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.33 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{3/2}} \, dx=\frac {\left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right ) \left (i (A+i B) \cosh \left (\frac {x}{2}\right )+(A+i B) \sinh \left (\frac {x}{2}\right )+(1+i) \sqrt [4]{-1} (A-3 i B) \arctan \left (\frac {i+\tanh \left (\frac {x}{4}\right )}{\sqrt {2}}\right ) (-i+\sinh (x))\right )}{2 (a+i a \sinh (x))^{3/2}} \] Input:
Integrate[(A + B*Sinh[x])/(a + I*a*Sinh[x])^(3/2),x]
Output:
((Cosh[x/2] + I*Sinh[x/2])*(I*(A + I*B)*Cosh[x/2] + (A + I*B)*Sinh[x/2] + (1 + I)*(-1)^(1/4)*(A - (3*I)*B)*ArcTan[(I + Tanh[x/4])/Sqrt[2]]*(-I + Sin h[x])))/(2*(a + I*a*Sinh[x])^(3/2))
Time = 0.32 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3229, 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.
| \(\displaystyle \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A-i B \sin (i x)}{(a+a \sin (i x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle \frac {(A-3 i B) \int \frac {1}{\sqrt {i \sinh (x) a+a}}dx}{4 a}+\frac {(-B+i A) \cosh (x)}{2 (a+i a \sinh (x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A-3 i B) \int \frac {1}{\sqrt {\sin (i x) a+a}}dx}{4 a}+\frac {(-B+i A) \cosh (x)}{2 (a+i a \sinh (x))^{3/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {i (A-3 i B) \int \frac {1}{2 a-\frac {a^2 \cosh ^2(x)}{i \sinh (x) a+a}}d\frac {a \cosh (x)}{\sqrt {i \sinh (x) a+a}}}{2 a}+\frac {(-B+i A) \cosh (x)}{2 (a+i a \sinh (x))^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {i (A-3 i B) \text {arctanh}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{2 \sqrt {2} a^{3/2}}+\frac {(-B+i A) \cosh (x)}{2 (a+i a \sinh (x))^{3/2}}\) |
Input:
Int[(A + B*Sinh[x])/(a + I*a*Sinh[x])^(3/2),x]
Output:
((I/2)*(A - (3*I)*B)*ArcTanh[(Sqrt[a]*Cosh[x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[ x]])])/(Sqrt[2]*a^(3/2)) + ((I*A - B)*Cosh[x])/(2*(a + I*a*Sinh[x])^(3/2))
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])
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]
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)]
\[\int \frac {A +B \sinh \left (x \right )}{\left (a +i a \sinh \left (x \right )\right )^{\frac {3}{2}}}d x\]
Input:
int((A+B*sinh(x))/(a+I*a*sinh(x))^(3/2),x)
Output:
int((A+B*sinh(x))/(a+I*a*sinh(x))^(3/2),x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (54) = 108\).
Time = 0.09 (sec) , antiderivative size = 264, normalized size of antiderivative = 3.34 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{3/2}} \, dx=\frac {\sqrt {\frac {1}{2}} {\left (a^{2} e^{\left (2 \, x\right )} - 2 i \, a^{2} e^{x} - a^{2}\right )} \sqrt {-\frac {A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}} \log \left (\frac {\sqrt {\frac {1}{2}} a^{2} \sqrt {-\frac {A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (i \, A + 3 \, B\right )}}{i \, A + 3 \, B}\right ) - \sqrt {\frac {1}{2}} {\left (a^{2} e^{\left (2 \, x\right )} - 2 i \, a^{2} e^{x} - a^{2}\right )} \sqrt {-\frac {A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}} \log \left (-\frac {\sqrt {\frac {1}{2}} a^{2} \sqrt {-\frac {A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}} - \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (i \, A + 3 \, B\right )}}{i \, A + 3 \, B}\right ) - 2 \, {\left ({\left (i \, A - B\right )} e^{\left (2 \, x\right )} - {\left (A + i \, B\right )} e^{x}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}}}{2 \, {\left (a^{2} e^{\left (2 \, x\right )} - 2 i \, a^{2} e^{x} - a^{2}\right )}} \] Input:
integrate((A+B*sinh(x))/(a+I*a*sinh(x))^(3/2),x, algorithm="fricas")
Output:
1/2*(sqrt(1/2)*(a^2*e^(2*x) - 2*I*a^2*e^x - a^2)*sqrt(-(A^2 - 6*I*A*B - 9* B^2)/a^3)*log((sqrt(1/2)*a^2*sqrt(-(A^2 - 6*I*A*B - 9*B^2)/a^3) + sqrt(1/2 *I*a*e^(-x))*(I*A + 3*B))/(I*A + 3*B)) - sqrt(1/2)*(a^2*e^(2*x) - 2*I*a^2* e^x - a^2)*sqrt(-(A^2 - 6*I*A*B - 9*B^2)/a^3)*log(-(sqrt(1/2)*a^2*sqrt(-(A ^2 - 6*I*A*B - 9*B^2)/a^3) - sqrt(1/2*I*a*e^(-x))*(I*A + 3*B))/(I*A + 3*B) ) - 2*((I*A - B)*e^(2*x) - (A + I*B)*e^x)*sqrt(1/2*I*a*e^(-x)))/(a^2*e^(2* x) - 2*I*a^2*e^x - a^2)
\[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{3/2}} \, dx=\int \frac {A + B \sinh {\left (x \right )}}{\left (i a \left (\sinh {\left (x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((A+B*sinh(x))/(a+I*a*sinh(x))**(3/2),x)
Output:
Integral((A + B*sinh(x))/(I*a*(sinh(x) - I))**(3/2), x)
\[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{3/2}} \, dx=\int { \frac {B \sinh \left (x\right ) + A}{{\left (i \, a \sinh \left (x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*sinh(x))/(a+I*a*sinh(x))^(3/2),x, algorithm="maxima")
Output:
integrate((B*sinh(x) + A)/(I*a*sinh(x) + a)^(3/2), x)
\[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{3/2}} \, dx=\int { \frac {B \sinh \left (x\right ) + A}{{\left (i \, a \sinh \left (x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+B*sinh(x))/(a+I*a*sinh(x))^(3/2),x, algorithm="giac")
Output:
integrate((B*sinh(x) + A)/(I*a*sinh(x) + a)^(3/2), x)
Timed out. \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{3/2}} \, dx=\int \frac {A+B\,\mathrm {sinh}\left (x\right )}{{\left (a+a\,\mathrm {sinh}\left (x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \] Input:
int((A + B*sinh(x))/(a + a*sinh(x)*1i)^(3/2),x)
Output:
int((A + B*sinh(x))/(a + a*sinh(x)*1i)^(3/2), x)
\[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sinh \left (x \right ) i +1}}{\sinh \left (x \right )^{3} i +\sinh \left (x \right )^{2}+\sinh \left (x \right ) i +1}d x \right ) a -\left (\int \frac {\sqrt {\sinh \left (x \right ) i +1}\, \sinh \left (x \right )^{2}}{\sinh \left (x \right )^{3} i +\sinh \left (x \right )^{2}+\sinh \left (x \right ) i +1}d x \right ) b i -\left (\int \frac {\sqrt {\sinh \left (x \right ) i +1}\, \sinh \left (x \right )}{\sinh \left (x \right )^{3} i +\sinh \left (x \right )^{2}+\sinh \left (x \right ) i +1}d x \right ) a i +\left (\int \frac {\sqrt {\sinh \left (x \right ) i +1}\, \sinh \left (x \right )}{\sinh \left (x \right )^{3} i +\sinh \left (x \right )^{2}+\sinh \left (x \right ) i +1}d x \right ) b \right )}{a^{2}} \] Input:
int((A+B*sinh(x))/(a+I*a*sinh(x))^(3/2),x)
Output:
(sqrt(a)*(int(sqrt(sinh(x)*i + 1)/(sinh(x)**3*i + sinh(x)**2 + sinh(x)*i + 1),x)*a - int((sqrt(sinh(x)*i + 1)*sinh(x)**2)/(sinh(x)**3*i + sinh(x)**2 + sinh(x)*i + 1),x)*b*i - int((sqrt(sinh(x)*i + 1)*sinh(x))/(sinh(x)**3*i + sinh(x)**2 + sinh(x)*i + 1),x)*a*i + int((sqrt(sinh(x)*i + 1)*sinh(x))/ (sinh(x)**3*i + sinh(x)**2 + sinh(x)*i + 1),x)*b))/a**2