Integrand size = 20, antiderivative size = 112 \[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\frac {64 a^3 (7 i A+5 B) \cosh (x)}{105 \sqrt {a+i a \sinh (x)}}+\frac {16}{105} a^2 (7 i A+5 B) \cosh (x) \sqrt {a+i a \sinh (x)}+\frac {2}{35} a (7 i A+5 B) \cosh (x) (a+i a \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2} \] Output:
64/105*a^3*(7*I*A+5*B)*cosh(x)/(a+I*a*sinh(x))^(1/2)+16/105*a^2*(7*I*A+5*B )*cosh(x)*(a+I*a*sinh(x))^(1/2)+2/35*a*(7*I*A+5*B)*cosh(x)*(a+I*a*sinh(x)) ^(3/2)+2/7*B*cosh(x)*(a+I*a*sinh(x))^(5/2)
Time = 2.68 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.89 \[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\frac {a^2 \left (\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )\right ) \sqrt {a+i a \sinh (x)} (1246 i A+1040 B+(-42 i A-120 B) \cosh (2 x)+(-392 A+505 i B) \sinh (x)-15 i B \sinh (3 x))}{210 \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )} \] Input:
Integrate[(a + I*a*Sinh[x])^(5/2)*(A + B*Sinh[x]),x]
Output:
(a^2*(Cosh[x/2] - I*Sinh[x/2])*Sqrt[a + I*a*Sinh[x]]*((1246*I)*A + 1040*B + ((-42*I)*A - 120*B)*Cosh[2*x] + (-392*A + (505*I)*B)*Sinh[x] - (15*I)*B* Sinh[3*x]))/(210*(Cosh[x/2] + I*Sinh[x/2]))
Time = 0.47 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3230, 3042, 3126, 3042, 3126, 3042, 3125}
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.
| \(\displaystyle \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+a \sin (i x))^{5/2} (A-i B \sin (i x))dx\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {1}{7} (7 A-5 i B) \int (i \sinh (x) a+a)^{5/2}dx+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 A-5 i B) \int (\sin (i x) a+a)^{5/2}dx+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {1}{7} (7 A-5 i B) \left (\frac {8}{5} a \int (i \sinh (x) a+a)^{3/2}dx+\frac {2}{5} i a \cosh (x) (a+i a \sinh (x))^{3/2}\right )+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 A-5 i B) \left (\frac {8}{5} a \int (\sin (i x) a+a)^{3/2}dx+\frac {2}{5} i a \cosh (x) (a+i a \sinh (x))^{3/2}\right )+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {1}{7} (7 A-5 i B) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {i \sinh (x) a+a}dx+\frac {2}{3} i a \cosh (x) \sqrt {a+i a \sinh (x)}\right )+\frac {2}{5} i a \cosh (x) (a+i a \sinh (x))^{3/2}\right )+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 A-5 i B) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin (i x) a+a}dx+\frac {2}{3} i a \cosh (x) \sqrt {a+i a \sinh (x)}\right )+\frac {2}{5} i a \cosh (x) (a+i a \sinh (x))^{3/2}\right )+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2}\) |
| \(\Big \downarrow \) 3125 |
| \(\displaystyle \frac {1}{7} (7 A-5 i B) \left (\frac {8}{5} a \left (\frac {8 i a^2 \cosh (x)}{3 \sqrt {a+i a \sinh (x)}}+\frac {2}{3} i a \cosh (x) \sqrt {a+i a \sinh (x)}\right )+\frac {2}{5} i a \cosh (x) (a+i a \sinh (x))^{3/2}\right )+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2}\) |
Input:
Int[(a + I*a*Sinh[x])^(5/2)*(A + B*Sinh[x]),x]
Output:
(2*B*Cosh[x]*(a + I*a*Sinh[x])^(5/2))/7 + ((7*A - (5*I)*B)*(((2*I)/5)*a*Co sh[x]*(a + I*a*Sinh[x])^(3/2) + (8*a*((((8*I)/3)*a^2*Cosh[x])/Sqrt[a + I*a *Sinh[x]] + ((2*I)/3)*a*Cosh[x]*Sqrt[a + I*a*Sinh[x]]))/5))/7
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[n - 1/2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
\[\int \left (a +i a \sinh \left (x \right )\right )^{\frac {5}{2}} \left (A +B \sinh \left (x \right )\right )d x\]
Input:
int((a+I*a*sinh(x))^(5/2)*(A+B*sinh(x)),x)
Output:
int((a+I*a*sinh(x))^(5/2)*(A+B*sinh(x)),x)
Time = 0.09 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.12 \[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=-\frac {1}{420} \, {\left (15 \, B a^{2} e^{\left (7 \, x\right )} + 21 \, {\left (2 \, A - 5 i \, B\right )} a^{2} e^{\left (6 \, x\right )} + 35 \, {\left (-10 i \, A - 11 \, B\right )} a^{2} e^{\left (5 \, x\right )} - 525 \, {\left (4 \, A - 3 i \, B\right )} a^{2} e^{\left (4 \, x\right )} + 525 \, {\left (-4 i \, A - 3 \, B\right )} a^{2} e^{\left (3 \, x\right )} - 35 \, {\left (10 \, A - 11 i \, B\right )} a^{2} e^{\left (2 \, x\right )} + 21 \, {\left (2 i \, A + 5 \, B\right )} a^{2} e^{x} - 15 i \, B a^{2}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} e^{\left (-3 \, x\right )} \] Input:
integrate((a+I*a*sinh(x))^(5/2)*(A+B*sinh(x)),x, algorithm="fricas")
Output:
-1/420*(15*B*a^2*e^(7*x) + 21*(2*A - 5*I*B)*a^2*e^(6*x) + 35*(-10*I*A - 11 *B)*a^2*e^(5*x) - 525*(4*A - 3*I*B)*a^2*e^(4*x) + 525*(-4*I*A - 3*B)*a^2*e ^(3*x) - 35*(10*A - 11*I*B)*a^2*e^(2*x) + 21*(2*I*A + 5*B)*a^2*e^x - 15*I* B*a^2)*sqrt(1/2*I*a*e^(-x))*e^(-3*x)
Timed out. \[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\text {Timed out} \] Input:
integrate((a+I*a*sinh(x))**(5/2)*(A+B*sinh(x)),x)
Output:
Timed out
\[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\int { {\left (B \sinh \left (x\right ) + A\right )} {\left (i \, a \sinh \left (x\right ) + a\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a+I*a*sinh(x))^(5/2)*(A+B*sinh(x)),x, algorithm="maxima")
Output:
integrate((B*sinh(x) + A)*(I*a*sinh(x) + a)^(5/2), x)
\[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\int { {\left (B \sinh \left (x\right ) + A\right )} {\left (i \, a \sinh \left (x\right ) + a\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a+I*a*sinh(x))^(5/2)*(A+B*sinh(x)),x, algorithm="giac")
Output:
integrate((B*sinh(x) + A)*(I*a*sinh(x) + a)^(5/2), x)
Timed out. \[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\int \left (A+B\,\mathrm {sinh}\left (x\right )\right )\,{\left (a+a\,\mathrm {sinh}\left (x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \] 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)
\[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sinh \left (x \right ) i +1}d x \right ) a -\left (\int \sqrt {\sinh \left (x \right ) i +1}\, \sinh \left (x \right )^{3}d x \right ) b -\left (\int \sqrt {\sinh \left (x \right ) i +1}\, \sinh \left (x \right )^{2}d x \right ) a +2 \left (\int \sqrt {\sinh \left (x \right ) i +1}\, \sinh \left (x \right )^{2}d x \right ) b i +2 \left (\int \sqrt {\sinh \left (x \right ) i +1}\, \sinh \left (x \right )d x \right ) a i +\left (\int \sqrt {\sinh \left (x \right ) i +1}\, \sinh \left (x \right )d x \right ) b \right ) \] Input:
int((a+I*a*sinh(x))^(5/2)*(A+B*sinh(x)),x)
Output:
sqrt(a)*a**2*(int(sqrt(sinh(x)*i + 1),x)*a - int(sqrt(sinh(x)*i + 1)*sinh( x)**3,x)*b - int(sqrt(sinh(x)*i + 1)*sinh(x)**2,x)*a + 2*int(sqrt(sinh(x)* i + 1)*sinh(x)**2,x)*b*i + 2*int(sqrt(sinh(x)*i + 1)*sinh(x),x)*a*i + int( sqrt(sinh(x)*i + 1)*sinh(x),x)*b)