\(\int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx\) [112]

Optimal result

Integrand size = 20, antiderivative size = 112 \[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\frac {64 a^3 (7 i A+5 B) \cosh (x)}{105 \sqrt {a+i a \sinh (x)}}+\frac {16}{105} a^2 (7 i A+5 B) \cosh (x) \sqrt {a+i a \sinh (x)}+\frac {2}{35} a (7 i A+5 B) \cosh (x) (a+i a \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2} \]

[Out]

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)+64/105*a^3*(7*I*A+5*B)*co
sh(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)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2830, 2726, 2725} \[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\frac {64 a^3 (5 B+7 i A) \cosh (x)}{105 \sqrt {a+i a \sinh (x)}}+\frac {16}{105} a^2 (5 B+7 i A) \cosh (x) \sqrt {a+i a \sinh (x)}+\frac {2}{35} a (5 B+7 i A) \cosh (x) (a+i a \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2} \]

[In]

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

[Out]

(64*a^3*((7*I)*A + 5*B)*Cosh[x])/(105*Sqrt[a + I*a*Sinh[x]]) + (16*a^2*((7*I)*A + 5*B)*Cosh[x]*Sqrt[a + I*a*Si
nh[x]])/105 + (2*a*((7*I)*A + 5*B)*Cosh[x]*(a + I*a*Sinh[x])^(3/2))/35 + (2*B*Cosh[x]*(a + I*a*Sinh[x])^(5/2))
/7

Rule 2725

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] && EqQ[a^2 - b^2, 0]

Rule 2726

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] + Dist[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]

Rule 2830

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] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[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)]

Rubi steps \begin{align*} \text {integral}& = \frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2}+\frac {1}{7} (7 A-5 i B) \int (a+i a \sinh (x))^{5/2} \, dx \\ & = \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}+\frac {1}{35} (8 a (7 A-5 i B)) \int (a+i a \sinh (x))^{3/2} \, dx \\ & = \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}+\frac {1}{105} \left (32 a^2 (7 A-5 i B)\right ) \int \sqrt {a+i a \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.89 \[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=\frac {a^2 \left (\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )\right ) \sqrt {a+i a \sinh (x)} (1246 i A+1040 B+(-42 i A-120 B) \cosh (2 x)+(-392 A+505 i B) \sinh (x)-15 i B \sinh (3 x))}{210 \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )} \]

[In]

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

[Out]

(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]))

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.12 \[ \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx=-\frac {1}{420} \, {\left (15 \, B a^{2} e^{\left (7 \, x\right )} + 21 \, {\left (2 \, A - 5 i \, B\right )} a^{2} e^{\left (6 \, x\right )} + 35 \, {\left (-10 i \, A - 11 \, B\right )} a^{2} e^{\left (5 \, x\right )} - 525 \, {\left (4 \, A - 3 i \, B\right )} a^{2} e^{\left (4 \, x\right )} + 525 \, {\left (-4 i \, A - 3 \, B\right )} a^{2} e^{\left (3 \, x\right )} - 35 \, {\left (10 \, A - 11 i \, B\right )} a^{2} e^{\left (2 \, x\right )} + 21 \, {\left (2 i \, A + 5 \, B\right )} a^{2} e^{x} - 15 i \, B a^{2}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} e^{\left (-3 \, x\right )} \]

[In]

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

[Out]

-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)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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