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

Optimal. Leaf size=112 \[ \frac {64 a^3 (5 B+7 i A) \cosh (x)}{105 \sqrt {a+i a \sinh (x)}}+\frac {16}{105} a^2 (5 B+7 i A) \cosh (x) \sqrt {a+i a \sinh (x)}+\frac {2}{35} a (5 B+7 i A) \cosh (x) (a+i a \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2} \]

[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)

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Rubi [A]  time = 0.10, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2751, 2647, 2646} \[ \frac {64 a^3 (5 B+7 i A) \cosh (x)}{105 \sqrt {a+i a \sinh (x)}}+\frac {16}{105} a^2 (5 B+7 i A) \cosh (x) \sqrt {a+i a \sinh (x)}+\frac {2}{35} a (5 B+7 i A) \cosh (x) (a+i a \sinh (x))^{3/2}+\frac {2}{7} B \cosh (x) (a+i a \sinh (x))^{5/2} \]

Antiderivative was successfully verified.

[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 2646

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 2647

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] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2751

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*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)]

Rubi steps

\begin {align*} \int (a+i a \sinh (x))^{5/2} (A+B \sinh (x)) \, dx &=\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*}

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Mathematica [A]  time = 0.36, size = 100, normalized size = 0.89 \[ \frac {a^2 \sqrt {a+i a \sinh (x)} \left (\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )\right ) ((-392 A+505 i B) \sinh (x)+(-120 B-42 i A) \cosh (2 x)+1246 i A-15 i B \sinh (3 x)+1040 B)}{210 \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )} \]

Antiderivative was successfully verified.

[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]))

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fricas [A]  time = 0.48, size = 125, normalized size = 1.12 \[ -\frac {1}{420} \, {\left (15 \, B a^{2} e^{\left (7 \, x\right )} + {\left (42 \, A - 105 i \, B\right )} a^{2} e^{\left (6 \, x\right )} + 35 \, {\left (-10 i \, A - 11 \, B\right )} a^{2} e^{\left (5 \, x\right )} - {\left (2100 \, A - 1575 i \, B\right )} a^{2} e^{\left (4 \, x\right )} + 525 \, {\left (-4 i \, A - 3 \, B\right )} a^{2} e^{\left (3 \, x\right )} - {\left (350 \, A - 385 i \, B\right )} a^{2} e^{\left (2 \, x\right )} + 21 \, {\left (2 i \, A + 5 \, B\right )} a^{2} e^{x} - 15 i \, B a^{2}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} e^{\left (-3 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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) + (42*A - 105*I*B)*a^2*e^(6*x) + 35*(-10*I*A - 11*B)*a^2*e^(5*x) - (2100*A - 1575*I*B
)*a^2*e^(4*x) + 525*(-4*I*A - 3*B)*a^2*e^(3*x) - (350*A - 385*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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sinh \relax (x) + A\right )} {\left (i \, a \sinh \relax (x) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \left (a +i a \sinh \relax (x )\right )^{\frac {5}{2}} \left (A +B \sinh \relax (x )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sinh \relax (x) + A\right )} {\left (i \, a \sinh \relax (x) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+B\,\mathrm {sinh}\relax (x)\right )\,{\left (a+a\,\mathrm {sinh}\relax (x)\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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