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]

(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

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Rubi [A]  time = 0.100143, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, 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 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)]

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 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]

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*}

Mathematica [A]  time = 0.348767, 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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Maple [F]  time = 0.121, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\sinh \left ( x \right ) \right ) ^{{\frac{5}{2}}} \left ( A+B\sinh \left ( x \right ) \right ) \, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sinh \left (x\right ) + A\right )}{\left (i \, a \sinh \left (x\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

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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Fricas [A]  time = 1.81776, size = 414, normalized size = 3.7 \begin{align*} -\frac{\sqrt{\frac{1}{2}}{\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 )} -{\left (350 i \, A + 385 \, B\right )} a^{2} e^{\left (5 \, x\right )} - 525 \,{\left (4 \, A - 3 i \, B\right )} a^{2} e^{\left (4 \, x\right )} -{\left (2100 i \, A + 1575 \, B\right )} a^{2} e^{\left (3 \, x\right )} - 35 \,{\left (10 \, A - 11 i \, B\right )} a^{2} e^{\left (2 \, x\right )} -{\left (-42 i \, A - 105 \, B\right )} a^{2} e^{x} - 15 i \, B a^{2}\right )} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a} e^{\left (-\frac{1}{2} \, x\right )}}{420 \,{\left (e^{\left (4 \, x\right )} - i \, e^{\left (3 \, x\right )}\right )}} \end{align*}

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*sqrt(1/2)*(15*B*a^2*e^(7*x) + 21*(2*A - 5*I*B)*a^2*e^(6*x) - (350*I*A + 385*B)*a^2*e^(5*x) - 525*(4*A -
3*I*B)*a^2*e^(4*x) - (2100*I*A + 1575*B)*a^2*e^(3*x) - 35*(10*A - 11*I*B)*a^2*e^(2*x) - (-42*I*A - 105*B)*a^2
*e^x - 15*I*B*a^2)*sqrt(I*a*e^(2*x) + 2*a*e^x - I*a)*e^(-1/2*x)/(e^(4*x) - I*e^(3*x))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sinh \left (x\right ) + A\right )}{\left (i \, a \sinh \left (x\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

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)