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

Optimal. Leaf size=110 \[ \frac{(5 B+3 i A) \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (x)}{\sqrt{2} \sqrt{a+i a \sinh (x)}}\right )}{16 \sqrt{2} a^{5/2}}+\frac{(5 B+3 i A) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac{(-B+i A) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}} \]

[Out]

(((3*I)*A + 5*B)*ArcTanh[(Sqrt[a]*Cosh[x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[x]])])/(16*Sqrt[2]*a^(5/2)) + ((I*A - B)
*Cosh[x])/(4*(a + I*a*Sinh[x])^(5/2)) + (((3*I)*A + 5*B)*Cosh[x])/(16*a*(a + I*a*Sinh[x])^(3/2))

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Rubi [A]  time = 0.100081, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2750, 2650, 2649, 206} \[ \frac{(5 B+3 i A) \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (x)}{\sqrt{2} \sqrt{a+i a \sinh (x)}}\right )}{16 \sqrt{2} a^{5/2}}+\frac{(5 B+3 i A) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac{(-B+i A) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)*A + 5*B)*ArcTanh[(Sqrt[a]*Cosh[x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[x]])])/(16*Sqrt[2]*a^(5/2)) + ((I*A - B)
*Cosh[x])/(4*(a + I*a*Sinh[x])^(5/2)) + (((3*I)*A + 5*B)*Cosh[x])/(16*a*(a + I*a*Sinh[x])^(3/2))

Rule 2750

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] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(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)]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx &=\frac{(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac{(3 A-5 i B) \int \frac{1}{(a+i a \sinh (x))^{3/2}} \, dx}{8 a}\\ &=\frac{(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac{(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac{(3 A-5 i B) \int \frac{1}{\sqrt{a+i a \sinh (x)}} \, dx}{32 a^2}\\ &=\frac{(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac{(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac{(3 i A+5 B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cosh (x)}{\sqrt{a+i a \sinh (x)}}\right )}{16 a^2}\\ &=\frac{(3 i A+5 B) \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (x)}{\sqrt{2} \sqrt{a+i a \sinh (x)}}\right )}{16 \sqrt{2} a^{5/2}}+\frac{(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac{(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.211919, size = 184, normalized size = 1.67 \[ \frac{\left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right ) \left (8 (A+i B) \sinh \left (\frac{x}{2}\right )+2 (5 B+3 i A) \sinh \left (\frac{x}{2}\right ) (\sinh (x)-i)+(5 B+3 i A) \left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right )^3+4 i (A+i B) \left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right )+(1-i) \sqrt [4]{-1} (3 A-5 i B) \tan ^{-1}\left (\frac{\tanh \left (\frac{x}{4}\right )+i}{\sqrt{2}}\right ) \left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right )^4\right )}{16 (a+i a \sinh (x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cosh[x/2] + I*Sinh[x/2])*((4*I)*(A + I*B)*(Cosh[x/2] + I*Sinh[x/2]) + ((3*I)*A + 5*B)*(Cosh[x/2] + I*Sinh[x/
2])^3 + (1 - I)*(-1)^(1/4)*(3*A - (5*I)*B)*ArcTan[(I + Tanh[x/4])/Sqrt[2]]*(Cosh[x/2] + I*Sinh[x/2])^4 + 8*(A
+ I*B)*Sinh[x/2] + 2*((3*I)*A + 5*B)*Sinh[x/2]*(-I + Sinh[x])))/(16*(a + I*a*Sinh[x])^(5/2))

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Maple [F]  time = 0.092, size = 0, normalized size = 0. \begin{align*} \int{(A+B\sinh \left ( x \right ) ) \left ( a+ia\sinh \left ( x \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sinh \left (x\right ) + A}{{\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+B*sinh(x))/(a+I*a*sinh(x))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.94748, size = 1218, normalized size = 11.07 \begin{align*} \frac{8 \, \sqrt{\frac{1}{2}}{\left ({\left (-3 i \, A - 5 \, B\right )} e^{\left (4 \, x\right )} -{\left (11 \, A + 3 i \, B\right )} e^{\left (3 \, x\right )} +{\left (-11 i \, A + 3 \, B\right )} e^{\left (2 \, x\right )} -{\left (3 \, A - 5 i \, B\right )} e^{x}\right )} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a} e^{\left (-\frac{1}{2} \, x\right )} + \sqrt{\frac{1}{2}}{\left (4 \, a^{3} e^{\left (5 \, x\right )} - 20 i \, a^{3} e^{\left (4 \, x\right )} - 40 \, a^{3} e^{\left (3 \, x\right )} + 40 i \, a^{3} e^{\left (2 \, x\right )} + 20 \, a^{3} e^{x} - 4 i \, a^{3}\right )} \sqrt{-\frac{9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} \log \left (\frac{\sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a}{\left (3 i \, A + 5 \, B\right )} e^{\left (-\frac{1}{2} \, x\right )} + \sqrt{\frac{1}{2}}{\left (a^{3} e^{x} - i \, a^{3}\right )} \sqrt{-\frac{9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}}}{{\left (3 i \, A + 5 \, B\right )} e^{x} + 3 \, A - 5 i \, B}\right ) - \sqrt{\frac{1}{2}}{\left (4 \, a^{3} e^{\left (5 \, x\right )} - 20 i \, a^{3} e^{\left (4 \, x\right )} - 40 \, a^{3} e^{\left (3 \, x\right )} + 40 i \, a^{3} e^{\left (2 \, x\right )} + 20 \, a^{3} e^{x} - 4 i \, a^{3}\right )} \sqrt{-\frac{9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} \log \left (\frac{\sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a}{\left (3 i \, A + 5 \, B\right )} e^{\left (-\frac{1}{2} \, x\right )} - \sqrt{\frac{1}{2}}{\left (a^{3} e^{x} - i \, a^{3}\right )} \sqrt{-\frac{9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}}}{{\left (3 i \, A + 5 \, B\right )} e^{x} + 3 \, A - 5 i \, B}\right )}{8 \,{\left (8 \, a^{3} e^{\left (5 \, x\right )} - 40 i \, a^{3} e^{\left (4 \, x\right )} - 80 \, a^{3} e^{\left (3 \, x\right )} + 80 i \, a^{3} e^{\left (2 \, x\right )} + 40 \, a^{3} e^{x} - 8 i \, a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*sqrt(1/2)*((-3*I*A - 5*B)*e^(4*x) - (11*A + 3*I*B)*e^(3*x) + (-11*I*A + 3*B)*e^(2*x) - (3*A - 5*I*B)*e^
x)*sqrt(I*a*e^(2*x) + 2*a*e^x - I*a)*e^(-1/2*x) + sqrt(1/2)*(4*a^3*e^(5*x) - 20*I*a^3*e^(4*x) - 40*a^3*e^(3*x)
+ 40*I*a^3*e^(2*x) + 20*a^3*e^x - 4*I*a^3)*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)/a^5)*log((sqrt(1/2)*sqrt(I*a*e^(
2*x) + 2*a*e^x - I*a)*(3*I*A + 5*B)*e^(-1/2*x) + sqrt(1/2)*(a^3*e^x - I*a^3)*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)
/a^5))/((3*I*A + 5*B)*e^x + 3*A - 5*I*B)) - sqrt(1/2)*(4*a^3*e^(5*x) - 20*I*a^3*e^(4*x) - 40*a^3*e^(3*x) + 40*
I*a^3*e^(2*x) + 20*a^3*e^x - 4*I*a^3)*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)/a^5)*log((sqrt(1/2)*sqrt(I*a*e^(2*x) +
2*a*e^x - I*a)*(3*I*A + 5*B)*e^(-1/2*x) - sqrt(1/2)*(a^3*e^x - I*a^3)*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)/a^5))
/((3*I*A + 5*B)*e^x + 3*A - 5*I*B)))/(8*a^3*e^(5*x) - 40*I*a^3*e^(4*x) - 80*a^3*e^(3*x) + 80*I*a^3*e^(2*x) + 4
0*a^3*e^x - 8*I*a^3)

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

[Out]

Timed out

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

[Out]

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