3.124 \(\int \frac{A+B \sinh (x)}{(a+i a \sinh (x))^{3/2}} \, dx\)

Optimal. Leaf size=79 \[ \frac{(3 B+i A) \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (x)}{\sqrt{2} \sqrt{a+i a \sinh (x)}}\right )}{2 \sqrt{2} a^{3/2}}+\frac{(-B+i A) \cosh (x)}{2 (a+i a \sinh (x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0768225, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2750, 2649, 206} \[ \frac{(3 B+i A) \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (x)}{\sqrt{2} \sqrt{a+i a \sinh (x)}}\right )}{2 \sqrt{2} a^{3/2}}+\frac{(-B+i A) \cosh (x)}{2 (a+i a \sinh (x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sinh[x])/(a + I*a*Sinh[x])^(3/2),x]

[Out]

((I*A + 3*B)*ArcTanh[(Sqrt[a]*Cosh[x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[x]])])/(2*Sqrt[2]*a^(3/2)) + ((I*A - B)*Cosh
[x])/(2*(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 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))^{3/2}} \, dx &=\frac{(i A-B) \cosh (x)}{2 (a+i a \sinh (x))^{3/2}}+\frac{(A-3 i B) \int \frac{1}{\sqrt{a+i a \sinh (x)}} \, dx}{4 a}\\ &=\frac{(i A-B) \cosh (x)}{2 (a+i a \sinh (x))^{3/2}}+\frac{(i A+3 B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cosh (x)}{\sqrt{a+i a \sinh (x)}}\right )}{2 a}\\ &=\frac{(i A+3 B) \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (x)}{\sqrt{2} \sqrt{a+i a \sinh (x)}}\right )}{2 \sqrt{2} a^{3/2}}+\frac{(i A-B) \cosh (x)}{2 (a+i a \sinh (x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.230316, size = 105, normalized size = 1.33 \[ \frac{\left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right ) \left ((A+i B) \sinh \left (\frac{x}{2}\right )+i (A+i B) \cosh \left (\frac{x}{2}\right )+(1+i) \sqrt [4]{-1} (A-3 i B) (\sinh (x)-i) \tan ^{-1}\left (\frac{\tanh \left (\frac{x}{4}\right )+i}{\sqrt{2}}\right )\right )}{2 (a+i a \sinh (x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sinh[x])/(a + I*a*Sinh[x])^(3/2),x]

[Out]

((Cosh[x/2] + I*Sinh[x/2])*(I*(A + I*B)*Cosh[x/2] + (A + I*B)*Sinh[x/2] + (1 + I)*(-1)^(1/4)*(A - (3*I)*B)*Arc
Tan[(I + Tanh[x/4])/Sqrt[2]]*(-I + Sinh[x])))/(2*(a + I*a*Sinh[x])^(3/2))

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Maple [F]  time = 0.088, size = 0, normalized size = 0. \begin{align*} \int{(A+B\sinh \left ( x \right ) ) \left ( a+ia\sinh \left ( x \right ) \right ) ^{-{\frac{3}{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))^(3/2),x)

[Out]

int((A+B*sinh(x))/(a+I*a*sinh(x))^(3/2),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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sinh(x) + A)/(I*a*sinh(x) + a)^(3/2), x)

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Fricas [B]  time = 1.93149, size = 936, normalized size = 11.85 \begin{align*} \frac{\sqrt{\frac{1}{2}}{\left ({\left (-2 i \, A + 2 \, B\right )} e^{\left (2 \, x\right )} + 2 \,{\left (A + 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 (a^{2} e^{\left (3 \, x\right )} - 3 i \, a^{2} e^{\left (2 \, x\right )} - 3 \, a^{2} e^{x} + i \, a^{2}\right )} \sqrt{-\frac{A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}} \log \left (\frac{\sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a}{\left (i \, A + 3 \, B\right )} e^{\left (-\frac{1}{2} \, x\right )} + \sqrt{\frac{1}{2}}{\left (a^{2} e^{x} - i \, a^{2}\right )} \sqrt{-\frac{A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}}}{{\left (i \, A + 3 \, B\right )} e^{x} + A - 3 i \, B}\right ) - \sqrt{\frac{1}{2}}{\left (a^{2} e^{\left (3 \, x\right )} - 3 i \, a^{2} e^{\left (2 \, x\right )} - 3 \, a^{2} e^{x} + i \, a^{2}\right )} \sqrt{-\frac{A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}} \log \left (\frac{\sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a}{\left (i \, A + 3 \, B\right )} e^{\left (-\frac{1}{2} \, x\right )} - \sqrt{\frac{1}{2}}{\left (a^{2} e^{x} - i \, a^{2}\right )} \sqrt{-\frac{A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}}}{{\left (i \, A + 3 \, B\right )} e^{x} + A - 3 i \, B}\right )}{2 \, a^{2} e^{\left (3 \, x\right )} - 6 i \, a^{2} e^{\left (2 \, x\right )} - 6 \, a^{2} e^{x} + 2 i \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(a+I*a*sinh(x))^(3/2),x, algorithm="fricas")

[Out]

(sqrt(1/2)*((-2*I*A + 2*B)*e^(2*x) + 2*(A + I*B)*e^x)*sqrt(I*a*e^(2*x) + 2*a*e^x - I*a)*e^(-1/2*x) + sqrt(1/2)
*(a^2*e^(3*x) - 3*I*a^2*e^(2*x) - 3*a^2*e^x + I*a^2)*sqrt(-(A^2 - 6*I*A*B - 9*B^2)/a^3)*log((sqrt(1/2)*sqrt(I*
a*e^(2*x) + 2*a*e^x - I*a)*(I*A + 3*B)*e^(-1/2*x) + sqrt(1/2)*(a^2*e^x - I*a^2)*sqrt(-(A^2 - 6*I*A*B - 9*B^2)/
a^3))/((I*A + 3*B)*e^x + A - 3*I*B)) - sqrt(1/2)*(a^2*e^(3*x) - 3*I*a^2*e^(2*x) - 3*a^2*e^x + I*a^2)*sqrt(-(A^
2 - 6*I*A*B - 9*B^2)/a^3)*log((sqrt(1/2)*sqrt(I*a*e^(2*x) + 2*a*e^x - I*a)*(I*A + 3*B)*e^(-1/2*x) - sqrt(1/2)*
(a^2*e^x - I*a^2)*sqrt(-(A^2 - 6*I*A*B - 9*B^2)/a^3))/((I*A + 3*B)*e^x + A - 3*I*B)))/(2*a^2*e^(3*x) - 6*I*a^2
*e^(2*x) - 6*a^2*e^x + 2*I*a^2)

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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))**(3/2),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{3}{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))^(3/2),x, algorithm="giac")

[Out]

integrate((B*sinh(x) + A)/(I*a*sinh(x) + a)^(3/2), x)