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]

1/2*(I*A-B)*cosh(x)/(a+I*a*sinh(x))^(3/2)+1/4*(I*A+3*B)*arctanh(1/2*cosh(x)*a^(1/2)*2^(1/2)/(a+I*a*sinh(x))^(1
/2))/a^(3/2)*2^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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 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])

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 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)]

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*}

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Mathematica [A]  time = 0.25, 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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fricas [B]  time = 0.58, size = 267, normalized size = 3.38 \[ \frac {\sqrt {\frac {1}{2}} {\left (a^{2} e^{\left (2 \, x\right )} - 2 i \, a^{2} e^{x} - a^{2}\right )} \sqrt {-\frac {A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}} \log \left (\frac {\sqrt {\frac {1}{2}} a^{2} \sqrt {-\frac {A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (i \, A + 3 \, B\right )}}{i \, A + 3 \, B}\right ) - \sqrt {\frac {1}{2}} {\left (a^{2} e^{\left (2 \, x\right )} - 2 i \, a^{2} e^{x} - a^{2}\right )} \sqrt {-\frac {A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}} \log \left (-\frac {\sqrt {\frac {1}{2}} a^{2} \sqrt {-\frac {A^{2} - 6 i \, A B - 9 \, B^{2}}{a^{3}}} - \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (i \, A + 3 \, B\right )}}{i \, A + 3 \, B}\right ) - {\left (2 \, {\left (i \, A - B\right )} e^{\left (2 \, x\right )} - {\left (2 \, A + 2 i \, B\right )} e^{x}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}}}{2 \, {\left (a^{2} e^{\left (2 \, x\right )} - 2 i \, a^{2} e^{x} - a^{2}\right )}} \]

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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sinh \relax (x) + A}{{\left (i \, a \sinh \relax (x) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {A +B \sinh \relax (x )}{\left (a +i a \sinh \relax (x )\right )^{\frac {3}{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sinh \relax (x) + A}{{\left (i \, a \sinh \relax (x) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\mathrm {sinh}\relax (x)}{{\left (a+a\,\mathrm {sinh}\relax (x)\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sinh(x))/(a + a*sinh(x)*1i)^(3/2),x)

[Out]

int((A + B*sinh(x))/(a + a*sinh(x)*1i)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \sinh {\relax (x )}}{\left (i a \left (\sinh {\relax (x )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

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]

Integral((A + B*sinh(x))/(I*a*(sinh(x) - I))**(3/2), x)

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