3.116 \(\int \frac {A+B \sinh (x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=43 \[ -\frac {(A+2 i B) \cosh (x)}{3 (\sinh (x)+i)}-\frac {(B+i A) \cosh (x)}{3 (\sinh (x)+i)^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2750, 2648} \[ -\frac {(A+2 i B) \cosh (x)}{3 (\sinh (x)+i)}-\frac {(B+i A) \cosh (x)}{3 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*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)}{(i+\sinh (x))^2} \, dx &=-\frac {(i A+B) \cosh (x)}{3 (i+\sinh (x))^2}+\frac {1}{3} (-i A+2 B) \int \frac {1}{i+\sinh (x)} \, dx\\ &=-\frac {(i A+B) \cosh (x)}{3 (i+\sinh (x))^2}-\frac {(A+2 i B) \cosh (x)}{3 (i+\sinh (x))}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 32, normalized size = 0.74 \[ \frac {\cosh (x) (-(A+2 i B) \sinh (x)-2 i A+B)}{3 (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 46, normalized size = 1.07 \[ -\frac {6 \, B e^{\left (2 \, x\right )} + {\left (6 \, A + 6 i \, B\right )} e^{x} + 2 i \, A - 4 \, B}{3 \, e^{\left (3 \, x\right )} + 9 i \, e^{\left (2 \, x\right )} - 9 \, e^{x} - 3 i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(6*B*e^(2*x) + (6*A + 6*I*B)*e^x + 2*I*A - 4*B)/(3*e^(3*x) + 9*I*e^(2*x) - 9*e^x - 3*I)

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giac [A]  time = 0.18, size = 32, normalized size = 0.74 \[ -\frac {6 \, B e^{\left (2 \, x\right )} + 6 \, A e^{x} + 6 i \, B e^{x} + 2 i \, A - 4 \, B}{3 \, {\left (e^{x} + i\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*B*e^(2*x) + 6*A*e^x + 6*I*B*e^x + 2*I*A - 4*B)/(e^x + I)^3

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maple [A]  time = 0.04, size = 52, normalized size = 1.21 \[ -\frac {-2 i A -2 B}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\frac {2 A}{\tanh \left (\frac {x}{2}\right )+i}-\frac {2 \left (2 i B -2 A \right )}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-2*I*A-2*B)/(tanh(1/2*x)+I)^2-2*A/(tanh(1/2*x)+I)-2/3*(2*I*B-2*A)/(tanh(1/2*x)+I)^3

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maxima [B]  time = 0.33, size = 141, normalized size = 3.28 \[ -2 \, A {\left (\frac {3 \, e^{\left (-x\right )}}{9 \, e^{\left (-x\right )} + 9 i \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-3 \, x\right )} - 3 i} - \frac {i}{9 \, e^{\left (-x\right )} + 9 i \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-3 \, x\right )} - 3 i}\right )} + \frac {1}{2} \, B {\left (-\frac {12 i \, e^{\left (-x\right )}}{9 \, e^{\left (-x\right )} + 9 i \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-3 \, x\right )} - 3 i} + \frac {12 \, e^{\left (-2 \, x\right )}}{9 \, e^{\left (-x\right )} + 9 i \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-3 \, x\right )} - 3 i} - \frac {8}{9 \, e^{\left (-x\right )} + 9 i \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-3 \, x\right )} - 3 i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*A*(3*e^(-x)/(9*e^(-x) + 9*I*e^(-2*x) - 3*e^(-3*x) - 3*I) - I/(9*e^(-x) + 9*I*e^(-2*x) - 3*e^(-3*x) - 3*I))
+ 1/2*B*(-12*I*e^(-x)/(9*e^(-x) + 9*I*e^(-2*x) - 3*e^(-3*x) - 3*I) + 12*e^(-2*x)/(9*e^(-x) + 9*I*e^(-2*x) - 3*
e^(-3*x) - 3*I) - 8/(9*e^(-x) + 9*I*e^(-2*x) - 3*e^(-3*x) - 3*I))

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mupad [B]  time = 0.61, size = 39, normalized size = 0.91 \[ -\frac {\frac {2\,A}{3}+\frac {B\,4{}\mathrm {i}}{3}-{\mathrm {e}}^x\,\left (-2\,B+A\,2{}\mathrm {i}\right )-B\,{\mathrm {e}}^{2\,x}\,2{}\mathrm {i}}{{\left (-1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*A)/3 + (B*4i)/3 - exp(x)*(A*2i - 2*B) - B*exp(2*x)*2i)/(exp(x)*1i - 1)^3

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sympy [A]  time = 0.20, size = 51, normalized size = 1.19 \[ \frac {2 i A + 6 B e^{2 x} - 4 B + \left (6 A + 6 i B\right ) e^{x}}{- 3 e^{3 x} - 9 i e^{2 x} + 9 e^{x} + 3 i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*I*A + 6*B*exp(2*x) - 4*B + (6*A + 6*I*B)*exp(x))/(-3*exp(3*x) - 9*I*exp(2*x) + 9*exp(x) + 3*I)

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