3.115 \(\int \frac{A+B \sinh (x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=23 \[ B x-\frac{(B+i A) \cosh (x)}{\sinh (x)+i} \]

[Out]

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

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Rubi [A]  time = 0.037834, antiderivative size = 23, normalized size of antiderivative = 1., 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 = {2735, 2648} \[ B x-\frac{(B+i A) \cosh (x)}{\sinh (x)+i} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sinh[x])/(I + Sinh[x]),x]

[Out]

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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]

Rubi steps

\begin{align*} \int \frac{A+B \sinh (x)}{i+\sinh (x)} \, dx &=B x-(-A+i B) \int \frac{1}{i+\sinh (x)} \, dx\\ &=B x-\frac{(i A+B) \cosh (x)}{i+\sinh (x)}\\ \end{align*}

Mathematica [B]  time = 0.23284, size = 53, normalized size = 2.3 \[ \cosh (x) \left (\frac{2 i B \sin ^{-1}\left (\frac{\sqrt{1-i \sinh (x)}}{\sqrt{2}}\right )}{\sqrt{\cosh ^2(x)}}-\frac{B+i A}{\sinh (x)+i}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Cosh[x]*(((2*I)*B*ArcSin[Sqrt[1 - I*Sinh[x]]/Sqrt[2]])/Sqrt[Cosh[x]^2] - (I*A + B)/(I + Sinh[x]))

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Maple [B]  time = 0.03, size = 46, normalized size = 2. \begin{align*} B\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) -2\,{\frac{B}{\tanh \left ( x/2 \right ) +i}}-{2\,iA \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}-B\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.26283, size = 35, normalized size = 1.52 \begin{align*} B{\left (x + \frac{2 i}{e^{\left (-x\right )} - i}\right )} - \frac{2 \, A}{e^{\left (-x\right )} - i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*(x + 2*I/(e^(-x) - I)) - 2*A/(e^(-x) - I)

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Fricas [A]  time = 1.8033, size = 58, normalized size = 2.52 \begin{align*} \frac{B x e^{x} + i \, B x - 2 \, A + 2 i \, B}{e^{x} + i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*x*e^x + I*B*x - 2*A + 2*I*B)/(e^x + I)

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Sympy [A]  time = 0.217398, size = 15, normalized size = 0.65 \begin{align*} B x + \frac{- 2 A + 2 i B}{e^{x} + i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x + (-2*A + 2*I*B)/(exp(x) + I)

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Giac [A]  time = 1.3178, size = 23, normalized size = 1. \begin{align*} B x - \frac{2 \,{\left (A - i \, B\right )}}{e^{x} + i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x - 2*(A - I*B)/(e^x + I)