3.3.47 \(\int \frac {A+B \cosh (x)}{i+\sinh (x)} \, dx\) [247]

Optimal. Leaf size=25 \[ B \log (i+\sinh (x))-\frac {A \cosh (x)}{1-i \sinh (x)} \]

[Out]

B*ln(I+sinh(x))-A*cosh(x)/(1-I*sinh(x))

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Rubi [A]
time = 0.06, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4486, 2727, 2746, 31} \begin {gather*} B \log (\sinh (x)+i)-\frac {A \cosh (x)}{1-i \sinh (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Cosh[x])/(I + Sinh[x]),x]

[Out]

B*Log[I + Sinh[x]] - (A*Cosh[x])/(1 - I*Sinh[x])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2727

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 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 4486

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 48, normalized size = 1.92 \begin {gather*} -2 i B \text {ArcTan}\left (\tanh \left (\frac {x}{2}\right )\right )+B \log (\cosh (x))-\frac {2 i A \sinh \left (\frac {x}{2}\right )}{\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*I)*B*ArcTan[Tanh[x/2]] + B*Log[Cosh[x]] - ((2*I)*A*Sinh[x/2])/(Cosh[x/2] - I*Sinh[x/2])

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Maple [A]
time = 0.59, size = 46, normalized size = 1.84

method	result	size
risch	\(-B x -\frac {2 A}{{\mathrm e}^{x}+i}+2 B \ln \left ({\mathrm e}^{x}+i\right )\)	\(25\)
default	\(-B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+2 B \ln \left (\tanh \left (\frac {x}{2}\right )+i\right )-\frac {2 i A}{\tanh \left (\frac {x}{2}\right )+i}-B \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cosh(x))/(I+sinh(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 19, normalized size = 0.76 \begin {gather*} B \log \left (\sinh \left (x\right ) + i\right ) - \frac {2 \, A}{e^{\left (-x\right )} - i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*log(sinh(x) + I) - 2*A/(e^(-x) - I)

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Fricas [A]
time = 0.41, size = 36, normalized size = 1.44 \begin {gather*} -\frac {B x e^{x} + i \, B x - 2 \, {\left (B e^{x} + i \, B\right )} \log \left (e^{x} + i\right ) + 2 \, A}{e^{x} + i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.08, size = 20, normalized size = 0.80 \begin {gather*} - \frac {2 A}{e^{x} + i} - B x + 2 B \log {\left (e^{x} + i \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*A/(exp(x) + I) - B*x + 2*B*log(exp(x) + I)

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Giac [A]
time = 0.41, size = 22, normalized size = 0.88 \begin {gather*} -B x + 2 \, B \log \left (e^{x} + i\right ) - \frac {2 \, A}{e^{x} + i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.14, size = 24, normalized size = 0.96 \begin {gather*} -B\,x-\frac {2\,A}{{\mathrm {e}}^x+1{}\mathrm {i}}+2\,B\,\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cosh(x))/(sinh(x) + 1i),x)

[Out]

2*B*log(exp(x) + 1i) - (2*A)/(exp(x) + 1i) - B*x

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