### 3.802 $$\int \frac{A+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx$$

Optimal. Leaf size=71 $\frac{x (2 a A+b C)}{2 a^2}-\frac{1}{2} \left (\frac{b C}{a^2}+\frac{2 A}{a}-\frac{C}{b}\right ) \log (a+b \sinh (x)+b \cosh (x))-\frac{C \sinh (x)}{2 a}+\frac{C \cosh (x)}{2 a}$

[Out]

((2*a*A + b*C)*x)/(2*a^2) + (C*Cosh[x])/(2*a) - (((2*A)/a - C/b + (b*C)/a^2)*Log[a + b*Cosh[x] + b*Sinh[x]])/2
- (C*Sinh[x])/(2*a)

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Rubi [A]  time = 0.0560363, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.053, Rules used = {3131} $\frac{x (2 a A+b C)}{2 a^2}-\frac{1}{2} \left (\frac{b C}{a^2}+\frac{2 A}{a}-\frac{C}{b}\right ) \log (a+b \sinh (x)+b \cosh (x))-\frac{C \sinh (x)}{2 a}+\frac{C \cosh (x)}{2 a}$

Antiderivative was successfully veriﬁed.

[In]

Int[(A + C*Sinh[x])/(a + b*Cosh[x] + b*Sinh[x]),x]

[Out]

((2*a*A + b*C)*x)/(2*a^2) + (C*Cosh[x])/(2*a) - (((2*A)/a - C/b + (b*C)/a^2)*Log[a + b*Cosh[x] + b*Sinh[x]])/2
- (C*Sinh[x])/(2*a)

Rule 3131

Int[((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x
_)]), x_Symbol] :> Simp[((2*a*A - c*C)*x)/(2*a^2), x] + (-Simp[(C*Cos[d + e*x])/(2*a*e), x] + Simp[(c*C*Sin[d
+ e*x])/(2*a*b*e), x] + Simp[((-(a^2*C) + 2*a*c*A + b^2*C)*Log[RemoveContent[a + b*Cos[d + e*x] + c*Sin[d + e*
x], x]])/(2*a^2*b*e), x]) /; FreeQ[{a, b, c, d, e, A, C}, x] && EqQ[b^2 + c^2, 0]

Rubi steps

\begin{align*} \int \frac{A+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx &=\frac{(2 a A+b C) x}{2 a^2}+\frac{C \cosh (x)}{2 a}-\frac{1}{2} \left (\frac{2 A}{a}-\frac{C}{b}+\frac{b C}{a^2}\right ) \log (a+b \cosh (x)+b \sinh (x))-\frac{C \sinh (x)}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.210844, size = 86, normalized size = 1.21 $\frac{x \left (a^2 C+2 a A b+b^2 C\right )+2 \left (a^2 C-2 a A b-b^2 C\right ) \log \left ((b-a) \sinh \left (\frac{x}{2}\right )+(a+b) \cosh \left (\frac{x}{2}\right )\right )-2 a b C \sinh (x)+2 a b C \cosh (x)}{4 a^2 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(A + C*Sinh[x])/(a + b*Cosh[x] + b*Sinh[x]),x]

[Out]

((2*a*A*b + a^2*C + b^2*C)*x + 2*a*b*C*Cosh[x] + 2*(-2*a*A*b + a^2*C - b^2*C)*Log[(a + b)*Cosh[x/2] + (-a + b)
*Sinh[x/2]] - 2*a*b*C*Sinh[x])/(4*a^2*b)

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Maple [B]  time = 0.053, size = 136, normalized size = 1.9 \begin{align*}{\frac{C}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{A}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{bC}{2\,{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{C}{2\,b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{A}{a}\ln \left ( a\tanh \left ({\frac{x}{2}} \right ) -\tanh \left ({\frac{x}{2}} \right ) b-a-b \right ) }+{\frac{C}{2\,b}\ln \left ( a\tanh \left ({\frac{x}{2}} \right ) -\tanh \left ({\frac{x}{2}} \right ) b-a-b \right ) }-{\frac{bC}{2\,{a}^{2}}\ln \left ( a\tanh \left ({\frac{x}{2}} \right ) -\tanh \left ({\frac{x}{2}} \right ) b-a-b \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sinh(x))/(a+b*cosh(x)+b*sinh(x)),x)

[Out]

C/a/(tanh(1/2*x)+1)+1/a*ln(tanh(1/2*x)+1)*A+1/2/a^2*ln(tanh(1/2*x)+1)*b*C-1/2*C/b*ln(tanh(1/2*x)-1)-1/a*ln(a*t
anh(1/2*x)-tanh(1/2*x)*b-a-b)*A+1/2/b*ln(a*tanh(1/2*x)-tanh(1/2*x)*b-a-b)*C-1/2/a^2*b*ln(a*tanh(1/2*x)-tanh(1/
2*x)*b-a-b)*C

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Maxima [A]  time = 1.12256, size = 78, normalized size = 1.1 \begin{align*} \frac{1}{2} \, C{\left (\frac{x}{b} + \frac{e^{\left (-x\right )}}{a} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (a e^{\left (-x\right )} + b\right )}{a^{2} b}\right )} - \frac{A \log \left (a e^{\left (-x\right )} + b\right )}{a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sinh(x))/(a+b*cosh(x)+b*sinh(x)),x, algorithm="maxima")

[Out]

1/2*C*(x/b + e^(-x)/a + (a^2 - b^2)*log(a*e^(-x) + b)/(a^2*b)) - A*log(a*e^(-x) + b)/a

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Fricas [A]  time = 2.29658, size = 284, normalized size = 4. \begin{align*} \frac{C a b +{\left (2 \, A a b + C b^{2}\right )} x \cosh \left (x\right ) +{\left (2 \, A a b + C b^{2}\right )} x \sinh \left (x\right ) +{\left ({\left (C a^{2} - 2 \, A a b - C b^{2}\right )} \cosh \left (x\right ) +{\left (C a^{2} - 2 \, A a b - C b^{2}\right )} \sinh \left (x\right )\right )} \log \left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}{2 \,{\left (a^{2} b \cosh \left (x\right ) + a^{2} b \sinh \left (x\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sinh(x))/(a+b*cosh(x)+b*sinh(x)),x, algorithm="fricas")

[Out]

1/2*(C*a*b + (2*A*a*b + C*b^2)*x*cosh(x) + (2*A*a*b + C*b^2)*x*sinh(x) + ((C*a^2 - 2*A*a*b - C*b^2)*cosh(x) +
(C*a^2 - 2*A*a*b - C*b^2)*sinh(x))*log(b*cosh(x) + b*sinh(x) + a))/(a^2*b*cosh(x) + a^2*b*sinh(x))

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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+C*sinh(x))/(a+b*cosh(x)+b*sinh(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.13367, size = 78, normalized size = 1.1 \begin{align*} \frac{C e^{\left (-x\right )}}{2 \, a} + \frac{{\left (2 \, A a + C b\right )} x}{2 \, a^{2}} + \frac{{\left (C a^{2} - 2 \, A a b - C b^{2}\right )} \log \left ({\left | b e^{x} + a \right |}\right )}{2 \, a^{2} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sinh(x))/(a+b*cosh(x)+b*sinh(x)),x, algorithm="giac")

[Out]

1/2*C*e^(-x)/a + 1/2*(2*A*a + C*b)*x/a^2 + 1/2*(C*a^2 - 2*A*a*b - C*b^2)*log(abs(b*e^x + a))/(a^2*b)