### 3.789 $$\int \frac{A+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx$$

Optimal. Leaf size=120 $-\frac{2 \left (a c C+A \left (b^2-c^2\right )\right ) \tanh ^{-1}\left (\frac{c-(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2-b^2+c^2}}+\frac{b C \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}-\frac{c C x}{b^2-c^2}$

[Out]

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

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Rubi [A]  time = 0.11779, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.21, Rules used = {3137, 3124, 618, 206} $-\frac{2 \left (a c C+A \left (b^2-c^2\right )\right ) \tanh ^{-1}\left (\frac{c-(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2-b^2+c^2}}+\frac{b C \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}-\frac{c C x}{b^2-c^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3137

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

Rule 3124

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[(2*f)/e, Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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])

Rubi steps

\begin{align*} \int \frac{A+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx &=-\frac{c C x}{b^2-c^2}+\frac{b C \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\left (A+\frac{a c C}{b^2-c^2}\right ) \int \frac{1}{a+b \cosh (x)+c \sinh (x)} \, dx\\ &=-\frac{c C x}{b^2-c^2}+\frac{b C \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\left (2 \left (A+\frac{a c C}{b^2-c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+2 c x-(a-b) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )\\ &=-\frac{c C x}{b^2-c^2}+\frac{b C \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}-\left (4 \left (A+\frac{a c C}{b^2-c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2-b^2+c^2\right )-x^2} \, dx,x,2 c+2 (-a+b) \tanh \left (\frac{x}{2}\right )\right )\\ &=-\frac{c C x}{b^2-c^2}-\frac{2 \left (A+\frac{a c C}{b^2-c^2}\right ) \tanh ^{-1}\left (\frac{c-(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2+c^2}}\right )}{\sqrt{a^2-b^2+c^2}}+\frac{b C \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}\\ \end{align*}

Mathematica [A]  time = 0.200776, size = 104, normalized size = 0.87 $\frac{\frac{2 \left (a c C+A \left (b^2-c^2\right )\right ) \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{x}{2}\right )+c}{\sqrt{-a^2+b^2-c^2}}\right )}{\sqrt{-a^2+b^2-c^2}}+C (b \log (a+b \cosh (x)+c \sinh (x))-c x)}{(b-c) (b+c)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.053, size = 573, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.38677, size = 1212, normalized size = 10.1 \begin{align*} \left [-\frac{{\left (A b^{2} + C a c - A c^{2}\right )} \sqrt{a^{2} - b^{2} + c^{2}} \log \left (\frac{{\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \left (x\right )^{2} +{\left (b^{2} + 2 \, b c + c^{2}\right )} \sinh \left (x\right )^{2} + 2 \, a^{2} - b^{2} + c^{2} + 2 \,{\left (a b + a c\right )} \cosh \left (x\right ) + 2 \,{\left (a b + a c +{\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + 2 \, \sqrt{a^{2} - b^{2} + c^{2}}{\left ({\left (b + c\right )} \cosh \left (x\right ) +{\left (b + c\right )} \sinh \left (x\right ) + a\right )}}{{\left (b + c\right )} \cosh \left (x\right )^{2} +{\left (b + c\right )} \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left ({\left (b + c\right )} \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) + b - c}\right ) +{\left (C a^{2} b - C b^{3} + C b c^{2} + C c^{3} +{\left (C a^{2} - C b^{2}\right )} c\right )} x -{\left (C a^{2} b - C b^{3} + C b c^{2}\right )} \log \left (\frac{2 \,{\left (b \cosh \left (x\right ) + c \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b^{2} - b^{4} - c^{4} -{\left (a^{2} - 2 \, b^{2}\right )} c^{2}}, \frac{2 \,{\left (A b^{2} + C a c - A c^{2}\right )} \sqrt{-a^{2} + b^{2} - c^{2}} \arctan \left (\frac{\sqrt{-a^{2} + b^{2} - c^{2}}{\left ({\left (b + c\right )} \cosh \left (x\right ) +{\left (b + c\right )} \sinh \left (x\right ) + a\right )}}{a^{2} - b^{2} + c^{2}}\right ) -{\left (C a^{2} b - C b^{3} + C b c^{2} + C c^{3} +{\left (C a^{2} - C b^{2}\right )} c\right )} x +{\left (C a^{2} b - C b^{3} + C b c^{2}\right )} \log \left (\frac{2 \,{\left (b \cosh \left (x\right ) + c \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} b^{2} - b^{4} - c^{4} -{\left (a^{2} - 2 \, b^{2}\right )} c^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-((A*b^2 + C*a*c - A*c^2)*sqrt(a^2 - b^2 + c^2)*log(((b^2 + 2*b*c + c^2)*cosh(x)^2 + (b^2 + 2*b*c + c^2)*sinh
(x)^2 + 2*a^2 - b^2 + c^2 + 2*(a*b + a*c)*cosh(x) + 2*(a*b + a*c + (b^2 + 2*b*c + c^2)*cosh(x))*sinh(x) + 2*sq
rt(a^2 - b^2 + c^2)*((b + c)*cosh(x) + (b + c)*sinh(x) + a))/((b + c)*cosh(x)^2 + (b + c)*sinh(x)^2 + 2*a*cosh
(x) + 2*((b + c)*cosh(x) + a)*sinh(x) + b - c)) + (C*a^2*b - C*b^3 + C*b*c^2 + C*c^3 + (C*a^2 - C*b^2)*c)*x -
(C*a^2*b - C*b^3 + C*b*c^2)*log(2*(b*cosh(x) + c*sinh(x) + a)/(cosh(x) - sinh(x))))/(a^2*b^2 - b^4 - c^4 - (a^
2 - 2*b^2)*c^2), (2*(A*b^2 + C*a*c - A*c^2)*sqrt(-a^2 + b^2 - c^2)*arctan(sqrt(-a^2 + b^2 - c^2)*((b + c)*cosh
(x) + (b + c)*sinh(x) + a)/(a^2 - b^2 + c^2)) - (C*a^2*b - C*b^3 + C*b*c^2 + C*c^3 + (C*a^2 - C*b^2)*c)*x + (C
*a^2*b - C*b^3 + C*b*c^2)*log(2*(b*cosh(x) + c*sinh(x) + a)/(cosh(x) - sinh(x))))/(a^2*b^2 - b^4 - c^4 - (a^2
- 2*b^2)*c^2)]

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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)+c*sinh(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.15594, size = 165, normalized size = 1.38 \begin{align*} \frac{C b \log \left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + 2 \, a e^{x} + b - c\right )}{b^{2} - c^{2}} - \frac{C x}{b - c} + \frac{2 \,{\left (A b^{2} + C a c - A c^{2}\right )} \arctan \left (\frac{b e^{x} + c e^{x} + a}{\sqrt{-a^{2} + b^{2} - c^{2}}}\right )}{\sqrt{-a^{2} + b^{2} - c^{2}}{\left (b^{2} - c^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

C*b*log(b*e^(2*x) + c*e^(2*x) + 2*a*e^x + b - c)/(b^2 - c^2) - C*x/(b - c) + 2*(A*b^2 + C*a*c - A*c^2)*arctan(
(b*e^x + c*e^x + a)/sqrt(-a^2 + b^2 - c^2))/(sqrt(-a^2 + b^2 - c^2)*(b^2 - c^2))