3.795 \(\int \frac{B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx\)

Optimal. Leaf size=125 \[ \frac{2 a (b B-c C) \tanh ^{-1}\left (\frac{c-(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2-b^2+c^2}}-\frac{(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac{x (b B-c C)}{b^2-c^2} \]

[Out]

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

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Rubi [A]  time = 0.142734, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3136, 3124, 618, 206} \[ \frac{2 a (b B-c C) \tanh ^{-1}\left (\frac{c-(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2-b^2+c^2}}-\frac{(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac{x (b B-c C)}{b^2-c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3136

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + (Dist[(A*(b^2 + c^2
) - a*(b*B + c*C))/(b^2 + c^2), Int[1/(a + b*Cos[d + e*x] + c*Sin[d + e*x]), x], x] + Simp[((c*B - 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, B, C}, x] && NeQ[b^2 + c^
2, 0] && NeQ[A*(b^2 + c^2) - a*(b*B + 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{B \cosh (x)+C \sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx &=\frac{(b B-c C) x}{b^2-c^2}-\frac{(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}-\frac{(a (b B-c C)) \int \frac{1}{a+b \cosh (x)+c \sinh (x)} \, dx}{b^2-c^2}\\ &=\frac{(b B-c C) x}{b^2-c^2}-\frac{(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}-\frac{(2 a (b B-c C)) \operatorname{Subst}\left (\int \frac{1}{a+b+2 c x-(a-b) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{b^2-c^2}\\ &=\frac{(b B-c C) x}{b^2-c^2}-\frac{(B c-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac{(4 a (b B-c C)) \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 )}{b^2-c^2}\\ &=\frac{(b B-c C) x}{b^2-c^2}+\frac{2 a (b B-c C) \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-b C) \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}\\ \end{align*}

Mathematica [A]  time = 0.262715, size = 107, normalized size = 0.86 \[ \frac{-\frac{2 a (b B-c C) \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{x}{2}\right )+c}{\sqrt{-a^2+b^2-c^2}}\right )}{\sqrt{-a^2+b^2-c^2}}+(b C-B c) \log (a+b \cosh (x)+c \sinh (x))+x (b B-c C)}{(b-c) (b+c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.056, size = 873, normalized size = 7. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B/(2*b-2*c)*ln(tanh(1/2*x)+1)-2*C/(2*b-2*c)*ln(tanh(1/2*x)+1)-2*B/(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)*b*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)*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*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))*B*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^2/(a-b)*a*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^2/(a-b)*b*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((B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.30363, size = 1364, normalized size = 10.91 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((B*a*b - C*a*c)*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*sqrt(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)) + ((B - C)*a^2*b - (B - C)*b^3 + (B - C)*b*c^2 + (B - C)*c^3 + ((B - C)
*a^2 - (B - C)*b^2)*c)*x + (C*a^2*b - C*b^3 + C*b*c^2 - B*c^3 - (B*a^2 - B*b^2)*c)*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*(B*a*b - C*a*c)*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)) - ((B - C)*a^2*b
- (B - C)*b^3 + (B - C)*b*c^2 + (B - C)*c^3 + ((B - C)*a^2 - (B - C)*b^2)*c)*x - (C*a^2*b - C*b^3 + C*b*c^2 -
B*c^3 - (B*a^2 - B*b^2)*c)*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((B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.18471, size = 169, normalized size = 1.35 \begin{align*} \frac{{\left (B - C\right )} x}{b - c} + \frac{{\left (C b - B c\right )} \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{2 \,{\left (B a b - C a c\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((B*cosh(x)+C*sinh(x))/(a+b*cosh(x)+c*sinh(x)),x, algorithm="giac")

[Out]

(B - C)*x/(b - c) + (C*b - B*c)*log(b*e^(2*x) + c*e^(2*x) + 2*a*e^x + b - c)/(b^2 - c^2) - 2*(B*a*b - C*a*c)*a
rctan((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))