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

Optimal. Leaf size=104 \[ -\frac {2 a c \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}+\frac {b \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}-\frac {c x}{b^2-c^2} \]

[Out]

-c*x/(b^2-c^2)+b*ln(a+b*cosh(x)+c*sinh(x))/(b^2-c^2)-2*a*c*arctanh((c-(a-b)*tanh(1/2*x))/(a^2-b^2+c^2)^(1/2))/
(b^2-c^2)/(a^2-b^2+c^2)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3137, 3124, 618, 206} \[ -\frac {2 a c \tanh ^{-1}\left (\frac {c-(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2+c^2}}\right )}{\left (b^2-c^2\right ) \sqrt {a^2-b^2+c^2}}+\frac {b \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}-\frac {c x}{b^2-c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 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]

Rubi steps

\begin {align*} \int \frac {\sinh (x)}{a+b \cosh (x)+c \sinh (x)} \, dx &=-\frac {c x}{b^2-c^2}+\frac {b \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {(a c) \int \frac {1}{a+b \cosh (x)+c \sinh (x)} \, dx}{b^2-c^2}\\ &=-\frac {c x}{b^2-c^2}+\frac {b \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}+\frac {(2 a 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 {c x}{b^2-c^2}+\frac {b \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}-\frac {(4 a 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 {c x}{b^2-c^2}-\frac {2 a 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 \log (a+b \cosh (x)+c \sinh (x))}{b^2-c^2}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 86, normalized size = 0.83 \[ \frac {\frac {2 a c \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )+c}{\sqrt {-a^2+b^2-c^2}}\right )}{\sqrt {-a^2+b^2-c^2}}+b \log (a+b \cosh (x)+c \sinh (x))-c x}{b^2-c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 455, normalized size = 4.38 \[ \left [-\frac {\sqrt {a^{2} - b^{2} + c^{2}} a c \log \left (\frac {{\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \relax (x)^{2} + {\left (b^{2} + 2 \, b c + c^{2}\right )} \sinh \relax (x)^{2} + 2 \, a^{2} - b^{2} + c^{2} + 2 \, {\left (a b + a c\right )} \cosh \relax (x) + 2 \, {\left (a b + a c + {\left (b^{2} + 2 \, b c + c^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} - b^{2} + c^{2}} {\left ({\left (b + c\right )} \cosh \relax (x) + {\left (b + c\right )} \sinh \relax (x) + a\right )}}{{\left (b + c\right )} \cosh \relax (x)^{2} + {\left (b + c\right )} \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left ({\left (b + c\right )} \cosh \relax (x) + a\right )} \sinh \relax (x) + b - c}\right ) + {\left (a^{2} b - b^{3} + b c^{2} + c^{3} + {\left (a^{2} - b^{2}\right )} c\right )} x - {\left (a^{2} b - b^{3} + b c^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + c \sinh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{2} b^{2} - b^{4} - c^{4} - {\left (a^{2} - 2 \, b^{2}\right )} c^{2}}, \frac {2 \, \sqrt {-a^{2} + b^{2} - c^{2}} a c \arctan \left (\frac {\sqrt {-a^{2} + b^{2} - c^{2}} {\left ({\left (b + c\right )} \cosh \relax (x) + {\left (b + c\right )} \sinh \relax (x) + a\right )}}{a^{2} - b^{2} + c^{2}}\right ) - {\left (a^{2} b - b^{3} + b c^{2} + c^{3} + {\left (a^{2} - b^{2}\right )} c\right )} x + {\left (a^{2} b - b^{3} + b c^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + c \sinh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{2} b^{2} - b^{4} - c^{4} - {\left (a^{2} - 2 \, b^{2}\right )} c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(sqrt(a^2 - b^2 + c^2)*a*c*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)*cos
h(x) + a)*sinh(x) + b - c)) + (a^2*b - b^3 + b*c^2 + c^3 + (a^2 - b^2)*c)*x - (a^2*b - b^3 + b*c^2)*log(2*(b*c
osh(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*sqrt(-a^2 + b^2 -
c^2)*a*c*arctan(sqrt(-a^2 + b^2 - c^2)*((b + c)*cosh(x) + (b + c)*sinh(x) + a)/(a^2 - b^2 + c^2)) - (a^2*b - b
^3 + b*c^2 + c^3 + (a^2 - b^2)*c)*x + (a^2*b - b^3 + 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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giac [A]  time = 0.12, size = 106, normalized size = 1.02 \[ \frac {2 \, a c \arctan \left (\frac {b e^{x} + c e^{x} + a}{\sqrt {-a^{2} + b^{2} - c^{2}}}\right )}{\sqrt {-a^{2} + b^{2} - c^{2}} {\left (b^{2} - c^{2}\right )}} + \frac {b \log \left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + 2 \, a e^{x} + b - c\right )}{b^{2} - c^{2}} - \frac {x}{b - c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.20, size = 429, normalized size = 4.12 \[ -\frac {4 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{4 b +4 c}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 c \tanh \left (\frac {x}{2}\right )-a -b \right ) a b}{\left (b -c \right ) \left (b +c \right ) \left (a -b \right )}-\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 c \tanh \left (\frac {x}{2}\right )-a -b \right ) b^{2}}{\left (b -c \right ) \left (b +c \right ) \left (a -b \right )}-\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) a c}{\left (b -c \right ) \left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}}-\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) c b}{\left (b -c \right ) \left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}}+\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) c a b}{\left (b -c \right ) \left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}\, \left (a -b \right )}-\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tanh \left (\frac {x}{2}\right )-2 c}{2 \sqrt {-a^{2}+b^{2}-c^{2}}}\right ) c \,b^{2}}{\left (b -c \right ) \left (b +c \right ) \sqrt {-a^{2}+b^{2}-c^{2}}\, \left (a -b \right )}-\frac {4 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{4 b -4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(4*b+4*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
-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^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/(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*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-2/(b-c)/(b+c)/(-a^2+b^2-c^2)^(1/2)*arctan(1/2*(2*(a-b)*ta
nh(1/2*x)-2*c)/(-a^2+b^2-c^2)^(1/2))*c/(a-b)*b^2-4/(4*b-4*c)*ln(tanh(1/2*x)+1)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c^2-b^2+a^2>0)', see `assume?`
for more details)Is c^2-b^2+a^2 positive or negative?

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mupad [B]  time = 2.16, size = 325, normalized size = 3.12 \[ -\frac {x}{b-c}-\frac {\ln \left (\frac {2\,\left (b+a\,{\mathrm {e}}^x\right )}{{\left (b+c\right )}^2}-\frac {2\,\left (b-c+a\,{\mathrm {e}}^x\right )\,\left (a^2\,b+b\,c^2-b^3+a\,c\,\sqrt {a^2-b^2+c^2}\right )}{\left (b+c\right )\,\left (b^2-c^2\right )\,\left (a^2-b^2+c^2\right )}\right )\,\left (a^2\,b+b\,c^2-b^3+a\,c\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4}-\frac {\ln \left (\frac {2\,\left (b+a\,{\mathrm {e}}^x\right )}{{\left (b+c\right )}^2}-\frac {2\,\left (b-c+a\,{\mathrm {e}}^x\right )\,\left (a^2\,b+b\,c^2-b^3-a\,c\,\sqrt {a^2-b^2+c^2}\right )}{\left (b+c\right )\,\left (b^2-c^2\right )\,\left (a^2-b^2+c^2\right )}\right )\,\left (a^2\,b+b\,c^2-b^3-a\,c\,\sqrt {a^2-b^2+c^2}\right )}{-a^2\,b^2+a^2\,c^2+b^4-2\,b^2\,c^2+c^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- x/(b - c) - (log((2*(b + a*exp(x)))/(b + c)^2 - (2*(b - c + a*exp(x))*(a^2*b + b*c^2 - b^3 + a*c*(a^2 - b^2
+ c^2)^(1/2)))/((b + c)*(b^2 - c^2)*(a^2 - b^2 + c^2)))*(a^2*b + b*c^2 - b^3 + a*c*(a^2 - b^2 + c^2)^(1/2)))/(
b^4 + c^4 - a^2*b^2 + a^2*c^2 - 2*b^2*c^2) - (log((2*(b + a*exp(x)))/(b + c)^2 - (2*(b - c + a*exp(x))*(a^2*b
+ b*c^2 - b^3 - a*c*(a^2 - b^2 + c^2)^(1/2)))/((b + c)*(b^2 - c^2)*(a^2 - b^2 + c^2)))*(a^2*b + b*c^2 - b^3 -
a*c*(a^2 - b^2 + c^2)^(1/2)))/(b^4 + c^4 - a^2*b^2 + a^2*c^2 - 2*b^2*c^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(a+b*cosh(x)+c*sinh(x)),x)

[Out]

Timed out

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