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

Optimal. Leaf size=186 \[ \frac {\text {Li}_2\left (-\frac {b e^{2 x}}{2 a-\sqrt {4 a^2+b^2}}\right )}{2 \sqrt {4 a^2+b^2}}-\frac {\text {Li}_2\left (-\frac {b e^{2 x}}{2 a+\sqrt {4 a^2+b^2}}\right )}{2 \sqrt {4 a^2+b^2}}+\frac {x \log \left (\frac {b e^{2 x}}{2 a-\sqrt {4 a^2+b^2}}+1\right )}{\sqrt {4 a^2+b^2}}-\frac {x \log \left (\frac {b e^{2 x}}{\sqrt {4 a^2+b^2}+2 a}+1\right )}{\sqrt {4 a^2+b^2}} \]

[Out]

x*ln(1+b*exp(2*x)/(2*a-(4*a^2+b^2)^(1/2)))/(4*a^2+b^2)^(1/2)-x*ln(1+b*exp(2*x)/(2*a+(4*a^2+b^2)^(1/2)))/(4*a^2
+b^2)^(1/2)+1/2*polylog(2,-b*exp(2*x)/(2*a-(4*a^2+b^2)^(1/2)))/(4*a^2+b^2)^(1/2)-1/2*polylog(2,-b*exp(2*x)/(2*
a+(4*a^2+b^2)^(1/2)))/(4*a^2+b^2)^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5628, 3322, 2264, 2190, 2279, 2391} \[ \frac {\text {PolyLog}\left (2,-\frac {b e^{2 x}}{2 a-\sqrt {4 a^2+b^2}}\right )}{2 \sqrt {4 a^2+b^2}}-\frac {\text {PolyLog}\left (2,-\frac {b e^{2 x}}{\sqrt {4 a^2+b^2}+2 a}\right )}{2 \sqrt {4 a^2+b^2}}+\frac {x \log \left (\frac {b e^{2 x}}{2 a-\sqrt {4 a^2+b^2}}+1\right )}{\sqrt {4 a^2+b^2}}-\frac {x \log \left (\frac {b e^{2 x}}{\sqrt {4 a^2+b^2}+2 a}+1\right )}{\sqrt {4 a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Cosh[x]*Sinh[x]),x]

[Out]

(x*Log[1 + (b*E^(2*x))/(2*a - Sqrt[4*a^2 + b^2])])/Sqrt[4*a^2 + b^2] - (x*Log[1 + (b*E^(2*x))/(2*a + Sqrt[4*a^
2 + b^2])])/Sqrt[4*a^2 + b^2] + PolyLog[2, -((b*E^(2*x))/(2*a - Sqrt[4*a^2 + b^2]))]/(2*Sqrt[4*a^2 + b^2]) - P
olyLog[2, -((b*E^(2*x))/(2*a + Sqrt[4*a^2 + b^2]))]/(2*Sqrt[4*a^2 + b^2])

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[2,
Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(-(I*b) + 2*a*E^(-(I*e) + f*fz*x) + I*b*E^(2*(-(I*e) + f*fz*x))), x], x]
/; FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 5628

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + Cosh[(c_.) + (d_.)*(x_)]*(b_.)*Sinh[(c_.) + (d_.)*(x_)])^(n_.), x_Symbo
l] :> Int[(e + f*x)^m*(a + (b*Sinh[2*c + 2*d*x])/2)^n, x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int \frac {x}{a+b \cosh (x) \sinh (x)} \, dx &=\int \frac {x}{a+\frac {1}{2} b \sinh (2 x)} \, dx\\ &=2 \int \frac {e^{2 x} x}{-\frac {b}{2}+2 a e^{2 x}+\frac {1}{2} b e^{4 x}} \, dx\\ &=\frac {(2 b) \int \frac {e^{2 x} x}{2 a-\sqrt {4 a^2+b^2}+b e^{2 x}} \, dx}{\sqrt {4 a^2+b^2}}-\frac {(2 b) \int \frac {e^{2 x} x}{2 a+\sqrt {4 a^2+b^2}+b e^{2 x}} \, dx}{\sqrt {4 a^2+b^2}}\\ &=\frac {x \log \left (1+\frac {b e^{2 x}}{2 a-\sqrt {4 a^2+b^2}}\right )}{\sqrt {4 a^2+b^2}}-\frac {x \log \left (1+\frac {b e^{2 x}}{2 a+\sqrt {4 a^2+b^2}}\right )}{\sqrt {4 a^2+b^2}}-\frac {\int \log \left (1+\frac {b e^{2 x}}{2 a-\sqrt {4 a^2+b^2}}\right ) \, dx}{\sqrt {4 a^2+b^2}}+\frac {\int \log \left (1+\frac {b e^{2 x}}{2 a+\sqrt {4 a^2+b^2}}\right ) \, dx}{\sqrt {4 a^2+b^2}}\\ &=\frac {x \log \left (1+\frac {b e^{2 x}}{2 a-\sqrt {4 a^2+b^2}}\right )}{\sqrt {4 a^2+b^2}}-\frac {x \log \left (1+\frac {b e^{2 x}}{2 a+\sqrt {4 a^2+b^2}}\right )}{\sqrt {4 a^2+b^2}}-\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{2 a-\sqrt {4 a^2+b^2}}\right )}{x} \, dx,x,e^{2 x}\right )}{2 \sqrt {4 a^2+b^2}}+\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{2 a+\sqrt {4 a^2+b^2}}\right )}{x} \, dx,x,e^{2 x}\right )}{2 \sqrt {4 a^2+b^2}}\\ &=\frac {x \log \left (1+\frac {b e^{2 x}}{2 a-\sqrt {4 a^2+b^2}}\right )}{\sqrt {4 a^2+b^2}}-\frac {x \log \left (1+\frac {b e^{2 x}}{2 a+\sqrt {4 a^2+b^2}}\right )}{\sqrt {4 a^2+b^2}}+\frac {\text {Li}_2\left (-\frac {b e^{2 x}}{2 a-\sqrt {4 a^2+b^2}}\right )}{2 \sqrt {4 a^2+b^2}}-\frac {\text {Li}_2\left (-\frac {b e^{2 x}}{2 a+\sqrt {4 a^2+b^2}}\right )}{2 \sqrt {4 a^2+b^2}}\\ \end {align*}

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Mathematica [C]  time = 1.87, size = 956, normalized size = 5.14 \[ \frac {1}{2} \left (-\frac {i \pi \tanh ^{-1}\left (\frac {2 a \tanh (x)-b}{\sqrt {4 a^2+b^2}}\right )}{\sqrt {4 a^2+b^2}}-\frac {2 \cos ^{-1}\left (-\frac {2 i a}{b}\right ) \tanh ^{-1}\left (\frac {(2 a+i b) \cot \left (\frac {1}{4} (4 i x+\pi )\right )}{\sqrt {-4 a^2-b^2}}\right )+(\pi -4 i x) \tanh ^{-1}\left (\frac {(2 a-i b) \tan \left (\frac {1}{4} (4 i x+\pi )\right )}{\sqrt {-4 a^2-b^2}}\right )-\left (\cos ^{-1}\left (-\frac {2 i a}{b}\right )+2 i \tanh ^{-1}\left (\frac {(2 a+i b) \cot \left (\frac {1}{4} (4 i x+\pi )\right )}{\sqrt {-4 a^2-b^2}}\right )\right ) \log \left (\frac {(2 i a+b) \left (-2 i a+b+\sqrt {-4 a^2-b^2}\right ) \left (i \cot \left (\frac {1}{4} (4 i x+\pi )\right )+1\right )}{b \left (2 i a+b+i \sqrt {-4 a^2-b^2} \cot \left (\frac {1}{4} (4 i x+\pi )\right )\right )}\right )-\left (\cos ^{-1}\left (-\frac {2 i a}{b}\right )-2 i \tanh ^{-1}\left (\frac {(2 a+i b) \cot \left (\frac {1}{4} (4 i x+\pi )\right )}{\sqrt {-4 a^2-b^2}}\right )\right ) \log \left (\frac {(2 i a+b) \left (2 i a-b+\sqrt {-4 a^2-b^2}\right ) \left (\cot \left (\frac {1}{4} (4 i x+\pi )\right )+i\right )}{b \left (2 a-i b+\sqrt {-4 a^2-b^2} \cot \left (\frac {1}{4} (4 i x+\pi )\right )\right )}\right )+\left (\cos ^{-1}\left (-\frac {2 i a}{b}\right )-2 i \tanh ^{-1}\left (\frac {(2 a+i b) \cot \left (\frac {1}{4} (4 i x+\pi )\right )}{\sqrt {-4 a^2-b^2}}\right )-2 i \tanh ^{-1}\left (\frac {(2 a-i b) \tan \left (\frac {1}{4} (4 i x+\pi )\right )}{\sqrt {-4 a^2-b^2}}\right )\right ) \log \left (-\frac {(-1)^{3/4} \sqrt {-4 a^2-b^2} e^{-x}}{2 \sqrt {-i b} \sqrt {a+b \cosh (x) \sinh (x)}}\right )+\left (\cos ^{-1}\left (-\frac {2 i a}{b}\right )+2 i \left (\tanh ^{-1}\left (\frac {(2 a+i b) \cot \left (\frac {1}{4} (4 i x+\pi )\right )}{\sqrt {-4 a^2-b^2}}\right )+\tanh ^{-1}\left (\frac {(2 a-i b) \tan \left (\frac {1}{4} (4 i x+\pi )\right )}{\sqrt {-4 a^2-b^2}}\right )\right )\right ) \log \left (\frac {\sqrt [4]{-1} \sqrt {-4 a^2-b^2} e^x}{2 \sqrt {-i b} \sqrt {a+b \cosh (x) \sinh (x)}}\right )+i \left (\text {Li}_2\left (\frac {\left (2 i a+\sqrt {-4 a^2-b^2}\right ) \left (2 i a+b-i \sqrt {-4 a^2-b^2} \cot \left (\frac {1}{4} (4 i x+\pi )\right )\right )}{b \left (2 i a+b+i \sqrt {-4 a^2-b^2} \cot \left (\frac {1}{4} (4 i x+\pi )\right )\right )}\right )-\text {Li}_2\left (\frac {\left (2 a+i \sqrt {-4 a^2-b^2}\right ) \left (-2 a+i b+\sqrt {-4 a^2-b^2} \cot \left (\frac {1}{4} (4 i x+\pi )\right )\right )}{b \left (2 i a+b+i \sqrt {-4 a^2-b^2} \cot \left (\frac {1}{4} (4 i x+\pi )\right )\right )}\right )\right )}{\sqrt {-4 a^2-b^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*Pi*ArcTanh[(-b + 2*a*Tanh[x])/Sqrt[4*a^2 + b^2]])/Sqrt[4*a^2 + b^2] - (2*ArcCos[((-2*I)*a)/b]*ArcTanh[(
(2*a + I*b)*Cot[(Pi + (4*I)*x)/4])/Sqrt[-4*a^2 - b^2]] + (Pi - (4*I)*x)*ArcTanh[((2*a - I*b)*Tan[(Pi + (4*I)*x
)/4])/Sqrt[-4*a^2 - b^2]] - (ArcCos[((-2*I)*a)/b] + (2*I)*ArcTanh[((2*a + I*b)*Cot[(Pi + (4*I)*x)/4])/Sqrt[-4*
a^2 - b^2]])*Log[(((2*I)*a + b)*((-2*I)*a + b + Sqrt[-4*a^2 - b^2])*(1 + I*Cot[(Pi + (4*I)*x)/4]))/(b*((2*I)*a
+ b + I*Sqrt[-4*a^2 - b^2]*Cot[(Pi + (4*I)*x)/4]))] - (ArcCos[((-2*I)*a)/b] - (2*I)*ArcTanh[((2*a + I*b)*Cot[
(Pi + (4*I)*x)/4])/Sqrt[-4*a^2 - b^2]])*Log[(((2*I)*a + b)*((2*I)*a - b + Sqrt[-4*a^2 - b^2])*(I + Cot[(Pi + (
4*I)*x)/4]))/(b*(2*a - I*b + Sqrt[-4*a^2 - b^2]*Cot[(Pi + (4*I)*x)/4]))] + (ArcCos[((-2*I)*a)/b] - (2*I)*ArcTa
nh[((2*a + I*b)*Cot[(Pi + (4*I)*x)/4])/Sqrt[-4*a^2 - b^2]] - (2*I)*ArcTanh[((2*a - I*b)*Tan[(Pi + (4*I)*x)/4])
/Sqrt[-4*a^2 - b^2]])*Log[-1/2*((-1)^(3/4)*Sqrt[-4*a^2 - b^2])/(Sqrt[(-I)*b]*E^x*Sqrt[a + b*Cosh[x]*Sinh[x]])]
+ (ArcCos[((-2*I)*a)/b] + (2*I)*(ArcTanh[((2*a + I*b)*Cot[(Pi + (4*I)*x)/4])/Sqrt[-4*a^2 - b^2]] + ArcTanh[((
2*a - I*b)*Tan[(Pi + (4*I)*x)/4])/Sqrt[-4*a^2 - b^2]]))*Log[((-1)^(1/4)*Sqrt[-4*a^2 - b^2]*E^x)/(2*Sqrt[(-I)*b
]*Sqrt[a + b*Cosh[x]*Sinh[x]])] + I*(PolyLog[2, (((2*I)*a + Sqrt[-4*a^2 - b^2])*((2*I)*a + b - I*Sqrt[-4*a^2 -
b^2]*Cot[(Pi + (4*I)*x)/4]))/(b*((2*I)*a + b + I*Sqrt[-4*a^2 - b^2]*Cot[(Pi + (4*I)*x)/4]))] - PolyLog[2, ((2
*a + I*Sqrt[-4*a^2 - b^2])*(-2*a + I*b + Sqrt[-4*a^2 - b^2]*Cot[(Pi + (4*I)*x)/4]))/(b*((2*I)*a + b + I*Sqrt[-
4*a^2 - b^2]*Cot[(Pi + (4*I)*x)/4]))]))/Sqrt[-4*a^2 - b^2])/2

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fricas [B]  time = 0.48, size = 754, normalized size = 4.05 \[ -\frac {b x \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} \log \left (\frac {{\left (2 \, a \cosh \relax (x) + 2 \, a \sinh \relax (x) - {\left (b \cosh \relax (x) + b \sinh \relax (x)\right )} \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}}\right )} \sqrt {-\frac {b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} + 2 \, a}{b}} + b}{b}\right ) + b x \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {{\left (2 \, a \cosh \relax (x) + 2 \, a \sinh \relax (x) - {\left (b \cosh \relax (x) + b \sinh \relax (x)\right )} \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}}\right )} \sqrt {-\frac {b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} + 2 \, a}{b}} - b}{b}\right ) - b x \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} \log \left (\frac {{\left (2 \, a \cosh \relax (x) + 2 \, a \sinh \relax (x) + {\left (b \cosh \relax (x) + b \sinh \relax (x)\right )} \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}}\right )} \sqrt {\frac {b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} - 2 \, a}{b}} + b}{b}\right ) - b x \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} \log \left (-\frac {{\left (2 \, a \cosh \relax (x) + 2 \, a \sinh \relax (x) + {\left (b \cosh \relax (x) + b \sinh \relax (x)\right )} \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}}\right )} \sqrt {\frac {b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} - 2 \, a}{b}} - b}{b}\right ) + b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (-\frac {{\left (2 \, a \cosh \relax (x) + 2 \, a \sinh \relax (x) - {\left (b \cosh \relax (x) + b \sinh \relax (x)\right )} \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}}\right )} \sqrt {-\frac {b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} + 2 \, a}{b}} + b}{b} + 1\right ) + b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {{\left (2 \, a \cosh \relax (x) + 2 \, a \sinh \relax (x) - {\left (b \cosh \relax (x) + b \sinh \relax (x)\right )} \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}}\right )} \sqrt {-\frac {b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} + 2 \, a}{b}} - b}{b} + 1\right ) - b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (-\frac {{\left (2 \, a \cosh \relax (x) + 2 \, a \sinh \relax (x) + {\left (b \cosh \relax (x) + b \sinh \relax (x)\right )} \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}}\right )} \sqrt {\frac {b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} - 2 \, a}{b}} + b}{b} + 1\right ) - b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} {\rm Li}_2\left (\frac {{\left (2 \, a \cosh \relax (x) + 2 \, a \sinh \relax (x) + {\left (b \cosh \relax (x) + b \sinh \relax (x)\right )} \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}}\right )} \sqrt {\frac {b \sqrt {\frac {4 \, a^{2} + b^{2}}{b^{2}}} - 2 \, a}{b}} - b}{b} + 1\right )}{4 \, a^{2} + b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x*sqrt((4*a^2 + b^2)/b^2)*log(((2*a*cosh(x) + 2*a*sinh(x) - (b*cosh(x) + b*sinh(x))*sqrt((4*a^2 + b^2)/b^2
))*sqrt(-(b*sqrt((4*a^2 + b^2)/b^2) + 2*a)/b) + b)/b) + b*x*sqrt((4*a^2 + b^2)/b^2)*log(-((2*a*cosh(x) + 2*a*s
inh(x) - (b*cosh(x) + b*sinh(x))*sqrt((4*a^2 + b^2)/b^2))*sqrt(-(b*sqrt((4*a^2 + b^2)/b^2) + 2*a)/b) - b)/b) -
b*x*sqrt((4*a^2 + b^2)/b^2)*log(((2*a*cosh(x) + 2*a*sinh(x) + (b*cosh(x) + b*sinh(x))*sqrt((4*a^2 + b^2)/b^2)
)*sqrt((b*sqrt((4*a^2 + b^2)/b^2) - 2*a)/b) + b)/b) - b*x*sqrt((4*a^2 + b^2)/b^2)*log(-((2*a*cosh(x) + 2*a*sin
h(x) + (b*cosh(x) + b*sinh(x))*sqrt((4*a^2 + b^2)/b^2))*sqrt((b*sqrt((4*a^2 + b^2)/b^2) - 2*a)/b) - b)/b) + b*
sqrt((4*a^2 + b^2)/b^2)*dilog(-((2*a*cosh(x) + 2*a*sinh(x) - (b*cosh(x) + b*sinh(x))*sqrt((4*a^2 + b^2)/b^2))*
sqrt(-(b*sqrt((4*a^2 + b^2)/b^2) + 2*a)/b) + b)/b + 1) + b*sqrt((4*a^2 + b^2)/b^2)*dilog(((2*a*cosh(x) + 2*a*s
inh(x) - (b*cosh(x) + b*sinh(x))*sqrt((4*a^2 + b^2)/b^2))*sqrt(-(b*sqrt((4*a^2 + b^2)/b^2) + 2*a)/b) - b)/b +
1) - b*sqrt((4*a^2 + b^2)/b^2)*dilog(-((2*a*cosh(x) + 2*a*sinh(x) + (b*cosh(x) + b*sinh(x))*sqrt((4*a^2 + b^2)
/b^2))*sqrt((b*sqrt((4*a^2 + b^2)/b^2) - 2*a)/b) + b)/b + 1) - b*sqrt((4*a^2 + b^2)/b^2)*dilog(((2*a*cosh(x) +
2*a*sinh(x) + (b*cosh(x) + b*sinh(x))*sqrt((4*a^2 + b^2)/b^2))*sqrt((b*sqrt((4*a^2 + b^2)/b^2) - 2*a)/b) - b)
/b + 1))/(4*a^2 + b^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{b \cosh \relax (x) \sinh \relax (x) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.30, size = 376, normalized size = 2.02 \[ \frac {\ln \left (1-\frac {b \,{\mathrm e}^{2 x}}{-2 a -\sqrt {4 a^{2}+b^{2}}}\right ) x}{-2 a -\sqrt {4 a^{2}+b^{2}}}+\frac {2 \ln \left (1-\frac {b \,{\mathrm e}^{2 x}}{-2 a -\sqrt {4 a^{2}+b^{2}}}\right ) a x}{\sqrt {4 a^{2}+b^{2}}\, \left (-2 a -\sqrt {4 a^{2}+b^{2}}\right )}-\frac {x^{2}}{-2 a -\sqrt {4 a^{2}+b^{2}}}-\frac {2 a \,x^{2}}{\sqrt {4 a^{2}+b^{2}}\, \left (-2 a -\sqrt {4 a^{2}+b^{2}}\right )}+\frac {\polylog \left (2, \frac {b \,{\mathrm e}^{2 x}}{-2 a -\sqrt {4 a^{2}+b^{2}}}\right )}{-4 a -2 \sqrt {4 a^{2}+b^{2}}}+\frac {\polylog \left (2, \frac {b \,{\mathrm e}^{2 x}}{-2 a -\sqrt {4 a^{2}+b^{2}}}\right ) a}{\sqrt {4 a^{2}+b^{2}}\, \left (-2 a -\sqrt {4 a^{2}+b^{2}}\right )}+\frac {x \ln \left (1-\frac {b \,{\mathrm e}^{2 x}}{\sqrt {4 a^{2}+b^{2}}-2 a}\right )}{\sqrt {4 a^{2}+b^{2}}}-\frac {x^{2}}{\sqrt {4 a^{2}+b^{2}}}+\frac {\polylog \left (2, \frac {b \,{\mathrm e}^{2 x}}{\sqrt {4 a^{2}+b^{2}}-2 a}\right )}{2 \sqrt {4 a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(-2*a-(4*a^2+b^2)^(1/2))*ln(1-b*exp(2*x)/(-2*a-(4*a^2+b^2)^(1/2)))*x+2/(4*a^2+b^2)^(1/2)/(-2*a-(4*a^2+b^2)^(
1/2))*ln(1-b*exp(2*x)/(-2*a-(4*a^2+b^2)^(1/2)))*a*x-1/(-2*a-(4*a^2+b^2)^(1/2))*x^2-2/(4*a^2+b^2)^(1/2)/(-2*a-(
4*a^2+b^2)^(1/2))*a*x^2+1/2/(-2*a-(4*a^2+b^2)^(1/2))*polylog(2,b*exp(2*x)/(-2*a-(4*a^2+b^2)^(1/2)))+1/(4*a^2+b
^2)^(1/2)/(-2*a-(4*a^2+b^2)^(1/2))*polylog(2,b*exp(2*x)/(-2*a-(4*a^2+b^2)^(1/2)))*a+1/(4*a^2+b^2)^(1/2)*x*ln(1
-b*exp(2*x)/((4*a^2+b^2)^(1/2)-2*a))-1/(4*a^2+b^2)^(1/2)*x^2+1/2/(4*a^2+b^2)^(1/2)*polylog(2,b*exp(2*x)/((4*a^
2+b^2)^(1/2)-2*a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{b \cosh \relax (x) \sinh \relax (x) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{a+b\,\mathrm {cosh}\relax (x)\,\mathrm {sinh}\relax (x)} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*cosh(x)*sinh(x)),x)

[Out]

int(x/(a + b*cosh(x)*sinh(x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{a + b \sinh {\relax (x )} \cosh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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