3.82 \(\int \frac {\sinh ^2(x)}{a+b \tanh (x)} \, dx\)

Optimal. Leaf size=84 \[ \frac {a^2 b \log (a+b \tanh (x))}{\left (a^2-b^2\right )^2}-\frac {\cosh ^2(x) (b-a \tanh (x))}{2 \left (a^2-b^2\right )}+\frac {a \log (1-\tanh (x))}{4 (a+b)^2}-\frac {a \log (\tanh (x)+1)}{4 (a-b)^2} \]

[Out]

1/4*a*ln(1-tanh(x))/(a+b)^2-1/4*a*ln(1+tanh(x))/(a-b)^2+a^2*b*ln(a+b*tanh(x))/(a^2-b^2)^2-1/2*cosh(x)^2*(b-a*t
anh(x))/(a^2-b^2)

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Rubi [A]  time = 0.16, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3516, 1647, 801} \[ \frac {a^2 b \log (a+b \tanh (x))}{\left (a^2-b^2\right )^2}-\frac {\cosh ^2(x) (b-a \tanh (x))}{2 \left (a^2-b^2\right )}+\frac {a \log (1-\tanh (x))}{4 (a+b)^2}-\frac {a \log (\tanh (x)+1)}{4 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(a + b*Tanh[x]),x]

[Out]

(a*Log[1 - Tanh[x]])/(4*(a + b)^2) - (a*Log[1 + Tanh[x]])/(4*(a - b)^2) + (a^2*b*Log[a + b*Tanh[x]])/(a^2 - b^
2)^2 - (Cosh[x]^2*(b - a*Tanh[x]))/(2*(a^2 - b^2))

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 73, normalized size = 0.87 \[ \frac {\left (b^3-a^2 b\right ) \cosh (2 x)+a \left (-2 x \left (a^2+b^2\right )+\left (a^2-b^2\right ) \sinh (2 x)+4 a b \log (a \cosh (x)+b \sinh (x))\right )}{4 (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^2/(a + b*Tanh[x]),x]

[Out]

((-(a^2*b) + b^3)*Cosh[2*x] + a*(-2*(a^2 + b^2)*x + 4*a*b*Log[a*Cosh[x] + b*Sinh[x]] + (a^2 - b^2)*Sinh[2*x]))
/(4*(a - b)^2*(a + b)^2)

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fricas [B]  time = 0.81, size = 334, normalized size = 3.98 \[ \frac {{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x)^{4} + 4 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \relax (x)^{4} - 4 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} x \cosh \relax (x)^{2} - a^{3} - a^{2} b + a b^{2} + b^{3} + 2 \, {\left (3 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x)^{2} - 2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} x\right )} \sinh \relax (x)^{2} + 8 \, {\left (a^{2} b \cosh \relax (x)^{2} + 2 \, a^{2} b \cosh \relax (x) \sinh \relax (x) + a^{2} b \sinh \relax (x)^{2}\right )} \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 4 \, {\left ({\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \relax (x)^{3} - 2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} x \cosh \relax (x)\right )} \sinh \relax (x)}{8 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(a+b*tanh(x)),x, algorithm="fricas")

[Out]

1/8*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^4 + 4*(a^3 - a^2*b - a*b^2 + b^3)*cosh(x)*sinh(x)^3 + (a^3 - a^2*b -
a*b^2 + b^3)*sinh(x)^4 - 4*(a^3 + 2*a^2*b + a*b^2)*x*cosh(x)^2 - a^3 - a^2*b + a*b^2 + b^3 + 2*(3*(a^3 - a^2*b
- a*b^2 + b^3)*cosh(x)^2 - 2*(a^3 + 2*a^2*b + a*b^2)*x)*sinh(x)^2 + 8*(a^2*b*cosh(x)^2 + 2*a^2*b*cosh(x)*sinh
(x) + a^2*b*sinh(x)^2)*log(2*(a*cosh(x) + b*sinh(x))/(cosh(x) - sinh(x))) + 4*((a^3 - a^2*b - a*b^2 + b^3)*cos
h(x)^3 - 2*(a^3 + 2*a^2*b + a*b^2)*x*cosh(x))*sinh(x))/((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 + 2*(a^4 - 2*a^2*b^2
+ b^4)*cosh(x)*sinh(x) + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^2)

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giac [A]  time = 0.13, size = 101, normalized size = 1.20 \[ \frac {a^{2} b \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {a x}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {{\left (2 \, a e^{\left (2 \, x\right )} - a + b\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {e^{\left (2 \, x\right )}}{8 \, {\left (a + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(a+b*tanh(x)),x, algorithm="giac")

[Out]

a^2*b*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^4 - 2*a^2*b^2 + b^4) - 1/2*a*x/(a^2 - 2*a*b + b^2) + 1/8*(2*a
*e^(2*x) - a + b)*e^(-2*x)/(a^2 - 2*a*b + b^2) + 1/8*e^(2*x)/(a + b)

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maple [A]  time = 0.12, size = 145, normalized size = 1.73 \[ \frac {a^{2} b \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \tanh \left (\frac {x}{2}\right ) b +a \right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}+\frac {4}{\left (8 a +8 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {4}{\left (8 a -8 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {8}{\left (16 a -16 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 \left (a -b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(a+b*tanh(x)),x)

[Out]

a^2*b/(a-b)^2/(a+b)^2*ln(a*tanh(1/2*x)^2+2*tanh(1/2*x)*b+a)+4/(8*a+8*b)/(tanh(1/2*x)-1)^2+8/(16*a+16*b)/(tanh(
1/2*x)-1)+1/2*a/(a+b)^2*ln(tanh(1/2*x)-1)-4/(8*a-8*b)/(tanh(1/2*x)+1)^2+8/(16*a-16*b)/(tanh(1/2*x)+1)-1/2*a/(a
-b)^2*ln(tanh(1/2*x)+1)

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maxima [A]  time = 0.33, size = 83, normalized size = 0.99 \[ \frac {a^{2} b \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {a x}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {e^{\left (2 \, x\right )}}{8 \, {\left (a + b\right )}} - \frac {e^{\left (-2 \, x\right )}}{8 \, {\left (a - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^2/(a+b*tanh(x)),x, algorithm="maxima")

[Out]

a^2*b*log(-(a - b)*e^(-2*x) - a - b)/(a^4 - 2*a^2*b^2 + b^4) - 1/2*a*x/(a^2 + 2*a*b + b^2) + 1/8*e^(2*x)/(a +
b) - 1/8*e^(-2*x)/(a - b)

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mupad [B]  time = 1.43, size = 81, normalized size = 0.96 \[ \frac {{\mathrm {e}}^{2\,x}}{8\,a+8\,b}-\frac {{\mathrm {e}}^{-2\,x}}{8\,a-8\,b}-\frac {a\,x}{2\,{\left (a-b\right )}^2}+\frac {a^2\,b\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^4-2\,a^2\,b^2+b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^2/(a + b*tanh(x)),x)

[Out]

exp(2*x)/(8*a + 8*b) - exp(-2*x)/(8*a - 8*b) - (a*x)/(2*(a - b)^2) + (a^2*b*log(a - b + a*exp(2*x) + b*exp(2*x
)))/(a^4 + b^4 - 2*a^2*b^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{2}{\relax (x )}}{a + b \tanh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**2/(a+b*tanh(x)),x)

[Out]

Integral(sinh(x)**2/(a + b*tanh(x)), x)

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