3.711 \(\int \frac {\cosh ^2(x) \sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx\)

Optimal. Leaf size=194 \[ \frac {b x}{8 \left (a^2-b^2\right )}-\frac {a^2 b x}{2 \left (a^2-b^2\right )^2}+\frac {a \sinh ^4(x)}{4 \left (a^2-b^2\right )}-\frac {a b^2 \sinh ^2(x)}{2 \left (a^2-b^2\right )^2}-\frac {b \sinh (x) \cosh ^3(x)}{4 \left (a^2-b^2\right )}+\frac {b \sinh (x) \cosh (x)}{8 \left (a^2-b^2\right )}+\frac {a^2 b \sinh (x) \cosh (x)}{2 \left (a^2-b^2\right )^2}-\frac {a^2 b^3 x}{\left (a^2-b^2\right )^3}+\frac {a^3 b^2 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3109, 2568, 2635, 8, 2564, 30, 3097, 3133} \[ \frac {b x}{8 \left (a^2-b^2\right )}-\frac {a^2 b x}{2 \left (a^2-b^2\right )^2}-\frac {a^2 b^3 x}{\left (a^2-b^2\right )^3}+\frac {a \sinh ^4(x)}{4 \left (a^2-b^2\right )}-\frac {a b^2 \sinh ^2(x)}{2 \left (a^2-b^2\right )^2}-\frac {b \sinh (x) \cosh ^3(x)}{4 \left (a^2-b^2\right )}+\frac {b \sinh (x) \cosh (x)}{8 \left (a^2-b^2\right )}+\frac {a^2 b \sinh (x) \cosh (x)}{2 \left (a^2-b^2\right )^2}+\frac {a^3 b^2 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cosh[x]^2*Sinh[x]^3)/(a*Cosh[x] + b*Sinh[x]),x]

[Out]

-((a^2*b^3*x)/(a^2 - b^2)^3) - (a^2*b*x)/(2*(a^2 - b^2)^2) + (b*x)/(8*(a^2 - b^2)) + (a^3*b^2*Log[a*Cosh[x] +
b*Sinh[x]])/(a^2 - b^2)^3 + (a^2*b*Cosh[x]*Sinh[x])/(2*(a^2 - b^2)^2) + (b*Cosh[x]*Sinh[x])/(8*(a^2 - b^2)) -
(b*Cosh[x]^3*Sinh[x])/(4*(a^2 - b^2)) - (a*b^2*Sinh[x]^2)/(2*(a^2 - b^2)^2) + (a*Sinh[x]^4)/(4*(a^2 - b^2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3097

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

Rule 3109

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Dist[
a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Dist[(a*b)/(a^2 + b^2), Int[(Cos[c + d*x]^(m
- 1)*Sin[c + d*x]^(n - 1))/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3133

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] + Simp[((c*B - b*C)*L
og[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] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

Rubi steps

\begin {align*} \int \frac {\cosh ^2(x) \sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx &=\frac {a \int \cosh (x) \sinh ^3(x) \, dx}{a^2-b^2}-\frac {b \int \cosh ^2(x) \sinh ^2(x) \, dx}{a^2-b^2}+\frac {(a b) \int \frac {\cosh (x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx}{a^2-b^2}\\ &=-\frac {b \cosh ^3(x) \sinh (x)}{4 \left (a^2-b^2\right )}+\frac {\left (a^2 b\right ) \int \sinh ^2(x) \, dx}{\left (a^2-b^2\right )^2}-\frac {\left (a b^2\right ) \int \cosh (x) \sinh (x) \, dx}{\left (a^2-b^2\right )^2}+\frac {\left (a^2 b^2\right ) \int \frac {\sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^2}+\frac {a \operatorname {Subst}\left (\int x^3 \, dx,x,i \sinh (x)\right )}{a^2-b^2}+\frac {b \int \cosh ^2(x) \, dx}{4 \left (a^2-b^2\right )}\\ &=-\frac {a^2 b^3 x}{\left (a^2-b^2\right )^3}+\frac {a^2 b \cosh (x) \sinh (x)}{2 \left (a^2-b^2\right )^2}+\frac {b \cosh (x) \sinh (x)}{8 \left (a^2-b^2\right )}-\frac {b \cosh ^3(x) \sinh (x)}{4 \left (a^2-b^2\right )}+\frac {a \sinh ^4(x)}{4 \left (a^2-b^2\right )}+\frac {\left (i a^3 b^2\right ) \int \frac {-i b \cosh (x)-i a \sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^3}-\frac {\left (a^2 b\right ) \int 1 \, dx}{2 \left (a^2-b^2\right )^2}+\frac {\left (a b^2\right ) \operatorname {Subst}(\int x \, dx,x,i \sinh (x))}{\left (a^2-b^2\right )^2}+\frac {b \int 1 \, dx}{8 \left (a^2-b^2\right )}\\ &=-\frac {a^2 b^3 x}{\left (a^2-b^2\right )^3}-\frac {a^2 b x}{2 \left (a^2-b^2\right )^2}+\frac {b x}{8 \left (a^2-b^2\right )}+\frac {a^3 b^2 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3}+\frac {a^2 b \cosh (x) \sinh (x)}{2 \left (a^2-b^2\right )^2}+\frac {b \cosh (x) \sinh (x)}{8 \left (a^2-b^2\right )}-\frac {b \cosh ^3(x) \sinh (x)}{4 \left (a^2-b^2\right )}-\frac {a b^2 \sinh ^2(x)}{2 \left (a^2-b^2\right )^2}+\frac {a \sinh ^4(x)}{4 \left (a^2-b^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.64, size = 128, normalized size = 0.66 \[ \frac {-4 a \left (a^4-b^4\right ) \cosh (2 x)+a \left (a^2-b^2\right )^2 \cosh (4 x)-b \left (-8 a^2 \left (a^2-b^2\right ) \sinh (2 x)+\left (a^2-b^2\right )^2 \sinh (4 x)+4 \left (3 a^4 x-8 a^3 b \log (a \cosh (x)+b \sinh (x))+6 a^2 b^2 x-b^4 x\right )\right )}{32 (a-b)^3 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cosh[x]^2*Sinh[x]^3)/(a*Cosh[x] + b*Sinh[x]),x]

[Out]

(-4*a*(a^4 - b^4)*Cosh[2*x] + a*(a^2 - b^2)^2*Cosh[4*x] - b*(4*(3*a^4*x + 6*a^2*b^2*x - b^4*x - 8*a^3*b*Log[a*
Cosh[x] + b*Sinh[x]]) - 8*a^2*(a^2 - b^2)*Sinh[2*x] + (a^2 - b^2)^2*Sinh[4*x]))/(32*(a - b)^3*(a + b)^3)

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fricas [B]  time = 0.46, size = 1158, normalized size = 5.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2*sinh(x)^3/(a*cosh(x)+b*sinh(x)),x, algorithm="fricas")

[Out]

1/64*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^8 + 8*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 +
a*b^4 - b^5)*cosh(x)*sinh(x)^7 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^8 - 4*(a^5 - 2*a
^4*b + 2*a^2*b^3 - a*b^4)*cosh(x)^6 - 4*(a^5 - 2*a^4*b + 2*a^2*b^3 - a*b^4 - 7*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^
2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^6 - 8*(3*a^4*b + 8*a^3*b^2 + 6*a^2*b^3 - b^5)*x*cosh(x)^4 + 8*(7*(a^5
- a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 - 3*(a^5 - 2*a^4*b + 2*a^2*b^3 - a*b^4)*cosh(x))*sinh
(x)^5 + a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 + 2*(35*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4
- b^5)*cosh(x)^4 - 30*(a^5 - 2*a^4*b + 2*a^2*b^3 - a*b^4)*cosh(x)^2 - 4*(3*a^4*b + 8*a^3*b^2 + 6*a^2*b^3 - b^
5)*x)*sinh(x)^4 + 8*(7*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^5 - 10*(a^5 - 2*a^4*b + 2*a
^2*b^3 - a*b^4)*cosh(x)^3 - 4*(3*a^4*b + 8*a^3*b^2 + 6*a^2*b^3 - b^5)*x*cosh(x))*sinh(x)^3 - 4*(a^5 + 2*a^4*b
- 2*a^2*b^3 - a*b^4)*cosh(x)^2 + 4*(7*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 - a^5 - 2*
a^4*b + 2*a^2*b^3 + a*b^4 - 15*(a^5 - 2*a^4*b + 2*a^2*b^3 - a*b^4)*cosh(x)^4 - 12*(3*a^4*b + 8*a^3*b^2 + 6*a^2
*b^3 - b^5)*x*cosh(x)^2)*sinh(x)^2 + 64*(a^3*b^2*cosh(x)^4 + 4*a^3*b^2*cosh(x)^3*sinh(x) + 6*a^3*b^2*cosh(x)^2
*sinh(x)^2 + 4*a^3*b^2*cosh(x)*sinh(x)^3 + a^3*b^2*sinh(x)^4)*log(2*(a*cosh(x) + b*sinh(x))/(cosh(x) - sinh(x)
)) + 8*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^7 - 3*(a^5 - 2*a^4*b + 2*a^2*b^3 - a*b^4)*
cosh(x)^5 - 4*(3*a^4*b + 8*a^3*b^2 + 6*a^2*b^3 - b^5)*x*cosh(x)^3 - (a^5 + 2*a^4*b - 2*a^2*b^3 - a*b^4)*cosh(x
))*sinh(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^4 + 4*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3*s
inh(x) + 6*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x)^2 + 4*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos
h(x)*sinh(x)^3 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^4)

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giac [A]  time = 0.12, size = 199, normalized size = 1.03 \[ \frac {a^{3} b^{2} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (3 \, a b - b^{2}\right )} x}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {{\left (18 \, a b e^{\left (4 \, x\right )} - 6 \, b^{2} e^{\left (4 \, x\right )} - 4 \, a^{2} e^{\left (2 \, x\right )} + 4 \, a b e^{\left (2 \, x\right )} + a^{2} - 2 \, a b + b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} - 4 \, a e^{\left (2 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2*sinh(x)^3/(a*cosh(x)+b*sinh(x)),x, algorithm="giac")

[Out]

a^3*b^2*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - 1/8*(3*a*b - b^2)*x/(a^3
- 3*a^2*b + 3*a*b^2 - b^3) + 1/64*(18*a*b*e^(4*x) - 6*b^2*e^(4*x) - 4*a^2*e^(2*x) + 4*a*b*e^(2*x) + a^2 - 2*a
*b + b^2)*e^(-4*x)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 1/64*(a*e^(4*x) + b*e^(4*x) - 4*a*e^(2*x))/(a^2 + 2*a*b +
b^2)

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maple [A]  time = 0.23, size = 321, normalized size = 1.65 \[ \frac {4}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {16}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}+\frac {a}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {a}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {3 b \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a}{8 \left (a +b \right )^{3}}+\frac {b^{2} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}+\frac {a^{3} b^{2} \ln \left (a +2 \tanh \left (\frac {x}{2}\right ) b +a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}+\frac {4}{\left (16 a -16 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {16}{\left (32 a -32 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {a}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {a}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 b \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a}{8 \left (a -b \right )^{3}}+\frac {b^{2} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2*sinh(x)^3/(a*cosh(x)+b*sinh(x)),x)

[Out]

4/(16*a+16*b)/(tanh(1/2*x)-1)^4+16/(32*a+32*b)/(tanh(1/2*x)-1)^3+1/8/(a+b)^2/(tanh(1/2*x)-1)^2*a+3/8/(a+b)^2/(
tanh(1/2*x)-1)^2*b-1/8*a/(a+b)^2/(tanh(1/2*x)-1)+1/8*b/(a+b)^2/(tanh(1/2*x)-1)+3/8*b/(a+b)^3*ln(tanh(1/2*x)-1)
*a+1/8*b^2/(a+b)^3*ln(tanh(1/2*x)-1)+a^3*b^2/(a-b)^3/(a+b)^3*ln(a+2*tanh(1/2*x)*b+a*tanh(1/2*x)^2)+4/(16*a-16*
b)/(tanh(1/2*x)+1)^4-16/(32*a-32*b)/(tanh(1/2*x)+1)^3+1/8*a/(a-b)^2/(tanh(1/2*x)+1)+1/8*b/(a-b)^2/(tanh(1/2*x)
+1)+1/8/(a-b)^2/(tanh(1/2*x)+1)^2*a-3/8/(a-b)^2/(tanh(1/2*x)+1)^2*b-3/8*b/(a-b)^3*ln(tanh(1/2*x)+1)*a+1/8*b^2/
(a-b)^3*ln(tanh(1/2*x)+1)

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maxima [A]  time = 0.50, size = 153, normalized size = 0.79 \[ \frac {a^{3} b^{2} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (3 \, a b + b^{2}\right )} x}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (4 \, a e^{\left (-2 \, x\right )} - a - b\right )} e^{\left (4 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {4 \, a e^{\left (-2 \, x\right )} - {\left (a - b\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2*sinh(x)^3/(a*cosh(x)+b*sinh(x)),x, algorithm="maxima")

[Out]

a^3*b^2*log(-(a - b)*e^(-2*x) - a - b)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) - 1/8*(3*a*b + b^2)*x/(a^3 + 3*a^2*
b + 3*a*b^2 + b^3) - 1/64*(4*a*e^(-2*x) - a - b)*e^(4*x)/(a^2 + 2*a*b + b^2) - 1/64*(4*a*e^(-2*x) - (a - b)*e^
(-4*x))/(a^2 - 2*a*b + b^2)

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mupad [B]  time = 1.87, size = 127, normalized size = 0.65 \[ \frac {{\mathrm {e}}^{-4\,x}}{64\,a-64\,b}+\frac {{\mathrm {e}}^{4\,x}}{64\,a+64\,b}-\frac {x\,\left (3\,a\,b-b^2\right )}{8\,{\left (a-b\right )}^3}-\frac {a\,{\mathrm {e}}^{2\,x}}{16\,{\left (a+b\right )}^2}-\frac {a\,{\mathrm {e}}^{-2\,x}}{16\,{\left (a-b\right )}^2}+\frac {a^3\,b^2\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(x)^2*sinh(x)^3)/(a*cosh(x) + b*sinh(x)),x)

[Out]

exp(-4*x)/(64*a - 64*b) + exp(4*x)/(64*a + 64*b) - (x*(3*a*b - b^2))/(8*(a - b)^3) - (a*exp(2*x))/(16*(a + b)^
2) - (a*exp(-2*x))/(16*(a - b)^2) + (a^3*b^2*log(a - b + a*exp(2*x) + b*exp(2*x)))/(a^6 - b^6 + 3*a^2*b^4 - 3*
a^4*b^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(cosh(x)**2*sinh(x)**3/(a*cosh(x)+b*sinh(x)),x)

[Out]

Timed out

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