3.61 \(\int \frac {\sinh ^3(x)}{a+b \text {sech}(x)} \, dx\)

Optimal. Leaf size=61 \[ -\frac {b \cosh ^2(x)}{2 a^2}+\frac {b \left (a^2-b^2\right ) \log (a \cosh (x)+b)}{a^4}-\frac {\left (a^2-b^2\right ) \cosh (x)}{a^3}+\frac {\cosh ^3(x)}{3 a} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3872, 2837, 12, 772} \[ -\frac {\left (a^2-b^2\right ) \cosh (x)}{a^3}+\frac {b \left (a^2-b^2\right ) \log (a \cosh (x)+b)}{a^4}-\frac {b \cosh ^2(x)}{2 a^2}+\frac {\cosh ^3(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(a + b*Sech[x]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 772

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

Rule 2837

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 66, normalized size = 1.08 \[ \frac {\left (12 a b^2-9 a^3\right ) \cosh (x)+a^3 \cosh (3 x)-3 a^2 b \cosh (2 x)+12 a^2 b \log (a \cosh (x)+b)-12 b^3 \log (a \cosh (x)+b)}{12 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-9*a^3 + 12*a*b^2)*Cosh[x] - 3*a^2*b*Cosh[2*x] + a^3*Cosh[3*x] + 12*a^2*b*Log[b + a*Cosh[x]] - 12*b^3*Log[b
+ a*Cosh[x]])/(12*a^4)

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fricas [B]  time = 0.41, size = 490, normalized size = 8.03 \[ \frac {a^{3} \cosh \relax (x)^{6} + a^{3} \sinh \relax (x)^{6} - 3 \, a^{2} b \cosh \relax (x)^{5} + 3 \, {\left (2 \, a^{3} \cosh \relax (x) - a^{2} b\right )} \sinh \relax (x)^{5} - 24 \, {\left (a^{2} b - b^{3}\right )} x \cosh \relax (x)^{3} - 3 \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} \cosh \relax (x)^{4} + 3 \, {\left (5 \, a^{3} \cosh \relax (x)^{2} - 5 \, a^{2} b \cosh \relax (x) - 3 \, a^{3} + 4 \, a b^{2}\right )} \sinh \relax (x)^{4} - 3 \, a^{2} b \cosh \relax (x) + 2 \, {\left (10 \, a^{3} \cosh \relax (x)^{3} - 15 \, a^{2} b \cosh \relax (x)^{2} - 12 \, {\left (a^{2} b - b^{3}\right )} x - 6 \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x)^{3} + a^{3} - 3 \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} \cosh \relax (x)^{2} + 3 \, {\left (5 \, a^{3} \cosh \relax (x)^{4} - 10 \, a^{2} b \cosh \relax (x)^{3} - 3 \, a^{3} + 4 \, a b^{2} - 24 \, {\left (a^{2} b - b^{3}\right )} x \cosh \relax (x) - 6 \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} \cosh \relax (x)^{2}\right )} \sinh \relax (x)^{2} + 24 \, {\left ({\left (a^{2} b - b^{3}\right )} \cosh \relax (x)^{3} + 3 \, {\left (a^{2} b - b^{3}\right )} \cosh \relax (x)^{2} \sinh \relax (x) + 3 \, {\left (a^{2} b - b^{3}\right )} \cosh \relax (x) \sinh \relax (x)^{2} + {\left (a^{2} b - b^{3}\right )} \sinh \relax (x)^{3}\right )} \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 3 \, {\left (2 \, a^{3} \cosh \relax (x)^{5} - 5 \, a^{2} b \cosh \relax (x)^{4} - 24 \, {\left (a^{2} b - b^{3}\right )} x \cosh \relax (x)^{2} - 4 \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} \cosh \relax (x)^{3} - a^{2} b - 2 \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}{24 \, {\left (a^{4} \cosh \relax (x)^{3} + 3 \, a^{4} \cosh \relax (x)^{2} \sinh \relax (x) + 3 \, a^{4} \cosh \relax (x) \sinh \relax (x)^{2} + a^{4} \sinh \relax (x)^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^3*cosh(x)^6 + a^3*sinh(x)^6 - 3*a^2*b*cosh(x)^5 + 3*(2*a^3*cosh(x) - a^2*b)*sinh(x)^5 - 24*(a^2*b - b^
3)*x*cosh(x)^3 - 3*(3*a^3 - 4*a*b^2)*cosh(x)^4 + 3*(5*a^3*cosh(x)^2 - 5*a^2*b*cosh(x) - 3*a^3 + 4*a*b^2)*sinh(
x)^4 - 3*a^2*b*cosh(x) + 2*(10*a^3*cosh(x)^3 - 15*a^2*b*cosh(x)^2 - 12*(a^2*b - b^3)*x - 6*(3*a^3 - 4*a*b^2)*c
osh(x))*sinh(x)^3 + a^3 - 3*(3*a^3 - 4*a*b^2)*cosh(x)^2 + 3*(5*a^3*cosh(x)^4 - 10*a^2*b*cosh(x)^3 - 3*a^3 + 4*
a*b^2 - 24*(a^2*b - b^3)*x*cosh(x) - 6*(3*a^3 - 4*a*b^2)*cosh(x)^2)*sinh(x)^2 + 24*((a^2*b - b^3)*cosh(x)^3 +
3*(a^2*b - b^3)*cosh(x)^2*sinh(x) + 3*(a^2*b - b^3)*cosh(x)*sinh(x)^2 + (a^2*b - b^3)*sinh(x)^3)*log(2*(a*cosh
(x) + b)/(cosh(x) - sinh(x))) + 3*(2*a^3*cosh(x)^5 - 5*a^2*b*cosh(x)^4 - 24*(a^2*b - b^3)*x*cosh(x)^2 - 4*(3*a
^3 - 4*a*b^2)*cosh(x)^3 - a^2*b - 2*(3*a^3 - 4*a*b^2)*cosh(x))*sinh(x))/(a^4*cosh(x)^3 + 3*a^4*cosh(x)^2*sinh(
x) + 3*a^4*cosh(x)*sinh(x)^2 + a^4*sinh(x)^3)

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giac [A]  time = 0.14, size = 87, normalized size = 1.43 \[ \frac {a^{2} {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 3 \, a b {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 12 \, a^{2} {\left (e^{\left (-x\right )} + e^{x}\right )} + 12 \, b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )}}{24 \, a^{3}} + \frac {{\left (a^{2} b - b^{3}\right )} \log \left ({\left | a {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^2*(e^(-x) + e^x)^3 - 3*a*b*(e^(-x) + e^x)^2 - 12*a^2*(e^(-x) + e^x) + 12*b^2*(e^(-x) + e^x))/a^3 + (a^
2*b - b^3)*log(abs(a*(e^(-x) + e^x) + 2*b))/a^4

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maple [B]  time = 0.12, size = 361, normalized size = 5.92 \[ -\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a^{2}}+\frac {b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a^{4}}-\frac {1}{3 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b^{2}}{a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b^{2}}{a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a^{2}}+\frac {b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a^{4}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{3 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {b \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +a +b \right )}{a \left (a -b \right )}-\frac {b^{2} \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +a +b \right )}{a^{2} \left (a -b \right )}-\frac {b^{3} \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +a +b \right )}{a^{3} \left (a -b \right )}+\frac {b^{4} \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +a +b \right )}{a^{4} \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b/a^2*ln(tanh(1/2*x)-1)+b^3/a^4*ln(tanh(1/2*x)-1)-1/3/a/(tanh(1/2*x)-1)^3-1/2/a/(tanh(1/2*x)-1)^2-1/2/a^2/(ta
nh(1/2*x)-1)^2*b+1/2/a/(tanh(1/2*x)-1)-1/2/a^2/(tanh(1/2*x)-1)*b-1/a^3/(tanh(1/2*x)-1)*b^2-1/2/a/(tanh(1/2*x)+
1)+1/2/a^2/(tanh(1/2*x)+1)*b+1/a^3/(tanh(1/2*x)+1)*b^2-b/a^2*ln(tanh(1/2*x)+1)+b^3/a^4*ln(tanh(1/2*x)+1)-1/2/a
/(tanh(1/2*x)+1)^2-1/2/a^2/(tanh(1/2*x)+1)^2*b+1/3/a/(tanh(1/2*x)+1)^3+b/a/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1/2*x
)^2*b+a+b)-b^2/a^2/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b+a+b)-b^3/a^3/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^
2*b+a+b)+b^4/a^4/(a-b)*ln(a*tanh(1/2*x)^2-tanh(1/2*x)^2*b+a+b)

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maxima [B]  time = 0.32, size = 128, normalized size = 2.10 \[ -\frac {{\left (3 \, a b e^{\left (-x\right )} - a^{2} + 3 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )} e^{\left (-2 \, x\right )}\right )} e^{\left (3 \, x\right )}}{24 \, a^{3}} - \frac {3 \, a b e^{\left (-2 \, x\right )} - a^{2} e^{\left (-3 \, x\right )} + 3 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )} e^{\left (-x\right )}}{24 \, a^{3}} + \frac {{\left (a^{2} b - b^{3}\right )} x}{a^{4}} + \frac {{\left (a^{2} b - b^{3}\right )} \log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*a*b*e^(-x) - a^2 + 3*(3*a^2 - 4*b^2)*e^(-2*x))*e^(3*x)/a^3 - 1/24*(3*a*b*e^(-2*x) - a^2*e^(-3*x) + 3*
(3*a^2 - 4*b^2)*e^(-x))/a^3 + (a^2*b - b^3)*x/a^4 + (a^2*b - b^3)*log(2*b*e^(-x) + a*e^(-2*x) + a)/a^4

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mupad [B]  time = 1.60, size = 123, normalized size = 2.02 \[ \frac {{\mathrm {e}}^{-3\,x}}{24\,a}+\frac {{\mathrm {e}}^{3\,x}}{24\,a}-\frac {x\,\left (a^2\,b-b^3\right )}{a^4}-\frac {{\mathrm {e}}^x\,\left (3\,a^2-4\,b^2\right )}{8\,a^3}-\frac {b\,{\mathrm {e}}^{-2\,x}}{8\,a^2}-\frac {b\,{\mathrm {e}}^{2\,x}}{8\,a^2}+\frac {\ln \left (a+2\,b\,{\mathrm {e}}^x+a\,{\mathrm {e}}^{2\,x}\right )\,\left (a^2\,b-b^3\right )}{a^4}-\frac {{\mathrm {e}}^{-x}\,\left (3\,a^2-4\,b^2\right )}{8\,a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(a + b/cosh(x)),x)

[Out]

exp(-3*x)/(24*a) + exp(3*x)/(24*a) - (x*(a^2*b - b^3))/a^4 - (exp(x)*(3*a^2 - 4*b^2))/(8*a^3) - (b*exp(-2*x))/
(8*a^2) - (b*exp(2*x))/(8*a^2) + (log(a + 2*b*exp(x) + a*exp(2*x))*(a^2*b - b^3))/a^4 - (exp(-x)*(3*a^2 - 4*b^
2))/(8*a^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{3}{\relax (x )}}{a + b \operatorname {sech}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**3/(a+b*sech(x)),x)

[Out]

Integral(sinh(x)**3/(a + b*sech(x)), x)

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