3.94 \(\int \frac{\cosh ^3(x)}{a+b \text{csch}(x)} \, dx\)

Optimal. Leaf size=57 \[ \frac{\left (a^2+b^2\right ) \sinh (x)}{a^3}-\frac{b \left (a^2+b^2\right ) \log (a \sinh (x)+b)}{a^4}-\frac{b \sinh ^2(x)}{2 a^2}+\frac{\sinh ^3(x)}{3 a} \]

[Out]

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

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Rubi [A]  time = 0.15953, antiderivative size = 57, normalized size of antiderivative = 1., 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} \[ \frac{\left (a^2+b^2\right ) \sinh (x)}{a^3}-\frac{b \left (a^2+b^2\right ) \log (a \sinh (x)+b)}{a^4}-\frac{b \sinh ^2(x)}{2 a^2}+\frac{\sinh ^3(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(a + b*Csch[x]),x]

[Out]

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

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.107951, size = 56, normalized size = 0.98 \[ \frac{6 a \left (a^2+b^2\right ) \sinh (x)-6 b \left (a^2+b^2\right ) \log (a \sinh (x)+b)-3 a^2 b \sinh ^2(x)+2 a^3 \sinh ^3(x)}{6 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.035, size = 274, normalized size = 4.8 \begin{align*} -{\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}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{b}{2\,{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{{b}^{2}}{{a}^{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{b}{{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{{b}^{3}}{{a}^{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{b}{{a}^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-2\,a\tanh \left ( x/2 \right ) -b \right ) }-{\frac{{b}^{3}}{{a}^{4}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-2\,a\tanh \left ( x/2 \right ) -b \right ) }-{\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}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{b}{2\,{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{{b}^{2}}{{a}^{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{b}{{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{{b}^{3}}{{a}^{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.00699, size = 171, normalized size = 3. \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(a+b*csch(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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Fricas [B]  time = 1.61395, size = 1251, normalized size = 21.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(a+b*csch(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*sinh
(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh ^{3}{\left (x \right )}}{a + b \operatorname{csch}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**3/(a+b*csch(x)),x)

[Out]

Integral(cosh(x)**3/(a + b*csch(x)), x)

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Giac [A]  time = 1.15823, size = 131, normalized size = 2.3 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(a+b*csch(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