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

Optimal. Leaf size=112 \[ \frac {2 b^4 \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b}}-\frac {b \sinh (x) \cosh (x)}{2 a^2}-\frac {b x \left (a^2+2 b^2\right )}{2 a^4}+\frac {\left (2 a^2+3 b^2\right ) \sinh (x)}{3 a^3}+\frac {\sinh (x) \cosh ^2(x)}{3 a} \]

[Out]

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

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Rubi [A]  time = 0.42, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3853, 4104, 3919, 3831, 2659, 205} \[ -\frac {b x \left (a^2+2 b^2\right )}{2 a^4}+\frac {\left (2 a^2+3 b^2\right ) \sinh (x)}{3 a^3}+\frac {2 b^4 \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b}}-\frac {b \sinh (x) \cosh (x)}{2 a^2}+\frac {\sinh (x) \cosh ^2(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(a + b*Sech[x]),x]

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3853

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(a*f*n), x] - Dist[1/(a*d*n), Int[((d*Csc[e + f*x])^(n + 1)*Simp[b*n - a*(n + 1)*Csc[e
+ f*x] - b*(n + 1)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 -
b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {\cosh ^3(x)}{a+b \text {sech}(x)} \, dx &=\frac {\cosh ^2(x) \sinh (x)}{3 a}+\frac {\int \frac {\cosh ^2(x) \left (-3 b+2 a \text {sech}(x)+2 b \text {sech}^2(x)\right )}{a+b \text {sech}(x)} \, dx}{3 a}\\ &=-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh ^2(x) \sinh (x)}{3 a}-\frac {\int \frac {\cosh (x) \left (-2 \left (2 a^2+3 b^2\right )-a b \text {sech}(x)+3 b^2 \text {sech}^2(x)\right )}{a+b \text {sech}(x)} \, dx}{6 a^2}\\ &=\frac {\left (2 a^2+3 b^2\right ) \sinh (x)}{3 a^3}-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh ^2(x) \sinh (x)}{3 a}+\frac {\int \frac {-3 b \left (a^2+2 b^2\right )-3 a b^2 \text {sech}(x)}{a+b \text {sech}(x)} \, dx}{6 a^3}\\ &=-\frac {b \left (a^2+2 b^2\right ) x}{2 a^4}+\frac {\left (2 a^2+3 b^2\right ) \sinh (x)}{3 a^3}-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh ^2(x) \sinh (x)}{3 a}+\frac {b^4 \int \frac {\text {sech}(x)}{a+b \text {sech}(x)} \, dx}{a^4}\\ &=-\frac {b \left (a^2+2 b^2\right ) x}{2 a^4}+\frac {\left (2 a^2+3 b^2\right ) \sinh (x)}{3 a^3}-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh ^2(x) \sinh (x)}{3 a}+\frac {b^3 \int \frac {1}{1+\frac {a \cosh (x)}{b}} \, dx}{a^4}\\ &=-\frac {b \left (a^2+2 b^2\right ) x}{2 a^4}+\frac {\left (2 a^2+3 b^2\right ) \sinh (x)}{3 a^3}-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh ^2(x) \sinh (x)}{3 a}+\frac {\left (2 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {a}{b}-\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^4}\\ &=-\frac {b \left (a^2+2 b^2\right ) x}{2 a^4}+\frac {2 b^4 \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b}}+\frac {\left (2 a^2+3 b^2\right ) \sinh (x)}{3 a^3}-\frac {b \cosh (x) \sinh (x)}{2 a^2}+\frac {\cosh ^2(x) \sinh (x)}{3 a}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 99, normalized size = 0.88 \[ \frac {a^3 \sinh (3 x)-6 b x \left (a^2+2 b^2\right )+3 a \left (3 a^2+4 b^2\right ) \sinh (x)-\frac {24 b^4 \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-3 a^2 b \sinh (2 x)}{12 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*b*(a^2 + 2*b^2)*x - (24*b^4*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 3*a*(3*a^2 + 4
*b^2)*Sinh[x] - 3*a^2*b*Sinh[2*x] + a^3*Sinh[3*x])/(12*a^4)

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fricas [B]  time = 0.46, size = 1562, normalized size = 13.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*((a^5 - a^3*b^2)*cosh(x)^6 + (a^5 - a^3*b^2)*sinh(x)^6 - 3*(a^4*b - a^2*b^3)*cosh(x)^5 - 3*(a^4*b - a^2*
b^3 - 2*(a^5 - a^3*b^2)*cosh(x))*sinh(x)^5 - a^5 + a^3*b^2 - 12*(a^4*b + a^2*b^3 - 2*b^5)*x*cosh(x)^3 + 3*(3*a
^5 + a^3*b^2 - 4*a*b^4)*cosh(x)^4 + 3*(3*a^5 + a^3*b^2 - 4*a*b^4 + 5*(a^5 - a^3*b^2)*cosh(x)^2 - 5*(a^4*b - a^
2*b^3)*cosh(x))*sinh(x)^4 + 2*(10*(a^5 - a^3*b^2)*cosh(x)^3 - 15*(a^4*b - a^2*b^3)*cosh(x)^2 - 6*(a^4*b + a^2*
b^3 - 2*b^5)*x + 6*(3*a^5 + a^3*b^2 - 4*a*b^4)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + a^3*b^2 - 4*a*b^4)*cosh(x)^2 -
3*(3*a^5 + a^3*b^2 - 4*a*b^4 - 5*(a^5 - a^3*b^2)*cosh(x)^4 + 10*(a^4*b - a^2*b^3)*cosh(x)^3 + 12*(a^4*b + a^2*
b^3 - 2*b^5)*x*cosh(x) - 6*(3*a^5 + a^3*b^2 - 4*a*b^4)*cosh(x)^2)*sinh(x)^2 - 24*(b^4*cosh(x)^3 + 3*b^4*cosh(x
)^2*sinh(x) + 3*b^4*cosh(x)*sinh(x)^2 + b^4*sinh(x)^3)*sqrt(-a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2
*a*b*cosh(x) - a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(-a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(
a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sinh(x) + a)) + 3*(a^4*b - a^2*b^3)*cosh(x) + 3*(2
*(a^5 - a^3*b^2)*cosh(x)^5 + a^4*b - a^2*b^3 - 5*(a^4*b - a^2*b^3)*cosh(x)^4 - 12*(a^4*b + a^2*b^3 - 2*b^5)*x*
cosh(x)^2 + 4*(3*a^5 + a^3*b^2 - 4*a*b^4)*cosh(x)^3 - 2*(3*a^5 + a^3*b^2 - 4*a*b^4)*cosh(x))*sinh(x))/((a^6 -
a^4*b^2)*cosh(x)^3 + 3*(a^6 - a^4*b^2)*cosh(x)^2*sinh(x) + 3*(a^6 - a^4*b^2)*cosh(x)*sinh(x)^2 + (a^6 - a^4*b^
2)*sinh(x)^3), 1/24*((a^5 - a^3*b^2)*cosh(x)^6 + (a^5 - a^3*b^2)*sinh(x)^6 - 3*(a^4*b - a^2*b^3)*cosh(x)^5 - 3
*(a^4*b - a^2*b^3 - 2*(a^5 - a^3*b^2)*cosh(x))*sinh(x)^5 - a^5 + a^3*b^2 - 12*(a^4*b + a^2*b^3 - 2*b^5)*x*cosh
(x)^3 + 3*(3*a^5 + a^3*b^2 - 4*a*b^4)*cosh(x)^4 + 3*(3*a^5 + a^3*b^2 - 4*a*b^4 + 5*(a^5 - a^3*b^2)*cosh(x)^2 -
5*(a^4*b - a^2*b^3)*cosh(x))*sinh(x)^4 + 2*(10*(a^5 - a^3*b^2)*cosh(x)^3 - 15*(a^4*b - a^2*b^3)*cosh(x)^2 - 6
*(a^4*b + a^2*b^3 - 2*b^5)*x + 6*(3*a^5 + a^3*b^2 - 4*a*b^4)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + a^3*b^2 - 4*a*b^4
)*cosh(x)^2 - 3*(3*a^5 + a^3*b^2 - 4*a*b^4 - 5*(a^5 - a^3*b^2)*cosh(x)^4 + 10*(a^4*b - a^2*b^3)*cosh(x)^3 + 12
*(a^4*b + a^2*b^3 - 2*b^5)*x*cosh(x) - 6*(3*a^5 + a^3*b^2 - 4*a*b^4)*cosh(x)^2)*sinh(x)^2 - 48*(b^4*cosh(x)^3
+ 3*b^4*cosh(x)^2*sinh(x) + 3*b^4*cosh(x)*sinh(x)^2 + b^4*sinh(x)^3)*sqrt(a^2 - b^2)*arctan(-(a*cosh(x) + a*si
nh(x) + b)/sqrt(a^2 - b^2)) + 3*(a^4*b - a^2*b^3)*cosh(x) + 3*(2*(a^5 - a^3*b^2)*cosh(x)^5 + a^4*b - a^2*b^3 -
5*(a^4*b - a^2*b^3)*cosh(x)^4 - 12*(a^4*b + a^2*b^3 - 2*b^5)*x*cosh(x)^2 + 4*(3*a^5 + a^3*b^2 - 4*a*b^4)*cosh
(x)^3 - 2*(3*a^5 + a^3*b^2 - 4*a*b^4)*cosh(x))*sinh(x))/((a^6 - a^4*b^2)*cosh(x)^3 + 3*(a^6 - a^4*b^2)*cosh(x)
^2*sinh(x) + 3*(a^6 - a^4*b^2)*cosh(x)*sinh(x)^2 + (a^6 - a^4*b^2)*sinh(x)^3)]

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giac [A]  time = 0.14, size = 133, normalized size = 1.19 \[ \frac {2 \, b^{4} \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} a^{4}} + \frac {a^{2} e^{\left (3 \, x\right )} - 3 \, a b e^{\left (2 \, x\right )} + 9 \, a^{2} e^{x} + 12 \, b^{2} e^{x}}{24 \, a^{3}} - \frac {{\left (a^{2} b + 2 \, b^{3}\right )} x}{2 \, a^{4}} + \frac {{\left (3 \, a^{2} b e^{x} - a^{3} - 3 \, {\left (3 \, a^{3} + 4 \, a b^{2}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{24 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^4*arctan((a*e^x + b)/sqrt(a^2 - b^2))/(sqrt(a^2 - b^2)*a^4) + 1/24*(a^2*e^(3*x) - 3*a*b*e^(2*x) + 9*a^2*e^
x + 12*b^2*e^x)/a^3 - 1/2*(a^2*b + 2*b^3)*x/a^4 + 1/24*(3*a^2*b*e^x - a^3 - 3*(3*a^3 + 4*a*b^2)*e^(2*x))*e^(-3
*x)/a^4

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maple [B]  time = 0.15, size = 264, normalized size = 2.36 \[ -\frac {1}{3 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {1}{a \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b^{2}}{a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 a^{2}}+\frac {b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a^{4}}+\frac {2 b^{4} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {1}{3 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {1}{a \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {b^{2}}{a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 a^{2}}-\frac {b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 1.71, size = 209, normalized size = 1.87 \[ \frac {{\mathrm {e}}^{3\,x}}{24\,a}-\frac {{\mathrm {e}}^{-3\,x}}{24\,a}-\frac {x\,\left (a^2\,b+2\,b^3\right )}{2\,a^4}+\frac {{\mathrm {e}}^x\,\left (3\,a^2+4\,b^2\right )}{8\,a^3}+\frac {b\,{\mathrm {e}}^{-2\,x}}{8\,a^2}-\frac {b\,{\mathrm {e}}^{2\,x}}{8\,a^2}-\frac {{\mathrm {e}}^{-x}\,\left (3\,a^2+4\,b^2\right )}{8\,a^3}+\frac {b^4\,\ln \left (-\frac {2\,b^4\,{\mathrm {e}}^x}{a^5}-\frac {2\,b^4\,\left (a+b\,{\mathrm {e}}^x\right )}{a^5\,\sqrt {a+b}\,\sqrt {b-a}}\right )}{a^4\,\sqrt {a+b}\,\sqrt {b-a}}-\frac {b^4\,\ln \left (\frac {2\,b^4\,\left (a+b\,{\mathrm {e}}^x\right )}{a^5\,\sqrt {a+b}\,\sqrt {b-a}}-\frac {2\,b^4\,{\mathrm {e}}^x}{a^5}\right )}{a^4\,\sqrt {a+b}\,\sqrt {b-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(a + b/cosh(x)),x)

[Out]

exp(3*x)/(24*a) - exp(-3*x)/(24*a) - (x*(a^2*b + 2*b^3))/(2*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) - (exp(-x)*(3*a^2 + 4*b^2))/(8*a^3) + (b^4*log(- (2*b^4*exp(x))/a^5 - (2*
b^4*(a + b*exp(x)))/(a^5*(a + b)^(1/2)*(b - a)^(1/2))))/(a^4*(a + b)^(1/2)*(b - a)^(1/2)) - (b^4*log((2*b^4*(a
+ b*exp(x)))/(a^5*(a + b)^(1/2)*(b - a)^(1/2)) - (2*b^4*exp(x))/a^5))/(a^4*(a + b)^(1/2)*(b - a)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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