3.114 \(\int \frac {\cosh ^3(x)}{a+b \tanh (x)} \, dx\)

Optimal. Leaf size=132 \[ \frac {a \sinh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {a b^2 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {a \sinh (x)}{a^2-b^2}-\frac {b \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {b^4 \tan ^{-1}\left (\frac {\cosh (x) (a \tanh (x)+b)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {b^3 \cosh (x)}{\left (a^2-b^2\right )^2} \]

[Out]

b^4*arctan(cosh(x)*(b+a*tanh(x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)+b^3*cosh(x)/(a^2-b^2)^2-1/3*b*cosh(x)^3/(a^2
-b^2)-a*b^2*sinh(x)/(a^2-b^2)^2+a*sinh(x)/(a^2-b^2)+1/3*a*sinh(x)^3/(a^2-b^2)

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Rubi [A]  time = 0.16, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3511, 3486, 2633, 2637, 3509, 206} \[ \frac {a \sinh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {a b^2 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {a \sinh (x)}{a^2-b^2}-\frac {b \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {b^3 \cosh (x)}{\left (a^2-b^2\right )^2}+\frac {b^4 \tan ^{-1}\left (\frac {\cosh (x) (a \tanh (x)+b)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(a + b*Tanh[x]),x]

[Out]

(b^4*ArcTan[(Cosh[x]*(b + a*Tanh[x]))/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (b^3*Cosh[x])/(a^2 - b^2)^2 - (b*C
osh[x]^3)/(3*(a^2 - b^2)) - (a*b^2*Sinh[x])/(a^2 - b^2)^2 + (a*Sinh[x])/(a^2 - b^2) + (a*Sinh[x]^3)/(3*(a^2 -
b^2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3509

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

Rule 3511

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^
2), Int[(d*Sec[e + f*x])^m*(a - b*Tan[e + f*x]), x], x] + Dist[b^2/(d^2*(a^2 + b^2)), Int[(d*Sec[e + f*x])^(m
+ 2)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 + b^2, 0] && ILtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\cosh ^3(x)}{a+b \tanh (x)} \, dx &=\frac {\int \cosh ^3(x) (a-b \tanh (x)) \, dx}{a^2-b^2}-\frac {b^2 \int \frac {\cosh (x)}{a+b \tanh (x)} \, dx}{a^2-b^2}\\ &=-\frac {b \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {b^2 \int \cosh (x) (a-b \tanh (x)) \, dx}{\left (a^2-b^2\right )^2}+\frac {b^4 \int \frac {\text {sech}(x)}{a+b \tanh (x)} \, dx}{\left (a^2-b^2\right )^2}+\frac {a \int \cosh ^3(x) \, dx}{a^2-b^2}\\ &=\frac {b^3 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {b \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {\left (a b^2\right ) \int \cosh (x) \, dx}{\left (a^2-b^2\right )^2}+\frac {\left (i b^4\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,\cosh (x) (-i b-i a \tanh (x))\right )}{\left (a^2-b^2\right )^2}+\frac {(i a) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-i \sinh (x)\right )}{a^2-b^2}\\ &=\frac {b^4 \tan ^{-1}\left (\frac {\cosh (x) (b+a \tanh (x))}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {b^3 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {b \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {a b^2 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {a \sinh (x)}{a^2-b^2}+\frac {a \sinh ^3(x)}{3 \left (a^2-b^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.48, size = 258, normalized size = 1.95 \[ \frac {9 a^4 \sqrt {a-b} \sinh (x)+a^4 \sqrt {a-b} \sinh (3 x)+9 a^3 b \sqrt {a-b} \sinh (x)+a^3 b \sqrt {a-b} \sinh (3 x)-21 a^2 b^2 \sqrt {a-b} \sinh (x)-a^2 b^2 \sqrt {a-b} \sinh (3 x)-3 b \sqrt {a-b} \left (a^3+a^2 b-5 a b^2-5 b^3\right ) \cosh (x)+24 b^4 \sqrt {a+b} \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )-21 a b^3 \sqrt {a-b} \sinh (x)-a b^3 \sqrt {a-b} \sinh (3 x)-b (a-b)^{3/2} (a+b)^2 \cosh (3 x)}{12 (a-b)^{5/2} (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*b^4*Sqrt[a + b]*ArcTan[(b + a*Tanh[x/2])/(Sqrt[a - b]*Sqrt[a + b])] - 3*Sqrt[a - b]*b*(a^3 + a^2*b - 5*a*b
^2 - 5*b^3)*Cosh[x] - (a - b)^(3/2)*b*(a + b)^2*Cosh[3*x] + 9*a^4*Sqrt[a - b]*Sinh[x] + 9*a^3*Sqrt[a - b]*b*Si
nh[x] - 21*a^2*Sqrt[a - b]*b^2*Sinh[x] - 21*a*Sqrt[a - b]*b^3*Sinh[x] + a^4*Sqrt[a - b]*Sinh[3*x] + a^3*Sqrt[a
- b]*b*Sinh[3*x] - a^2*Sqrt[a - b]*b^2*Sinh[3*x] - a*Sqrt[a - b]*b^3*Sinh[3*x])/(12*(a - b)^(5/2)*(a + b)^3)

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fricas [B]  time = 0.71, size = 1871, normalized size = 14.17 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 + 6*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3
+ a*b^4 - b^5)*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)*sinh(x)^6 - a^5 - a^4*b
+ 2*a^3*b^2 + 2*a^2*b^3 - a*b^4 - b^5 + 3*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*cosh(x)^
4 + 3*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5 + 5*(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)^4 + 4*(5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 + 3*(3*a^5
- a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + a^4*b - 10*a^3*b^2 - 6*a^
2*b^3 + 7*a*b^4 + 5*b^5)*cosh(x)^2 - 3*(3*a^5 + a^4*b - 10*a^3*b^2 - 6*a^2*b^3 + 7*a*b^4 + 5*b^5 - 5*(a^5 - a^
4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 - 6*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5
*b^5)*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 + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 - 2*sqrt(-a^2 +
b^2)*(cosh(x) + sinh(x)) - a + b)/((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + a - b))
+ 6*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^5 + 2*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^
3 + 7*a*b^4 - 5*b^5)*cosh(x)^3 - (3*a^5 + a^4*b - 10*a^3*b^2 - 6*a^2*b^3 + 7*a*b^4 + 5*b^5)*cosh(x))*sinh(x))/
((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x) + 3*(
a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^2 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^3), 1/24*((
a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 + 6*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4
- b^5)*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)*sinh(x)^6 - a^5 - a^4*b + 2*a^3
*b^2 + 2*a^2*b^3 - a*b^4 - b^5 + 3*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*cosh(x)^4 + 3*(3
*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5 + 5*(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)^4 + 4*(5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 + 3*(3*a^5 - a^4*b
- 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + a^4*b - 10*a^3*b^2 - 6*a^2*b^3 +
7*a*b^4 + 5*b^5)*cosh(x)^2 - 3*(3*a^5 + a^4*b - 10*a^3*b^2 - 6*a^2*b^3 + 7*a*b^4 + 5*b^5 - 5*(a^5 - a^4*b - 2*
a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 - 6*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*co
sh(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)*sq
rt(a^2 - b^2)*arctan(sqrt(a^2 - b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) + 6*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^
2*b^3 + a*b^4 - b^5)*cosh(x)^5 + 2*(3*a^5 - a^4*b - 10*a^3*b^2 + 6*a^2*b^3 + 7*a*b^4 - 5*b^5)*cosh(x)^3 - (3*a
^5 + a^4*b - 10*a^3*b^2 - 6*a^2*b^3 + 7*a*b^4 + 5*b^5)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*
cosh(x)^3 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x) + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*co
sh(x)*sinh(x)^2 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^3)]

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giac [A]  time = 0.14, size = 162, normalized size = 1.23 \[ \frac {2 \, b^{4} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {{\left (9 \, a e^{\left (2 \, x\right )} - 15 \, b e^{\left (2 \, x\right )} + a - b\right )} e^{\left (-3 \, x\right )}}{24 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {a^{2} e^{\left (3 \, x\right )} + 2 \, a b e^{\left (3 \, x\right )} + b^{2} e^{\left (3 \, x\right )} + 9 \, a^{2} e^{x} + 24 \, a b e^{x} + 15 \, b^{2} e^{x}}{24 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^4*arctan((a*e^x + b*e^x)/sqrt(a^2 - b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) - 1/24*(9*a*e^(2*x) -
15*b*e^(2*x) + a - b)*e^(-3*x)/(a^2 - 2*a*b + b^2) + 1/24*(a^2*e^(3*x) + 2*a*b*e^(3*x) + b^2*e^(3*x) + 9*a^2*e
^x + 24*a*b*e^x + 15*b^2*e^x)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3)

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maple [A]  time = 0.13, size = 198, normalized size = 1.50 \[ \frac {2 b^{4} \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {2}{3 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3} \left (2 b +2 a \right )}-\frac {1}{\left (2 b +2 a \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {a}{\left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {3 b}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {2}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3} \left (2 a -2 b \right )}+\frac {1}{\left (2 a -2 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a}{\left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {3 b}{2 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*tanh(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 = 2.25, size = 221, normalized size = 1.67 \[ \frac {{\mathrm {e}}^{3\,x}}{24\,a+24\,b}-\frac {{\mathrm {e}}^{-3\,x}}{24\,a-24\,b}-\frac {{\mathrm {e}}^{-x}\,\left (3\,a-5\,b\right )}{8\,{\left (a-b\right )}^2}+\frac {{\mathrm {e}}^x\,\left (3\,a+5\,b\right )}{8\,{\left (a+b\right )}^2}-\frac {b^4\,\ln \left (-\frac {2\,b^4\,{\mathrm {e}}^x}{a^5+a^4\,b-2\,a^3\,b^2-2\,a^2\,b^3+a\,b^4+b^5}-\frac {2\,b^4}{{\left (a+b\right )}^{7/2}\,{\left (b-a\right )}^{3/2}}\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}}+\frac {b^4\,\ln \left (\frac {2\,b^4}{{\left (a+b\right )}^{7/2}\,{\left (b-a\right )}^{3/2}}-\frac {2\,b^4\,{\mathrm {e}}^x}{a^5+a^4\,b-2\,a^3\,b^2-2\,a^2\,b^3+a\,b^4+b^5}\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(a + b*tanh(x)),x)

[Out]

exp(3*x)/(24*a + 24*b) - exp(-3*x)/(24*a - 24*b) - (exp(-x)*(3*a - 5*b))/(8*(a - b)^2) + (exp(x)*(3*a + 5*b))/
(8*(a + b)^2) - (b^4*log(- (2*b^4*exp(x))/(a*b^4 + a^4*b + a^5 + b^5 - 2*a^2*b^3 - 2*a^3*b^2) - (2*b^4)/((a +
b)^(7/2)*(b - a)^(3/2))))/((a + b)^(5/2)*(b - a)^(5/2)) + (b^4*log((2*b^4)/((a + b)^(7/2)*(b - a)^(3/2)) - (2*
b^4*exp(x))/(a*b^4 + a^4*b + a^5 + b^5 - 2*a^2*b^3 - 2*a^3*b^2)))/((a + b)^(5/2)*(b - a)^(5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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