3.114 \(\int \frac{\cosh ^3(x)}{a+b \tanh (x)} \, dx\)
Optimal. Leaf size=132 \[ \frac{a \sinh ^3(x)}{3 \left (a^2-b^2\right )}-\frac{a b^2 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac{a \sinh (x)}{a^2-b^2}-\frac{b \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac{b^3 \cosh (x)}{\left (a^2-b^2\right )^2}+\frac{b^4 \tan ^{-1}\left (\frac{\cosh (x) (a \tanh (x)+b)}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]
[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))
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Rubi [A] time = 0.15572, antiderivative size = 132, normalized size of antiderivative = 1.,
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 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]
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 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 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 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])
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*}
Mathematica [A] time = 0.45916, size = 258, normalized size = 1.95 \[ \frac{-21 a^2 b^2 \sqrt{a-b} \sinh (x)-a^2 b^2 \sqrt{a-b} \sinh (3 x)-3 b \sqrt{a-b} \left (a^2 b+a^3-5 a b^2-5 b^3\right ) \cosh (x)+9 a^4 \sqrt{a-b} \sinh (x)+a^4 \sqrt{a-b} \sinh (3 x)+9 a^3 b \sqrt{a-b} \sinh (x)+a^3 b \sqrt{a-b} \sinh (3 x)-21 a b^3 \sqrt{a-b} \sinh (x)-a b^3 \sqrt{a-b} \sinh (3 x)+24 b^4 \sqrt{a+b} \tan ^{-1}\left (\frac{a \tanh \left (\frac{x}{2}\right )+b}{\sqrt{a-b} \sqrt{a+b}}\right )-b (a-b)^{3/2} (a+b)^2 \cosh (3 x)}{12 (a-b)^{5/2} (a+b)^3} \]
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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Maple [A] time = 0.044, size = 198, normalized size = 1.5 \begin{align*} -{\frac{2}{6\,a-6\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{2\,a-2\,b} \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 ) ^{-1}}+{\frac{3\,b}{2\, \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+2\,{\frac{{b}^{4}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{2}{6\,a+6\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{2\,b+2\,a} \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 ) ^{-1}}-{\frac{3\,b}{2\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cosh(x)^3/(a+b*tanh(x)),x)
[Out]
-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+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
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
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
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Fricas [B] time = 2.56586, size = 4251, normalized size = 32.2 \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*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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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh ^{3}{\left (x \right )}}{a + b \tanh{\left (x \right )}}\, dx \end{align*}
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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Giac [A] time = 1.26922, size = 219, normalized size = 1.66 \begin{align*} \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 )}} \end{align*}
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)