Integrand size = 15, antiderivative size = 153 \[ \int \frac {\tanh ^3(x)}{a+b \cosh ^3(x)} \, dx=-\frac {b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \cosh (x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}+\frac {\log (\cosh (x))}{a}+\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \cosh (x)\right )}{3 a^{5/3}}-\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cosh (x)+b^{2/3} \cosh ^2(x)\right )}{6 a^{5/3}}-\frac {\log \left (a+b \cosh ^3(x)\right )}{3 a}+\frac {\text {sech}^2(x)}{2 a} \]
ln(cosh(x))/a+1/3*b^(2/3)*ln(a^(1/3)+b^(1/3)*cosh(x))/a^(5/3)-1/6*b^(2/3)* ln(a^(2/3)-a^(1/3)*b^(1/3)*cosh(x)+b^(2/3)*cosh(x)^2)/a^(5/3)-1/3*ln(a+b*c osh(x)^3)/a+1/2*sech(x)^2/a-1/3*b^(2/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*cosh (x))/a^(1/3)*3^(1/2))/a^(5/3)*3^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.35 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.95 \[ \int \frac {\tanh ^3(x)}{a+b \cosh ^3(x)} \, dx=\frac {-6 x+6 \log (\cosh (x))-2 \text {RootSum}\left [b+3 b \text {$\#$1}^2+8 a \text {$\#$1}^3+3 b \text {$\#$1}^4+b \text {$\#$1}^6\&,\frac {-b x+b \log \left (e^x-\text {$\#$1}\right )-4 a x \text {$\#$1}^3+4 a \log \left (e^x-\text {$\#$1}\right ) \text {$\#$1}^3-3 b x \text {$\#$1}^4+3 b \log \left (e^x-\text {$\#$1}\right ) \text {$\#$1}^4}{b+2 b \text {$\#$1}^2+4 a \text {$\#$1}^3+b \text {$\#$1}^4}\&\right ]+3 \text {sech}^2(x)}{6 a} \]
(-6*x + 6*Log[Cosh[x]] - 2*RootSum[b + 3*b*#1^2 + 8*a*#1^3 + 3*b*#1^4 + b* #1^6 & , (-(b*x) + b*Log[E^x - #1] - 4*a*x*#1^3 + 4*a*Log[E^x - #1]*#1^3 - 3*b*x*#1^4 + 3*b*Log[E^x - #1]*#1^4)/(b + 2*b*#1^2 + 4*a*#1^3 + b*#1^4) & ] + 3*Sech[x]^2)/(6*a)
Time = 0.45 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 26, 3709, 2373, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\tanh ^3(x)}{a+b \cosh ^3(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i}{\tan \left (\frac {\pi }{2}+i x\right )^3 \left (a+b \sin \left (\frac {\pi }{2}+i x\right )^3\right )}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {1}{\left (b \sin \left (i x+\frac {\pi }{2}\right )^3+a\right ) \tan \left (i x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 3709 |
\(\displaystyle -\int \frac {\left (1-\cosh ^2(x)\right ) \text {sech}^3(x)}{b \cosh ^3(x)+a}d\cosh (x)\) |
\(\Big \downarrow \) 2373 |
\(\displaystyle -\int \left (\frac {\text {sech}^3(x)}{a}-\frac {\text {sech}(x)}{a}+\frac {b \left (\cosh ^2(x)-1\right )}{a \left (b \cosh ^3(x)+a\right )}\right )d\cosh (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \cosh (x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}-\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cosh (x)+b^{2/3} \cosh ^2(x)\right )}{6 a^{5/3}}+\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \cosh (x)\right )}{3 a^{5/3}}-\frac {\log \left (a+b \cosh ^3(x)\right )}{3 a}+\frac {\text {sech}^2(x)}{2 a}+\frac {\log (\cosh (x))}{a}\) |
-((b^(2/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Cosh[x])/(Sqrt[3]*a^(1/3))])/(Sqrt[ 3]*a^(5/3))) + Log[Cosh[x]]/a + (b^(2/3)*Log[a^(1/3) + b^(1/3)*Cosh[x]])/( 3*a^(5/3)) - (b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Cosh[x] + b^(2/3)*Cosh [x]^2])/(6*a^(5/3)) - Log[a + b*Cosh[x]^3]/(3*a) + Sech[x]^2/(2*a)
3.1.81.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[E xpandIntegrand[(c*x)^m*(Pq/(a + b*x^n)), x], x] /; FreeQ[{a, b, c, m}, x] & & PolyQ[Pq, x] && IntegerQ[n] && !IGtQ[m, 0]
Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si mp[ff^(m + 1)/f Subst[Int[x^m*((a + b*(c*ff*x)^n)^p/(1 - ff^2*x^2)^((m + 1)/2)), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && ILtQ[(m - 1)/2, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.07 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.61
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2} a}+\frac {\ln \left (1+{\mathrm e}^{2 x}\right )}{a}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (27 a^{5} \textit {\_Z}^{3}+27 a^{4} \textit {\_Z}^{2}+9 a^{3} \textit {\_Z} +a^{2}-b^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 x}+\left (\frac {6 a^{2} \textit {\_R}}{b}+\frac {2 a}{b}\right ) {\mathrm e}^{x}+1\right )\right )\) | \(93\) |
default | \(\frac {\frac {2}{\left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )^{2}}+\ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )-\frac {2}{1+\tanh \left (\frac {x}{2}\right )^{2}}}{a}-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a -b \right ) \textit {\_Z}^{3}+\left (-3 a -3 b \right ) \textit {\_Z}^{2}+\left (3 a -3 b \right ) \textit {\_Z} -a -b \right )}{\sum }\frac {\left (\textit {\_R}^{2} a -\textit {\_R}^{2} b -2 \textit {\_R} a -4 \textit {\_R} b +a +b \right ) \ln \left (\tanh \left (\frac {x}{2}\right )^{2}-\textit {\_R} \right )}{\textit {\_R}^{2} a -\textit {\_R}^{2} b -2 \textit {\_R} a -2 \textit {\_R} b +a -b}}{3 a}\) | \(145\) |
2*exp(2*x)/(1+exp(2*x))^2/a+1/a*ln(1+exp(2*x))+sum(_R*ln(exp(2*x)+(6*a^2/b *_R+2*a/b)*exp(x)+1),_R=RootOf(27*_Z^3*a^5+27*_Z^2*a^4+9*_Z*a^3+a^2-b^2))
Result contains complex when optimal does not.
Time = 0.95 (sec) , antiderivative size = 1435, normalized size of antiderivative = 9.38 \[ \int \frac {\tanh ^3(x)}{a+b \cosh ^3(x)} \, dx=\text {Too large to display} \]
-1/12*(2*(a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 + 2*a*cosh(x)^ 2 + 2*(3*a*cosh(x)^2 + a)*sinh(x)^2 + 4*(a*cosh(x)^3 + a*cosh(x))*sinh(x) + a)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3 ) + 2/a)*log(b*cosh(x)^2 + b*sinh(x)^2 - (a^2*cosh(x) + a^2*sinh(x))*((1/2 )^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a) + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) + b) - 24*cosh(x)^2 + (6*cosh(x)^ 4 + 24*cosh(x)*sinh(x)^3 + 6*sinh(x)^4 + 12*(3*cosh(x)^2 + 1)*sinh(x)^2 - (a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 + 2*a*cosh(x)^2 + 2*(3* a*cosh(x)^2 + a)*sinh(x)^2 + 4*(a*cosh(x)^3 + a*cosh(x))*sinh(x) + a)*((1/ 2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a) - 3*sqrt(1/3)*(a*cosh(x)^4 + 4*a*cosh(x)*sinh(x)^3 + a*sinh(x)^4 + 2*a*cos h(x)^2 + 2*(3*a*cosh(x)^2 + a)*sinh(x)^2 + 4*(a*cosh(x)^3 + a*cosh(x))*sin h(x) + a)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^ 2)/a^5)^(1/3) + 2/a)^2*a^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a ^5 - (a^2 - b^2)/a^5)^(1/3) + 2/a)*a + 4)/a^2) + 12*cosh(x)^2 + 24*(cosh(x )^3 + cosh(x))*sinh(x) + 6)*log(b*cosh(x)^2 + b*sinh(x)^2 + 1/2*(a^2*cosh( x) + a^2*sinh(x))*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b ^2)/a^5)^(1/3) + 2/a) + 3/2*sqrt(1/3)*(a^2*cosh(x) + a^2*sinh(x))*sqrt(-(( (1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2)/a^5)^(1/3) + 2/ a)^2*a^2 - 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(1/a^3 + b^2/a^5 - (a^2 - b^2...
\[ \int \frac {\tanh ^3(x)}{a+b \cosh ^3(x)} \, dx=\int \frac {\tanh ^{3}{\left (x \right )}}{a + b \cosh ^{3}{\left (x \right )}}\, dx \]
\[ \int \frac {\tanh ^3(x)}{a+b \cosh ^3(x)} \, dx=\int { \frac {\tanh \left (x\right )^{3}}{b \cosh \left (x\right )^{3} + a} \,d x } \]
2*b*(x/(a*b) - integrate((b*e^(5*x) + 3*b*e^(3*x) + 8*a*e^(2*x) + 3*b*e^x) *e^x/(b*e^(6*x) + 3*b*e^(4*x) + 8*a*e^(3*x) + 3*b*e^(2*x) + b), x)/(a*b)) + 6*b*integrate(e^(4*x)/(b*e^(6*x) + 3*b*e^(4*x) + 8*a*e^(3*x) + 3*b*e^(2* x) + b), x)/a - 2*(x*e^(4*x) + (2*x - 1)*e^(2*x) + x)/(a*e^(4*x) + 2*a*e^( 2*x) + a) + log(e^(2*x) + 1)/a + 8*integrate(e^(3*x)/(b*e^(6*x) + 3*b*e^(4 *x) + 8*a*e^(3*x) + 3*b*e^(2*x) + b), x)
Time = 0.28 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.25 \[ \int \frac {\tanh ^3(x)}{a+b \cosh ^3(x)} \, dx=-\frac {b \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} + e^{\left (-x\right )} + e^{x} \right |}\right )}{3 \, a^{2}} + \frac {\log \left (e^{\left (-x\right )} + e^{x}\right )}{a} - \frac {\log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} + 8 \, a \right |}\right )}{3 \, a} + \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} + e^{\left (-x\right )} + e^{x}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{2}} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} {\left (e^{\left (-x\right )} + e^{x}\right )} + 4 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{2}} - \frac {3 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4}{2 \, a {\left (e^{\left (-x\right )} + e^{x}\right )}^{2}} \]
-1/3*b*(-a/b)^(1/3)*log(abs(-2*(-a/b)^(1/3) + e^(-x) + e^x))/a^2 + log(e^( -x) + e^x)/a - 1/3*log(abs(b*(e^(-x) + e^x)^3 + 8*a))/a + 1/3*sqrt(3)*(-a* b^2)^(1/3)*arctan(1/3*sqrt(3)*((-a/b)^(1/3) + e^(-x) + e^x)/(-a/b)^(1/3))/ a^2 + 1/6*(-a*b^2)^(1/3)*log((e^(-x) + e^x)^2 + 2*(-a/b)^(1/3)*(e^(-x) + e ^x) + 4*(-a/b)^(2/3))/a^2 - 1/2*(3*(e^(-x) + e^x)^2 - 4)/(a*(e^(-x) + e^x) ^2)
Time = 0.88 (sec) , antiderivative size = 1173, normalized size of antiderivative = 7.67 \[ \int \frac {\tanh ^3(x)}{a+b \cosh ^3(x)} \, dx=\text {Too large to display} \]
2/(a + a*exp(2*x)) - 2/(a + 2*a*exp(2*x) + a*exp(4*x)) + symsum(log(-(5033 1648*a^6*exp(2*x) - 786432*b^6*exp(2*x) + 452984832*root(27*a^5*z^3 + 27*a ^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^7 + 50331648*a^6 - 786432*b^6 + 1358 954496*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^2*a^8 + 1 358954496*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^3*a^9 + 50593792*a^2*b^4 - 102498304*a^4*b^2 + 1358954496*root(27*a^5*z^3 + 27*a ^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^2*a^8*exp(2*x) + 1358954496*root(27*a^ 5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^3*a^9*exp(2*x) + 50593792* a^2*b^4*exp(2*x) - 102498304*a^4*b^2*exp(2*x) + 7602176*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^3*b^4 - 465305600*root(27*a^5*z^ 3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^5*b^2 + 524288*a*b^5*exp(x) + 24379392*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^2*a^4 *b^4 - 1383333888*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k )^2*a^6*b^2 + 18874368*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^3*a^5*b^4 - 1370750976*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)^3*a^7*b^2 + 452984832*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^7*exp(2*x) - 5242880*a^3*b^3*exp(x) - 524288*root(27 *a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^2*b^5*exp(x) - 891289 6*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^4*b^3*exp(x) + 7602176*root(27*a^5*z^3 + 27*a^4*z^2 + 9*a^3*z + a^2 - b^2, z, k)*a^...