Integrand size = 13, antiderivative size = 132 \[ \int \frac {\cosh ^3(x)}{a+b \tanh (x)} \, dx=\frac {b^4 \arctan \left (\frac {\cosh (x) (b+a \tanh (x))}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {b^3 \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {b \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {a b^2 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {a \sinh (x)}{a^2-b^2}+\frac {a \sinh ^3(x)}{3 \left (a^2-b^2\right )} \]
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)
Time = 0.37 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.95 \[ \int \frac {\cosh ^3(x)}{a+b \tanh (x)} \, dx=\frac {24 b^4 \sqrt {a+b} \arctan \left (\frac {b+a \tanh \left (\frac {x}{2}\right )}{\sqrt {a-b} \sqrt {a+b}}\right )-3 \sqrt {a-b} b \left (a^3+a^2 b-5 a b^2-5 b^3\right ) \cosh (x)-(a-b)^{3/2} b (a+b)^2 \cosh (3 x)+9 a^4 \sqrt {a-b} \sinh (x)+9 a^3 \sqrt {a-b} b \sinh (x)-21 a^2 \sqrt {a-b} b^2 \sinh (x)-21 a \sqrt {a-b} b^3 \sinh (x)+a^4 \sqrt {a-b} \sinh (3 x)+a^3 \sqrt {a-b} b \sinh (3 x)-a^2 \sqrt {a-b} b^2 \sinh (3 x)-a \sqrt {a-b} b^3 \sinh (3 x)}{12 (a-b)^{5/2} (a+b)^3} \]
(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*Sinh [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)
Result contains complex when optimal does not.
Time = 0.88 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.077, Rules used = {3042, 3990, 3042, 3967, 3042, 3113, 2009, 3990, 3042, 3967, 3042, 3117, 3988, 219}
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.
\(\displaystyle \int \frac {\cosh ^3(x)}{a+b \tanh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sec (i x)^3 (a-i b \tan (i x))}dx\) |
\(\Big \downarrow \) 3990 |
\(\displaystyle \frac {\int \cosh ^3(x) (a-b \tanh (x))dx}{a^2-b^2}-\frac {b^2 \int \frac {\cosh (x)}{a+b \tanh (x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a+i b \tan (i x)}{\sec (i x)^3}dx}{a^2-b^2}-\frac {b^2 \int \frac {1}{\sec (i x) (a-i b \tan (i x))}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle \frac {a \int \cosh ^3(x)dx-\frac {1}{3} b \cosh ^3(x)}{a^2-b^2}-\frac {b^2 \int \frac {1}{\sec (i x) (a-i b \tan (i x))}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {1}{3} b \cosh ^3(x)+a \int \sin \left (i x+\frac {\pi }{2}\right )^3dx}{a^2-b^2}-\frac {b^2 \int \frac {1}{\sec (i x) (a-i b \tan (i x))}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle \frac {-\frac {1}{3} b \cosh ^3(x)+i a \int \left (\sinh ^2(x)+1\right )d(-i \sinh (x))}{a^2-b^2}-\frac {b^2 \int \frac {1}{\sec (i x) (a-i b \tan (i x))}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} b \cosh ^3(x)+i a \left (-\frac {1}{3} i \sinh ^3(x)-i \sinh (x)\right )}{a^2-b^2}-\frac {b^2 \int \frac {1}{\sec (i x) (a-i b \tan (i x))}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3990 |
\(\displaystyle -\frac {b^2 \left (\frac {\int \cosh (x) (a-b \tanh (x))dx}{a^2-b^2}-\frac {b^2 \int \frac {\text {sech}(x)}{a+b \tanh (x)}dx}{a^2-b^2}\right )}{a^2-b^2}+\frac {-\frac {1}{3} b \cosh ^3(x)+i a \left (-\frac {1}{3} i \sinh ^3(x)-i \sinh (x)\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {1}{3} b \cosh ^3(x)+i a \left (-\frac {1}{3} i \sinh ^3(x)-i \sinh (x)\right )}{a^2-b^2}-\frac {b^2 \left (\frac {\int \frac {a+i b \tan (i x)}{\sec (i x)}dx}{a^2-b^2}-\frac {b^2 \int \frac {\sec (i x)}{a-i b \tan (i x)}dx}{a^2-b^2}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle \frac {-\frac {1}{3} b \cosh ^3(x)+i a \left (-\frac {1}{3} i \sinh ^3(x)-i \sinh (x)\right )}{a^2-b^2}-\frac {b^2 \left (\frac {a \int \cosh (x)dx-b \cosh (x)}{a^2-b^2}-\frac {b^2 \int \frac {\sec (i x)}{a-i b \tan (i x)}dx}{a^2-b^2}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {1}{3} b \cosh ^3(x)+i a \left (-\frac {1}{3} i \sinh ^3(x)-i \sinh (x)\right )}{a^2-b^2}-\frac {b^2 \left (\frac {-b \cosh (x)+a \int \sin \left (i x+\frac {\pi }{2}\right )dx}{a^2-b^2}-\frac {b^2 \int \frac {\sec (i x)}{a-i b \tan (i x)}dx}{a^2-b^2}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {-\frac {1}{3} b \cosh ^3(x)+i a \left (-\frac {1}{3} i \sinh ^3(x)-i \sinh (x)\right )}{a^2-b^2}-\frac {b^2 \left (\frac {a \sinh (x)-b \cosh (x)}{a^2-b^2}-\frac {b^2 \int \frac {\sec (i x)}{a-i b \tan (i x)}dx}{a^2-b^2}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 3988 |
\(\displaystyle \frac {-\frac {1}{3} b \cosh ^3(x)+i a \left (-\frac {1}{3} i \sinh ^3(x)-i \sinh (x)\right )}{a^2-b^2}-\frac {b^2 \left (\frac {a \sinh (x)-b \cosh (x)}{a^2-b^2}-\frac {i b^2 \int \frac {1}{a^2-b^2+\cosh ^2(x) (b+a \tanh (x))^2}d(-i \cosh (x) (b+a \tanh (x)))}{a^2-b^2}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {b^2 \left (\frac {a \sinh (x)-b \cosh (x)}{a^2-b^2}-\frac {b^2 \arctan \left (\frac {\cosh (x) (a \tanh (x)+b)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}\right )}{a^2-b^2}+\frac {-\frac {1}{3} b \cosh ^3(x)+i a \left (-\frac {1}{3} i \sinh ^3(x)-i \sinh (x)\right )}{a^2-b^2}\) |
-((b^2*(-((b^2*ArcTan[(Cosh[x]*(b + a*Tanh[x]))/Sqrt[a^2 - b^2]])/(a^2 - b ^2)^(3/2)) + (-(b*Cosh[x]) + a*Sinh[x])/(a^2 - b^2)))/(a^2 - b^2)) + (-1/3 *(b*Cosh[x]^3) + I*a*((-I)*Sinh[x] - (I/3)*Sinh[x]^3))/(a^2 - b^2)
3.2.14.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
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] + Simp[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])
Int[sec[(e_.) + (f_.)*(x_)]/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbo l] :> Simp[-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]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_ Symbol] :> Simp[1/(a^2 + b^2) Int[Sec[e + f*x]^m*(a - b*Tan[e + f*x]), x] , x] + Simp[b^2/(a^2 + b^2) Int[Sec[e + f*x]^(m + 2)/(a + b*Tan[e + f*x]) , x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 0] && ILtQ[(m - 1)/2, 0]
Time = 2.05 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.30
method | result | size |
risch | \(\frac {{\mathrm e}^{3 x}}{24 a +24 b}+\frac {3 \,{\mathrm e}^{x} a}{8 \left (a +b \right )^{2}}+\frac {5 \,{\mathrm e}^{x} b}{8 \left (a +b \right )^{2}}-\frac {3 \,{\mathrm e}^{-x} a}{8 \left (a -b \right )^{2}}+\frac {5 \,{\mathrm e}^{-x} b}{8 \left (a -b \right )^{2}}-\frac {{\mathrm e}^{-3 x}}{24 \left (a -b \right )}-\frac {b^{4} \ln \left ({\mathrm e}^{x}-\frac {a -b}{\sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {b^{4} \ln \left ({\mathrm e}^{x}+\frac {a -b}{\sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}\) | \(172\) |
default | \(-\frac {2}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3} \left (2 a -2 b \right )}+\frac {1}{\left (2 a -2 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 a -3 b}{2 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {2}{3 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3} \left (2 a +2 b \right )}-\frac {1}{\left (2 a +2 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {2 a +3 b}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {2 b^{4} \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}\) | \(176\) |
1/24/(a+b)*exp(x)^3+3/8/(a+b)^2*exp(x)*a+5/8/(a+b)^2*exp(x)*b-3/8/(a-b)^2/ exp(x)*a+5/8/(a-b)^2/exp(x)*b-1/24/(a-b)/exp(x)^3-1/(-a^2+b^2)^(1/2)*b^4/( a+b)^2/(a-b)^2*ln(exp(x)-(a-b)/(-a^2+b^2)^(1/2))+1/(-a^2+b^2)^(1/2)*b^4/(a +b)^2/(a-b)^2*ln(exp(x)+(a-b)/(-a^2+b^2)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 908 vs. \(2 (124) = 248\).
Time = 0.29 (sec) , antiderivative size = 1871, normalized size of antiderivative = 14.17 \[ \int \frac {\cosh ^3(x)}{a+b \tanh (x)} \, dx=\text {Too large to display} \]
[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)*c osh(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 - 24*(b^4*cosh(x)^3 + 3*b^4*cosh(x)^2*sinh(x) + 3*b^4*c osh(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 ...
\[ \int \frac {\cosh ^3(x)}{a+b \tanh (x)} \, dx=\int \frac {\cosh ^{3}{\left (x \right )}}{a + b \tanh {\left (x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cosh ^3(x)}{a+b \tanh (x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.23 \[ \int \frac {\cosh ^3(x)}{a+b \tanh (x)} \, dx=\frac {2 \, b^{4} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {{\left (9 \, a e^{\left (2 \, x\right )} - 15 \, b e^{\left (2 \, x\right )} + a - b\right )} e^{\left (-3 \, x\right )}}{24 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {a^{2} e^{\left (3 \, x\right )} + 2 \, a b e^{\left (3 \, x\right )} + b^{2} e^{\left (3 \, x\right )} + 9 \, a^{2} e^{x} + 24 \, a b e^{x} + 15 \, b^{2} e^{x}}{24 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} \]
2*b^4*arctan((a*e^x + b*e^x)/sqrt(a^2 - b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqr t(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)
Time = 2.77 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.67 \[ \int \frac {\cosh ^3(x)}{a+b \tanh (x)} \, dx=\frac {{\mathrm {e}}^{3\,x}}{24\,a+24\,b}-\frac {{\mathrm {e}}^{-3\,x}}{24\,a-24\,b}-\frac {{\mathrm {e}}^{-x}\,\left (3\,a-5\,b\right )}{8\,{\left (a-b\right )}^2}+\frac {{\mathrm {e}}^x\,\left (3\,a+5\,b\right )}{8\,{\left (a+b\right )}^2}-\frac {b^4\,\ln \left (-\frac {2\,b^4\,{\mathrm {e}}^x}{a^5+a^4\,b-2\,a^3\,b^2-2\,a^2\,b^3+a\,b^4+b^5}-\frac {2\,b^4}{{\left (a+b\right )}^{7/2}\,{\left (b-a\right )}^{3/2}}\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}}+\frac {b^4\,\ln \left (\frac {2\,b^4}{{\left (a+b\right )}^{7/2}\,{\left (b-a\right )}^{3/2}}-\frac {2\,b^4\,{\mathrm {e}}^x}{a^5+a^4\,b-2\,a^3\,b^2-2\,a^2\,b^3+a\,b^4+b^5}\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}} \]
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))