Integrand size = 15, antiderivative size = 78 \[ \int \frac {\cosh ^7(x)}{a+b \cosh ^2(x)} \, dx=-\frac {a^3 \arctan \left (\frac {\sqrt {b} \sinh (x)}{\sqrt {a+b}}\right )}{b^{7/2} \sqrt {a+b}}+\frac {\left (a^2-a b+b^2\right ) \sinh (x)}{b^3}-\frac {(a-2 b) \sinh ^3(x)}{3 b^2}+\frac {\sinh ^5(x)}{5 b} \]
(a^2-a*b+b^2)*sinh(x)/b^3-1/3*(a-2*b)*sinh(x)^3/b^2+1/5*sinh(x)^5/b-a^3*ar ctan(sinh(x)*b^(1/2)/(a+b)^(1/2))/b^(7/2)/(a+b)^(1/2)
Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \frac {\cosh ^7(x)}{a+b \cosh ^2(x)} \, dx=\frac {a^3 \arctan \left (\frac {\sqrt {a+b} \text {csch}(x)}{\sqrt {b}}\right )}{b^{7/2} \sqrt {a+b}}+\frac {\left (8 a^2-6 a b+5 b^2\right ) \sinh (x)}{8 b^3}-\frac {(4 a-5 b) \sinh (3 x)}{48 b^2}+\frac {\sinh (5 x)}{80 b} \]
(a^3*ArcTan[(Sqrt[a + b]*Csch[x])/Sqrt[b]])/(b^(7/2)*Sqrt[a + b]) + ((8*a^ 2 - 6*a*b + 5*b^2)*Sinh[x])/(8*b^3) - ((4*a - 5*b)*Sinh[3*x])/(48*b^2) + S inh[5*x]/(80*b)
Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3665, 300, 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.
\(\displaystyle \int \frac {\cosh ^7(x)}{a+b \cosh ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (\frac {\pi }{2}+i x\right )^7}{a+b \sin \left (\frac {\pi }{2}+i x\right )^2}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle \int \frac {\left (\sinh ^2(x)+1\right )^3}{a+b \sinh ^2(x)+b}d\sinh (x)\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \int \left (-\frac {a^3}{b^3 \left (a+b \sinh ^2(x)+b\right )}+\frac {a^2-a b+b^2}{b^3}-\frac {(a-2 b) \sinh ^2(x)}{b^2}+\frac {\sinh ^4(x)}{b}\right )d\sinh (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 \arctan \left (\frac {\sqrt {b} \sinh (x)}{\sqrt {a+b}}\right )}{b^{7/2} \sqrt {a+b}}+\frac {\left (a^2-a b+b^2\right ) \sinh (x)}{b^3}-\frac {(a-2 b) \sinh ^3(x)}{3 b^2}+\frac {\sinh ^5(x)}{5 b}\) |
-((a^3*ArcTan[(Sqrt[b]*Sinh[x])/Sqrt[a + b]])/(b^(7/2)*Sqrt[a + b])) + ((a ^2 - a*b + b^2)*Sinh[x])/b^3 - ((a - 2*b)*Sinh[x]^3)/(3*b^2) + Sinh[x]^5/( 5*b)
3.1.20.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(205\) vs. \(2(66)=132\).
Time = 1.42 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.64
method | result | size |
risch | \(\frac {{\mathrm e}^{5 x}}{160 b}+\frac {5 \,{\mathrm e}^{3 x}}{96 b}-\frac {{\mathrm e}^{3 x} a}{24 b^{2}}+\frac {{\mathrm e}^{x} a^{2}}{2 b^{3}}-\frac {3 a \,{\mathrm e}^{x}}{8 b^{2}}+\frac {5 \,{\mathrm e}^{x}}{16 b}-\frac {{\mathrm e}^{-x} a^{2}}{2 b^{3}}+\frac {3 a \,{\mathrm e}^{-x}}{8 b^{2}}-\frac {5 \,{\mathrm e}^{-x}}{16 b}-\frac {5 \,{\mathrm e}^{-3 x}}{96 b}+\frac {{\mathrm e}^{-3 x} a}{24 b^{2}}-\frac {{\mathrm e}^{-5 x}}{160 b}-\frac {a^{3} \ln \left ({\mathrm e}^{2 x}+\frac {2 \left (a +b \right ) {\mathrm e}^{x}}{\sqrt {-a b -b^{2}}}-1\right )}{2 \sqrt {-a b -b^{2}}\, b^{3}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 x}-\frac {2 \left (a +b \right ) {\mathrm e}^{x}}{\sqrt {-a b -b^{2}}}-1\right )}{2 \sqrt {-a b -b^{2}}\, b^{3}}\) | \(206\) |
default | \(-\frac {2 a^{3} \left (\frac {\arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) \sqrt {a +b}+2 \sqrt {a}}{2 \sqrt {b}}\right )}{2 \sqrt {a +b}\, \sqrt {b}}+\frac {\arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) \sqrt {a +b}-2 \sqrt {a}}{2 \sqrt {b}}\right )}{2 \sqrt {a +b}\, \sqrt {b}}\right )}{b^{3}}-\frac {1}{5 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{5}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {-7 b +4 a}{8 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {11 b -4 a}{12 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {a^{2}-a b +b^{2}}{b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}-\frac {1}{5 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{5}}-\frac {7 b -4 a}{8 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {11 b -4 a}{12 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a^{2}-a b +b^{2}}{b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}\) | \(257\) |
1/160/b*exp(5*x)+5/96/b*exp(3*x)-1/24/b^2*exp(3*x)*a+1/2/b^3*exp(x)*a^2-3/ 8*a/b^2*exp(x)+5/16/b*exp(x)-1/2/b^3*exp(-x)*a^2+3/8*a/b^2*exp(-x)-5/16/b* exp(-x)-5/96/b*exp(-3*x)+1/24/b^2*exp(-3*x)*a-1/160/b*exp(-5*x)-1/2/(-a*b- b^2)^(1/2)*a^3/b^3*ln(exp(2*x)+2*(a+b)/(-a*b-b^2)^(1/2)*exp(x)-1)+1/2/(-a* b-b^2)^(1/2)*a^3/b^3*ln(exp(2*x)-2*(a+b)/(-a*b-b^2)^(1/2)*exp(x)-1)
Leaf count of result is larger than twice the leaf count of optimal. 1232 vs. \(2 (66) = 132\).
Time = 0.28 (sec) , antiderivative size = 2508, normalized size of antiderivative = 32.15 \[ \int \frac {\cosh ^7(x)}{a+b \cosh ^2(x)} \, dx=\text {Too large to display} \]
[1/480*(3*(a*b^3 + b^4)*cosh(x)^10 + 30*(a*b^3 + b^4)*cosh(x)*sinh(x)^9 + 3*(a*b^3 + b^4)*sinh(x)^10 - 5*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^8 - 5*( 4*a^2*b^2 - a*b^3 - 5*b^4 - 27*(a*b^3 + b^4)*cosh(x)^2)*sinh(x)^8 + 40*(9* (a*b^3 + b^4)*cosh(x)^3 - (4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x))*sinh(x)^7 + 30*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^6 + 10*(63*(a*b^3 + b^4) *cosh(x)^4 + 24*a^3*b + 6*a^2*b^2 - 3*a*b^3 + 15*b^4 - 14*(4*a^2*b^2 - a*b ^3 - 5*b^4)*cosh(x)^2)*sinh(x)^6 + 4*(189*(a*b^3 + b^4)*cosh(x)^5 - 70*(4* a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^3 + 45*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b ^4)*cosh(x))*sinh(x)^5 - 30*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^ 4 + 10*(63*(a*b^3 + b^4)*cosh(x)^6 - 35*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x )^4 - 24*a^3*b - 6*a^2*b^2 + 3*a*b^3 - 15*b^4 + 45*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^2)*sinh(x)^4 - 3*a*b^3 - 3*b^4 + 40*(9*(a*b^3 + b^4 )*cosh(x)^7 - 7*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^5 + 15*(8*a^3*b + 2*a^ 2*b^2 - a*b^3 + 5*b^4)*cosh(x)^3 - 3*(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4) *cosh(x))*sinh(x)^3 + 5*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^2 + 5*(27*(a*b ^3 + b^4)*cosh(x)^8 - 28*(4*a^2*b^2 - a*b^3 - 5*b^4)*cosh(x)^6 + 90*(8*a^3 *b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^4 + 4*a^2*b^2 - a*b^3 - 5*b^4 - 36 *(8*a^3*b + 2*a^2*b^2 - a*b^3 + 5*b^4)*cosh(x)^2)*sinh(x)^2 - 240*(a^3*cos h(x)^5 + 5*a^3*cosh(x)^4*sinh(x) + 10*a^3*cosh(x)^3*sinh(x)^2 + 10*a^3*cos h(x)^2*sinh(x)^3 + 5*a^3*cosh(x)*sinh(x)^4 + a^3*sinh(x)^5)*sqrt(-a*b -...
Timed out. \[ \int \frac {\cosh ^7(x)}{a+b \cosh ^2(x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cosh ^7(x)}{a+b \cosh ^2(x)} \, dx=\int { \frac {\cosh \left (x\right )^{7}}{b \cosh \left (x\right )^{2} + a} \,d x } \]
1/480*(3*b^2*e^(10*x) - 3*b^2 - 5*(4*a*b - 5*b^2)*e^(8*x) + 30*(8*a^2 - 6* a*b + 5*b^2)*e^(6*x) - 30*(8*a^2 - 6*a*b + 5*b^2)*e^(4*x) + 5*(4*a*b - 5*b ^2)*e^(2*x))*e^(-5*x)/b^3 - 1/128*integrate(256*(a^3*e^(3*x) + a^3*e^x)/(b ^4*e^(4*x) + b^4 + 2*(2*a*b^3 + b^4)*e^(2*x)), x)
\[ \int \frac {\cosh ^7(x)}{a+b \cosh ^2(x)} \, dx=\int { \frac {\cosh \left (x\right )^{7}}{b \cosh \left (x\right )^{2} + a} \,d x } \]
Time = 2.34 (sec) , antiderivative size = 293, normalized size of antiderivative = 3.76 \[ \int \frac {\cosh ^7(x)}{a+b \cosh ^2(x)} \, dx=\frac {{\mathrm {e}}^{5\,x}}{160\,b}-\frac {{\mathrm {e}}^{-5\,x}}{160\,b}-\frac {{\mathrm {e}}^{-x}\,\left (8\,a^2-6\,a\,b+5\,b^2\right )}{16\,b^3}+\frac {\left (2\,\mathrm {atan}\left (\frac {\left (b^9\,\sqrt {b^8+a\,b^7}+a\,b^8\,\sqrt {b^8+a\,b^7}\right )\,\left ({\mathrm {e}}^x\,\left (\frac {2\,a^7}{b^{11}\,{\left (a+b\right )}^2\,\sqrt {a^6}}-\frac {4\,\left (2\,a^4\,b^4\,\sqrt {a^6}+2\,a^5\,b^3\,\sqrt {a^6}\right )}{a^3\,b^8\,\left (a+b\right )\,\sqrt {b^7\,\left (a+b\right )}\,\sqrt {b^8+a\,b^7}}\right )-\frac {2\,a^7\,{\mathrm {e}}^{3\,x}}{b^{11}\,{\left (a+b\right )}^2\,\sqrt {a^6}}\right )}{4\,a^4}\right )-2\,\mathrm {atan}\left (\frac {a^3\,{\mathrm {e}}^x\,\sqrt {b^7\,\left (a+b\right )}}{2\,b^3\,\left (a+b\right )\,\sqrt {a^6}}\right )\right )\,\sqrt {a^6}}{2\,\sqrt {b^8+a\,b^7}}+\frac {{\mathrm {e}}^{-3\,x}\,\left (4\,a-5\,b\right )}{96\,b^2}-\frac {{\mathrm {e}}^{3\,x}\,\left (4\,a-5\,b\right )}{96\,b^2}+\frac {{\mathrm {e}}^x\,\left (8\,a^2-6\,a\,b+5\,b^2\right )}{16\,b^3} \]
exp(5*x)/(160*b) - exp(-5*x)/(160*b) - (exp(-x)*(8*a^2 - 6*a*b + 5*b^2))/( 16*b^3) + ((2*atan(((b^9*(a*b^7 + b^8)^(1/2) + a*b^8*(a*b^7 + b^8)^(1/2))* (exp(x)*((2*a^7)/(b^11*(a + b)^2*(a^6)^(1/2)) - (4*(2*a^4*b^4*(a^6)^(1/2) + 2*a^5*b^3*(a^6)^(1/2)))/(a^3*b^8*(a + b)*(b^7*(a + b))^(1/2)*(a*b^7 + b^ 8)^(1/2))) - (2*a^7*exp(3*x))/(b^11*(a + b)^2*(a^6)^(1/2))))/(4*a^4)) - 2* atan((a^3*exp(x)*(b^7*(a + b))^(1/2))/(2*b^3*(a + b)*(a^6)^(1/2))))*(a^6)^ (1/2))/(2*(a*b^7 + b^8)^(1/2)) + (exp(-3*x)*(4*a - 5*b))/(96*b^2) - (exp(3 *x)*(4*a - 5*b))/(96*b^2) + (exp(x)*(8*a^2 - 6*a*b + 5*b^2))/(16*b^3)