Integrand size = 25, antiderivative size = 134 \[ \int \frac {\cosh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {(a-b) \cosh ^2(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac {(a-b) (3 a+2 b) \sinh (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sinh ^2(e+f x)}} \]
arctanh(sinh(f*x+e)*b^(1/2)/(a+b*sinh(f*x+e)^2)^(1/2))/b^(5/2)/f-1/3*(a-b) *cosh(f*x+e)^2*sinh(f*x+e)/a/b/f/(a+b*sinh(f*x+e)^2)^(3/2)-1/3*(a-b)*(3*a+ 2*b)*sinh(f*x+e)/a^2/b^2/f/(a+b*sinh(f*x+e)^2)^(1/2)
Time = 0.65 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.94 \[ \int \frac {\cosh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {b} \sinh (e+f x)}{\sqrt {2 a-b+b \cosh (2 (e+f x))}}\right )}{b^{5/2}}+\frac {2 \sqrt {2} (-a+b) \left (3 a^2+a b-b^2+b (2 a+b) \cosh (2 (e+f x))\right ) \sinh (e+f x)}{3 a^2 b^2 (2 a-b+b \cosh (2 (e+f x)))^{3/2}}}{f} \]
(ArcTanh[(Sqrt[2]*Sqrt[b]*Sinh[e + f*x])/Sqrt[2*a - b + b*Cosh[2*(e + f*x) ]]]/b^(5/2) + (2*Sqrt[2]*(-a + b)*(3*a^2 + a*b - b^2 + b*(2*a + b)*Cosh[2* (e + f*x)])*Sinh[e + f*x])/(3*a^2*b^2*(2*a - b + b*Cosh[2*(e + f*x)])^(3/2 )))/f
Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3669, 315, 298, 224, 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.
\(\displaystyle \int \frac {\cosh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i e+i f x)^5}{\left (a-b \sin (i e+i f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \frac {\left (\sinh ^2(e+f x)+1\right )^2}{\left (b \sinh ^2(e+f x)+a\right )^{5/2}}d\sinh (e+f x)}{f}\) |
\(\Big \downarrow \) 315 |
\(\displaystyle \frac {\frac {\int \frac {3 a \sinh ^2(e+f x)+a+2 b}{\left (b \sinh ^2(e+f x)+a\right )^{3/2}}d\sinh (e+f x)}{3 a b}-\frac {(a-b) \sinh (e+f x) \left (\sinh ^2(e+f x)+1\right )}{3 a b \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {\frac {3 a \int \frac {1}{\sqrt {b \sinh ^2(e+f x)+a}}d\sinh (e+f x)}{b}+\frac {\left (-\frac {3 a}{b}+\frac {2 b}{a}+1\right ) \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}}{3 a b}-\frac {(a-b) \sinh (e+f x) \left (\sinh ^2(e+f x)+1\right )}{3 a b \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {\frac {3 a \int \frac {1}{1-\frac {b \sinh ^2(e+f x)}{b \sinh ^2(e+f x)+a}}d\frac {\sinh (e+f x)}{\sqrt {b \sinh ^2(e+f x)+a}}}{b}+\frac {\left (-\frac {3 a}{b}+\frac {2 b}{a}+1\right ) \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}}{3 a b}-\frac {(a-b) \sinh (e+f x) \left (\sinh ^2(e+f x)+1\right )}{3 a b \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{b^{3/2}}+\frac {\left (-\frac {3 a}{b}+\frac {2 b}{a}+1\right ) \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}}{3 a b}-\frac {(a-b) \sinh (e+f x) \left (\sinh ^2(e+f x)+1\right )}{3 a b \left (a+b \sinh ^2(e+f x)\right )^{3/2}}}{f}\) |
(-1/3*((a - b)*Sinh[e + f*x]*(1 + Sinh[e + f*x]^2))/(a*b*(a + b*Sinh[e + f *x]^2)^(3/2)) + ((3*a*ArcTanh[(Sqrt[b]*Sinh[e + f*x])/Sqrt[a + b*Sinh[e + f*x]^2]])/b^(3/2) + ((1 - (3*a)/b + (2*b)/a)*Sinh[e + f*x])/Sqrt[a + b*Sin h[e + f*x]^2])/(3*a*b))/f
3.4.91.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.20 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.55
method | result | size |
derivativedivides | \(\frac {\sinh \left (f x +e \right )}{3 a f \left (a +b \sinh \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \sinh \left (f x +e \right )}{3 a^{2} f \sqrt {a +b \sinh \left (f x +e \right )^{2}}}-\frac {\sinh \left (f x +e \right )^{3}}{3 f b \left (a +b \sinh \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {\sinh \left (f x +e \right )}{f \,b^{2} \sqrt {a +b \sinh \left (f x +e \right )^{2}}}+\frac {\ln \left (\sqrt {b}\, \sinh \left (f x +e \right )+\sqrt {a +b \sinh \left (f x +e \right )^{2}}\right )}{f \,b^{\frac {5}{2}}}-\frac {2 \sinh \left (f x +e \right )}{3 f b \left (a +b \sinh \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \sinh \left (f x +e \right )}{3 f a b \sqrt {a +b \sinh \left (f x +e \right )^{2}}}\) | \(208\) |
default | \(\frac {\sinh \left (f x +e \right )}{3 a f \left (a +b \sinh \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \sinh \left (f x +e \right )}{3 a^{2} f \sqrt {a +b \sinh \left (f x +e \right )^{2}}}-\frac {\sinh \left (f x +e \right )^{3}}{3 f b \left (a +b \sinh \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}-\frac {\sinh \left (f x +e \right )}{f \,b^{2} \sqrt {a +b \sinh \left (f x +e \right )^{2}}}+\frac {\ln \left (\sqrt {b}\, \sinh \left (f x +e \right )+\sqrt {a +b \sinh \left (f x +e \right )^{2}}\right )}{f \,b^{\frac {5}{2}}}-\frac {2 \sinh \left (f x +e \right )}{3 f b \left (a +b \sinh \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}+\frac {2 \sinh \left (f x +e \right )}{3 f a b \sqrt {a +b \sinh \left (f x +e \right )^{2}}}\) | \(208\) |
1/3*sinh(f*x+e)/a/f/(a+b*sinh(f*x+e)^2)^(3/2)+2/3*sinh(f*x+e)/a^2/f/(a+b*s inh(f*x+e)^2)^(1/2)-1/3/f*sinh(f*x+e)^3/b/(a+b*sinh(f*x+e)^2)^(3/2)-1/f/b^ 2*sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2)+1/f/b^(5/2)*ln(b^(1/2)*sinh(f*x+e) +(a+b*sinh(f*x+e)^2)^(1/2))-2/3/f/b*sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2)+ 2/3/f/a/b*sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 2932 vs. \(2 (120) = 240\).
Time = 0.62 (sec) , antiderivative size = 6774, normalized size of antiderivative = 50.55 \[ \int \frac {\cosh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\cosh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cosh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cosh \left (f x + e\right )^{5}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Exception generated. \[ \int \frac {\cosh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {\cosh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {cosh}\left (e+f\,x\right )}^5}{{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]