Integrand size = 25, antiderivative size = 143 \[ \int \frac {\sinh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \cosh (e+f x)}{\sqrt {a-b+b \cosh ^2(e+f x)}}\right )}{b^{5/2} f}-\frac {a (3 a-5 b) \cosh (e+f x)}{3 (a-b)^2 b^2 f \sqrt {a-b+b \cosh ^2(e+f x)}}-\frac {a \cosh (e+f x) \sinh ^2(e+f x)}{3 (a-b) b f \left (a-b+b \cosh ^2(e+f x)\right )^{3/2}} \]
arctanh(cosh(f*x+e)*b^(1/2)/(a-b+b*cosh(f*x+e)^2)^(1/2))/b^(5/2)/f-1/3*a*c osh(f*x+e)*sinh(f*x+e)^2/(a-b)/b/f/(a-b+b*cosh(f*x+e)^2)^(3/2)-1/3*a*(3*a- 5*b)*cosh(f*x+e)/(a-b)^2/b^2/f/(a-b+b*cosh(f*x+e)^2)^(1/2)
Time = 1.10 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.91 \[ \int \frac {\sinh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\frac {-\frac {2 \sqrt {2} a \cosh (e+f x) \left (3 a^2-7 a b+3 b^2+(2 a-3 b) b \cosh (2 (e+f x))\right )}{3 (a-b)^2 b^2 (2 a-b+b \cosh (2 (e+f x)))^{3/2}}+\frac {\log \left (\sqrt {2} \sqrt {b} \cosh (e+f x)+\sqrt {2 a-b+b \cosh (2 (e+f x))}\right )}{b^{5/2}}}{f} \]
((-2*Sqrt[2]*a*Cosh[e + f*x]*(3*a^2 - 7*a*b + 3*b^2 + (2*a - 3*b)*b*Cosh[2 *(e + f*x)]))/(3*(a - b)^2*b^2*(2*a - b + b*Cosh[2*(e + f*x)])^(3/2)) + Lo g[Sqrt[2]*Sqrt[b]*Cosh[e + f*x] + Sqrt[2*a - b + b*Cosh[2*(e + f*x)]]]/b^( 5/2))/f
Time = 0.35 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 26, 3665, 315, 25, 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.
\(\displaystyle \int \frac {\sinh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \sin (i e+i f x)^5}{\left (a-b \sin (i e+i f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\sin (i e+i f x)^5}{\left (a-b \sin (i e+i f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle \frac {\int \frac {\left (1-\cosh ^2(e+f x)\right )^2}{\left (b \cosh ^2(e+f x)+a-b\right )^{5/2}}d\cosh (e+f x)}{f}\) |
\(\Big \downarrow \) 315 |
\(\displaystyle \frac {\frac {\int -\frac {-3 (a-b) \cosh ^2(e+f x)+a-3 b}{\left (b \cosh ^2(e+f x)+a-b\right )^{3/2}}d\cosh (e+f x)}{3 b (a-b)}+\frac {a \cosh (e+f x) \left (1-\cosh ^2(e+f x)\right )}{3 b (a-b) \left (a+b \cosh ^2(e+f x)-b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {a \cosh (e+f x) \left (1-\cosh ^2(e+f x)\right )}{3 b (a-b) \left (a+b \cosh ^2(e+f x)-b\right )^{3/2}}-\frac {\int \frac {-3 (a-b) \cosh ^2(e+f x)+a-3 b}{\left (b \cosh ^2(e+f x)+a-b\right )^{3/2}}d\cosh (e+f x)}{3 b (a-b)}}{f}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {a \cosh (e+f x) \left (1-\cosh ^2(e+f x)\right )}{3 b (a-b) \left (a+b \cosh ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {a (3 a-5 b) \cosh (e+f x)}{b (a-b) \sqrt {a+b \cosh ^2(e+f x)-b}}-\frac {3 (a-b) \int \frac {1}{\sqrt {b \cosh ^2(e+f x)+a-b}}d\cosh (e+f x)}{b}}{3 b (a-b)}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {a \cosh (e+f x) \left (1-\cosh ^2(e+f x)\right )}{3 b (a-b) \left (a+b \cosh ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {a (3 a-5 b) \cosh (e+f x)}{b (a-b) \sqrt {a+b \cosh ^2(e+f x)-b}}-\frac {3 (a-b) \int \frac {1}{1-\frac {b \cosh ^2(e+f x)}{b \cosh ^2(e+f x)+a-b}}d\frac {\cosh (e+f x)}{\sqrt {b \cosh ^2(e+f x)+a-b}}}{b}}{3 b (a-b)}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {a \cosh (e+f x) \left (1-\cosh ^2(e+f x)\right )}{3 b (a-b) \left (a+b \cosh ^2(e+f x)-b\right )^{3/2}}-\frac {\frac {a (3 a-5 b) \cosh (e+f x)}{b (a-b) \sqrt {a+b \cosh ^2(e+f x)-b}}-\frac {3 (a-b) \text {arctanh}\left (\frac {\sqrt {b} \cosh (e+f x)}{\sqrt {a+b \cosh ^2(e+f x)-b}}\right )}{b^{3/2}}}{3 b (a-b)}}{f}\) |
((a*Cosh[e + f*x]*(1 - Cosh[e + f*x]^2))/(3*(a - b)*b*(a - b + b*Cosh[e + f*x]^2)^(3/2)) - ((-3*(a - b)*ArcTanh[(Sqrt[b]*Cosh[e + f*x])/Sqrt[a - b + b*Cosh[e + f*x]^2]])/b^(3/2) + (a*(3*a - 5*b)*Cosh[e + f*x])/((a - b)*b*S qrt[a - b + b*Cosh[e + f*x]^2]))/(3*(a - b)*b))/f
3.2.16.3.1 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[((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[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]
Time = 0.19 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.61
method | result | size |
default | \(\frac {\sqrt {\left (a +b \sinh \left (f x +e \right )^{2}\right ) \cosh \left (f x +e \right )^{2}}\, \left (\frac {\ln \left (\frac {\frac {a}{2}+\frac {b}{2}+b \sinh \left (f x +e \right )^{2}}{\sqrt {b}}+\sqrt {\left (a +b \sinh \left (f x +e \right )^{2}\right ) \cosh \left (f x +e \right )^{2}}\right )}{2 b^{\frac {5}{2}}}+\frac {a^{2} \left (2 b \sinh \left (f x +e \right )^{2}+3 a -b \right ) \cosh \left (f x +e \right )^{2}}{3 b^{2} \sqrt {\left (a +b \sinh \left (f x +e \right )^{2}\right ) \cosh \left (f x +e \right )^{2}}\, \left (a +b \sinh \left (f x +e \right )^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 a \cosh \left (f x +e \right )^{2}}{b^{2} \left (a -b \right ) \sqrt {\left (a +b \sinh \left (f x +e \right )^{2}\right ) \cosh \left (f x +e \right )^{2}}}\right )}{\cosh \left (f x +e \right ) \sqrt {a +b \sinh \left (f x +e \right )^{2}}\, f}\) | \(230\) |
((a+b*sinh(f*x+e)^2)*cosh(f*x+e)^2)^(1/2)*(1/2/b^(5/2)*ln((1/2*a+1/2*b+b*s inh(f*x+e)^2)/b^(1/2)+((a+b*sinh(f*x+e)^2)*cosh(f*x+e)^2)^(1/2))+1/3*a^2/b ^2*(2*b*sinh(f*x+e)^2+3*a-b)*cosh(f*x+e)^2/((a+b*sinh(f*x+e)^2)*cosh(f*x+e )^2)^(1/2)/(a+b*sinh(f*x+e)^2)/(a^2-2*a*b+b^2)-2*a/b^2*cosh(f*x+e)^2/(a-b) /((a+b*sinh(f*x+e)^2)*cosh(f*x+e)^2)^(1/2))/cosh(f*x+e)/(a+b*sinh(f*x+e)^2 )^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 3631 vs. \(2 (129) = 258\).
Time = 0.61 (sec) , antiderivative size = 7938, normalized size of antiderivative = 55.51 \[ \int \frac {\sinh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\sinh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sinh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sinh \left (f x + e\right )^{5}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Exception generated. \[ \int \frac {\sinh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {\sinh ^5(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {sinh}\left (e+f\,x\right )}^5}{{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]