Integrand size = 23, antiderivative size = 79 \[ \int \frac {\sinh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\frac {\cosh (e+f x)}{3 (a-b) f \left (a-b+b \cosh ^2(e+f x)\right )^{3/2}}+\frac {2 \cosh (e+f x)}{3 (a-b)^2 f \sqrt {a-b+b \cosh ^2(e+f x)}} \]
1/3*cosh(f*x+e)/(a-b)/f/(a-b+b*cosh(f*x+e)^2)^(3/2)+2/3*cosh(f*x+e)/(a-b)^ 2/f/(a-b+b*cosh(f*x+e)^2)^(1/2)
Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.80 \[ \int \frac {\sinh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\frac {2 \sqrt {2} \cosh (e+f x) (3 a-2 b+b \cosh (2 (e+f x)))}{3 (a-b)^2 f (2 a-b+b \cosh (2 (e+f x)))^{3/2}} \]
(2*Sqrt[2]*Cosh[e + f*x]*(3*a - 2*b + b*Cosh[2*(e + f*x)]))/(3*(a - b)^2*f *(2*a - b + b*Cosh[2*(e + f*x)])^(3/2))
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 26, 3665, 209, 208}
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.
\(\displaystyle \int \frac {\sinh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \sin (i e+i f x)}{\left (a-b \sin (i e+i f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\sin (i e+i f x)}{\left (a-b \sin (i e+i f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle \frac {\int \frac {1}{\left (b \cosh ^2(e+f x)+a-b\right )^{5/2}}d\cosh (e+f x)}{f}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{\left (b \cosh ^2(e+f x)+a-b\right )^{3/2}}d\cosh (e+f x)}{3 (a-b)}+\frac {\cosh (e+f x)}{3 (a-b) \left (a+b \cosh ^2(e+f x)-b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {2 \cosh (e+f x)}{3 (a-b)^2 \sqrt {a+b \cosh ^2(e+f x)-b}}+\frac {\cosh (e+f x)}{3 (a-b) \left (a+b \cosh ^2(e+f x)-b\right )^{3/2}}}{f}\) |
(Cosh[e + f*x]/(3*(a - b)*(a - b + b*Cosh[e + f*x]^2)^(3/2)) + (2*Cosh[e + f*x])/(3*(a - b)^2*Sqrt[a - b + b*Cosh[e + f*x]^2]))/f
3.2.18.3.1 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)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
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 = 0.37 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.72
method | result | size |
default | \(\frac {\left (2 b \sinh \left (f x +e \right )^{2}+3 a -b \right ) \cosh \left (f x +e \right )}{3 \left (a +b \sinh \left (f x +e \right )^{2}\right )^{\frac {3}{2}} \left (a^{2}-2 a b +b^{2}\right ) f}\) | \(57\) |
risch | \(\text {Expression too large to display}\) | \(161059\) |
Leaf count of result is larger than twice the leaf count of optimal. 1186 vs. \(2 (71) = 142\).
Time = 0.39 (sec) , antiderivative size = 1186, normalized size of antiderivative = 15.01 \[ \int \frac {\sinh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]
2/3*sqrt(2)*(b*cosh(f*x + e)^6 + 6*b*cosh(f*x + e)*sinh(f*x + e)^5 + b*sin h(f*x + e)^6 + 3*(2*a - b)*cosh(f*x + e)^4 + 3*(5*b*cosh(f*x + e)^2 + 2*a - b)*sinh(f*x + e)^4 + 4*(5*b*cosh(f*x + e)^3 + 3*(2*a - b)*cosh(f*x + e)) *sinh(f*x + e)^3 + 3*(2*a - b)*cosh(f*x + e)^2 + 3*(5*b*cosh(f*x + e)^4 + 6*(2*a - b)*cosh(f*x + e)^2 + 2*a - b)*sinh(f*x + e)^2 + 6*(b*cosh(f*x + e )^5 + 2*(2*a - b)*cosh(f*x + e)^3 + (2*a - b)*cosh(f*x + e))*sinh(f*x + e) + b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e )^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))/((a^2*b^2 - 2*a*b^ 3 + b^4)*f*cosh(f*x + e)^8 + 8*(a^2*b^2 - 2*a*b^3 + b^4)*f*cosh(f*x + e)*s inh(f*x + e)^7 + (a^2*b^2 - 2*a*b^3 + b^4)*f*sinh(f*x + e)^8 + 4*(2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f*cosh(f*x + e)^6 + 4*(7*(a^2*b^2 - 2*a*b^3 + b^4)*f*cosh(f*x + e)^2 + (2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f)*sinh(f* x + e)^6 + 2*(8*a^4 - 24*a^3*b + 27*a^2*b^2 - 14*a*b^3 + 3*b^4)*f*cosh(f*x + e)^4 + 8*(7*(a^2*b^2 - 2*a*b^3 + b^4)*f*cosh(f*x + e)^3 + 3*(2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(35*(a^2*b ^2 - 2*a*b^3 + b^4)*f*cosh(f*x + e)^4 + 30*(2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f*cosh(f*x + e)^2 + (8*a^4 - 24*a^3*b + 27*a^2*b^2 - 14*a*b^3 + 3*b ^4)*f)*sinh(f*x + e)^4 + 4*(2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f*cosh(f* x + e)^2 + 8*(7*(a^2*b^2 - 2*a*b^3 + b^4)*f*cosh(f*x + e)^5 + 10*(2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f*cosh(f*x + e)^3 + (8*a^4 - 24*a^3*b + 27...
Timed out. \[ \int \frac {\sinh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 485 vs. \(2 (71) = 142\).
Time = 0.32 (sec) , antiderivative size = 485, normalized size of antiderivative = 6.14 \[ \int \frac {\sinh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\frac {2 \, a^{2} b^{2} - 2 \, a b^{3} + b^{4} + 5 \, {\left (4 \, a^{3} b - 6 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, f x - 2 \, e\right )} + 2 \, {\left (24 \, a^{4} - 48 \, a^{3} b + 49 \, a^{2} b^{2} - 25 \, a b^{3} + 5 \, b^{4}\right )} e^{\left (-4 \, f x - 4 \, e\right )} + 10 \, {\left (6 \, a^{3} b - 9 \, a^{2} b^{2} + 5 \, a b^{3} - b^{4}\right )} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \, {\left (4 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} e^{\left (-8 \, f x - 8 \, e\right )} + {\left (2 \, a b^{3} - b^{4}\right )} e^{\left (-10 \, f x - 10 \, e\right )}}{3 \, {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (2 \, {\left (2 \, a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} + b e^{\left (-4 \, f x - 4 \, e\right )} + b\right )}^{\frac {5}{2}} f} + \frac {2 \, a b^{3} - b^{4} + 5 \, {\left (4 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} e^{\left (-2 \, f x - 2 \, e\right )} + 10 \, {\left (6 \, a^{3} b - 9 \, a^{2} b^{2} + 5 \, a b^{3} - b^{4}\right )} e^{\left (-4 \, f x - 4 \, e\right )} + 2 \, {\left (24 \, a^{4} - 48 \, a^{3} b + 49 \, a^{2} b^{2} - 25 \, a b^{3} + 5 \, b^{4}\right )} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \, {\left (4 \, a^{3} b - 6 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} e^{\left (-8 \, f x - 8 \, e\right )} + {\left (2 \, a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} e^{\left (-10 \, f x - 10 \, e\right )}}{3 \, {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (2 \, {\left (2 \, a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} + b e^{\left (-4 \, f x - 4 \, e\right )} + b\right )}^{\frac {5}{2}} f} \]
1/3*(2*a^2*b^2 - 2*a*b^3 + b^4 + 5*(4*a^3*b - 6*a^2*b^2 + 4*a*b^3 - b^4)*e ^(-2*f*x - 2*e) + 2*(24*a^4 - 48*a^3*b + 49*a^2*b^2 - 25*a*b^3 + 5*b^4)*e^ (-4*f*x - 4*e) + 10*(6*a^3*b - 9*a^2*b^2 + 5*a*b^3 - b^4)*e^(-6*f*x - 6*e) + 5*(4*a^2*b^2 - 4*a*b^3 + b^4)*e^(-8*f*x - 8*e) + (2*a*b^3 - b^4)*e^(-10 *f*x - 10*e))/((a^4 - 2*a^3*b + a^2*b^2)*(2*(2*a - b)*e^(-2*f*x - 2*e) + b *e^(-4*f*x - 4*e) + b)^(5/2)*f) + 1/3*(2*a*b^3 - b^4 + 5*(4*a^2*b^2 - 4*a* b^3 + b^4)*e^(-2*f*x - 2*e) + 10*(6*a^3*b - 9*a^2*b^2 + 5*a*b^3 - b^4)*e^( -4*f*x - 4*e) + 2*(24*a^4 - 48*a^3*b + 49*a^2*b^2 - 25*a*b^3 + 5*b^4)*e^(- 6*f*x - 6*e) + 5*(4*a^3*b - 6*a^2*b^2 + 4*a*b^3 - b^4)*e^(-8*f*x - 8*e) + (2*a^2*b^2 - 2*a*b^3 + b^4)*e^(-10*f*x - 10*e))/((a^4 - 2*a^3*b + a^2*b^2) *(2*(2*a - b)*e^(-2*f*x - 2*e) + b*e^(-4*f*x - 4*e) + b)^(5/2)*f)
Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (71) = 142\).
Time = 0.84 (sec) , antiderivative size = 254, normalized size of antiderivative = 3.22 \[ \int \frac {\sinh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\frac {2 \, {\left (\frac {a^{2} b e^{\left (6 \, e\right )}}{a^{4} e^{\left (6 \, e\right )} - 2 \, a^{3} b e^{\left (6 \, e\right )} + a^{2} b^{2} e^{\left (6 \, e\right )}} + {\left ({\left (\frac {a^{2} b e^{\left (2 \, f x + 12 \, e\right )}}{a^{4} e^{\left (6 \, e\right )} - 2 \, a^{3} b e^{\left (6 \, e\right )} + a^{2} b^{2} e^{\left (6 \, e\right )}} + \frac {3 \, {\left (2 \, a^{3} e^{\left (10 \, e\right )} - a^{2} b e^{\left (10 \, e\right )}\right )}}{a^{4} e^{\left (6 \, e\right )} - 2 \, a^{3} b e^{\left (6 \, e\right )} + a^{2} b^{2} e^{\left (6 \, e\right )}}\right )} e^{\left (2 \, f x\right )} + \frac {3 \, {\left (2 \, a^{3} e^{\left (8 \, e\right )} - a^{2} b e^{\left (8 \, e\right )}\right )}}{a^{4} e^{\left (6 \, e\right )} - 2 \, a^{3} b e^{\left (6 \, e\right )} + a^{2} b^{2} e^{\left (6 \, e\right )}}\right )} e^{\left (2 \, f x\right )}\right )}}{3 \, {\left (b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b\right )}^{\frac {3}{2}} f} \]
2/3*(a^2*b*e^(6*e)/(a^4*e^(6*e) - 2*a^3*b*e^(6*e) + a^2*b^2*e^(6*e)) + ((a ^2*b*e^(2*f*x + 12*e)/(a^4*e^(6*e) - 2*a^3*b*e^(6*e) + a^2*b^2*e^(6*e)) + 3*(2*a^3*e^(10*e) - a^2*b*e^(10*e))/(a^4*e^(6*e) - 2*a^3*b*e^(6*e) + a^2*b ^2*e^(6*e)))*e^(2*f*x) + 3*(2*a^3*e^(8*e) - a^2*b*e^(8*e))/(a^4*e^(6*e) - 2*a^3*b*e^(6*e) + a^2*b^2*e^(6*e)))*e^(2*f*x))/((b*e^(4*f*x + 4*e) + 4*a*e ^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b)^(3/2)*f)
Time = 1.99 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.68 \[ \int \frac {\sinh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx=\frac {4\,{\mathrm {e}}^{e+f\,x}\,\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )\,\sqrt {a+b\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}\,\left (b+6\,a\,{\mathrm {e}}^{2\,e+2\,f\,x}-4\,b\,{\mathrm {e}}^{2\,e+2\,f\,x}+b\,{\mathrm {e}}^{4\,e+4\,f\,x}\right )}{3\,f\,{\left (a-b\right )}^2\,{\left (b+4\,a\,{\mathrm {e}}^{2\,e+2\,f\,x}-2\,b\,{\mathrm {e}}^{2\,e+2\,f\,x}+b\,{\mathrm {e}}^{4\,e+4\,f\,x}\right )}^2} \]