Integrand size = 25, antiderivative size = 83 \[ \int \frac {\sinh ^3(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=-\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {b} \cosh (e+f x)}{\sqrt {a-b+b \cosh ^2(e+f x)}}\right )}{2 b^{3/2} f}+\frac {\cosh (e+f x) \sqrt {a-b+b \cosh ^2(e+f x)}}{2 b f} \]
-1/2*(a+b)*arctanh(cosh(f*x+e)*b^(1/2)/(a-b+b*cosh(f*x+e)^2)^(1/2))/b^(3/2 )/f+1/2*cosh(f*x+e)*(a-b+b*cosh(f*x+e)^2)^(1/2)/b/f
Time = 0.51 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.18 \[ \int \frac {\sinh ^3(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=\frac {\cosh (e+f x) \sqrt {2 a-b+b \cosh (2 (e+f x))}}{2 \sqrt {2} b f}-\frac {(a+b) \log \left (\sqrt {2} \sqrt {b} \cosh (e+f x)+\sqrt {2 a-b+b \cosh (2 (e+f x))}\right )}{2 b^{3/2} f} \]
(Cosh[e + f*x]*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])/(2*Sqrt[2]*b*f) - ((a + b)*Log[Sqrt[2]*Sqrt[b]*Cosh[e + f*x] + Sqrt[2*a - b + b*Cosh[2*(e + f*x) ]]])/(2*b^(3/2)*f)
Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 26, 3665, 299, 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.
\(\displaystyle \int \frac {\sinh ^3(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i \sin (i e+i f x)^3}{\sqrt {a-b \sin (i e+i f x)^2}}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\sin (i e+i f x)^3}{\sqrt {a-b \sin (i e+i f x)^2}}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle -\frac {\int \frac {1-\cosh ^2(e+f x)}{\sqrt {b \cosh ^2(e+f x)+a-b}}d\cosh (e+f x)}{f}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle -\frac {\frac {(a+b) \int \frac {1}{\sqrt {b \cosh ^2(e+f x)+a-b}}d\cosh (e+f x)}{2 b}-\frac {\cosh (e+f x) \sqrt {a+b \cosh ^2(e+f x)-b}}{2 b}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {\frac {(a+b) \int \frac {1}{1-\frac {b \cosh ^2(e+f x)}{b \cosh ^2(e+f x)+a-b}}d\frac {\cosh (e+f x)}{\sqrt {b \cosh ^2(e+f x)+a-b}}}{2 b}-\frac {\cosh (e+f x) \sqrt {a+b \cosh ^2(e+f x)-b}}{2 b}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {b} \cosh (e+f x)}{\sqrt {a+b \cosh ^2(e+f x)-b}}\right )}{2 b^{3/2}}-\frac {\cosh (e+f x) \sqrt {a+b \cosh ^2(e+f x)-b}}{2 b}}{f}\) |
-((((a + b)*ArcTanh[(Sqrt[b]*Cosh[e + f*x])/Sqrt[a - b + b*Cosh[e + f*x]^2 ]])/(2*b^(3/2)) - (Cosh[e + f*x]*Sqrt[a - b + b*Cosh[e + f*x]^2])/(2*b))/f )
3.1.98.3.1 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[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 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 = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.60
method | result | size |
default | \(\frac {\sqrt {\left (a +b \sinh \left (f x +e \right )^{2}\right ) \cosh \left (f x +e \right )^{2}}\, \left (\frac {\sqrt {\left (a +b \sinh \left (f x +e \right )^{2}\right ) \cosh \left (f x +e \right )^{2}}}{2 b}-\frac {\left (a +b \right ) \ln \left (\frac {\frac {a}{2}+\frac {b}{2}+b \sinh \left (f x +e \right )^{2}}{\sqrt {b}}+\sqrt {\left (a +b \sinh \left (f x +e \right )^{2}\right ) \cosh \left (f x +e \right )^{2}}\right )}{4 b^{\frac {3}{2}}}\right )}{\cosh \left (f x +e \right ) \sqrt {a +b \sinh \left (f x +e \right )^{2}}\, f}\) | \(133\) |
((a+b*sinh(f*x+e)^2)*cosh(f*x+e)^2)^(1/2)*(1/2/b*((a+b*sinh(f*x+e)^2)*cosh (f*x+e)^2)^(1/2)-1/4*(a+b)/b^(3/2)*ln((1/2*a+1/2*b+b*sinh(f*x+e)^2)/b^(1/2 )+((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. 720 vs. \(2 (71) = 142\).
Time = 0.33 (sec) , antiderivative size = 2116, normalized size of antiderivative = 25.49 \[ \int \frac {\sinh ^3(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=\text {Too large to display} \]
[1/8*(((a + b)*cosh(f*x + e)^2 + 2*(a + b)*cosh(f*x + e)*sinh(f*x + e) + ( a + b)*sinh(f*x + e)^2)*sqrt(b)*log((a^2*b*cosh(f*x + e)^8 + 8*a^2*b*cosh( f*x + e)*sinh(f*x + e)^7 + a^2*b*sinh(f*x + e)^8 + 2*(a^3 + a^2*b)*cosh(f* x + e)^6 + 2*(14*a^2*b*cosh(f*x + e)^2 + a^3 + a^2*b)*sinh(f*x + e)^6 + 4* (14*a^2*b*cosh(f*x + e)^3 + 3*(a^3 + a^2*b)*cosh(f*x + e))*sinh(f*x + e)^5 + (9*a^2*b - 4*a*b^2 + b^3)*cosh(f*x + e)^4 + (70*a^2*b*cosh(f*x + e)^4 + 9*a^2*b - 4*a*b^2 + b^3 + 30*(a^3 + a^2*b)*cosh(f*x + e)^2)*sinh(f*x + e) ^4 + 4*(14*a^2*b*cosh(f*x + e)^5 + 10*(a^3 + a^2*b)*cosh(f*x + e)^3 + (9*a ^2*b - 4*a*b^2 + b^3)*cosh(f*x + e))*sinh(f*x + e)^3 + b^3 + 2*(3*a*b^2 - b^3)*cosh(f*x + e)^2 + 2*(14*a^2*b*cosh(f*x + e)^6 + 15*(a^3 + a^2*b)*cosh (f*x + e)^4 + 3*a*b^2 - b^3 + 3*(9*a^2*b - 4*a*b^2 + b^3)*cosh(f*x + e)^2) *sinh(f*x + e)^2 - sqrt(2)*(a^2*cosh(f*x + e)^6 + 6*a^2*cosh(f*x + e)*sinh (f*x + e)^5 + a^2*sinh(f*x + e)^6 + 3*a^2*cosh(f*x + e)^4 + 3*(5*a^2*cosh( f*x + e)^2 + a^2)*sinh(f*x + e)^4 + 4*(5*a^2*cosh(f*x + e)^3 + 3*a^2*cosh( f*x + e))*sinh(f*x + e)^3 + (4*a*b - b^2)*cosh(f*x + e)^2 + (15*a^2*cosh(f *x + e)^4 + 18*a^2*cosh(f*x + e)^2 + 4*a*b - b^2)*sinh(f*x + e)^2 + b^2 + 2*(3*a^2*cosh(f*x + e)^5 + 6*a^2*cosh(f*x + e)^3 + (4*a*b - b^2)*cosh(f*x + e))*sinh(f*x + e))*sqrt(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)) + 4*(2*a^2*b*cosh(f*x + e)^7 + 3*(a^3 + a^2*b)*cosh(f*x + e)^5 + (...
Timed out. \[ \int \frac {\sinh ^3(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\sinh ^3(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=\int { \frac {\sinh \left (f x + e\right )^{3}}{\sqrt {b \sinh \left (f x + e\right )^{2} + a}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 430 vs. \(2 (71) = 142\).
Time = 0.44 (sec) , antiderivative size = 430, normalized size of antiderivative = 5.18 \[ \int \frac {\sinh ^3(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=\frac {{\left (\frac {4 \, {\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {\sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b} e^{\left (2 \, e\right )}}{b} + \frac {2 \, {\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} \log \left ({\left | {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} \sqrt {b} + 2 \, a - b \right |}\right )}{b^{\frac {3}{2}}} - \frac {2 \, {\left (2 \, {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} a e^{\left (2 \, e\right )} - {\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )} b e^{\left (2 \, e\right )} + b^{\frac {3}{2}} e^{\left (2 \, e\right )}\right )}}{{\left ({\left (\sqrt {b} e^{\left (2 \, f x + 2 \, e\right )} - \sqrt {b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}\right )}^{2} - b\right )} b}\right )} e^{\left (-2 \, e\right )}}{8 \, f} \]
1/8*(4*(a*e^(2*e) + b*e^(2*e))*arctan(-(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e ^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))/sqrt(-b)) /(sqrt(-b)*b) + sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f* x + 2*e) + b)*e^(2*e)/b + 2*(a*e^(2*e) + b*e^(2*e))*log(abs((sqrt(b)*e^(2* f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*sqrt(b) + 2*a - b))/b^(3/2) - 2*(2*(sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*a *e^(2*e) - (sqrt(b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f* x + 2*e) - 2*b*e^(2*f*x + 2*e) + b))*b*e^(2*e) + b^(3/2)*e^(2*e))/(((sqrt( b)*e^(2*f*x + 2*e) - sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^ (2*f*x + 2*e) + b))^2 - b)*b))*e^(-2*e)/f
Timed out. \[ \int \frac {\sinh ^3(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=\int \frac {{\mathrm {sinh}\left (e+f\,x\right )}^3}{\sqrt {b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}} \,d x \]