Integrand size = 25, antiderivative size = 77 \[ \int \frac {\cosh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{b^{3/2} f}-\frac {(a-b) \sinh (e+f x)}{a b f \sqrt {a+b \sinh ^2(e+f x)}} \] Output:
arctanh(b^(1/2)*sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2))/b^(3/2)/f-(a-b)*sin h(f*x+e)/a/b/f/(a+b*sinh(f*x+e)^2)^(1/2)
Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.16 \[ \int \frac {\cosh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sqrt {b} (-a+b) \sinh (e+f x)+a^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} \sinh (e+f x)}{\sqrt {a}}\right ) \sqrt {1+\frac {b \sinh ^2(e+f x)}{a}}}{a b^{3/2} f \sqrt {a+b \sinh ^2(e+f x)}} \] Input:
Integrate[Cosh[e + f*x]^3/(a + b*Sinh[e + f*x]^2)^(3/2),x]
Output:
(Sqrt[b]*(-a + b)*Sinh[e + f*x] + a^(3/2)*ArcSinh[(Sqrt[b]*Sinh[e + f*x])/ Sqrt[a]]*Sqrt[1 + (b*Sinh[e + f*x]^2)/a])/(a*b^(3/2)*f*Sqrt[a + b*Sinh[e + f*x]^2])
Time = 0.33 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3669, 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.
\(\displaystyle \int \frac {\cosh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i e+i f x)^3}{\left (a-b \sin (i e+i f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \frac {\sinh ^2(e+f x)+1}{\left (b \sinh ^2(e+f x)+a\right )^{3/2}}d\sinh (e+f x)}{f}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {\int \frac {1}{\sqrt {b \sinh ^2(e+f x)+a}}d\sinh (e+f x)}{b}-\frac {(a-b) \sinh (e+f x)}{a b \sqrt {a+b \sinh ^2(e+f x)}}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {\int \frac {1}{1-\frac {b \sinh ^2(e+f x)}{b \sinh ^2(e+f x)+a}}d\frac {\sinh (e+f x)}{\sqrt {b \sinh ^2(e+f x)+a}}}{b}-\frac {(a-b) \sinh (e+f x)}{a b \sqrt {a+b \sinh ^2(e+f x)}}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {b} \sinh (e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}}\right )}{b^{3/2}}-\frac {(a-b) \sinh (e+f x)}{a b \sqrt {a+b \sinh ^2(e+f x)}}}{f}\) |
Input:
Int[Cosh[e + f*x]^3/(a + b*Sinh[e + f*x]^2)^(3/2),x]
Output:
(ArcTanh[(Sqrt[b]*Sinh[e + f*x])/Sqrt[a + b*Sinh[e + f*x]^2]]/b^(3/2) - (( a - b)*Sinh[e + f*x])/(a*b*Sqrt[a + b*Sinh[e + f*x]^2]))/f
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[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 = 0.32 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {\frac {\sinh \left (f x +e \right )}{a \sqrt {a +b \sinh \left (f x +e \right )^{2}}}-\frac {\sinh \left (f x +e \right )}{b \sqrt {a +b \sinh \left (f x +e \right )^{2}}}+\frac {\ln \left (\sqrt {b}\, \sinh \left (f x +e \right )+\sqrt {a +b \sinh \left (f x +e \right )^{2}}\right )}{b^{\frac {3}{2}}}}{f}\) | \(85\) |
default | \(\frac {\frac {\sinh \left (f x +e \right )}{a \sqrt {a +b \sinh \left (f x +e \right )^{2}}}-\frac {\sinh \left (f x +e \right )}{b \sqrt {a +b \sinh \left (f x +e \right )^{2}}}+\frac {\ln \left (\sqrt {b}\, \sinh \left (f x +e \right )+\sqrt {a +b \sinh \left (f x +e \right )^{2}}\right )}{b^{\frac {3}{2}}}}{f}\) | \(85\) |
Input:
int(cosh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/f*(sinh(f*x+e)/a/(a+b*sinh(f*x+e)^2)^(1/2)-sinh(f*x+e)/b/(a+b*sinh(f*x+e )^2)^(1/2)+1/b^(3/2)*ln(b^(1/2)*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. 996 vs. \(2 (69) = 138\).
Time = 0.21 (sec) , antiderivative size = 3014, normalized size of antiderivative = 39.14 \[ \int \frac {\cosh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(cosh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\cosh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(cosh(f*x+e)**3/(a+b*sinh(f*x+e)**2)**(3/2),x)
Output:
Timed out
\[ \int \frac {\cosh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cosh \left (f x + e\right )^{3}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(cosh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")
Output:
integrate(cosh(f*x + e)^3/(b*sinh(f*x + e)^2 + a)^(3/2), x)
Exception generated. \[ \int \frac {\cosh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cosh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio n over extensionNot implemented, e.g. for multivariate mod/approx polynomi alsError:
Timed out. \[ \int \frac {\cosh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {cosh}\left (e+f\,x\right )}^3}{{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \] Input:
int(cosh(e + f*x)^3/(a + b*sinh(e + f*x)^2)^(3/2),x)
Output:
int(cosh(e + f*x)^3/(a + b*sinh(e + f*x)^2)^(3/2), x)
\[ \int \frac {\cosh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sinh \left (f x +e \right )^{2} b +a}\, \cosh \left (f x +e \right )^{3}}{\sinh \left (f x +e \right )^{4} b^{2}+2 \sinh \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:
int(cosh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(3/2),x)
Output:
int((sqrt(sinh(e + f*x)**2*b + a)*cosh(e + f*x)**3)/(sinh(e + f*x)**4*b**2 + 2*sinh(e + f*x)**2*a*b + a**2),x)