Integrand size = 25, antiderivative size = 140 \[ \int \coth ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=-\frac {\sqrt {a} (2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )}{2 f}+\frac {(2 a+3 b) \sqrt {a+b \sinh ^2(e+f x)}}{2 f}+\frac {(2 a+3 b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}{6 a f}-\frac {\text {csch}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{2 a f} \]
1/6*(2*a+3*b)*(a+b*sinh(f*x+e)^2)^(3/2)/a/f-1/2*csch(f*x+e)^2*(a+b*sinh(f* x+e)^2)^(5/2)/a/f-1/2*(2*a+3*b)*arctanh((a+b*sinh(f*x+e)^2)^(1/2)/a^(1/2)) *a^(1/2)/f+1/2*(2*a+3*b)*(a+b*sinh(f*x+e)^2)^(1/2)/f
Time = 0.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.64 \[ \int \coth ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\frac {-3 \sqrt {a} (2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )+\left (8 a+5 b+b \cosh (2 (e+f x))-3 a \text {csch}^2(e+f x)\right ) \sqrt {a+b \sinh ^2(e+f x)}}{6 f} \]
(-3*Sqrt[a]*(2*a + 3*b)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a]] + (8* a + 5*b + b*Cosh[2*(e + f*x)] - 3*a*Csch[e + f*x]^2)*Sqrt[a + b*Sinh[e + f *x]^2])/(6*f)
Time = 0.31 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 26, 3673, 87, 60, 60, 73, 221}
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.
\(\displaystyle \int \coth ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \left (a-b \sin (i e+i f x)^2\right )^{3/2}}{\tan (i e+i f x)^3}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\left (a-b \sin (i e+i f x)^2\right )^{3/2}}{\tan (i e+i f x)^3}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \text {csch}^4(e+f x) \left (\sinh ^2(e+f x)+1\right ) \left (b \sinh ^2(e+f x)+a\right )^{3/2}d\sinh ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {\frac {(2 a+3 b) \int \text {csch}^2(e+f x) \left (b \sinh ^2(e+f x)+a\right )^{3/2}d\sinh ^2(e+f x)}{2 a}-\frac {\text {csch}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{a}}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\frac {(2 a+3 b) \left (a \int \text {csch}^2(e+f x) \sqrt {b \sinh ^2(e+f x)+a}d\sinh ^2(e+f x)+\frac {2}{3} \left (a+b \sinh ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {\text {csch}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{a}}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\frac {(2 a+3 b) \left (a \left (a \int \frac {\text {csch}^2(e+f x)}{\sqrt {b \sinh ^2(e+f x)+a}}d\sinh ^2(e+f x)+2 \sqrt {a+b \sinh ^2(e+f x)}\right )+\frac {2}{3} \left (a+b \sinh ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {\text {csch}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{a}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {(2 a+3 b) \left (a \left (\frac {2 a \int \frac {1}{\frac {\sinh ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sinh ^2(e+f x)+a}}{b}+2 \sqrt {a+b \sinh ^2(e+f x)}\right )+\frac {2}{3} \left (a+b \sinh ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {\text {csch}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{a}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {(2 a+3 b) \left (a \left (2 \sqrt {a+b \sinh ^2(e+f x)}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sinh ^2(e+f x)}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b \sinh ^2(e+f x)\right )^{3/2}\right )}{2 a}-\frac {\text {csch}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{5/2}}{a}}{2 f}\) |
(-((Csch[e + f*x]^2*(a + b*Sinh[e + f*x]^2)^(5/2))/a) + ((2*a + 3*b)*((2*( a + b*Sinh[e + f*x]^2)^(3/2))/3 + a*(-2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a]] + 2*Sqrt[a + b*Sinh[e + f*x]^2])))/(2*a))/(2*f)
3.5.72.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m + 1)/2)/(2*f) Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 )/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ erQ[(m - 1)/2]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.60
method | result | size |
default | \(\frac {\operatorname {`\,int/indef0`\,}\left (\frac {\sinh \left (f x +e \right )^{3} b^{2}+\left (2 a b +b^{2}\right ) \sinh \left (f x +e \right )+\frac {a^{2}+2 a b}{\sinh \left (f x +e \right )}+\frac {a^{2}}{\sinh \left (f x +e \right )^{3}}}{\sqrt {a +b \sinh \left (f x +e \right )^{2}}}, \sinh \left (f x +e \right )\right )}{f}\) | \(84\) |
`int/indef0`((sinh(f*x+e)^3*b^2+(2*a*b+b^2)*sinh(f*x+e)+(a^2+2*a*b)/sinh(f *x+e)+a^2/sinh(f*x+e)^3)/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/f
Leaf count of result is larger than twice the leaf count of optimal. 1102 vs. \(2 (120) = 240\).
Time = 0.54 (sec) , antiderivative size = 2406, normalized size of antiderivative = 17.19 \[ \int \coth ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
[1/24*(6*((2*a + 3*b)*cosh(f*x + e)^7 + 7*(2*a + 3*b)*cosh(f*x + e)*sinh(f *x + e)^6 + (2*a + 3*b)*sinh(f*x + e)^7 - 2*(2*a + 3*b)*cosh(f*x + e)^5 + (21*(2*a + 3*b)*cosh(f*x + e)^2 - 4*a - 6*b)*sinh(f*x + e)^5 + 5*(7*(2*a + 3*b)*cosh(f*x + e)^3 - 2*(2*a + 3*b)*cosh(f*x + e))*sinh(f*x + e)^4 + (2* a + 3*b)*cosh(f*x + e)^3 + (35*(2*a + 3*b)*cosh(f*x + e)^4 - 20*(2*a + 3*b )*cosh(f*x + e)^2 + 2*a + 3*b)*sinh(f*x + e)^3 + (21*(2*a + 3*b)*cosh(f*x + e)^5 - 20*(2*a + 3*b)*cosh(f*x + e)^3 + 3*(2*a + 3*b)*cosh(f*x + e))*sin h(f*x + e)^2 + (7*(2*a + 3*b)*cosh(f*x + e)^6 - 10*(2*a + 3*b)*cosh(f*x + e)^4 + 3*(2*a + 3*b)*cosh(f*x + e)^2)*sinh(f*x + e))*sqrt(a)*log((b*cosh(f *x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(4*a - b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 4*a - b)*sinh(f*x + e)^2 - 4*sqrt(2)*sqrt(a)*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))*(cos h(f*x + e) + sinh(f*x + e)) + 4*(b*cosh(f*x + e)^3 + (4*a - b)*cosh(f*x + e))*sinh(f*x + e) + b)/(cosh(f*x + e)^4 + 4*cosh(f*x + e)*sinh(f*x + e)^3 + sinh(f*x + e)^4 + 2*(3*cosh(f*x + e)^2 - 1)*sinh(f*x + e)^2 - 2*cosh(f*x + e)^2 + 4*(cosh(f*x + e)^3 - cosh(f*x + e))*sinh(f*x + e) + 1)) + sqrt(2 )*(b*cosh(f*x + e)^8 + 8*b*cosh(f*x + e)*sinh(f*x + e)^7 + b*sinh(f*x + e) ^8 + 8*(2*a + b)*cosh(f*x + e)^6 + 4*(7*b*cosh(f*x + e)^2 + 4*a + 2*b)*sin h(f*x + e)^6 + 8*(7*b*cosh(f*x + e)^3 + 6*(2*a + b)*cosh(f*x + e))*sinh...
Timed out. \[ \int \coth ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
\[ \int \coth ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \coth \left (f x + e\right )^{3} \,d x } \]
Exception generated. \[ \int \coth ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{%%%{256,[1]%%%},[6,14,6]%%%}+%%%{%%%{-768,[2]%%%},[6,14,5] %%%}+%%%{
Timed out. \[ \int \coth ^3(e+f x) \left (a+b \sinh ^2(e+f x)\right )^{3/2} \, dx=\int {\mathrm {coth}\left (e+f\,x\right )}^3\,{\left (b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]