Integrand size = 24, antiderivative size = 141 \[ \int x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\frac {3 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 d^2}+\frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \left (c+d x^2\right )}{2 d}-\frac {3 \sqrt {b} (b c-a d) e^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 d^{5/2}} \]
1/2*(e*(b*x^2+a)/(d*x^2+c))^(3/2)*(d*x^2+c)/d-3/2*(-a*d+b*c)*e^(3/2)*arcta nh(d^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^(1/2)/e^(1/2))*b^(1/2)/d^(5/2)+ 3/2*(-a*d+b*c)*e*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/d^2
Time = 1.89 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.96 \[ \int x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {d} \sqrt {a+b x^2} \left (3 b c-2 a d+b d x^2\right )-3 \sqrt {b} (b c-a d) \sqrt {c+d x^2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )\right )}{2 d^{5/2} \sqrt {a+b x^2}} \]
(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[d]*Sqrt[a + b*x^2]*(3*b*c - 2*a *d + b*d*x^2) - 3*Sqrt[b]*(b*c - a*d)*Sqrt[c + d*x^2]*ArcTanh[(Sqrt[d]*Sqr t[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])]))/(2*d^(5/2)*Sqrt[a + b*x^2])
Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2053, 2051, 252, 262, 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.
\(\displaystyle \int x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 2053 |
\(\displaystyle \frac {1}{2} \int \left (\frac {e \left (b x^2+a\right )}{d x^2+c}\right )^{3/2}dx^2\) |
\(\Big \downarrow \) 2051 |
\(\displaystyle e (b c-a d) \int \frac {x^8}{\left (b e-d x^4\right )^2}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle e (b c-a d) \left (\frac {x^6}{2 d \left (b e-d x^4\right )}-\frac {3 \int \frac {x^4}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{2 d}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle e (b c-a d) \left (\frac {x^6}{2 d \left (b e-d x^4\right )}-\frac {3 \left (\frac {b e \int \frac {1}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{d}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d}\right )}{2 d}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle e (b c-a d) \left (\frac {x^6}{2 d \left (b e-d x^4\right )}-\frac {3 \left (\frac {\sqrt {b} \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{d^{3/2}}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{d}\right )}{2 d}\right )\) |
(b*c - a*d)*e*(x^6/(2*d*(b*e - d*x^4)) - (3*(-(Sqrt[(e*(a + b*x^2))/(c + d *x^2)]/d) + (Sqrt[b]*Sqrt[e]*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d* x^2)])/(Sqrt[b]*Sqrt[e])])/d^(3/2)))/(2*d))
3.3.78.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_ Symbol] :> With[{q = Denominator[p]}, Simp[q*e*((b*c - a*d)/n) Subst[Int[ x^(q*(p + 1) - 1)*(((-a)*e + c*x^q)^(1/n - 1)/(b*e - d*x^q)^(1/n + 1)), x], x, (e*((a + b*x^n)/(c + d*x^n)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e}, x] && FractionQ[p] && IntegerQ[1/n]
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.) ))^(p_), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(e*( (a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.16 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.64
method | result | size |
risch | \(\frac {\left (d \,x^{2}+c \right ) b e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{2 d^{2}}+\frac {\left (\frac {3 b \left (a d -b c \right ) \ln \left (\frac {\frac {1}{2} e d a +\frac {1}{2} e b c +b d e \,x^{2}}{\sqrt {b d e}}+\sqrt {b d e \,x^{4}+\left (e d a +e b c \right ) x^{2}+a c e}\right )}{2 \sqrt {b d e}}-\frac {\left (2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right ) \left (b \,x^{2}+a \right )}{\left (a d -b c \right ) \sqrt {b d e \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{2 d^{2} \left (b \,x^{2}+a \right )}\) | \(231\) |
default | \(-\frac {\left (-3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,d^{2} x^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c d \,x^{2}-2 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b d \,x^{2}-3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c d +3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2}+4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, a d -4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, b c -2 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b c \right ) \left (d \,x^{2}+c \right ) {\left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )}^{\frac {3}{2}}}{4 d^{2} \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (b \,x^{2}+a \right )}\) | \(432\) |
1/2/d^2*(d*x^2+c)*b*e*(e*(b*x^2+a)/(d*x^2+c))^(1/2)+1/2/d^2*(3/2*b*(a*d-b* c)*ln((1/2*e*d*a+1/2*e*b*c+b*d*e*x^2)/(b*d*e)^(1/2)+(b*d*e*x^4+(a*d*e+b*c* e)*x^2+a*c*e)^(1/2))/(b*d*e)^(1/2)-(2*a^2*d^2-4*a*b*c*d+2*b^2*c^2)*(b*x^2+ a)/(a*d-b*c)/(b*d*e*x^4+a*d*e*x^2+b*c*e*x^2+a*c*e)^(1/2))*e/(b*x^2+a)*(e*( b*x^2+a)/(d*x^2+c))^(1/2)*((d*x^2+c)*e*(b*x^2+a))^(1/2)
Time = 0.46 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.33 \[ \int x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\left [-\frac {3 \, {\left (b c - a d\right )} \sqrt {\frac {b e}{d}} e \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \, {\left (2 \, b d^{3} x^{4} + b c^{2} d + a c d^{2} + {\left (3 \, b c d^{2} + a d^{3}\right )} x^{2}\right )} \sqrt {\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ) - 4 \, {\left (b d e x^{2} + {\left (3 \, b c - 2 \, a d\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, d^{2}}, \frac {3 \, {\left (b c - a d\right )} \sqrt {-\frac {b e}{d}} e \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-\frac {b e}{d}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} e x^{2} + a b e\right )}}\right ) + 2 \, {\left (b d e x^{2} + {\left (3 \, b c - 2 \, a d\right )} e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, d^{2}}\right ] \]
[-1/8*(3*(b*c - a*d)*sqrt(b*e/d)*e*log(8*b^2*d^2*e*x^4 + 8*(b^2*c*d + a*b* d^2)*e*x^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*e + 4*(2*b*d^3*x^4 + b*c^2*d + a*c*d^2 + (3*b*c*d^2 + a*d^3)*x^2)*sqrt(b*e/d)*sqrt((b*e*x^2 + a*e)/(d*x ^2 + c))) - 4*(b*d*e*x^2 + (3*b*c - 2*a*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d^2, 1/4*(3*(b*c - a*d)*sqrt(-b*e/d)*e*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(-b*e/d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(b^2*e*x^2 + a*b*e)) + 2*(b*d*e*x^2 + (3*b*c - 2*a*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d^ 2]
Timed out. \[ \int x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\text {Timed out} \]
Exception generated. \[ \int x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Exception generated. \[ \int x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]
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%%%{%%%{2,[0,3,0]%%%},[2,0,0,0]%%%}+%%%{%%{[%%%{-4,[0,2,0]%%%} ,0]:[1,0,
Timed out. \[ \int x \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx=\int x\,{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2} \,d x \]