Integrand size = 24, antiderivative size = 146 \[ \int \frac {x}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=-\frac {3 (b c-a d)}{2 b^2 e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {c+d x^2}{2 b e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {3 \sqrt {d} (b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 b^{5/2} e^{3/2}} \]
3/2*(-a*d+b*c)*arctanh(d^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b^(1/2)/e^(1/ 2))*d^(1/2)/b^(5/2)/e^(3/2)-3/2*(-a*d+b*c)/b^2/e/(e*(b*x^2+a)/(d*x^2+c))^( 1/2)+1/2*(d*x^2+c)/b/e/(e*(b*x^2+a)/(d*x^2+c))^(1/2)
Time = 1.86 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95 \[ \int \frac {x}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\frac {\sqrt {b} \sqrt {c+d x^2} \left (-2 b c+3 a d+b d x^2\right )+3 \sqrt {d} (b c-a d) \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 b^{5/2} e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}} \]
(Sqrt[b]*Sqrt[c + d*x^2]*(-2*b*c + 3*a*d + b*d*x^2) + 3*Sqrt[d]*(b*c - a*d )*Sqrt[a + b*x^2]*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^ 2])])/(2*b^(5/2)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*Sqrt[c + d*x^2])
Time = 0.25 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2053, 2051, 253, 264, 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.
\(\displaystyle \int \frac {x}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2053 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\left (\frac {e \left (b x^2+a\right )}{d x^2+c}\right )^{3/2}}dx^2\) |
\(\Big \downarrow \) 2051 |
\(\displaystyle e (b c-a d) \int \frac {1}{x^4 \left (b e-d x^4\right )^2}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle e (b c-a d) \left (\frac {3 \int \frac {1}{x^4 \left (b e-d x^4\right )}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{2 b e}+\frac {1}{2 b e x^2 \left (b e-d x^4\right )}\right )\) |
\(\Big \downarrow \) 264 |
\(\displaystyle e (b c-a d) \left (\frac {3 \left (\frac {d \int \frac {1}{b e-d x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{b e}-\frac {1}{b e x^2}\right )}{2 b e}+\frac {1}{2 b e x^2 \left (b e-d x^4\right )}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle e (b c-a d) \left (\frac {3 \left (\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{b^{3/2} e^{3/2}}-\frac {1}{b e x^2}\right )}{2 b e}+\frac {1}{2 b e x^2 \left (b e-d x^4\right )}\right )\) |
(b*c - a*d)*e*(1/(2*b*e*x^2*(b*e - d*x^4)) + (3*(-(1/(b*e*x^2)) + (Sqrt[d] *ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/( b^(3/2)*e^(3/2))))/(2*b*e))
3.4.9.3.1 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*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && 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 = 0.18 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.60
method | result | size |
risch | \(\frac {\left (b \,x^{2}+a \right ) d}{2 b^{2} e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}-\frac {\left (\frac {3 d \left (a d -b c \right ) \ln \left (\frac {\frac {1}{2} e d a +\frac {1}{2} e b c +b d e \,x^{2}}{\sqrt {b d e}}+\sqrt {b d e \,x^{4}+\left (e d a +e b c \right ) x^{2}+a c e}\right )}{2 \sqrt {b d e}}+\frac {\left (-2 a^{2} d^{2}+4 a b c d -2 b^{2} c^{2}\right ) \left (d \,x^{2}+c \right )}{\left (a d -b c \right ) \sqrt {b d e \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{2 b^{2} e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\) | \(234\) |
default | \(-\frac {\left (3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b \,d^{2} x^{2}-3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c d \,x^{2}-2 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b d \,x^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{2}-3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c d -2 \sqrt {b d}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, a d -4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, a d +4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, b c \right ) \left (b \,x^{2}+a \right )}{4 b^{2} \sqrt {b d}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \left (d \,x^{2}+c \right ) {\left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )}^{\frac {3}{2}}}\) | \(432\) |
1/2/b^2*(b*x^2+a)*d/e/(e*(b*x^2+a)/(d*x^2+c))^(1/2)-1/2/b^2*(3/2*d*(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)*(d*x^2 +c)/(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/(e*(b*x^2+a)/ (d*x^2+c))^(1/2)*((d*x^2+c)*e*(b*x^2+a))^(1/2)/(d*x^2+c)
Time = 0.54 (sec) , antiderivative size = 443, normalized size of antiderivative = 3.03 \[ \int \frac {x}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (b^{2} c - a b d\right )} e x^{2} + {\left (a b c - a^{2} d\right )} e\right )} \sqrt {\frac {d}{b e}} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} - 4 \, {\left (2 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + a b c d + {\left (3 \, b^{2} c d + a b d^{2}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {\frac {d}{b e}}\right ) - 4 \, {\left (b d^{2} x^{4} - 2 \, b c^{2} + 3 \, a c d - {\left (b c d - 3 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, {\left (b^{3} e^{2} x^{2} + a b^{2} e^{2}\right )}}, -\frac {3 \, {\left ({\left (b^{2} c - a b d\right )} e x^{2} + {\left (a b c - a^{2} d\right )} e\right )} \sqrt {-\frac {d}{b e}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}} \sqrt {-\frac {d}{b e}}}{2 \, {\left (b d x^{2} + a d\right )}}\right ) - 2 \, {\left (b d^{2} x^{4} - 2 \, b c^{2} + 3 \, a c d - {\left (b c d - 3 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, {\left (b^{3} e^{2} x^{2} + a b^{2} e^{2}\right )}}\right ] \]
[-1/8*(3*((b^2*c - a*b*d)*e*x^2 + (a*b*c - a^2*d)*e)*sqrt(d/(b*e))*log(8*b ^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 - 4 *(2*b^2*d^2*x^4 + b^2*c^2 + a*b*c*d + (3*b^2*c*d + a*b*d^2)*x^2)*sqrt((b*e *x^2 + a*e)/(d*x^2 + c))*sqrt(d/(b*e))) - 4*(b*d^2*x^4 - 2*b*c^2 + 3*a*c*d - (b*c*d - 3*a*d^2)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b^3*e^2*x^2 + a*b^2*e^2), -1/4*(3*((b^2*c - a*b*d)*e*x^2 + (a*b*c - a^2*d)*e)*sqrt(-d/ (b*e))*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c) )*sqrt(-d/(b*e))/(b*d*x^2 + a*d)) - 2*(b*d^2*x^4 - 2*b*c^2 + 3*a*c*d - (b* c*d - 3*a*d^2)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b^3*e^2*x^2 + a*b^ 2*e^2)]
Timed out. \[ \int \frac {x}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int \frac {x}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]
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,[1,0,0]%%%},[2,1,0]%%%}+%%%{%%{[-4,0]:[1,0,%%%{-1,[1 ,1,1]%%%}
Timed out. \[ \int \frac {x}{\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx=\int \frac {x}{{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2}} \,d x \]