Integrand size = 26, antiderivative size = 165 \[ \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx=\frac {3 (b c-a d) e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 c^2}+\frac {(b c-a d) \left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 c \left (a-\frac {c \left (a+b x^2\right )}{c+d x^2}\right )}-\frac {3 \sqrt {a} (b c-a d) e^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{2 c^{5/2}} \]
1/2*(-a*d+b*c)*(e*(b*x^2+a)/(d*x^2+c))^(3/2)/c/(a-c*(b*x^2+a)/(d*x^2+c))-3 /2*(-a*d+b*c)*e^(3/2)*arctanh(c^(1/2)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/a^(1/2 )/e^(1/2))*a^(1/2)/c^(5/2)+3/2*(-a*d+b*c)*e*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/ c^2
Time = 3.80 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.88 \[ \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx=\frac {e \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt {c} \sqrt {a+b x^2} \left (2 b c x^2-a \left (c+3 d x^2\right )\right )-3 \sqrt {a} (b c-a d) x^2 \sqrt {c+d x^2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )\right )}{2 c^{5/2} x^2 \sqrt {a+b x^2}} \]
(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[c]*Sqrt[a + b*x^2]*(2*b*c*x^2 - a*(c + 3*d*x^2)) - 3*Sqrt[a]*(b*c - a*d)*x^2*Sqrt[c + d*x^2]*ArcTanh[(Sqr t[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])]))/(2*c^(5/2)*x^2*Sqrt[a + b*x^2])
Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.75, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2053, 2052, 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.
\(\displaystyle \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 2053 |
\(\displaystyle \frac {1}{2} \int \frac {\left (\frac {e \left (b x^2+a\right )}{d x^2+c}\right )^{3/2}}{x^4}dx^2\) |
\(\Big \downarrow \) 2052 |
\(\displaystyle e (b c-a d) \int \frac {x^8}{\left (a e-c x^4\right )^2}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle e (b c-a d) \left (\frac {x^6}{2 c \left (a e-c x^4\right )}-\frac {3 \int \frac {x^4}{a e-c x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{2 c}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle e (b c-a d) \left (\frac {x^6}{2 c \left (a e-c x^4\right )}-\frac {3 \left (\frac {a e \int \frac {1}{a e-c x^4}d\sqrt {\frac {e \left (b x^2+a\right )}{d x^2+c}}}{c}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c}\right )}{2 c}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle e (b c-a d) \left (\frac {x^6}{2 c \left (a e-c x^4\right )}-\frac {3 \left (\frac {\sqrt {a} \sqrt {e} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{c^{3/2}}-\frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{c}\right )}{2 c}\right )\) |
(b*c - a*d)*e*(x^6/(2*c*(a*e - c*x^4)) - (3*(-(Sqrt[(e*(a + b*x^2))/(c + d *x^2)]/c) + (Sqrt[a]*Sqrt[e]*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c + d* x^2)])/(Sqrt[a]*Sqrt[e])])/c^(3/2)))/(2*c))
3.3.80.3.1 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[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_S ymbol] :> With[{q = Denominator[p]}, Simp[q*e*(b*c - a*d) Subst[Int[x^(q* (p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a + b* x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]
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 = 0.21 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.45
method | result | size |
risch | \(-\frac {a \left (d \,x^{2}+c \right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}}{2 c^{2} x^{2}}-\frac {\left (-\frac {3 a \left (a d -b c \right ) \ln \left (\frac {2 a c e +\left (e d a +e b c \right ) x^{2}+2 \sqrt {a c e}\, \sqrt {b d e \,x^{4}+\left (e d a +e b c \right ) x^{2}+a c e}}{x^{2}}\right )}{2 \sqrt {a c e}}-\frac {\left (-2 a^{2} d^{2}+4 a b c d -2 b^{2} c^{2}\right ) \left (b \,x^{2}+a \right )}{\left (a d -b c \right ) \sqrt {b d e \,x^{4}+a d e \,x^{2}+b c e \,x^{2}+a c e}}\right ) e \sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) e \left (b \,x^{2}+a \right )}}{2 c^{2} \left (b \,x^{2}+a \right )}\) | \(239\) |
default | \(-\frac {\left (-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b \,d^{2} x^{6}-3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{2} c \,d^{2} x^{4}+3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a b \,c^{2} d \,x^{4}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a \,d^{2} x^{4}-4 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b c d \,x^{4}-3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a^{2} c^{2} d \,x^{2}+3 \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) a b \,c^{3} x^{2}+4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, a c d \,x^{2}-4 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {a c}\, b \,c^{2} x^{2}+2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, d \,x^{2}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a c d \,x^{2}-2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b \,c^{2} x^{2}+2 \left (b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c \right )^{\frac {3}{2}} \sqrt {a c}\, c \right ) \left (d \,x^{2}+c \right ) {\left (\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}\right )}^{\frac {3}{2}}}{4 \sqrt {a c}\, x^{2} c^{3} \left (b \,x^{2}+a \right ) \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}}\) | \(641\) |
-1/2*a/c^2*(d*x^2+c)/x^2*e*(e*(b*x^2+a)/(d*x^2+c))^(1/2)-1/2/c^2*(-3/2*a*( a*d-b*c)/(a*c*e)^(1/2)*ln((2*a*c*e+(a*d*e+b*c*e)*x^2+2*(a*c*e)^(1/2)*(b*d* e*x^4+(a*d*e+b*c*e)*x^2+a*c*e)^(1/2))/x^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 = 0.91 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.12 \[ \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx=\left [-\frac {3 \, {\left (b c - a d\right )} \sqrt {\frac {a e}{c}} e x^{2} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \, {\left (a b c^{2} + a^{2} c d\right )} e x^{2} + 4 \, {\left ({\left (b c^{2} d + a c d^{2}\right )} x^{4} + 2 \, a c^{3} + {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2}\right )} \sqrt {\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{x^{4}}\right ) - 4 \, {\left ({\left (2 \, b c - 3 \, a d\right )} e x^{2} - a c e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, c^{2} x^{2}}, \frac {3 \, {\left (b c - a d\right )} \sqrt {-\frac {a e}{c}} e x^{2} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {-\frac {a e}{c}} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (a b e x^{2} + a^{2} e\right )}}\right ) + 2 \, {\left ({\left (2 \, b c - 3 \, a d\right )} e x^{2} - a c e\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, c^{2} x^{2}}\right ] \]
[-1/8*(3*(b*c - a*d)*sqrt(a*e/c)*e*x^2*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2 )*e*x^4 + 8*a^2*c^2*e + 8*(a*b*c^2 + a^2*c*d)*e*x^2 + 4*((b*c^2*d + a*c*d^ 2)*x^4 + 2*a*c^3 + (b*c^3 + 3*a*c^2*d)*x^2)*sqrt(a*e/c)*sqrt((b*e*x^2 + a* e)/(d*x^2 + c)))/x^4) - 4*((2*b*c - 3*a*d)*e*x^2 - a*c*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(c^2*x^2), 1/4*(3*(b*c - a*d)*sqrt(-a*e/c)*e*x^2*arctan (1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(-a*e/c)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(a*b*e*x^2 + a^2*e)) + 2*((2*b*c - 3*a*d)*e*x^2 - a*c*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(c^2*x^2)]
Timed out. \[ \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx=\text {Exception raised: TypeError} \]
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,[0,1,0]%%%},[6,0,0]%%%}+%%%{%%{[-4,0]:[1,0,%%%{-1,[1 ,1,1]%%%}
Timed out. \[ \int \frac {\left (\frac {e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx=\int \frac {{\left (\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}\right )}^{3/2}}{x^3} \,d x \]