Integrand size = 48, antiderivative size = 154 \[ \int \frac {1}{x^2 \sqrt {-b x+a^2 x^2} \left (a x^2+x \sqrt {-b x+a^2 x^2}\right )^{3/2}} \, dx=-\frac {4 \sqrt {-b x+a^2 x^2} \left (3003 b^3-4032 a^2 b^2 x-5120 a^4 b x^2-8192 a^6 x^3\right ) \sqrt {x \left (a x+\sqrt {-b x+a^2 x^2}\right )}}{45045 b^5 x^5}+\frac {4 \left (6699 a b^3+448 a^3 b^2 x+1024 a^5 b x^2+8192 a^7 x^3\right ) \sqrt {x \left (a x+\sqrt {-b x+a^2 x^2}\right )}}{45045 b^5 x^4} \]
-4/45045*(a^2*x^2-b*x)^(1/2)*(-8192*a^6*x^3-5120*a^4*b*x^2-4032*a^2*b^2*x+ 3003*b^3)*(x*(a*x+(a^2*x^2-b*x)^(1/2)))^(1/2)/b^5/x^5+4/45045*(8192*a^7*x^ 3+1024*a^5*b*x^2+448*a^3*b^2*x+6699*a*b^3)*(x*(a*x+(a^2*x^2-b*x)^(1/2)))^( 1/2)/b^5/x^4
Time = 3.54 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^2 \sqrt {-b x+a^2 x^2} \left (a x^2+x \sqrt {-b x+a^2 x^2}\right )^{3/2}} \, dx=\frac {4 \sqrt {x \left (a x+\sqrt {x \left (-b+a^2 x\right )}\right )} \left (3003 b^4+1024 a^5 b x^2 \left (-3 a x+\sqrt {x \left (-b+a^2 x\right )}\right )+8192 a^7 x^3 \left (a x+\sqrt {x \left (-b+a^2 x\right )}\right )+64 a^3 b^2 x \left (-17 a x+7 \sqrt {x \left (-b+a^2 x\right )}\right )+21 a b^3 \left (-335 a x+319 \sqrt {x \left (-b+a^2 x\right )}\right )\right )}{45045 b^5 x^4 \sqrt {x \left (-b+a^2 x\right )}} \]
(4*Sqrt[x*(a*x + Sqrt[x*(-b + a^2*x)])]*(3003*b^4 + 1024*a^5*b*x^2*(-3*a*x + Sqrt[x*(-b + a^2*x)]) + 8192*a^7*x^3*(a*x + Sqrt[x*(-b + a^2*x)]) + 64* a^3*b^2*x*(-17*a*x + 7*Sqrt[x*(-b + a^2*x)]) + 21*a*b^3*(-335*a*x + 319*Sq rt[x*(-b + a^2*x)])))/(45045*b^5*x^4*Sqrt[x*(-b + a^2*x)])
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.
\(\displaystyle \int \frac {1}{x^2 \sqrt {a^2 x^2-b x} \left (x \sqrt {a^2 x^2-b x}+a x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt {a^2 x-b} \int \frac {1}{x^{5/2} \sqrt {a^2 x-b} \left (a x^2+\sqrt {a^2 x^2-b x} x\right )^{3/2}}dx}{\sqrt {a^2 x^2-b x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {2 \sqrt {x} \sqrt {a^2 x-b} \int \frac {1}{x^2 \sqrt {a^2 x-b} \left (a x^2+\sqrt {a^2 x^2-b x} x\right )^{3/2}}d\sqrt {x}}{\sqrt {a^2 x^2-b x}}\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \frac {2 \sqrt {x} \sqrt {a^2 x-b} \int \frac {1}{x^2 \sqrt {a^2 x-b} \left (a x^2+\sqrt {a^2 x^2-b x} x\right )^{3/2}}d\sqrt {x}}{\sqrt {a^2 x^2-b x}}\) |
3.22.31.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
\[\int \frac {1}{x^{2} \sqrt {a^{2} x^{2}-b x}\, \left (a \,x^{2}+x \sqrt {a^{2} x^{2}-b x}\right )^{\frac {3}{2}}}d x\]
Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^2 \sqrt {-b x+a^2 x^2} \left (a x^2+x \sqrt {-b x+a^2 x^2}\right )^{3/2}} \, dx=\frac {4 \, {\left (8192 \, a^{7} x^{4} + 1024 \, a^{5} b x^{3} + 448 \, a^{3} b^{2} x^{2} + 6699 \, a b^{3} x + {\left (8192 \, a^{6} x^{3} + 5120 \, a^{4} b x^{2} + 4032 \, a^{2} b^{2} x - 3003 \, b^{3}\right )} \sqrt {a^{2} x^{2} - b x}\right )} \sqrt {a x^{2} + \sqrt {a^{2} x^{2} - b x} x}}{45045 \, b^{5} x^{5}} \]
integrate(1/x^2/(a^2*x^2-b*x)^(1/2)/(a*x^2+x*(a^2*x^2-b*x)^(1/2))^(3/2),x, algorithm="fricas")
4/45045*(8192*a^7*x^4 + 1024*a^5*b*x^3 + 448*a^3*b^2*x^2 + 6699*a*b^3*x + (8192*a^6*x^3 + 5120*a^4*b*x^2 + 4032*a^2*b^2*x - 3003*b^3)*sqrt(a^2*x^2 - b*x))*sqrt(a*x^2 + sqrt(a^2*x^2 - b*x)*x)/(b^5*x^5)
\[ \int \frac {1}{x^2 \sqrt {-b x+a^2 x^2} \left (a x^2+x \sqrt {-b x+a^2 x^2}\right )^{3/2}} \, dx=\int \frac {1}{x^{2} \left (x \left (a x + \sqrt {a^{2} x^{2} - b x}\right )\right )^{\frac {3}{2}} \sqrt {x \left (a^{2} x - b\right )}}\, dx \]
\[ \int \frac {1}{x^2 \sqrt {-b x+a^2 x^2} \left (a x^2+x \sqrt {-b x+a^2 x^2}\right )^{3/2}} \, dx=\int { \frac {1}{\sqrt {a^{2} x^{2} - b x} {\left (a x^{2} + \sqrt {a^{2} x^{2} - b x} x\right )}^{\frac {3}{2}} x^{2}} \,d x } \]
integrate(1/x^2/(a^2*x^2-b*x)^(1/2)/(a*x^2+x*(a^2*x^2-b*x)^(1/2))^(3/2),x, algorithm="maxima")
\[ \int \frac {1}{x^2 \sqrt {-b x+a^2 x^2} \left (a x^2+x \sqrt {-b x+a^2 x^2}\right )^{3/2}} \, dx=\int { \frac {1}{\sqrt {a^{2} x^{2} - b x} {\left (a x^{2} + \sqrt {a^{2} x^{2} - b x} x\right )}^{\frac {3}{2}} x^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{x^2 \sqrt {-b x+a^2 x^2} \left (a x^2+x \sqrt {-b x+a^2 x^2}\right )^{3/2}} \, dx=\int \frac {1}{x^2\,\sqrt {a^2\,x^2-b\,x}\,{\left (a\,x^2+x\,\sqrt {a^2\,x^2-b\,x}\right )}^{3/2}} \,d x \]