Integrand size = 43, antiderivative size = 64 \[ \int \frac {-b x^2+a x^3}{\left (b x^2+a x^3\right ) \sqrt {b^2 x+a^2 x^3}} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} \sqrt {b^2 x+a^2 x^3}}{b^2+a^2 x^2}\right )}{\sqrt {a} \sqrt {b}} \]
-2^(1/2)*arctan(2^(1/2)*a^(1/2)*b^(1/2)*(a^2*x^3+b^2*x)^(1/2)/(a^2*x^2+b^2 ))/a^(1/2)/b^(1/2)
Time = 0.00 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.42 \[ \int \frac {-b x^2+a x^3}{\left (b x^2+a x^3\right ) \sqrt {b^2 x+a^2 x^3}} \, dx=-\frac {\sqrt {2} \sqrt {x} \sqrt {b^2+a^2 x^2} \arctan \left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} \sqrt {x}}{\sqrt {b^2+a^2 x^2}}\right )}{\sqrt {a} \sqrt {b} \sqrt {x \left (b^2+a^2 x^2\right )}} \]
-((Sqrt[2]*Sqrt[x]*Sqrt[b^2 + a^2*x^2]*ArcTan[(Sqrt[2]*Sqrt[a]*Sqrt[b]*Sqr t[x])/Sqrt[b^2 + a^2*x^2]])/(Sqrt[a]*Sqrt[b]*Sqrt[x*(b^2 + a^2*x^2)]))
Time = 0.53 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.42, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {2026, 9, 25, 2467, 2035, 2213, 218}
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.
\(\displaystyle \int \frac {a x^3-b x^2}{\left (a x^3+b x^2\right ) \sqrt {a^2 x^3+b^2 x}} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {a x^3-b x^2}{x^2 (a x+b) \sqrt {a^2 x^3+b^2 x}}dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int -\frac {b-a x}{(a x+b) \sqrt {a^2 x^3+b^2 x}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {b-a x}{(b+a x) \sqrt {a^2 x^3+b^2 x}}dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle -\frac {\sqrt {x} \sqrt {a^2 x^2+b^2} \int \frac {b-a x}{\sqrt {x} (b+a x) \sqrt {b^2+a^2 x^2}}dx}{\sqrt {a^2 x^3+b^2 x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {a^2 x^2+b^2} \int \frac {b-a x}{(b+a x) \sqrt {b^2+a^2 x^2}}d\sqrt {x}}{\sqrt {a^2 x^3+b^2 x}}\) |
\(\Big \downarrow \) 2213 |
\(\displaystyle -\frac {2 b \sqrt {x} \sqrt {a^2 x^2+b^2} \int \frac {1}{2 a x b^2+b}d\frac {\sqrt {x}}{\sqrt {b^2+a^2 x^2}}}{\sqrt {a^2 x^3+b^2 x}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {\sqrt {2} \sqrt {x} \sqrt {a^2 x^2+b^2} \arctan \left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} \sqrt {x}}{\sqrt {a^2 x^2+b^2}}\right )}{\sqrt {a} \sqrt {b} \sqrt {a^2 x^3+b^2 x}}\) |
-((Sqrt[2]*Sqrt[x]*Sqrt[b^2 + a^2*x^2]*ArcTan[(Sqrt[2]*Sqrt[a]*Sqrt[b]*Sqr t[x])/Sqrt[b^2 + a^2*x^2]])/(Sqrt[a]*Sqrt[b]*Sqrt[b^2*x + a^2*x^3]))
3.9.46.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
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[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]) , x_Symbol] :> Simp[A Subst[Int[1/(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^ 4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]
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]
Time = 1.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.61
method | result | size |
default | \(\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {x \left (a^{2} x^{2}+b^{2}\right )}\, \sqrt {2}}{2 x \sqrt {a b}}\right )}{\sqrt {a b}}\) | \(39\) |
pseudoelliptic | \(\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {x \left (a^{2} x^{2}+b^{2}\right )}\, \sqrt {2}}{2 x \sqrt {a b}}\right )}{\sqrt {a b}}\) | \(39\) |
elliptic | \(\frac {i b \sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}\, \sqrt {2}\, \sqrt {\frac {i \left (x -\frac {i b}{a}\right ) a}{b}}\, \sqrt {\frac {i a x}{b}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}, \frac {\sqrt {2}}{2}\right )}{a \sqrt {a^{2} x^{3}+b^{2} x}}-\frac {2 i b^{2} \sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}\, \sqrt {2}\, \sqrt {\frac {i \left (x -\frac {i b}{a}\right ) a}{b}}\, \sqrt {\frac {i a x}{b}}\, \operatorname {EllipticPi}\left (\sqrt {-\frac {i \left (x +\frac {i b}{a}\right ) a}{b}}, -\frac {i b}{a \left (-\frac {i b}{a}+\frac {b}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{a^{2} \sqrt {a^{2} x^{3}+b^{2} x}\, \left (-\frac {i b}{a}+\frac {b}{a}\right )}\) | \(231\) |
Time = 0.28 (sec) , antiderivative size = 213, normalized size of antiderivative = 3.33 \[ \int \frac {-b x^2+a x^3}{\left (b x^2+a x^3\right ) \sqrt {b^2 x+a^2 x^3}} \, dx=\left [\frac {1}{4} \, \sqrt {2} \sqrt {-\frac {1}{a b}} \log \left (\frac {a^{4} x^{4} - 12 \, a^{3} b x^{3} + 6 \, a^{2} b^{2} x^{2} - 12 \, a b^{3} x + b^{4} + 4 \, \sqrt {2} {\left (a^{3} b x^{2} - 2 \, a^{2} b^{2} x + a b^{3}\right )} \sqrt {a^{2} x^{3} + b^{2} x} \sqrt {-\frac {1}{a b}}}{a^{4} x^{4} + 4 \, a^{3} b x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a b^{3} x + b^{4}}\right ), -\frac {1}{2} \, \sqrt {2} \sqrt {\frac {1}{a b}} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {a^{2} x^{3} + b^{2} x} a b \sqrt {\frac {1}{a b}}}{a^{2} x^{2} - 2 \, a b x + b^{2}}\right )\right ] \]
[1/4*sqrt(2)*sqrt(-1/(a*b))*log((a^4*x^4 - 12*a^3*b*x^3 + 6*a^2*b^2*x^2 - 12*a*b^3*x + b^4 + 4*sqrt(2)*(a^3*b*x^2 - 2*a^2*b^2*x + a*b^3)*sqrt(a^2*x^ 3 + b^2*x)*sqrt(-1/(a*b)))/(a^4*x^4 + 4*a^3*b*x^3 + 6*a^2*b^2*x^2 + 4*a*b^ 3*x + b^4)), -1/2*sqrt(2)*sqrt(1/(a*b))*arctan(2*sqrt(2)*sqrt(a^2*x^3 + b^ 2*x)*a*b*sqrt(1/(a*b))/(a^2*x^2 - 2*a*b*x + b^2))]
\[ \int \frac {-b x^2+a x^3}{\left (b x^2+a x^3\right ) \sqrt {b^2 x+a^2 x^3}} \, dx=\int \frac {a x - b}{\sqrt {x \left (a^{2} x^{2} + b^{2}\right )} \left (a x + b\right )}\, dx \]
\[ \int \frac {-b x^2+a x^3}{\left (b x^2+a x^3\right ) \sqrt {b^2 x+a^2 x^3}} \, dx=\int { \frac {a x^{3} - b x^{2}}{\sqrt {a^{2} x^{3} + b^{2} x} {\left (a x^{3} + b x^{2}\right )}} \,d x } \]
\[ \int \frac {-b x^2+a x^3}{\left (b x^2+a x^3\right ) \sqrt {b^2 x+a^2 x^3}} \, dx=\int { \frac {a x^{3} - b x^{2}}{\sqrt {a^{2} x^{3} + b^{2} x} {\left (a x^{3} + b x^{2}\right )}} \,d x } \]
Timed out. \[ \int \frac {-b x^2+a x^3}{\left (b x^2+a x^3\right ) \sqrt {b^2 x+a^2 x^3}} \, dx=\text {Hanged} \]