\(\int \frac {1}{(a x^2+b x^3)^{3/4}} \, dx\) [117]

Optimal result

Integrand size = 15, antiderivative size = 91 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/4}} \, dx=-\frac {2 \sqrt [4]{a x^2+b x^3}}{a x}-\frac {2 \sqrt {b} x^{3/2} \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right ),2\right )}{\sqrt {a} \left (a x^2+b x^3\right )^{3/4}} \] Output:

-2*(b*x^3+a*x^2)^(1/4)/a/x-2*b^(1/2)*x^(3/2)*(1+b*x/a)^(3/4)*InverseJacobi 
AM(1/2*arctan(b^(1/2)*x^(1/2)/a^(1/2)),2^(1/2))/a^(1/2)/(b*x^3+a*x^2)^(3/4 
)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.49 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/4}} \, dx=-\frac {2 x \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {1}{2},-\frac {b x}{a}\right )}{\left (x^2 (a+b x)\right )^{3/4}} \] Input:

Integrate[(a*x^2 + b*x^3)^(-3/4),x]
 

Output:

(-2*x*(1 + (b*x)/a)^(3/4)*Hypergeometric2F1[-1/2, 3/4, 1/2, -((b*x)/a)])/( 
x^2*(a + b*x))^(3/4)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.22, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1917, 61, 73, 765, 762}

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.

\(\displaystyle \int \frac {1}{\left (a x^2+b x^3\right )^{3/4}} \, dx\)

\(\Big \downarrow \) 1917

\(\displaystyle \frac {x^{3/2} (a+b x)^{3/4} \int \frac {1}{x^{3/2} (a+b x)^{3/4}}dx}{\left (a x^2+b x^3\right )^{3/4}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {x^{3/2} (a+b x)^{3/4} \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)^{3/4}}dx}{2 a}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{\left (a x^2+b x^3\right )^{3/4}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {x^{3/2} (a+b x)^{3/4} \left (-\frac {2 \int \frac {1}{\sqrt {\frac {a+b x}{b}-\frac {a}{b}}}d\sqrt [4]{a+b x}}{a}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{\left (a x^2+b x^3\right )^{3/4}}\)

\(\Big \downarrow \) 765

\(\displaystyle \frac {x^{3/2} (a+b x)^{3/4} \left (-\frac {2 \sqrt {1-\frac {a+b x}{a}} \int \frac {1}{\sqrt {1-\frac {a+b x}{a}}}d\sqrt [4]{a+b x}}{a \sqrt {\frac {a+b x}{b}-\frac {a}{b}}}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{\left (a x^2+b x^3\right )^{3/4}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {x^{3/2} (a+b x)^{3/4} \left (-\frac {2 \sqrt {1-\frac {a+b x}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{a+b x}}{\sqrt [4]{a}}\right ),-1\right )}{a^{3/4} \sqrt {\frac {a+b x}{b}-\frac {a}{b}}}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{\left (a x^2+b x^3\right )^{3/4}}\)

Input:

Int[(a*x^2 + b*x^3)^(-3/4),x]
 

Output:

(x^(3/2)*(a + b*x)^(3/4)*((-2*(a + b*x)^(1/4))/(a*Sqrt[x]) - (2*Sqrt[1 - ( 
a + b*x)/a]*EllipticF[ArcSin[(a + b*x)^(1/4)/a^(1/4)], -1])/(a^(3/4)*Sqrt[ 
-(a/b) + (a + b*x)/b])))/(a*x^2 + b*x^3)^(3/4)
 

Defintions of rubi rules used

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 765

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt 
[a + b*x^4]   Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ 
[b/a] &&  !GtQ[a, 0]
 

rule 1917

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + 
b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])   Int[ 
x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !Integ 
erQ[p] && NeQ[n, j] && PosQ[n - j]
 

Maple [F]

\[\int \frac {1}{\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{4}}}d x\]

Input:

int(1/(b*x^3+a*x^2)^(3/4),x)
 

Output:

int(1/(b*x^3+a*x^2)^(3/4),x)
 

Fricas [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(1/(b*x^3+a*x^2)^(3/4),x, algorithm="fricas")
 

Output:

integral((b*x^3 + a*x^2)^(-3/4), x)
 

Sympy [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/4}} \, dx=\int \frac {1}{\left (a x^{2} + b x^{3}\right )^{\frac {3}{4}}}\, dx \] Input:

integrate(1/(b*x**3+a*x**2)**(3/4),x)
 

Output:

Integral((a*x**2 + b*x**3)**(-3/4), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(1/(b*x^3+a*x^2)^(3/4),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(-3/4), x)
 

Giac [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate(1/(b*x^3+a*x^2)^(3/4),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a*x^2)^(-3/4), x)
 

Mupad [B] (verification not implemented)

Time = 10.14 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/4}} \, dx=-\frac {2\,x\,{\left (\frac {b\,x}{a}+1\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {3}{4};\ \frac {1}{2};\ -\frac {b\,x}{a}\right )}{{\left (b\,x^3+a\,x^2\right )}^{3/4}} \] Input:

int(1/(a*x^2 + b*x^3)^(3/4),x)
 

Output:

-(2*x*((b*x)/a + 1)^(3/4)*hypergeom([-1/2, 3/4], 1/2, -(b*x)/a))/(a*x^2 + 
b*x^3)^(3/4)
 

Reduce [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/4}} \, dx=\int \frac {\sqrt {x}\, \left (b x +a \right )^{\frac {3}{4}}}{\sqrt {b x +a}\, a \,x^{2}+\sqrt {b x +a}\, b \,x^{3}}d x \] Input:

int(1/(b*x^3+a*x^2)^(3/4),x)
 

Output:

int((sqrt(x)*(a + b*x)**(3/4))/(sqrt(a + b*x)*a*x**2 + sqrt(a + b*x)*b*x** 
3),x)