\(\int \frac {1}{x (a x^2+b x^3)^{3/4}} \, dx\) [398]

Optimal result

Integrand size = 19, antiderivative size = 121 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^{3/4}} \, dx=-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}+\frac {5 b \sqrt [4]{a x^2+b x^3}}{3 a^2 x}+\frac {5 b^{3/2} x^{3/2} \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right ),2\right )}{3 a^{3/2} \left (a x^2+b x^3\right )^{3/4}} \] Output:

-2/3*(b*x^3+a*x^2)^(1/4)/a/x^2+5/3*b*(b*x^3+a*x^2)^(1/4)/a^2/x+5/3*b^(3/2) 
*x^(3/2)*(1+b*x/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x^(1/2)/a^(1/2 
)),2^(1/2))/a^(3/2)/(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 = 46, normalized size of antiderivative = 0.38 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^{3/4}} \, dx=-\frac {2 \left (1+\frac {b x}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},-\frac {1}{2},-\frac {b x}{a}\right )}{3 \left (x^2 (a+b x)\right )^{3/4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.19, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1931, 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.

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

\(\Big \downarrow \) 1931

\(\displaystyle -\frac {5 b \int \frac {1}{\left (b x^3+a x^2\right )^{3/4}}dx}{6 a}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\)

\(\Big \downarrow \) 1917

\(\displaystyle -\frac {5 b x^{3/2} (a+b x)^{3/4} \int \frac {1}{x^{3/2} (a+b x)^{3/4}}dx}{6 a \left (a x^2+b x^3\right )^{3/4}}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {5 b x^{3/2} (a+b x)^{3/4} \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)^{3/4}}dx}{2 a}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a \left (a x^2+b x^3\right )^{3/4}}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {5 b x^{3/2} (a+b x)^{3/4} \left (-\frac {2 \int \frac {1}{\sqrt {\frac {a+b x}{b}-\frac {a}{b}}}d\sqrt [4]{a+b x}}{a}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a \left (a x^2+b x^3\right )^{3/4}}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\)

\(\Big \downarrow \) 765

\(\displaystyle -\frac {5 b x^{3/2} (a+b x)^{3/4} \left (-\frac {2 \sqrt {1-\frac {a+b x}{a}} \int \frac {1}{\sqrt {1-\frac {a+b x}{a}}}d\sqrt [4]{a+b x}}{a \sqrt {\frac {a+b x}{b}-\frac {a}{b}}}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a \left (a x^2+b x^3\right )^{3/4}}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\)

\(\Big \downarrow \) 762

\(\displaystyle -\frac {5 b x^{3/2} (a+b x)^{3/4} \left (-\frac {2 \sqrt {1-\frac {a+b x}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{a+b x}}{\sqrt [4]{a}}\right ),-1\right )}{a^{3/4} \sqrt {\frac {a+b x}{b}-\frac {a}{b}}}-\frac {2 \sqrt [4]{a+b x}}{a \sqrt {x}}\right )}{6 a \left (a x^2+b x^3\right )^{3/4}}-\frac {2 \sqrt [4]{a x^2+b x^3}}{3 a x^2}\)

Input:

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

Output:

(-2*(a*x^2 + b*x^3)^(1/4))/(3*a*x^2) - (5*b*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])))/(6*a*( 
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]
 

rule 1931

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p 
+ 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1)))   I 
nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] 
 &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[ 
m + j*p + 1, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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