\(\int \frac {1}{(a x+b x^2)^{5/3}} \, dx\) [57]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int \frac {1}{\left (a x+b x^2\right )^{5/3}} \, dx=-\frac {3 \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},1,\frac {1}{3},-\frac {b x}{a}\right )}{2 a \left (a x+b x^2\right )^{2/3}} \] Output:

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

Mathematica [A] (verified)

Time = 10.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.34 \[ \int \frac {1}{\left (a x+b x^2\right )^{5/3}} \, dx=-\frac {3 \left (1+\frac {b x}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {5}{3},\frac {1}{3},-\frac {b x}{a}\right )}{2 a (x (a+b x))^{2/3}} \] Input:

Integrate[(a*x + b*x^2)^(-5/3),x]
 

Output:

(-3*(1 + (b*x)/a)^(2/3)*Hypergeometric2F1[-2/3, 5/3, 1/3, -((b*x)/a)])/(2* 
a*(x*(a + b*x))^(2/3))
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(308\) vs. \(2(35)=70\).

Time = 0.50 (sec) , antiderivative size = 308, normalized size of antiderivative = 8.80, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1089, 1093, 1090, 234, 760}

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.

\(\displaystyle \int \frac {1}{\left (a x+b x^2\right )^{5/3}} \, dx\)

\(\Big \downarrow \) 1089

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

\(\Big \downarrow \) 1093

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

\(\Big \downarrow \) 1090

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

\(\Big \downarrow \) 234

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

\(\Big \downarrow \) 760

\(\displaystyle \frac {\sqrt [3]{2} 3^{3/4} \sqrt {2-\sqrt {3}} b \left (\frac {2 b^2 x}{a^2}+\frac {b}{a}+1\right ) \left (-\frac {b \left (a x+b x^2\right )}{a^2}\right )^{2/3} \sqrt {\frac {\left (-\frac {2 b^2 x}{a^2}-\frac {b}{a}\right )^2+\sqrt [3]{1-\frac {a^2 \left (-\frac {2 b^2 x}{a^2}-\frac {b}{a}\right )^2}{b^2}}+1}{\left (\frac {2 b^2 x}{a^2}+\frac {b}{a}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\frac {2 x b^2}{a^2}+\frac {b}{a}+\sqrt {3}+1}{\frac {2 x b^2}{a^2}+\frac {b}{a}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{a^2 \left (-\frac {2 b^2 x}{a^2}-\frac {b}{a}\right ) \sqrt {-\frac {\frac {2 b^2 x}{a^2}+\frac {b}{a}+1}{\left (\frac {2 b^2 x}{a^2}+\frac {b}{a}-\sqrt {3}+1\right )^2}} \left (a x+b x^2\right )^{2/3}}-\frac {3 (a+2 b x)}{2 a^2 \left (a x+b x^2\right )^{2/3}}\)

Input:

Int[(a*x + b*x^2)^(-5/3),x]
 

Output:

(-3*(a + 2*b*x))/(2*a^2*(a*x + b*x^2)^(2/3)) + (2^(1/3)*3^(3/4)*Sqrt[2 - S 
qrt[3]]*b*(1 + b/a + (2*b^2*x)/a^2)*(-((b*(a*x + b*x^2))/a^2))^(2/3)*Sqrt[ 
(1 + (-(b/a) - (2*b^2*x)/a^2)^2 + (1 - (a^2*(-(b/a) - (2*b^2*x)/a^2)^2)/b^ 
2)^(1/3))/(1 - Sqrt[3] + b/a + (2*b^2*x)/a^2)^2]*EllipticF[ArcSin[(1 + Sqr 
t[3] + b/a + (2*b^2*x)/a^2)/(1 - Sqrt[3] + b/a + (2*b^2*x)/a^2)], -7 + 4*S 
qrt[3]])/(a^2*(-(b/a) - (2*b^2*x)/a^2)*Sqrt[-((1 + b/a + (2*b^2*x)/a^2)/(1 
 - Sqrt[3] + b/a + (2*b^2*x)/a^2)^2)]*(a*x + b*x^2)^(2/3))
 

Defintions of rubi rules used

rule 234

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) 
   Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b 
}, x]
 

rule 760

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- 
s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) 
*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x 
] && NegQ[a]
 

rule 1089

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 
 3)/((p + 1)*(b^2 - 4*a*c)))   Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre 
eQ[{a, b, c}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1090

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* 
(c/(b^2 - 4*a*c)))^p)   Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, 
b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
 

rule 1093

Int[((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b*x + c*x^2)^p/((- 
c)*((b*x + c*x^2)/b^2))^p   Int[((-c)*(x/b) - c^2*(x^2/b^2))^p, x], x] /; F 
reeQ[{b, c}, x] && (IntegerQ[4*p] || IntegerQ[3*p])
 

Maple [F]

\[\int \frac {1}{\left (b \,x^{2}+a x \right )^{\frac {5}{3}}}d x\]

Input:

int(1/(b*x^2+a*x)^(5/3),x)
 

Output:

int(1/(b*x^2+a*x)^(5/3),x)
 

Fricas [F]

\[ \int \frac {1}{\left (a x+b x^2\right )^{5/3}} \, dx=\int { \frac {1}{{\left (b x^{2} + a x\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate(1/(b*x^2+a*x)^(5/3),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{\left (a x+b x^2\right )^{5/3}} \, dx=\int \frac {1}{\left (a x + b x^{2}\right )^{\frac {5}{3}}}\, dx \] Input:

integrate(1/(b*x**2+a*x)**(5/3),x)
 

Output:

Integral((a*x + b*x**2)**(-5/3), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a x+b x^2\right )^{5/3}} \, dx=\int { \frac {1}{{\left (b x^{2} + a x\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate(1/(b*x^2+a*x)^(5/3),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a*x)^(-5/3), x)
 

Giac [F]

\[ \int \frac {1}{\left (a x+b x^2\right )^{5/3}} \, dx=\int { \frac {1}{{\left (b x^{2} + a x\right )}^{\frac {5}{3}}} \,d x } \] Input:

integrate(1/(b*x^2+a*x)^(5/3),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a*x)^(-5/3), x)
 

Mupad [B] (verification not implemented)

Time = 9.72 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (a x+b x^2\right )^{5/3}} \, dx=-\frac {3\,x\,{\left (\frac {b\,x}{a}+1\right )}^{5/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {2}{3},\frac {5}{3};\ \frac {1}{3};\ -\frac {b\,x}{a}\right )}{2\,{\left (b\,x^2+a\,x\right )}^{5/3}} \] Input:

int(1/(a*x + b*x^2)^(5/3),x)
 

Output:

-(3*x*((b*x)/a + 1)^(5/3)*hypergeom([-2/3, 5/3], 1/3, -(b*x)/a))/(2*(a*x + 
 b*x^2)^(5/3))
 

Reduce [F]

\[ \int \frac {1}{\left (a x+b x^2\right )^{5/3}} \, dx=\int \frac {1}{x^{\frac {5}{3}} \left (b x +a \right )^{\frac {2}{3}} a +x^{\frac {8}{3}} \left (b x +a \right )^{\frac {2}{3}} b}d x \] Input:

int(1/(b*x^2+a*x)^(5/3),x)
 

Output:

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