\(\int \frac {1}{(c x)^{4/3} (a x+b x^2)^{2/3}} \, dx\) [143]

Optimal result

Integrand size = 21, antiderivative size = 202 \[ \int \frac {1}{(c x)^{4/3} \left (a x+b x^2\right )^{2/3}} \, dx=-\frac {\sqrt [3]{a x+b x^2}}{a (c x)^{4/3}}+\frac {2 b (c x)^{2/3} (a+b x)^{2/3} \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3} c^2 \left (a x+b x^2\right )^{2/3}}+\frac {b (c x)^{2/3} (a+b x)^{2/3} \log (x)}{3 a^{5/3} c^2 \left (a x+b x^2\right )^{2/3}}-\frac {b (c x)^{2/3} (a+b x)^{2/3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{a^{5/3} c^2 \left (a x+b x^2\right )^{2/3}} \] Output:

-(b*x^2+a*x)^(1/3)/a/(c*x)^(4/3)+2/3*b*(c*x)^(2/3)*(b*x+a)^(2/3)*arctan(1/ 
3*(a^(1/3)+2*(b*x+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)/a^(5/3)/c^2/(b*x^2+a* 
x)^(2/3)+1/3*b*(c*x)^(2/3)*(b*x+a)^(2/3)*ln(x)/a^(5/3)/c^2/(b*x^2+a*x)^(2/ 
3)-b*(c*x)^(2/3)*(b*x+a)^(2/3)*ln(a^(1/3)-(b*x+a)^(1/3))/a^(5/3)/c^2/(b*x^ 
2+a*x)^(2/3)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(c x)^{4/3} \left (a x+b x^2\right )^{2/3}} \, dx=\frac {x \left (-3 a^{5/3}-3 a^{2/3} b x+2 \sqrt {3} b x (a+b x)^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 b x (a+b x)^{2/3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )+b x (a+b x)^{2/3} \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )\right )}{3 a^{5/3} (c x)^{4/3} (x (a+b x))^{2/3}} \] Input:

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

Output:

(x*(-3*a^(5/3) - 3*a^(2/3)*b*x + 2*Sqrt[3]*b*x*(a + b*x)^(2/3)*ArcTan[(1 + 
 (2*(a + b*x)^(1/3))/a^(1/3))/Sqrt[3]] - 2*b*x*(a + b*x)^(2/3)*Log[a^(1/3) 
 - (a + b*x)^(1/3)] + b*x*(a + b*x)^(2/3)*Log[a^(2/3) + a^(1/3)*(a + b*x)^ 
(1/3) + (a + b*x)^(2/3)]))/(3*a^(5/3)*(c*x)^(4/3)*(x*(a + b*x))^(2/3))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.67, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1137, 52, 69, 16, 1082, 217}

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.

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

\(\Big \downarrow \) 1137

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

(x^2*(a + b*x)^(2/3)*(-((a + b*x)^(1/3)/(a*x)) - (2*b*(-((Sqrt[3]*ArcTan[( 
1 + (2*(a + b*x)^(1/3))/a^(1/3))/Sqrt[3]])/a^(2/3)) - Log[x]/(2*a^(2/3)) + 
 (3*Log[a^(1/3) - (a + b*x)^(1/3)])/(2*a^(2/3))))/(3*a)))/((c*x)^(4/3)*(a* 
x + b*x^2)^(2/3))
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 52

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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[3/(2*b*q^2)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1137

Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[( 
e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p))   Int[x^(m + p)*(b + c*x)^ 
p, x], x] /; FreeQ[{b, c, e, m}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 567, normalized size of antiderivative = 2.81 \[ \int \frac {1}{(c x)^{4/3} \left (a x+b x^2\right )^{2/3}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/3*(3*sqrt(1/3)*a*b*c*x^2*sqrt((-a^2*c)^(1/3)/c)*log((2*a*b*c*x^2 + 3*a^ 
2*c*x + 3*(-a^2*c)^(1/3)*(b*x^2 + a*x)^(1/3)*(c*x)^(2/3)*a + 3*sqrt(1/3)*( 
(-a^2*c)^(1/3)*a*c*x + 2*(b*x^2 + a*x)^(2/3)*(c*x)^(1/3)*a*c - (-a^2*c)^(2 
/3)*(b*x^2 + a*x)^(1/3)*(c*x)^(2/3))*sqrt((-a^2*c)^(1/3)/c))/x^2) + (-a^2* 
c)^(2/3)*b*x^2*log(-((-a^2*c)^(1/3)*a*c*x - (b*x^2 + a*x)^(2/3)*(c*x)^(1/3 
)*a*c - (-a^2*c)^(2/3)*(b*x^2 + a*x)^(1/3)*(c*x)^(2/3))/x) - 2*(-a^2*c)^(2 
/3)*b*x^2*log(((b*x^2 + a*x)^(1/3)*(c*x)^(2/3)*a - (-a^2*c)^(2/3)*x)/x) - 
3*(b*x^2 + a*x)^(1/3)*(c*x)^(2/3)*a^2)/(a^3*c^2*x^2), 1/3*(6*sqrt(1/3)*a*b 
*c*x^2*sqrt(-(-a^2*c)^(1/3)/c)*arctan(-sqrt(1/3)*((-a^2*c)^(1/3)*a*c*x - 2 
*(-a^2*c)^(2/3)*(b*x^2 + a*x)^(1/3)*(c*x)^(2/3))*sqrt(-(-a^2*c)^(1/3)/c)/( 
a^2*c*x)) + (-a^2*c)^(2/3)*b*x^2*log(-((-a^2*c)^(1/3)*a*c*x - (b*x^2 + a*x 
)^(2/3)*(c*x)^(1/3)*a*c - (-a^2*c)^(2/3)*(b*x^2 + a*x)^(1/3)*(c*x)^(2/3))/ 
x) - 2*(-a^2*c)^(2/3)*b*x^2*log(((b*x^2 + a*x)^(1/3)*(c*x)^(2/3)*a - (-a^2 
*c)^(2/3)*x)/x) - 3*(b*x^2 + a*x)^(1/3)*(c*x)^(2/3)*a^2)/(a^3*c^2*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(c x)^{4/3} \left (a x+b x^2\right )^{2/3}} \, dx=\frac {-2 \sqrt {3}\, \mathit {atan} \left (\frac {2 \left (b x +a \right )^{\frac {1}{6}}+a^{\frac {1}{6}}}{a^{\frac {1}{6}} \sqrt {3}}\right ) b x +2 \sqrt {3}\, \mathit {atan} \left (\frac {2 \left (b x +a \right )^{\frac {1}{6}}-a^{\frac {1}{6}}}{a^{\frac {1}{6}} \sqrt {3}}\right ) b x -3 a^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}}-2 \,\mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}+a^{\frac {1}{6}}\right ) b x -2 \,\mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}-a^{\frac {1}{6}}\right ) b x +\mathrm {log}\left (-a^{\frac {1}{6}} \left (b x +a \right )^{\frac {1}{6}}+\left (b x +a \right )^{\frac {1}{3}}+a^{\frac {1}{3}}\right ) b x +\mathrm {log}\left (a^{\frac {1}{6}} \left (b x +a \right )^{\frac {1}{6}}+\left (b x +a \right )^{\frac {1}{3}}+a^{\frac {1}{3}}\right ) b x}{3 c^{\frac {4}{3}} a^{\frac {5}{3}} x} \] Input:

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

Output:

( - 2*sqrt(3)*atan((2*(a + b*x)**(1/6) + a**(1/6))/(a**(1/6)*sqrt(3)))*b*x 
 + 2*sqrt(3)*atan((2*(a + b*x)**(1/6) - a**(1/6))/(a**(1/6)*sqrt(3)))*b*x 
- 3*a**(2/3)*(a + b*x)**(1/3) - 2*log((a + b*x)**(1/6) + a**(1/6))*b*x - 2 
*log((a + b*x)**(1/6) - a**(1/6))*b*x + log( - a**(1/6)*(a + b*x)**(1/6) + 
 (a + b*x)**(1/3) + a**(1/3))*b*x + log(a**(1/6)*(a + b*x)**(1/6) + (a + b 
*x)**(1/3) + a**(1/3))*b*x)/(3*c**(1/3)*a**(2/3)*a*c*x)