\(\int \frac {(c x)^{7/3}}{(a x+b x^2)^{2/3}} \, dx\) [139]

Optimal result

Integrand size = 21, antiderivative size = 280 \[ \int \frac {(c x)^{7/3}}{\left (a x+b x^2\right )^{2/3}} \, dx=-\frac {5 a c^2 \sqrt [3]{c x} \sqrt [3]{a x+b x^2}}{6 b^2}+\frac {c (c x)^{4/3} \sqrt [3]{a x+b x^2}}{2 b}-\frac {5 a^2 c^{5/3} (c x)^{2/3} (a+b x)^{2/3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{c x}}{\sqrt {3} \sqrt [3]{c} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} b^{8/3} \left (a x+b x^2\right )^{2/3}}-\frac {5 a^2 c^{5/3} (c x)^{2/3} (a+b x)^{2/3} \log (a+b x)}{18 b^{8/3} \left (a x+b x^2\right )^{2/3}}-\frac {5 a^2 c^{5/3} (c x)^{2/3} (a+b x)^{2/3} \log \left (1-\frac {\sqrt [3]{b} \sqrt [3]{c x}}{\sqrt [3]{c} \sqrt [3]{a+b x}}\right )}{6 b^{8/3} \left (a x+b x^2\right )^{2/3}} \] Output:

-5/6*a*c^2*(c*x)^(1/3)*(b*x^2+a*x)^(1/3)/b^2+1/2*c*(c*x)^(4/3)*(b*x^2+a*x) 
^(1/3)/b-5/9*a^2*c^(5/3)*(c*x)^(2/3)*(b*x+a)^(2/3)*arctan(1/3*3^(1/2)+2/3* 
b^(1/3)*(c*x)^(1/3)*3^(1/2)/c^(1/3)/(b*x+a)^(1/3))*3^(1/2)/b^(8/3)/(b*x^2+ 
a*x)^(2/3)-5/18*a^2*c^(5/3)*(c*x)^(2/3)*(b*x+a)^(2/3)*ln(b*x+a)/b^(8/3)/(b 
*x^2+a*x)^(2/3)-5/6*a^2*c^(5/3)*(c*x)^(2/3)*(b*x+a)^(2/3)*ln(1-b^(1/3)*(c* 
x)^(1/3)/c^(1/3)/(b*x+a)^(1/3))/b^(8/3)/(b*x^2+a*x)^(2/3)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.81 \[ \int \frac {(c x)^{7/3}}{\left (a x+b x^2\right )^{2/3}} \, dx=\frac {(c x)^{7/3} \left (-15 a^2 b^{2/3} x^{2/3}-6 a b^{5/3} x^{5/3}+9 b^{8/3} x^{8/3}-10 \sqrt {3} a^2 (a+b x)^{2/3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{b} \sqrt [3]{x}+2 \sqrt [3]{a+b x}}\right )-10 a^2 (a+b x)^{2/3} \log \left (-\sqrt [3]{b} \sqrt [3]{x}+\sqrt [3]{a+b x}\right )+5 a^2 (a+b x)^{2/3} \log \left (b^{2/3} x^{2/3}+\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )\right )}{18 b^{8/3} x^{5/3} (x (a+b x))^{2/3}} \] Input:

Integrate[(c*x)^(7/3)/(a*x + b*x^2)^(2/3),x]
 

Output:

((c*x)^(7/3)*(-15*a^2*b^(2/3)*x^(2/3) - 6*a*b^(5/3)*x^(5/3) + 9*b^(8/3)*x^ 
(8/3) - 10*Sqrt[3]*a^2*(a + b*x)^(2/3)*ArcTan[(Sqrt[3]*b^(1/3)*x^(1/3))/(b 
^(1/3)*x^(1/3) + 2*(a + b*x)^(1/3))] - 10*a^2*(a + b*x)^(2/3)*Log[-(b^(1/3 
)*x^(1/3)) + (a + b*x)^(1/3)] + 5*a^2*(a + b*x)^(2/3)*Log[b^(2/3)*x^(2/3) 
+ b^(1/3)*x^(1/3)*(a + b*x)^(1/3) + (a + b*x)^(2/3)]))/(18*b^(8/3)*x^(5/3) 
*(x*(a + b*x))^(2/3))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.66, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1137, 60, 60, 71}

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.

\(\displaystyle \int \frac {(c x)^{7/3}}{\left (a x+b x^2\right )^{2/3}} \, dx\)

\(\Big \downarrow \) 1137

\(\displaystyle \frac {(c x)^{7/3} (a+b x)^{2/3} \int \frac {x^{5/3}}{(a+b x)^{2/3}}dx}{x^{5/3} \left (a x+b x^2\right )^{2/3}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(c x)^{7/3} (a+b x)^{2/3} \left (\frac {x^{5/3} \sqrt [3]{a+b x}}{2 b}-\frac {5 a \int \frac {x^{2/3}}{(a+b x)^{2/3}}dx}{6 b}\right )}{x^{5/3} \left (a x+b x^2\right )^{2/3}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(c x)^{7/3} (a+b x)^{2/3} \left (\frac {x^{5/3} \sqrt [3]{a+b x}}{2 b}-\frac {5 a \left (\frac {x^{2/3} \sqrt [3]{a+b x}}{b}-\frac {2 a \int \frac {1}{\sqrt [3]{x} (a+b x)^{2/3}}dx}{3 b}\right )}{6 b}\right )}{x^{5/3} \left (a x+b x^2\right )^{2/3}}\)

\(\Big \downarrow \) 71

\(\displaystyle \frac {(c x)^{7/3} (a+b x)^{2/3} \left (\frac {x^{5/3} \sqrt [3]{a+b x}}{2 b}-\frac {5 a \left (\frac {x^{2/3} \sqrt [3]{a+b x}}{b}-\frac {2 a \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{b^{2/3}}-\frac {\log (a+b x)}{2 b^{2/3}}-\frac {3 \log \left (\frac {\sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a+b x}}-1\right )}{2 b^{2/3}}\right )}{3 b}\right )}{6 b}\right )}{x^{5/3} \left (a x+b x^2\right )^{2/3}}\)

Input:

Int[(c*x)^(7/3)/(a*x + b*x^2)^(2/3),x]
 

Output:

((c*x)^(7/3)*(a + b*x)^(2/3)*((x^(5/3)*(a + b*x)^(1/3))/(2*b) - (5*a*((x^( 
2/3)*(a + b*x)^(1/3))/b - (2*a*(-((Sqrt[3]*ArcTan[1/Sqrt[3] + (2*b^(1/3)*x 
^(1/3))/(Sqrt[3]*(a + b*x)^(1/3))])/b^(2/3)) - Log[a + b*x]/(2*b^(2/3)) - 
(3*Log[-1 + (b^(1/3)*x^(1/3))/(a + b*x)^(1/3)])/(2*b^(2/3))))/(3*b)))/(6*b 
)))/(x^(5/3)*(a*x + b*x^2)^(2/3))
 

Defintions of rubi rules used

rule 60

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

rule 71

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

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]

\[\int \frac {\left (c x \right )^{\frac {7}{3}}}{\left (b \,x^{2}+a x \right )^{\frac {2}{3}}}d x\]

Input:

int((c*x)^(7/3)/(b*x^2+a*x)^(2/3),x)
 

Output:

int((c*x)^(7/3)/(b*x^2+a*x)^(2/3),x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.84 \[ \int \frac {(c x)^{7/3}}{\left (a x+b x^2\right )^{2/3}} \, dx=-\frac {10 \, \sqrt {3} a^{2} c^{2} \left (-\frac {c}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x^{2} + a x\right )}^{\frac {1}{3}} \left (c x\right )^{\frac {1}{3}} b \left (-\frac {c}{b^{2}}\right )^{\frac {2}{3}} + \sqrt {3} c x}{3 \, c x}\right ) - 10 \, a^{2} c^{2} \left (-\frac {c}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {b x \left (-\frac {c}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{2} + a x\right )}^{\frac {1}{3}} \left (c x\right )^{\frac {1}{3}}}{x}\right ) + 5 \, a^{2} c^{2} \left (-\frac {c}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {b^{2} x^{2} \left (-\frac {c}{b^{2}}\right )^{\frac {2}{3}} - {\left (b x^{2} + a x\right )}^{\frac {1}{3}} \left (c x\right )^{\frac {1}{3}} b x \left (-\frac {c}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{2} + a x\right )}^{\frac {2}{3}} \left (c x\right )^{\frac {2}{3}}}{x^{2}}\right ) - 3 \, {\left (3 \, b c^{2} x - 5 \, a c^{2}\right )} {\left (b x^{2} + a x\right )}^{\frac {1}{3}} \left (c x\right )^{\frac {1}{3}}}{18 \, b^{2}} \] Input:

integrate((c*x)^(7/3)/(b*x^2+a*x)^(2/3),x, algorithm="fricas")
 

Output:

-1/18*(10*sqrt(3)*a^2*c^2*(-c/b^2)^(1/3)*arctan(1/3*(2*sqrt(3)*(b*x^2 + a* 
x)^(1/3)*(c*x)^(1/3)*b*(-c/b^2)^(2/3) + sqrt(3)*c*x)/(c*x)) - 10*a^2*c^2*( 
-c/b^2)^(1/3)*log((b*x*(-c/b^2)^(1/3) + (b*x^2 + a*x)^(1/3)*(c*x)^(1/3))/x 
) + 5*a^2*c^2*(-c/b^2)^(1/3)*log((b^2*x^2*(-c/b^2)^(2/3) - (b*x^2 + a*x)^( 
1/3)*(c*x)^(1/3)*b*x*(-c/b^2)^(1/3) + (b*x^2 + a*x)^(2/3)*(c*x)^(2/3))/x^2 
) - 3*(3*b*c^2*x - 5*a*c^2)*(b*x^2 + a*x)^(1/3)*(c*x)^(1/3))/b^2
 

Sympy [F]

\[ \int \frac {(c x)^{7/3}}{\left (a x+b x^2\right )^{2/3}} \, dx=\int \frac {\left (c x\right )^{\frac {7}{3}}}{\left (x \left (a + b x\right )\right )^{\frac {2}{3}}}\, dx \] Input:

integrate((c*x)**(7/3)/(b*x**2+a*x)**(2/3),x)
 

Output:

Integral((c*x)**(7/3)/(x*(a + b*x))**(2/3), x)
 

Maxima [F]

\[ \int \frac {(c x)^{7/3}}{\left (a x+b x^2\right )^{2/3}} \, dx=\int { \frac {\left (c x\right )^{\frac {7}{3}}}{{\left (b x^{2} + a x\right )}^{\frac {2}{3}}} \,d x } \] Input:

integrate((c*x)^(7/3)/(b*x^2+a*x)^(2/3),x, algorithm="maxima")
 

Output:

integrate((c*x)^(7/3)/(b*x^2 + a*x)^(2/3), x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.74 \[ \int \frac {(c x)^{7/3}}{\left (a x+b x^2\right )^{2/3}} \, dx=\frac {\frac {10 \, \sqrt {3} \left (b c\right )^{\frac {1}{3}} a^{3} c^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b c + \frac {a c}{x}\right )}^{\frac {1}{3}} + \left (b c\right )^{\frac {1}{3}}\right )}}{3 \, \left (b c\right )^{\frac {1}{3}}}\right )}{b^{3}} + \frac {5 \, \left (b c\right )^{\frac {1}{3}} a^{3} c^{3} \log \left ({\left (b c + \frac {a c}{x}\right )}^{\frac {2}{3}} + {\left (b c + \frac {a c}{x}\right )}^{\frac {1}{3}} \left (b c\right )^{\frac {1}{3}} + \left (b c\right )^{\frac {2}{3}}\right )}{b^{3}} - \frac {10 \, \left (b c\right )^{\frac {1}{3}} a^{3} c^{3} \log \left ({\left | {\left (b c + \frac {a c}{x}\right )}^{\frac {1}{3}} - \left (b c\right )^{\frac {1}{3}} \right |}\right )}{b^{3}} + \frac {3 \, {\left (8 \, {\left (b c + \frac {a c}{x}\right )}^{\frac {1}{3}} a^{3} b c^{5} - 5 \, {\left (b c + \frac {a c}{x}\right )}^{\frac {4}{3}} a^{3} c^{4}\right )} x^{2}}{a^{2} b^{2} c^{2}}}{18 \, a c} \] Input:

integrate((c*x)^(7/3)/(b*x^2+a*x)^(2/3),x, algorithm="giac")
 

Output:

1/18*(10*sqrt(3)*(b*c)^(1/3)*a^3*c^3*arctan(1/3*sqrt(3)*(2*(b*c + a*c/x)^( 
1/3) + (b*c)^(1/3))/(b*c)^(1/3))/b^3 + 5*(b*c)^(1/3)*a^3*c^3*log((b*c + a* 
c/x)^(2/3) + (b*c + a*c/x)^(1/3)*(b*c)^(1/3) + (b*c)^(2/3))/b^3 - 10*(b*c) 
^(1/3)*a^3*c^3*log(abs((b*c + a*c/x)^(1/3) - (b*c)^(1/3)))/b^3 + 3*(8*(b*c 
 + a*c/x)^(1/3)*a^3*b*c^5 - 5*(b*c + a*c/x)^(4/3)*a^3*c^4)*x^2/(a^2*b^2*c^ 
2))/(a*c)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{7/3}}{\left (a x+b x^2\right )^{2/3}} \, dx=\int \frac {{\left (c\,x\right )}^{7/3}}{{\left (b\,x^2+a\,x\right )}^{2/3}} \,d x \] Input:

int((c*x)^(7/3)/(a*x + b*x^2)^(2/3),x)
 

Output:

int((c*x)^(7/3)/(a*x + b*x^2)^(2/3), x)
 

Reduce [F]

\[ \int \frac {(c x)^{7/3}}{\left (a x+b x^2\right )^{2/3}} \, dx=c^{\frac {7}{3}} \left (\int \frac {x^{\frac {5}{3}}}{\left (b x +a \right )^{\frac {2}{3}}}d x \right ) \] Input:

int((c*x)^(7/3)/(b*x^2+a*x)^(2/3),x)
 

Output:

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