\(\int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx\) [3006]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 26, antiderivative size = 325 \[ \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx=\frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (d e-c f) (e+f x)^2}-\frac {(b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{6 (b e-a f) (d e-c f) (e+f x)}+\frac {(b c-a d)^2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d e-c f} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} (b e-a f)^{5/3} (d e-c f)^{4/3}}-\frac {(b c-a d)^2 \log (e+f x)}{18 (b e-a f)^{5/3} (d e-c f)^{4/3}}+\frac {(b c-a d)^2 \log \left (-\sqrt [3]{a+b x}+\frac {\sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt [3]{d e-c f}}\right )}{6 (b e-a f)^{5/3} (d e-c f)^{4/3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {96, 93} \[ \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx=\frac {(b c-a d)^2 \arctan \left (\frac {2 \sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt {3} \sqrt [3]{a+b x} \sqrt [3]{d e-c f}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3} (b e-a f)^{5/3} (d e-c f)^{4/3}}-\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d)}{6 (e+f x) (b e-a f) (d e-c f)}+\frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (e+f x)^2 (d e-c f)}-\frac {(b c-a d)^2 \log (e+f x)}{18 (b e-a f)^{5/3} (d e-c f)^{4/3}}+\frac {(b c-a d)^2 \log \left (\frac {\sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt [3]{d e-c f}}-\sqrt [3]{a+b x}\right )}{6 (b e-a f)^{5/3} (d e-c f)^{4/3}} \]

[In]

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

[Out]

((a + b*x)^(1/3)*(c + d*x)^(5/3))/(2*(d*e - c*f)*(e + f*x)^2) - ((b*c - a*d)*(a + b*x)^(1/3)*(c + d*x)^(2/3))/
(6*(b*e - a*f)*(d*e - c*f)*(e + f*x)) + ((b*c - a*d)^2*ArcTan[1/Sqrt[3] + (2*(b*e - a*f)^(1/3)*(c + d*x)^(1/3)
)/(Sqrt[3]*(d*e - c*f)^(1/3)*(a + b*x)^(1/3))])/(3*Sqrt[3]*(b*e - a*f)^(5/3)*(d*e - c*f)^(4/3)) - ((b*c - a*d)
^2*Log[e + f*x])/(18*(b*e - a*f)^(5/3)*(d*e - c*f)^(4/3)) + ((b*c - a*d)^2*Log[-(a + b*x)^(1/3) + ((b*e - a*f)
^(1/3)*(c + d*x)^(1/3))/(d*e - c*f)^(1/3)])/(6*(b*e - a*f)^(5/3)*(d*e - c*f)^(4/3))

Rule 93

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

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Dist[n*((d*e - c*f)/((m + 1)*(b*e - a*f
))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (d e-c f) (e+f x)^2}-\frac {(b c-a d) \int \frac {(c+d x)^{2/3}}{(a+b x)^{2/3} (e+f x)^2} \, dx}{6 (d e-c f)} \\ & = \frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (d e-c f) (e+f x)^2}-\frac {(b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{6 (b e-a f) (d e-c f) (e+f x)}-\frac {(b c-a d)^2 \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)} \, dx}{9 (b e-a f) (d e-c f)} \\ & = \frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (d e-c f) (e+f x)^2}-\frac {(b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{6 (b e-a f) (d e-c f) (e+f x)}+\frac {(b c-a d)^2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d e-c f} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} (b e-a f)^{5/3} (d e-c f)^{4/3}}-\frac {(b c-a d)^2 \log (e+f x)}{18 (b e-a f)^{5/3} (d e-c f)^{4/3}}+\frac {(b c-a d)^2 \log \left (-\sqrt [3]{a+b x}+\frac {\sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt [3]{d e-c f}}\right )}{6 (b e-a f)^{5/3} (d e-c f)^{4/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.17 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx=\frac {1}{18} (b c-a d)^2 \left (\frac {3 \sqrt [3]{a+b x} (c+d x)^{2/3} (b (-2 c e-3 d e x+c f x)+a (-d e+3 c f+2 d f x))}{(b c-a d)^2 (b e-a f) (-d e+c f) (e+f x)^2}-\frac {2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{-d e+c f} \sqrt [3]{a+b x}}{\sqrt [3]{b e-a f} \sqrt [3]{c+d x}}}{\sqrt {3}}\right )}{(b e-a f)^{5/3} (-d e+c f)^{4/3}}+\frac {2 \log \left (\sqrt [3]{b e-a f}+\frac {\sqrt [3]{-d e+c f} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )}{(b e-a f)^{5/3} (-d e+c f)^{4/3}}-\frac {\log \left ((b e-a f)^{2/3}+\frac {(-d e+c f)^{2/3} (a+b x)^{2/3}}{(c+d x)^{2/3}}-\frac {\sqrt [3]{b e-a f} \sqrt [3]{-d e+c f} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )}{(b e-a f)^{5/3} (-d e+c f)^{4/3}}\right ) \]

[In]

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

[Out]

((b*c - a*d)^2*((3*(a + b*x)^(1/3)*(c + d*x)^(2/3)*(b*(-2*c*e - 3*d*e*x + c*f*x) + a*(-(d*e) + 3*c*f + 2*d*f*x
)))/((b*c - a*d)^2*(b*e - a*f)*(-(d*e) + c*f)*(e + f*x)^2) - (2*Sqrt[3]*ArcTan[(1 - (2*(-(d*e) + c*f)^(1/3)*(a
 + b*x)^(1/3))/((b*e - a*f)^(1/3)*(c + d*x)^(1/3)))/Sqrt[3]])/((b*e - a*f)^(5/3)*(-(d*e) + c*f)^(4/3)) + (2*Lo
g[(b*e - a*f)^(1/3) + ((-(d*e) + c*f)^(1/3)*(a + b*x)^(1/3))/(c + d*x)^(1/3)])/((b*e - a*f)^(5/3)*(-(d*e) + c*
f)^(4/3)) - Log[(b*e - a*f)^(2/3) + ((-(d*e) + c*f)^(2/3)*(a + b*x)^(2/3))/(c + d*x)^(2/3) - ((b*e - a*f)^(1/3
)*(-(d*e) + c*f)^(1/3)*(a + b*x)^(1/3))/(c + d*x)^(1/3)]/((b*e - a*f)^(5/3)*(-(d*e) + c*f)^(4/3))))/18

Maple [F]

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

[In]

int((b*x+a)^(1/3)*(d*x+c)^(2/3)/(f*x+e)^3,x)

[Out]

int((b*x+a)^(1/3)*(d*x+c)^(2/3)/(f*x+e)^3,x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1918 vs. \(2 (275) = 550\).

Time = 0.40 (sec) , antiderivative size = 3992, normalized size of antiderivative = 12.28 \[ \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/18*(3*sqrt(1/3)*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^4 - (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d
^3)*e^3*f + (a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e^2*f^2 + ((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^2*f^2
 - (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e*f^3 + (a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*f^4)*x^2 +
2*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^3*f - (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e^2*f^2 + (
a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e*f^3)*x)*sqrt((-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a
*b*c + a^2*d)*e*f^2)^(1/3)/(d*e - c*f))*log(-(3*a^2*c*f^2 + (b^2*c + 2*a*b*d)*e^2 - 2*(2*a*b*c + a^2*d)*e*f +
3*(-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)*(b*e - a*f)*(b*x + a)^(1/
3)*(d*x + c)^(2/3) + (3*b^2*d*e^2 - 2*(b^2*c + 2*a*b*d)*e*f + (2*a*b*c + a^2*d)*f^2)*x - 3*sqrt(1/3)*(2*(b*d*e
^2 + a*c*f^2 - (b*c + a*d)*e*f)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*
e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*
a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)*(b*c*e - a*c*f + (b*d*e - a*d*f)*x))*sqrt((-b^2*d*e^3 + a^2*c*f^
3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)/(d*e - c*f)))/(f*x + e)) - (-b^2*d*e^3 + a^2*c*f^
3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*x^2 + 2*(b^2
*c^2 - 2*a*b*c*d + a^2*d^2)*e*f*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)
*e*f)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*
e*f^2)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) - (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c +
a^2*d)*e*f^2)^(1/3)*(b*c*e - a*c*f + (b*d*e - a*d*f)*x))/(d*x + c)) + 2*(-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a
*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*x^2 + 2*(b^2*c^2 - 2*a*b*c*d
 + a^2*d^2)*e*f*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)*e*f)*(b*x + a)^
(1/3)*(d*x + c)^(2/3) - (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*(d*
x + c))/(d*x + c)) + 3*(3*a^3*c^2*f^4 + (2*b^3*c*d + a*b^2*d^2)*e^4 - 2*(b^3*c^2 + 4*a*b^2*c*d + a^2*b*d^2)*e^
3*f + (7*a*b^2*c^2 + 10*a^2*b*c*d + a^3*d^2)*e^2*f^2 - 4*(2*a^2*b*c^2 + a^3*c*d)*e*f^3 + (3*b^3*d^2*e^4 - 4*(b
^3*c*d + 2*a*b^2*d^2)*e^3*f + (b^3*c^2 + 10*a*b^2*c*d + 7*a^2*b*d^2)*e^2*f^2 - 2*(a*b^2*c^2 + 4*a^2*b*c*d + a^
3*d^2)*e*f^3 + (a^2*b*c^2 + 2*a^3*c*d)*f^4)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^3*d^2*e^7 - a^3*c^2*e^2*f^5
 - (2*b^3*c*d + 3*a*b^2*d^2)*e^6*f + (b^3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*e^5*f^2 - (3*a*b^2*c^2 + 6*a^2*b*c*
d + a^3*d^2)*e^4*f^3 + (3*a^2*b*c^2 + 2*a^3*c*d)*e^3*f^4 + (b^3*d^2*e^5*f^2 - a^3*c^2*f^7 - (2*b^3*c*d + 3*a*b
^2*d^2)*e^4*f^3 + (b^3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*e^3*f^4 - (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*e^2*f^
5 + (3*a^2*b*c^2 + 2*a^3*c*d)*e*f^6)*x^2 + 2*(b^3*d^2*e^6*f - a^3*c^2*e*f^6 - (2*b^3*c*d + 3*a*b^2*d^2)*e^5*f^
2 + (b^3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*e^4*f^3 - (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*e^3*f^4 + (3*a^2*b*c
^2 + 2*a^3*c*d)*e^2*f^5)*x), -1/18*(6*sqrt(1/3)*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^4 - (b^3*c^3 - a*b^
2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e^3*f + (a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e^2*f^2 + ((b^3*c^2*d - 2*a*b
^2*c*d^2 + a^2*b*d^3)*e^2*f^2 - (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e*f^3 + (a*b^2*c^3 - 2*a^2*b*c
^2*d + a^3*c*d^2)*f^4)*x^2 + 2*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^3*f - (b^3*c^3 - a*b^2*c^2*d - a^2*b
*c*d^2 + a^3*d^3)*e^2*f^2 + (a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e*f^3)*x)*sqrt(-(-b^2*d*e^3 + a^2*c*f^3 +
(b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)/(d*e - c*f))*arctan(sqrt(1/3)*(2*(-b^2*d*e^3 + a^2*c*
f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) - (-b^2*d*e^3 +
 a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)*(b*c*e - a*c*f + (b*d*e - a*d*f)*x))*sqr
t(-(-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)/(d*e - c*f))/(b^2*c*e^2
- 2*a*b*c*e*f + a^2*c*f^2 + (b^2*d*e^2 - 2*a*b*d*e*f + a^2*d*f^2)*x)) + (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a
*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*x^2 + 2*(b^2*c^2 - 2*a*b*c*d
 + a^2*d^2)*e*f*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)*e*f)*(b*x + a)^
(2/3)*(d*x + c)^(1/3) + (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*(b*
x + a)^(1/3)*(d*x + c)^(2/3) - (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1
/3)*(b*c*e - a*c*f + (b*d*e - a*d*f)*x))/(d*x + c)) - 2*(-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2
*a*b*c + a^2*d)*e*f^2)^(2/3)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*x^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e*f*
x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2)*log(((b*d*e^2 + a*c*f^2 - (b*c + a*d)*e*f)*(b*x + a)^(1/3)*(d*x + c)^
(2/3) - (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*(d*x + c))/(d*x + c
)) - 3*(3*a^3*c^2*f^4 + (2*b^3*c*d + a*b^2*d^2)*e^4 - 2*(b^3*c^2 + 4*a*b^2*c*d + a^2*b*d^2)*e^3*f + (7*a*b^2*c
^2 + 10*a^2*b*c*d + a^3*d^2)*e^2*f^2 - 4*(2*a^2*b*c^2 + a^3*c*d)*e*f^3 + (3*b^3*d^2*e^4 - 4*(b^3*c*d + 2*a*b^2
*d^2)*e^3*f + (b^3*c^2 + 10*a*b^2*c*d + 7*a^2*b*d^2)*e^2*f^2 - 2*(a*b^2*c^2 + 4*a^2*b*c*d + a^3*d^2)*e*f^3 + (
a^2*b*c^2 + 2*a^3*c*d)*f^4)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^3*d^2*e^7 - a^3*c^2*e^2*f^5 - (2*b^3*c*d +
3*a*b^2*d^2)*e^6*f + (b^3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*e^5*f^2 - (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*e^4
*f^3 + (3*a^2*b*c^2 + 2*a^3*c*d)*e^3*f^4 + (b^3*d^2*e^5*f^2 - a^3*c^2*f^7 - (2*b^3*c*d + 3*a*b^2*d^2)*e^4*f^3
+ (b^3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*e^3*f^4 - (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*e^2*f^5 + (3*a^2*b*c^2
 + 2*a^3*c*d)*e*f^6)*x^2 + 2*(b^3*d^2*e^6*f - a^3*c^2*e*f^6 - (2*b^3*c*d + 3*a*b^2*d^2)*e^5*f^2 + (b^3*c^2 + 6
*a*b^2*c*d + 3*a^2*b*d^2)*e^4*f^3 - (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*e^3*f^4 + (3*a^2*b*c^2 + 2*a^3*c*d)*
e^2*f^5)*x)]

Sympy [F]

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

[In]

integrate((b*x+a)**(1/3)*(d*x+c)**(2/3)/(f*x+e)**3,x)

[Out]

Integral((a + b*x)**(1/3)*(c + d*x)**(2/3)/(e + f*x)**3, x)

Maxima [F]

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

[In]

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

[Out]

integrate((b*x + a)^(1/3)*(d*x + c)^(2/3)/(f*x + e)^3, x)

Giac [F]

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

[In]

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

[Out]

integrate((b*x + a)^(1/3)*(d*x + c)^(2/3)/(f*x + e)^3, x)

Mupad [F(-1)]

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

[In]

int(((a + b*x)^(1/3)*(c + d*x)^(2/3))/(e + f*x)^3,x)

[Out]

int(((a + b*x)^(1/3)*(c + d*x)^(2/3))/(e + f*x)^3, x)