3.31.16 \(\int \frac {(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx\) [3016]

Optimal. Leaf size=369 \[ \frac {f (9 b d e-5 b c f-4 a d f) (a+b x)^{2/3} \sqrt [3]{c+d x}}{6 b^2 d^2}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}-\frac {\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{3 \sqrt {3} b^{7/3} d^{8/3}}-\frac {\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log (c+d x)}{18 b^{7/3} d^{8/3}}-\frac {\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log \left (-1+\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{6 b^{7/3} d^{8/3}} \]

[Out]

1/6*f*(-4*a*d*f-5*b*c*f+9*b*d*e)*(b*x+a)^(2/3)*(d*x+c)^(1/3)/b^2/d^2+1/2*f*(b*x+a)^(2/3)*(d*x+c)^(1/3)*(f*x+e)
/b/d-1/18*(2*a^2*d^2*f^2-2*a*b*d*f*(-c*f+3*d*e)+b^2*(5*c^2*f^2-12*c*d*e*f+9*d^2*e^2))*ln(d*x+c)/b^(7/3)/d^(8/3
)-1/6*(2*a^2*d^2*f^2-2*a*b*d*f*(-c*f+3*d*e)+b^2*(5*c^2*f^2-12*c*d*e*f+9*d^2*e^2))*ln(-1+d^(1/3)*(b*x+a)^(1/3)/
b^(1/3)/(d*x+c)^(1/3))/b^(7/3)/d^(8/3)-1/9*(2*a^2*d^2*f^2-2*a*b*d*f*(-c*f+3*d*e)+b^2*(5*c^2*f^2-12*c*d*e*f+9*d
^2*e^2))*arctan(1/3*3^(1/2)+2/3*d^(1/3)*(b*x+a)^(1/3)/b^(1/3)/(d*x+c)^(1/3)*3^(1/2))/b^(7/3)/d^(8/3)*3^(1/2)

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Rubi [A]
time = 0.19, antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {92, 81, 61} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac {1}{\sqrt {3}}\right ) \left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right )}{3 \sqrt {3} b^{7/3} d^{8/3}}-\frac {\log (c+d x) \left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right )}{18 b^{7/3} d^{8/3}}-\frac {\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right ) \log \left (\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{6 b^{7/3} d^{8/3}}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (-4 a d f-5 b c f+9 b d e)}{6 b^2 d^2}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(9*b*d*e - 5*b*c*f - 4*a*d*f)*(a + b*x)^(2/3)*(c + d*x)^(1/3))/(6*b^2*d^2) + (f*(a + b*x)^(2/3)*(c + d*x)^(
1/3)*(e + f*x))/(2*b*d) - ((2*a^2*d^2*f^2 - 2*a*b*d*f*(3*d*e - c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2)
)*ArcTan[1/Sqrt[3] + (2*d^(1/3)*(a + b*x)^(1/3))/(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))])/(3*Sqrt[3]*b^(7/3)*d^(8/3
)) - ((2*a^2*d^2*f^2 - 2*a*b*d*f*(3*d*e - c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*Log[c + d*x])/(18*b
^(7/3)*d^(8/3)) - ((2*a^2*d^2*f^2 - 2*a*b*d*f*(3*d*e - c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*Log[-1
+ (d^(1/3)*(a + b*x)^(1/3))/(b^(1/3)*(c + d*x)^(1/3))])/(6*b^(7/3)*d^(8/3))

Rule 61

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] && Ne
Q[b*c - a*d, 0] && PosQ[d/b]

Rule 81

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

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rubi steps

\begin {align*} \int \frac {(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx &=\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}+\frac {\int \frac {\frac {1}{3} \left (6 b d e^2-f (2 b c e+a d e+3 a c f)\right )+\frac {1}{3} f (9 b d e-5 b c f-4 a d f) x}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx}{2 b d}\\ &=\frac {f (9 b d e-5 b c f-4 a d f) (a+b x)^{2/3} \sqrt [3]{c+d x}}{6 b^2 d^2}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}+\frac {\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \int \frac {1}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx}{9 b^2 d^2}\\ &=\frac {f (9 b d e-5 b c f-4 a d f) (a+b x)^{2/3} \sqrt [3]{c+d x}}{6 b^2 d^2}+\frac {f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}-\frac {\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{3 \sqrt {3} b^{7/3} d^{8/3}}-\frac {\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log (c+d x)}{18 b^{7/3} d^{8/3}}-\frac {\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log \left (-1+\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{6 b^{7/3} d^{8/3}}\\ \end {align*}

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Mathematica [A]
time = 0.75, size = 368, normalized size = 1.00 \begin {gather*} \frac {3 \sqrt [3]{b} d^{2/3} f (a+b x)^{2/3} \sqrt [3]{c+d x} (-5 b c f-4 a d f+3 b d (4 e+f x))+2 \sqrt {3} \left (2 a^2 d^2 f^2+2 a b d f (-3 d e+c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}}{\sqrt {3}}\right )-2 \left (2 a^2 d^2 f^2+2 a b d f (-3 d e+c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log \left (\sqrt [3]{d}-\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{a+b x}}\right )+\left (2 a^2 d^2 f^2+2 a b d f (-3 d e+c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log \left (d^{2/3}+\frac {\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{c+d x}}{\sqrt [3]{a+b x}}+\frac {b^{2/3} (c+d x)^{2/3}}{(a+b x)^{2/3}}\right )}{18 b^{7/3} d^{8/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^(1/3)*d^(2/3)*f*(a + b*x)^(2/3)*(c + d*x)^(1/3)*(-5*b*c*f - 4*a*d*f + 3*b*d*(4*e + f*x)) + 2*Sqrt[3]*(2*a
^2*d^2*f^2 + 2*a*b*d*f*(-3*d*e + c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*ArcTan[(1 + (2*b^(1/3)*(c +
d*x)^(1/3))/(d^(1/3)*(a + b*x)^(1/3)))/Sqrt[3]] - 2*(2*a^2*d^2*f^2 + 2*a*b*d*f*(-3*d*e + c*f) + b^2*(9*d^2*e^2
- 12*c*d*e*f + 5*c^2*f^2))*Log[d^(1/3) - (b^(1/3)*(c + d*x)^(1/3))/(a + b*x)^(1/3)] + (2*a^2*d^2*f^2 + 2*a*b*
d*f*(-3*d*e + c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*Log[d^(2/3) + (b^(1/3)*d^(1/3)*(c + d*x)^(1/3))
/(a + b*x)^(1/3) + (b^(2/3)*(c + d*x)^(2/3))/(a + b*x)^(2/3)])/(18*b^(7/3)*d^(8/3))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{2}}{\left (b x +a \right )^{\frac {1}{3}} \left (d x +c \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.85, size = 1003, normalized size = 2.72 \begin {gather*} \left [\frac {3 \, \sqrt {\frac {1}{3}} {\left (9 \, b^{3} d^{3} e^{2} + {\left (5 \, b^{3} c^{2} d + 2 \, a b^{2} c d^{2} + 2 \, a^{2} b d^{3}\right )} f^{2} - 6 \, {\left (2 \, b^{3} c d^{2} + a b^{2} d^{3}\right )} f e\right )} \sqrt {\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}} \log \left (-3 \, b d^{2} x - 2 \, b c d - a d^{2} - 3 \, \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} d - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d - \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} + \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}\right )} \sqrt {\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}}\right ) - 2 \, {\left (9 \, b^{2} d^{2} e^{2} + {\left (5 \, b^{2} c^{2} + 2 \, a b c d + 2 \, a^{2} d^{2}\right )} f^{2} - 6 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} f e\right )} \left (-b d^{2}\right )^{\frac {2}{3}} \log \left (\frac {{\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d - \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}}{b x + a}\right ) + {\left (9 \, b^{2} d^{2} e^{2} + {\left (5 \, b^{2} c^{2} + 2 \, a b c d + 2 \, a^{2} d^{2}\right )} f^{2} - 6 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} f e\right )} \left (-b d^{2}\right )^{\frac {2}{3}} \log \left (\frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d + \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} - \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}}{b x + a}\right ) + 3 \, {\left (3 \, b^{2} d^{3} f^{2} x + 12 \, b^{2} d^{3} f e - {\left (5 \, b^{2} c d^{2} + 4 \, a b d^{3}\right )} f^{2}\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{18 \, b^{3} d^{4}}, \frac {6 \, \sqrt {\frac {1}{3}} {\left (9 \, b^{3} d^{3} e^{2} + {\left (5 \, b^{3} c^{2} d + 2 \, a b^{2} c d^{2} + 2 \, a^{2} b d^{3}\right )} f^{2} - 6 \, {\left (2 \, b^{3} c d^{2} + a b^{2} d^{3}\right )} f e\right )} \sqrt {-\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} - \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}\right )} \sqrt {-\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}}}{b d^{2} x + a d^{2}}\right ) - 2 \, {\left (9 \, b^{2} d^{2} e^{2} + {\left (5 \, b^{2} c^{2} + 2 \, a b c d + 2 \, a^{2} d^{2}\right )} f^{2} - 6 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} f e\right )} \left (-b d^{2}\right )^{\frac {2}{3}} \log \left (\frac {{\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d - \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}}{b x + a}\right ) + {\left (9 \, b^{2} d^{2} e^{2} + {\left (5 \, b^{2} c^{2} + 2 \, a b c d + 2 \, a^{2} d^{2}\right )} f^{2} - 6 \, {\left (2 \, b^{2} c d + a b d^{2}\right )} f e\right )} \left (-b d^{2}\right )^{\frac {2}{3}} \log \left (\frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d + \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} - \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}}{b x + a}\right ) + 3 \, {\left (3 \, b^{2} d^{3} f^{2} x + 12 \, b^{2} d^{3} f e - {\left (5 \, b^{2} c d^{2} + 4 \, a b d^{3}\right )} f^{2}\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{18 \, b^{3} d^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(3*sqrt(1/3)*(9*b^3*d^3*e^2 + (5*b^3*c^2*d + 2*a*b^2*c*d^2 + 2*a^2*b*d^3)*f^2 - 6*(2*b^3*c*d^2 + a*b^2*d
^3)*f*e)*sqrt((-b*d^2)^(1/3)/b)*log(-3*b*d^2*x - 2*b*c*d - a*d^2 - 3*(-b*d^2)^(1/3)*(b*x + a)^(2/3)*(d*x + c)^
(1/3)*d - 3*sqrt(1/3)*(2*(b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - (-b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3)
+ (-b*d^2)^(1/3)*(b*d*x + a*d))*sqrt((-b*d^2)^(1/3)/b)) - 2*(9*b^2*d^2*e^2 + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*d^
2)*f^2 - 6*(2*b^2*c*d + a*b*d^2)*f*e)*(-b*d^2)^(2/3)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (-b*d^2)^(2/3)
*(b*x + a))/(b*x + a)) + (9*b^2*d^2*e^2 + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^2 - 6*(2*b^2*c*d + a*b*d^2)*f*
e)*(-b*d^2)^(2/3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d + (-b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) -
(-b*d^2)^(1/3)*(b*d*x + a*d))/(b*x + a)) + 3*(3*b^2*d^3*f^2*x + 12*b^2*d^3*f*e - (5*b^2*c*d^2 + 4*a*b*d^3)*f^2
)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b^3*d^4), 1/18*(6*sqrt(1/3)*(9*b^3*d^3*e^2 + (5*b^3*c^2*d + 2*a*b^2*c*d^2
+ 2*a^2*b*d^3)*f^2 - 6*(2*b^3*c*d^2 + a*b^2*d^3)*f*e)*sqrt(-(-b*d^2)^(1/3)/b)*arctan(sqrt(1/3)*(2*(-b*d^2)^(2/
3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2)^(1/3)*(b*d*x + a*d))*sqrt(-(-b*d^2)^(1/3)/b)/(b*d^2*x + a*d^2))
- 2*(9*b^2*d^2*e^2 + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^2 - 6*(2*b^2*c*d + a*b*d^2)*f*e)*(-b*d^2)^(2/3)*log
(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (-b*d^2)^(2/3)*(b*x + a))/(b*x + a)) + (9*b^2*d^2*e^2 + (5*b^2*c^2 + 2
*a*b*c*d + 2*a^2*d^2)*f^2 - 6*(2*b^2*c*d + a*b*d^2)*f*e)*(-b*d^2)^(2/3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b
*d + (-b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2)^(1/3)*(b*d*x + a*d))/(b*x + a)) + 3*(3*b^2*d^3*
f^2*x + 12*b^2*d^3*f*e - (5*b^2*c*d^2 + 4*a*b*d^3)*f^2)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b^3*d^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e + f x\right )^{2}}{\sqrt [3]{a + b x} \left (c + d x\right )^{\frac {2}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e+f\,x\right )}^2}{{\left (a+b\,x\right )}^{1/3}\,{\left (c+d\,x\right )}^{2/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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