3.2999 \(\int \frac{(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx\)

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

[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))

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Rubi [A]  time = 0.829032, antiderivative size = 369, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115 \[ -\frac{\log (c+d x) \left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right )}{18 b^{7/3} d^{8/3}}-\frac{\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right ) \log \left (\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{6 b^{7/3} d^{8/3}}-\frac{\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right ) \tan ^{-1}\left (\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt{3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac{1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{7/3} d^{8/3}}+\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x} (-4 a d f-5 b c f+9 b d e)}{6 b^2 d^2}+\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d} \]

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))

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Rubi in Sympy [A]  time = 45.6673, size = 386, normalized size = 1.05 \[ \frac{f \left (a + b x\right )^{\frac{2}{3}} \sqrt [3]{c + d x} \left (e + f x\right )}{2 b d} - \frac{f \left (a + b x\right )^{\frac{2}{3}} \sqrt [3]{c + d x} \left (4 a d f + 5 b c f - 9 b d e\right )}{6 b^{2} d^{2}} + \frac{\left (3 b d \left (- 6 b d e^{2} + f \left (3 a c f + e \left (a d + 2 b c\right )\right )\right ) - f \left (a d + 2 b c\right ) \left (4 a d f + 5 b c f - 9 b d e\right )\right ) \log{\left (-1 + \frac{\sqrt [3]{d} \sqrt [3]{a + b x}}{\sqrt [3]{b} \sqrt [3]{c + d x}} \right )}}{12 b^{\frac{7}{3}} d^{\frac{8}{3}}} + \frac{\left (3 b d \left (- 6 b d e^{2} + f \left (3 a c f + e \left (a d + 2 b c\right )\right )\right ) - f \left (a d + 2 b c\right ) \left (4 a d f + 5 b c f - 9 b d e\right )\right ) \log{\left (c + d x \right )}}{36 b^{\frac{7}{3}} d^{\frac{8}{3}}} + \frac{\sqrt{3} \left (3 b d \left (- 6 b d e^{2} + f \left (3 a c f + e \left (a d + 2 b c\right )\right )\right ) - f \left (a d + 2 b c\right ) \left (4 a d f + 5 b c f - 9 b d e\right )\right ) \operatorname{atan}{\left (\frac{\sqrt{3}}{3} + \frac{2 \sqrt{3} \sqrt [3]{d} \sqrt [3]{a + b x}}{3 \sqrt [3]{b} \sqrt [3]{c + d x}} \right )}}{18 b^{\frac{7}{3}} d^{\frac{8}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

f*(a + b*x)**(2/3)*(c + d*x)**(1/3)*(e + f*x)/(2*b*d) - f*(a + b*x)**(2/3)*(c +
d*x)**(1/3)*(4*a*d*f + 5*b*c*f - 9*b*d*e)/(6*b**2*d**2) + (3*b*d*(-6*b*d*e**2 +
f*(3*a*c*f + e*(a*d + 2*b*c))) - f*(a*d + 2*b*c)*(4*a*d*f + 5*b*c*f - 9*b*d*e))*
log(-1 + d**(1/3)*(a + b*x)**(1/3)/(b**(1/3)*(c + d*x)**(1/3)))/(12*b**(7/3)*d**
(8/3)) + (3*b*d*(-6*b*d*e**2 + f*(3*a*c*f + e*(a*d + 2*b*c))) - f*(a*d + 2*b*c)*
(4*a*d*f + 5*b*c*f - 9*b*d*e))*log(c + d*x)/(36*b**(7/3)*d**(8/3)) + sqrt(3)*(3*
b*d*(-6*b*d*e**2 + f*(3*a*c*f + e*(a*d + 2*b*c))) - f*(a*d + 2*b*c)*(4*a*d*f + 5
*b*c*f - 9*b*d*e))*atan(sqrt(3)/3 + 2*sqrt(3)*d**(1/3)*(a + b*x)**(1/3)/(3*b**(1
/3)*(c + d*x)**(1/3)))/(18*b**(7/3)*d**(8/3))

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Mathematica [C]  time = 0.30363, size = 162, normalized size = 0.44 \[ \frac{\sqrt [3]{c+d x} \left (2 \sqrt [3]{\frac{d (a+b x)}{a d-b c}} \left (2 a^2 d^2 f^2+2 a b d f (c f-3 d e)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right ) \, _2F_1\left (\frac{1}{3},\frac{1}{3};\frac{4}{3};\frac{b (c+d x)}{b c-a d}\right )+d f (a+b x) (-4 a d f-5 b c f+3 b d (4 e+f x))\right )}{6 b^2 d^3 \sqrt [3]{a+b x}} \]

Antiderivative was successfully verified.

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

[Out]

((c + d*x)^(1/3)*(d*f*(a + b*x)*(-5*b*c*f - 4*a*d*f + 3*b*d*(4*e + f*x)) + 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
))*((d*(a + b*x))/(-(b*c) + a*d))^(1/3)*Hypergeometric2F1[1/3, 1/3, 4/3, (b*(c +
d*x))/(b*c - a*d)]))/(6*b^2*d^3*(a + b*x)^(1/3))

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \[ \int{ \left ( fx+e \right ) ^{2}{\frac{1}{\sqrt [3]{bx+a}}} \left ( dx+c \right ) ^{-{\frac{2}{3}}}}\, dx \]

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

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 = 0.264257, size = 570, normalized size = 1.54 \[ \frac{\sqrt{3}{\left (3 \, \sqrt{3}{\left (3 \, b d f^{2} x + 12 \, b d e f -{\left (5 \, b c + 4 \, a d\right )} f^{2}\right )} \left (-b d^{2}\right )^{\frac{1}{3}}{\left (b x + a\right )}^{\frac{2}{3}}{\left (d x + c\right )}^{\frac{1}{3}} - \sqrt{3}{\left (9 \, b^{2} d^{2} e^{2} - 6 \,{\left (2 \, b^{2} c d + a b d^{2}\right )} e f +{\left (5 \, b^{2} c^{2} + 2 \, a b c d + 2 \, a^{2} d^{2}\right )} f^{2}\right )} \log \left (\frac{b d^{2} x + a d^{2} - \left (-b d^{2}\right )^{\frac{1}{3}}{\left (b x + a\right )}^{\frac{2}{3}}{\left (d x + c\right )}^{\frac{1}{3}} d + \left (-b d^{2}\right )^{\frac{2}{3}}{\left (b x + a\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{2}{3}}}{b x + a}\right ) + 2 \, \sqrt{3}{\left (9 \, b^{2} d^{2} e^{2} - 6 \,{\left (2 \, b^{2} c d + a b d^{2}\right )} e f +{\left (5 \, b^{2} c^{2} + 2 \, a b c d + 2 \, a^{2} d^{2}\right )} f^{2}\right )} \log \left (\frac{b d x + a d + \left (-b d^{2}\right )^{\frac{1}{3}}{\left (b x + a\right )}^{\frac{2}{3}}{\left (d x + c\right )}^{\frac{1}{3}}}{b x + a}\right ) + 6 \,{\left (9 \, b^{2} d^{2} e^{2} - 6 \,{\left (2 \, b^{2} c d + a b d^{2}\right )} e f +{\left (5 \, b^{2} c^{2} + 2 \, a b c d + 2 \, a^{2} d^{2}\right )} f^{2}\right )} \arctan \left (\frac{2 \, \sqrt{3} \left (-b d^{2}\right )^{\frac{1}{3}}{\left (b x + a\right )}^{\frac{2}{3}}{\left (d x + c\right )}^{\frac{1}{3}} - \sqrt{3}{\left (b d x + a d\right )}}{3 \,{\left (b d x + a d\right )}}\right )\right )}}{54 \, \left (-b d^{2}\right )^{\frac{1}{3}} b^{2} d^{2}} \]

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]

1/54*sqrt(3)*(3*sqrt(3)*(3*b*d*f^2*x + 12*b*d*e*f - (5*b*c + 4*a*d)*f^2)*(-b*d^2
)^(1/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - sqrt(3)*(9*b^2*d^2*e^2 - 6*(2*b^2*c*d
+ a*b*d^2)*e*f + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^2)*log((b*d^2*x + a*d^2 -
(-b*d^2)^(1/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3)*d + (-b*d^2)^(2/3)*(b*x + a)^(1/
3)*(d*x + c)^(2/3))/(b*x + a)) + 2*sqrt(3)*(9*b^2*d^2*e^2 - 6*(2*b^2*c*d + a*b*d
^2)*e*f + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^2)*log((b*d*x + a*d + (-b*d^2)^(
1/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b*x + a)) + 6*(9*b^2*d^2*e^2 - 6*(2*b^2*c
*d + a*b*d^2)*e*f + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^2)*arctan(1/3*(2*sqrt(
3)*(-b*d^2)^(1/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - sqrt(3)*(b*d*x + a*d))/(b*d*
x + a*d)))/((-b*d^2)^(1/3)*b^2*d^2)

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

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

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)