3.3004 \(\int \frac{1}{\sqrt [3]{a+b x} (c+d x)^{2/3} (e+f x)^3} \, dx\)

Optimal. Leaf size=477 \[ \frac{\log (e+f x) \left (5 a^2 d^2 f^2-2 a b d f (6 d e-c f)+b^2 \left (2 c^2 f^2-6 c d e f+9 d^2 e^2\right )\right )}{18 (b e-a f)^{7/3} (d e-c f)^{8/3}}-\frac{\left (5 a^2 d^2 f^2-2 a b d f (6 d e-c f)+b^2 \left (2 c^2 f^2-6 c d e f+9 d^2 e^2\right )\right ) \log \left (\frac{\sqrt [3]{a+b x} \sqrt [3]{d e-c f}}{\sqrt [3]{b e-a f}}-\sqrt [3]{c+d x}\right )}{6 (b e-a f)^{7/3} (d e-c f)^{8/3}}-\frac{\left (5 a^2 d^2 f^2-2 a b d f (6 d e-c f)+b^2 \left (2 c^2 f^2-6 c d e f+9 d^2 e^2\right )\right ) \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x} \sqrt [3]{d e-c f}}{\sqrt{3} \sqrt [3]{c+d x} \sqrt [3]{b e-a f}}+\frac{1}{\sqrt{3}}\right )}{3 \sqrt{3} (b e-a f)^{7/3} (d e-c f)^{8/3}}-\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x} (-5 a d f-4 b c f+9 b d e)}{6 (e+f x) (b e-a f)^2 (d e-c f)^2}-\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x}}{2 (e+f x)^2 (b e-a f) (d e-c f)} \]

[Out]

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

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Rubi [A]  time = 1.75664, antiderivative size = 477, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154 \[ \frac{\log (e+f x) \left (5 a^2 d^2 f^2-2 a b d f (6 d e-c f)+b^2 \left (2 c^2 f^2-6 c d e f+9 d^2 e^2\right )\right )}{18 (b e-a f)^{7/3} (d e-c f)^{8/3}}-\frac{\left (5 a^2 d^2 f^2-2 a b d f (6 d e-c f)+b^2 \left (2 c^2 f^2-6 c d e f+9 d^2 e^2\right )\right ) \log \left (\frac{\sqrt [3]{a+b x} \sqrt [3]{d e-c f}}{\sqrt [3]{b e-a f}}-\sqrt [3]{c+d x}\right )}{6 (b e-a f)^{7/3} (d e-c f)^{8/3}}-\frac{\left (5 a^2 d^2 f^2-2 a b d f (6 d e-c f)+b^2 \left (2 c^2 f^2-6 c d e f+9 d^2 e^2\right )\right ) \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x} \sqrt [3]{d e-c f}}{\sqrt{3} \sqrt [3]{c+d x} \sqrt [3]{b e-a f}}+\frac{1}{\sqrt{3}}\right )}{3 \sqrt{3} (b e-a f)^{7/3} (d e-c f)^{8/3}}-\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x} (-5 a d f-4 b c f+9 b d e)}{6 (e+f x) (b e-a f)^2 (d e-c f)^2}-\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x}}{2 (e+f x)^2 (b e-a f) (d e-c f)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out

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Mathematica [C]  time = 1.2835, size = 244, normalized size = 0.51 \[ \frac{(a+b x)^{2/3} \left ((e+f x)^2 \left (5 a^2 d^2 f^2+2 a b d f (c f-6 d e)+b^2 \left (2 c^2 f^2-6 c d e f+9 d^2 e^2\right )\right ) \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{2/3} \, _2F_1\left (\frac{2}{3},\frac{2}{3};\frac{5}{3};\frac{(c f-d e) (a+b x)}{(b c-a d) (e+f x)}\right )-f (c+d x) (b e-a f) ((e+f x) (-5 a d f-4 b c f+9 b d e)+3 (b e-a f) (d e-c f))\right )}{6 (c+d x)^{2/3} (e+f x)^2 (b e-a f)^3 (d e-c f)^2} \]

Antiderivative was successfully verified.

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

[Out]

((a + b*x)^(2/3)*(-(f*(b*e - a*f)*(c + d*x)*(3*(b*e - a*f)*(d*e - c*f) + (9*b*d*
e - 4*b*c*f - 5*a*d*f)*(e + f*x))) + (5*a^2*d^2*f^2 + 2*a*b*d*f*(-6*d*e + c*f) +
b^2*(9*d^2*e^2 - 6*c*d*e*f + 2*c^2*f^2))*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*
(e + f*x)))^(2/3)*(e + f*x)^2*Hypergeometric2F1[2/3, 2/3, 5/3, ((-(d*e) + c*f)*(
a + b*x))/((b*c - a*d)*(e + f*x))]))/(6*(b*e - a*f)^3*(d*e - c*f)^2*(c + d*x)^(2
/3)*(e + f*x)^2)

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Maple [F]  time = 0.112, size = 0, normalized size = 0. \[ \int{\frac{1}{ \left ( fx+e \right ) ^{3}}{\frac{1}{\sqrt [3]{bx+a}}} \left ( dx+c \right ) ^{-{\frac{2}{3}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [A]  time = 0.304093, size = 2020, normalized size = 4.23 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/54*sqrt(3)*(3*sqrt(3)*(-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*
c^2 + 2*a*c*d)*e*f^2)^(1/3)*(12*b*d*e^2*f + 3*a*c*f^3 - (7*b*c + 8*a*d)*e*f^2 +
(9*b*d*e*f^2 - (4*b*c + 5*a*d)*f^3)*x)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + sqrt(3)
*(9*b^2*d^2*e^4 - 6*(b^2*c*d + 2*a*b*d^2)*e^3*f + (2*b^2*c^2 + 2*a*b*c*d + 5*a^2
*d^2)*e^2*f^2 + (9*b^2*d^2*e^2*f^2 - 6*(b^2*c*d + 2*a*b*d^2)*e*f^3 + (2*b^2*c^2
+ 2*a*b*c*d + 5*a^2*d^2)*f^4)*x^2 + 2*(9*b^2*d^2*e^3*f - 6*(b^2*c*d + 2*a*b*d^2)
*e^2*f^2 + (2*b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*e*f^3)*x)*log((a*d^2*e^2 - 2*a*c*
d*e*f + a*c^2*f^2 - (-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 +
2*a*c*d)*e*f^2)^(1/3)*(d*e - c*f)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (b*d^2*e^2
- 2*b*c*d*e*f + b*c^2*f^2)*x + (-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f
- (b*c^2 + 2*a*c*d)*e*f^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^4 - 6*(b^2*c*d + 2*a*b*d^2)*e^3*f + (2*b^2*c^2 + 2*a*b*c*
d + 5*a^2*d^2)*e^2*f^2 + (9*b^2*d^2*e^2*f^2 - 6*(b^2*c*d + 2*a*b*d^2)*e*f^3 + (2
*b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*f^4)*x^2 + 2*(9*b^2*d^2*e^3*f - 6*(b^2*c*d + 2
*a*b*d^2)*e^2*f^2 + (2*b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*e*f^3)*x)*log((a*d*e - a
*c*f + (b*d*e - b*c*f)*x + (-b*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (
b*c^2 + 2*a*c*d)*e*f^2)^(1/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b*x + a)) + 6*(9
*b^2*d^2*e^4 - 6*(b^2*c*d + 2*a*b*d^2)*e^3*f + (2*b^2*c^2 + 2*a*b*c*d + 5*a^2*d^
2)*e^2*f^2 + (9*b^2*d^2*e^2*f^2 - 6*(b^2*c*d + 2*a*b*d^2)*e*f^3 + (2*b^2*c^2 + 2
*a*b*c*d + 5*a^2*d^2)*f^4)*x^2 + 2*(9*b^2*d^2*e^3*f - 6*(b^2*c*d + 2*a*b*d^2)*e^
2*f^2 + (2*b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*e*f^3)*x)*arctan(-1/3*(2*sqrt(3)*(-b
*d^2*e^3 + a*c^2*f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(1/3)*
(b*x + a)^(2/3)*(d*x + c)^(1/3) - sqrt(3)*(a*d*e - a*c*f + (b*d*e - b*c*f)*x))/(
a*d*e - a*c*f + (b*d*e - b*c*f)*x)))/((b^2*d^2*e^6 + a^2*c^2*e^2*f^4 - 2*(b^2*c*
d + a*b*d^2)*e^5*f + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*e^4*f^2 - 2*(a*b*c^2 + a^2*
c*d)*e^3*f^3 + (b^2*d^2*e^4*f^2 + a^2*c^2*f^6 - 2*(b^2*c*d + a*b*d^2)*e^3*f^3 +
(b^2*c^2 + 4*a*b*c*d + a^2*d^2)*e^2*f^4 - 2*(a*b*c^2 + a^2*c*d)*e*f^5)*x^2 + 2*(
b^2*d^2*e^5*f + a^2*c^2*e*f^5 - 2*(b^2*c*d + a*b*d^2)*e^4*f^2 + (b^2*c^2 + 4*a*b
*c*d + a^2*d^2)*e^3*f^3 - 2*(a*b*c^2 + a^2*c*d)*e^2*f^4)*x)*(-b*d^2*e^3 + a*c^2*
f^3 + (2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)^(1/3))

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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