3.1411 \(\int \frac{\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{16/3}} \, dx\)

Optimal. Leaf size=99 \[ \frac{3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \, _2F_1\left (-\frac{13}{6},-\frac{4}{3};-\frac{7}{6};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{104 c^2 d \sqrt [3]{1-\frac{(b+2 c x)^2}{b^2-4 a c}} (d (b+2 c x))^{13/3}} \]

[Out]

(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/3)*Hypergeometric2F1[-13/6, -4/3, -7/6, (b
+ 2*c*x)^2/(b^2 - 4*a*c)])/(104*c^2*d*(d*(b + 2*c*x))^(13/3)*(1 - (b + 2*c*x)^2
/(b^2 - 4*a*c))^(1/3))

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Rubi [A]  time = 0.296324, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107 \[ \frac{3 \left (b^2-4 a c\right ) \sqrt [3]{a+b x+c x^2} \, _2F_1\left (-\frac{13}{6},-\frac{4}{3};-\frac{7}{6};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{104 c^2 d \sqrt [3]{1-\frac{(b+2 c x)^2}{b^2-4 a c}} (d (b+2 c x))^{13/3}} \]

Antiderivative was successfully verified.

[In]  Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(16/3),x]

[Out]

(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/3)*Hypergeometric2F1[-13/6, -4/3, -7/6, (b
+ 2*c*x)^2/(b^2 - 4*a*c)])/(104*c^2*d*(d*(b + 2*c*x))^(13/3)*(1 - (b + 2*c*x)^2
/(b^2 - 4*a*c))^(1/3))

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Rubi in Sympy [A]  time = 31.3049, size = 105, normalized size = 1.06 \[ \frac{3 \left (- 4 a c + b^{2}\right ) \sqrt [3]{a - \frac{b^{2}}{4 c} + \frac{\left (b + 2 c x\right )^{2}}{4 c}}{{}_{2}F_{1}\left (\begin{matrix} - \frac{4}{3}, - \frac{13}{6} \\ - \frac{7}{6} \end{matrix}\middle |{- \frac{\left (b + 2 c x\right )^{2}}{4 a c - b^{2}}} \right )}}{104 c^{2} d \left (b d + 2 c d x\right )^{\frac{13}{3}} \sqrt [3]{\frac{\left (b + 2 c x\right )^{2}}{4 a c - b^{2}} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((c*x**2+b*x+a)**(4/3)/(2*c*d*x+b*d)**(16/3),x)

[Out]

3*(-4*a*c + b**2)*(a - b**2/(4*c) + (b + 2*c*x)**2/(4*c))**(1/3)*hyper((-4/3, -1
3/6), (-7/6,), -(b + 2*c*x)**2/(4*a*c - b**2))/(104*c**2*d*(b*d + 2*c*d*x)**(13/
3)*((b + 2*c*x)**2/(4*a*c - b**2) + 1)**(1/3))

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Mathematica [A]  time = 0.525644, size = 171, normalized size = 1.73 \[ \frac{3 \left (5 c (a+x (b+c x)) \left (-15 \left (b^2-4 a c\right ) (b+2 c x)^2+7 \left (b^2-4 a c\right )^2+16 (b+2 c x)^4\right )-8 \sqrt [3]{2} (b+2 c x)^6 \left (-\frac{c (a+x (b+c x))}{b^2-4 a c}\right )^{2/3} \, _2F_1\left (\frac{2}{3},\frac{5}{6};\frac{11}{6};\frac{(b+2 c x)^2}{b^2-4 a c}\right )\right )}{3640 c^3 d \left (b^2-4 a c\right ) (a+x (b+c x))^{2/3} (d (b+2 c x))^{13/3}} \]

Antiderivative was successfully verified.

[In]  Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(16/3),x]

[Out]

(3*(5*c*(a + x*(b + c*x))*(7*(b^2 - 4*a*c)^2 - 15*(b^2 - 4*a*c)*(b + 2*c*x)^2 +
16*(b + 2*c*x)^4) - 8*2^(1/3)*(b + 2*c*x)^6*(-((c*(a + x*(b + c*x)))/(b^2 - 4*a*
c)))^(2/3)*HypergeometricPFQ[{2/3, 5/6}, {11/6}, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/
(3640*c^3*(b^2 - 4*a*c)*d*(d*(b + 2*c*x))^(13/3)*(a + x*(b + c*x))^(2/3))

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Maple [F]  time = 0.156, size = 0, normalized size = 0. \[ \int{1 \left ( c{x}^{2}+bx+a \right ) ^{{\frac{4}{3}}} \left ( 2\,cdx+bd \right ) ^{-{\frac{16}{3}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(16/3),x)

[Out]

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(16/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac{16}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(16/3),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(16/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{{\left (32 \, c^{5} d^{5} x^{5} + 80 \, b c^{4} d^{5} x^{4} + 80 \, b^{2} c^{3} d^{5} x^{3} + 40 \, b^{3} c^{2} d^{5} x^{2} + 10 \, b^{4} c d^{5} x + b^{5} d^{5}\right )}{\left (2 \, c d x + b d\right )}^{\frac{1}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(16/3),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^(4/3)/((32*c^5*d^5*x^5 + 80*b*c^4*d^5*x^4 + 80*b^2*c^
3*d^5*x^3 + 40*b^3*c^2*d^5*x^2 + 10*b^4*c*d^5*x + b^5*d^5)*(2*c*d*x + b*d)^(1/3)
), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x**2+b*x+a)**(4/3)/(2*c*d*x+b*d)**(16/3),x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 1.83554, size = 1, normalized size = 0.01 \[ \mathit{Done} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(16/3),x, algorithm="giac")

[Out]

Done