3.1401 \(\int \frac{\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{11/3}} \, dx\)

Optimal. Leaf size=468 \[ -\frac{3 \sqrt [3]{a+b x+c x^2} (d (b+2 c x))^{4/3}}{16 c^2 d^5 \left (b^2-4 a c\right )}+\frac{9 \left (a+b x+c x^2\right )^{4/3} (d (b+2 c x))^{4/3}}{16 c d^5 \left (b^2-4 a c\right )^2}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{2/3}}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac{\log \left (-\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}-2 \sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}}{\sqrt [3]{a+b x+c x^2}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}+\frac{\log \left (\frac{2 \sqrt [3]{2} c^{2/3} d^{4/3} \left (a+b x+c x^2\right )^{2/3}+2^{2/3} \sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2} (d (b+2 c x))^{2/3}+(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}+\sqrt [3]{c} d^{2/3}}{\sqrt{3} \sqrt [3]{c} d^{2/3}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}} \]

[Out]

(-3*(d*(b + 2*c*x))^(4/3)*(a + b*x + c*x^2)^(1/3))/(16*c^2*(b^2 - 4*a*c)*d^5) +
(9*(d*(b + 2*c*x))^(4/3)*(a + b*x + c*x^2)^(4/3))/(16*c*(b^2 - 4*a*c)^2*d^5) + (
3*(a + b*x + c*x^2)^(7/3))/(4*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(8/3)) - (9*(a + b
*x + c*x^2)^(7/3))/(4*(b^2 - 4*a*c)^2*d^3*(b*d + 2*c*d*x)^(2/3)) - (Sqrt[3]*ArcT
an[(c^(1/3)*d^(2/3) + (2^(1/3)*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))/(
Sqrt[3]*c^(1/3)*d^(2/3))])/(16*2^(2/3)*c^(7/3)*d^(11/3)) - Log[-((2^(1/3)*(d*(b
+ 2*c*x))^(2/3) - 2*c^(1/3)*d^(2/3)*(a + b*x + c*x^2)^(1/3))/(a + b*x + c*x^2)^(
1/3))]/(16*2^(2/3)*c^(7/3)*d^(11/3)) + Log[((d*(b + 2*c*x))^(4/3) + 2^(2/3)*c^(1
/3)*d^(2/3)*(d*(b + 2*c*x))^(2/3)*(a + b*x + c*x^2)^(1/3) + 2*2^(1/3)*c^(2/3)*d^
(4/3)*(a + b*x + c*x^2)^(2/3))/(a + b*x + c*x^2)^(2/3)]/(32*2^(2/3)*c^(7/3)*d^(1
1/3))

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Rubi [A]  time = 2.75073, antiderivative size = 468, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429 \[ -\frac{3 \sqrt [3]{a+b x+c x^2} (d (b+2 c x))^{4/3}}{16 c^2 d^5 \left (b^2-4 a c\right )}+\frac{9 \left (a+b x+c x^2\right )^{4/3} (d (b+2 c x))^{4/3}}{16 c d^5 \left (b^2-4 a c\right )^2}-\frac{9 \left (a+b x+c x^2\right )^{7/3}}{4 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{2/3}}+\frac{3 \left (a+b x+c x^2\right )^{7/3}}{4 d \left (b^2-4 a c\right ) (b d+2 c d x)^{8/3}}-\frac{\log \left (-\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}-2 \sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2}}{\sqrt [3]{a+b x+c x^2}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}}+\frac{\log \left (\frac{2 \sqrt [3]{2} c^{2/3} d^{4/3} \left (a+b x+c x^2\right )^{2/3}+2^{2/3} \sqrt [3]{c} d^{2/3} \sqrt [3]{a+b x+c x^2} (d (b+2 c x))^{2/3}+(d (b+2 c x))^{4/3}}{\left (a+b x+c x^2\right )^{2/3}}\right )}{32\ 2^{2/3} c^{7/3} d^{11/3}}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{\sqrt [3]{2} (d (b+2 c x))^{2/3}}{\sqrt [3]{a+b x+c x^2}}+\sqrt [3]{c} d^{2/3}}{\sqrt{3} \sqrt [3]{c} d^{2/3}}\right )}{16\ 2^{2/3} c^{7/3} d^{11/3}} \]

Antiderivative was successfully verified.

[In]  Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(11/3),x]

[Out]

(-3*(d*(b + 2*c*x))^(4/3)*(a + b*x + c*x^2)^(1/3))/(16*c^2*(b^2 - 4*a*c)*d^5) +
(9*(d*(b + 2*c*x))^(4/3)*(a + b*x + c*x^2)^(4/3))/(16*c*(b^2 - 4*a*c)^2*d^5) + (
3*(a + b*x + c*x^2)^(7/3))/(4*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(8/3)) - (9*(a + b
*x + c*x^2)^(7/3))/(4*(b^2 - 4*a*c)^2*d^3*(b*d + 2*c*d*x)^(2/3)) - (Sqrt[3]*ArcT
an[(c^(1/3)*d^(2/3) + (2^(1/3)*(d*(b + 2*c*x))^(2/3))/(a + b*x + c*x^2)^(1/3))/(
Sqrt[3]*c^(1/3)*d^(2/3))])/(16*2^(2/3)*c^(7/3)*d^(11/3)) - Log[-((2^(1/3)*(d*(b
+ 2*c*x))^(2/3) - 2*c^(1/3)*d^(2/3)*(a + b*x + c*x^2)^(1/3))/(a + b*x + c*x^2)^(
1/3))]/(16*2^(2/3)*c^(7/3)*d^(11/3)) + Log[((d*(b + 2*c*x))^(4/3) + 2^(2/3)*c^(1
/3)*d^(2/3)*(d*(b + 2*c*x))^(2/3)*(a + b*x + c*x^2)^(1/3) + 2*2^(1/3)*c^(2/3)*d^
(4/3)*(a + b*x + c*x^2)^(2/3))/(a + b*x + c*x^2)^(2/3)]/(32*2^(2/3)*c^(7/3)*d^(1
1/3))

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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((c*x**2+b*x+a)**(4/3)/(2*c*d*x+b*d)**(11/3),x)

[Out]

Timed out

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Mathematica [C]  time = 0.642007, size = 167, normalized size = 0.36 \[ \frac{6 \sqrt [3]{2} (b+2 c x)^4 \left (\frac{c (a+x (b+c x))}{4 a c-b^2}\right )^{2/3} \, _2F_1\left (\frac{2}{3},\frac{2}{3};\frac{5}{3};\frac{(b+2 c x)^2}{b^2-4 a c}\right )-24 c \left (a^2 c+a \left (b^2+6 b c x+6 c^2 x^2\right )+x \left (b^3+6 b^2 c x+10 b c^2 x^2+5 c^3 x^3\right )\right )}{128 c^3 d (a+x (b+c x))^{2/3} (d (b+2 c x))^{8/3}} \]

Antiderivative was successfully verified.

[In]  Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(11/3),x]

[Out]

(-24*c*(a^2*c + a*(b^2 + 6*b*c*x + 6*c^2*x^2) + x*(b^3 + 6*b^2*c*x + 10*b*c^2*x^
2 + 5*c^3*x^3)) + 6*2^(1/3)*(b + 2*c*x)^4*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))
^(2/3)*HypergeometricPFQ[{2/3, 2/3}, {5/3}, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(128*c
^3*d*(d*(b + 2*c*x))^(8/3)*(a + x*(b + c*x))^(2/3))

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Maple [F]  time = 0.196, size = 0, normalized size = 0. \[ \int{1 \left ( c{x}^{2}+bx+a \right ) ^{{\frac{4}{3}}} \left ( 2\,cdx+bd \right ) ^{-{\frac{11}{3}}}}\, dx \]

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)^(11/3),x)

[Out]

int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(11/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac{11}{3}}}\,{d x} \]

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)^(11/3),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(11/3), x)

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Fricas [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)^(11/3),x, algorithm="fricas")

[Out]

Timed out

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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)**(11/3),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (c x^{2} + b x + a\right )}^{\frac{4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac{11}{3}}}\,{d x} \]

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)^(11/3),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(11/3), x)