3.1430 \(\int \frac{\left (a+b x+c x^2\right )^p}{b d+2 c d x} \, dx\)

Optimal. Leaf size=63 \[ \frac{\left (a+b x+c x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;1-\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d (p+1) \left (b^2-4 a c\right )} \]

[Out]

((a + b*x + c*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 - (b + 2*c*x)^2/
(b^2 - 4*a*c)])/((b^2 - 4*a*c)*d*(1 + p))

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Rubi [A]  time = 0.214855, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125 \[ \frac{\left (a+b x+c x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;1-\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d (p+1) \left (b^2-4 a c\right )} \]

Antiderivative was successfully verified.

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

[Out]

((a + b*x + c*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 - (b + 2*c*x)^2/
(b^2 - 4*a*c)])/((b^2 - 4*a*c)*d*(1 + p))

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Rubi in Sympy [A]  time = 32.9909, size = 60, normalized size = 0.95 \[ \frac{\left (a - \frac{b^{2}}{4 c} + \frac{\left (b + 2 c x\right )^{2}}{4 c}\right )^{p + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, p + 1 \\ p + 2 \end{matrix}\middle |{\frac{\left (b + 2 c x\right )^{2}}{4 a c - b^{2}} + 1} \right )}}{d \left (p + 1\right ) \left (- 4 a c + b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

(a - b**2/(4*c) + (b + 2*c*x)**2/(4*c))**(p + 1)*hyper((1, p + 1), (p + 2,), (b
+ 2*c*x)**2/(4*a*c - b**2) + 1)/(d*(p + 1)*(-4*a*c + b**2))

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Mathematica [A]  time = 0.243584, size = 82, normalized size = 1.3 \[ \frac{4^{-p-1} (a+x (b+c x))^p \left (\frac{c (a+x (b+c x))}{(b+2 c x)^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac{b^2-4 a c}{(b+2 c x)^2}\right )}{c d p} \]

Antiderivative was successfully verified.

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

[Out]

(4^(-1 - p)*(a + x*(b + c*x))^p*HypergeometricPFQ[{-p, -p}, {1 - p}, (b^2 - 4*a*
c)/(b + 2*c*x)^2])/(c*d*p*((c*(a + x*(b + c*x)))/(b + 2*c*x)^2)^p)

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Maple [F]  time = 0.144, size = 0, normalized size = 0. \[ \int{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{p}}{2\,cdx+bd}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

int((c*x^2+b*x+a)^p/(2*c*d*x+b*d),x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((c*x^2 + b*x + a)^p/(2*c*d*x + b*d), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (c x^{2} + b x + a\right )}^{p}}{2 \, c d x + b d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((c*x^2 + b*x + a)^p/(2*c*d*x + b*d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Integral((a + b*x + c*x**2)**p/(b + 2*c*x), x)/d

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integrate((c*x^2 + b*x + a)^p/(2*c*d*x + b*d), x)