3.1444 \(\int \frac{(a+b x+c x^2)^p}{(b d+2 c d x)^4} \, dx\)

Optimal. Leaf size=90 \[ \frac{2 \left (\frac{1}{4} \left (4 a-\frac{b^2}{c}\right )+\frac{(b+2 c x)^2}{4 c}\right )^{p+1} \, _2F_1\left (1,p-\frac{1}{2};-\frac{1}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{3 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3} \]

[Out]

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

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Rubi [A]  time = 0.0682533, antiderivative size = 85, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {694, 365, 364} \[ -\frac{\left (a+b x+c x^2\right )^p \left (1-\frac{(b+2 c x)^2}{b^2-4 a c}\right )^{-p} \, _2F_1\left (-\frac{3}{2},-p;-\frac{1}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{6 c d^4 (b+2 c x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}\right )^p}{x^4} \, dx,x,b d+2 c d x\right )}{2 c d}\\ &=\frac{\left (2^{-1+2 p} \left (a+b x+c x^2\right )^p \left (4+\frac{(b d+2 c d x)^2}{\left (a-\frac{b^2}{4 c}\right ) c d^2}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{x^2}{4 \left (a-\frac{b^2}{4 c}\right ) c d^2}\right )^p}{x^4} \, dx,x,b d+2 c d x\right )}{c d}\\ &=-\frac{2^{-1+2 p} \left (a+b x+c x^2\right )^p \left (4-\frac{4 (b+2 c x)^2}{b^2-4 a c}\right )^{-p} \, _2F_1\left (-\frac{3}{2},-p;-\frac{1}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{3 c d^4 (b+2 c x)^3}\\ \end{align*}

Mathematica [A]  time = 0.0436174, size = 92, normalized size = 1.02 \[ -\frac{2^{-2 p-1} (a+x (b+c x))^p \left (\frac{c (a+x (b+c x))}{4 a c-b^2}\right )^{-p} \, _2F_1\left (-\frac{3}{2},-p;-\frac{1}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{3 c d^4 (b+2 c x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.174, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{p}}{ \left ( 2\,cdx+bd \right ) ^{4}}}\, dx \end{align*}

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)^4,x)

[Out]

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

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

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)^4,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{2} + b x + a\right )}^{p}}{16 \, c^{4} d^{4} x^{4} + 32 \, b c^{3} d^{4} x^{3} + 24 \, b^{2} c^{2} d^{4} x^{2} + 8 \, b^{3} c d^{4} x + b^{4} d^{4}}, x\right ) \end{align*}

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)^4,x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^p/(16*c^4*d^4*x^4 + 32*b*c^3*d^4*x^3 + 24*b^2*c^2*d^4*x^2 + 8*b^3*c*d^4*x + b^4*d^4
), x)

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

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)**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{p}}{{\left (2 \, c d x + b d\right )}^{4}}\,{d x} \end{align*}

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)^4,x, algorithm="giac")

[Out]

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