3.2569 $$\int \frac{(a+b x+c x^2)^p}{(d+e x)^3} \, dx$$

Optimal. Leaf size=200 $-\frac{2^{2 p-1} \left (a+b x+c x^2\right )^p \left (\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} F_1\left (2 (1-p);-p,-p;3-2 p;\frac{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e (1-p) (d+e x)^2}$

[Out]

-((2^(-1 + 2*p)*(a + b*x + c*x^2)^p*AppellF1[2*(1 - p), -p, -p, 3 - 2*p, (2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)/(
2*c*(d + e*x)), (2*d - ((b + Sqrt[b^2 - 4*a*c])*e)/c)/(2*(d + e*x))])/(e*(1 - p)*((e*(b - Sqrt[b^2 - 4*a*c] +
2*c*x))/(c*(d + e*x)))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^p*(d + e*x)^2))

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Rubi [A]  time = 0.114028, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.1, Rules used = {758, 133} $-\frac{2^{2 p-1} \left (a+b x+c x^2\right )^p \left (\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} F_1\left (2 (1-p);-p,-p;3-2 p;\frac{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e (1-p) (d+e x)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((2^(-1 + 2*p)*(a + b*x + c*x^2)^p*AppellF1[2*(1 - p), -p, -p, 3 - 2*p, (2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)/(
2*c*(d + e*x)), (2*d - ((b + Sqrt[b^2 - 4*a*c])*e)/c)/(2*(d + e*x))])/(e*(1 - p)*((e*(b - Sqrt[b^2 - 4*a*c] +
2*c*x))/(c*(d + e*x)))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^p*(d + e*x)^2))

Rule 758

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

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^p}{(d+e x)^3} \, dx &=-\frac{\left (2^{2 p} \left (\frac{1}{d+e x}\right )^{2 p} \left (\frac{e \left (b-\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac{e \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (a+b x+c x^2\right )^p\right ) \operatorname{Subst}\left (\int x^{3-2 (1+p)} \left (1-\frac{1}{2} \left (2 d-\frac{\left (b-\sqrt{b^2-4 a c}\right ) e}{c}\right ) x\right )^p \left (1-\frac{1}{2} \left (2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}\right ) x\right )^p \, dx,x,\frac{1}{d+e x}\right )}{e}\\ &=-\frac{2^{-1+2 p} \left (\frac{e \left (b-\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac{e \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (2 (1-p);-p,-p;3-2 p;\frac{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e (1-p) (d+e x)^2}\\ \end{align*}

Mathematica [A]  time = 0.385815, size = 193, normalized size = 0.96 $\frac{2^{2 p-1} (a+x (b+c x))^p \left (\frac{e \left (-\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} F_1\left (2-2 p;-p,-p;3-2 p;\frac{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c (d+e x)},\frac{2 c d-b e+\sqrt{b^2-4 a c} e}{2 c d+2 c e x}\right )}{e (p-1) (d+e x)^2}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(-1 + 2*p)*(a + x*(b + c*x))^p*AppellF1[2 - 2*p, -p, -p, 3 - 2*p, (2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)/(2*c*
(d + e*x)), (2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)/(2*c*d + 2*c*e*x)])/(e*(-1 + p)*((e*(b - Sqrt[b^2 - 4*a*c] + 2
*c*x))/(c*(d + e*x)))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^p*(d + e*x)^2)

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Maple [F]  time = 1.32, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{p}}{ \left ( ex+d \right ) ^{3}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((c*x^2+b*x+a)^p/(e*x+d)^3,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 (e x + d\right )}^{3}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^p/(e*x+d)^3,x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^p/(e*x + d)^3, 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}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^p/(e*x+d)^3,x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^p/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), 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/(e*x+d)**3,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 (e x + d\right )}^{3}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^p/(e*x+d)^3,x, algorithm="giac")

[Out]

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