### 3.1441 $$\int \frac{(a+b x+c x^2)^p}{b d+2 c d x} \, dx$$

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 = 0.102314, 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, Rules used = {694, 266, 65} $\frac{\left (a+b x+c x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;1-\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d (p+1) \left (b^2-4 a c\right )}$

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))

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

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

Mathematica [A]  time = 0.0269272, size = 64, normalized size = 1.02 $\frac{(a+x (b+c x))^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{d (p+1) \left (b^2-4 a c\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [F]  time = 1.157, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{p}}{2\,cdx+bd}}\, 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),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. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}^{p}}{2 \, c d x + b d}\,{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),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. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{2} + b x + a\right )}^{p}}{2 \, c d x + b d}, 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),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. \begin{align*} \frac{\int \frac{\left (a + b x + c x^{2}\right )^{p}}{b + 2 c x}\, dx}{d} \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),x)

[Out]

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

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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}}{2 \, c d x + b d}\,{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),x, algorithm="giac")

[Out]

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