\(\int \frac {(a+b x+c x^2)^p}{b d+2 c d x} \, dx\) [1441]

Optimal result

Integrand size = 24, antiderivative size = 63 \[ \int \frac {\left (a+b x+c x^2\right )^p}{b d+2 c d x} \, dx=\frac {\left (a+b x+c x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1-\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {708, 272, 67} \[ \int \frac {\left (a+b x+c x^2\right )^p}{b d+2 c d x} \, dx=\frac {\left (a+b x+c x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,1-\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{d (p+1) \left (b^2-4 a c\right )} \]

[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 67

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

Rule 272

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 708

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {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 {\text {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] (verified)

Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x+c x^2\right )^p}{b d+2 c d x} \, dx=\frac {(a+x (b+c x))^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {4 c (a+x (b+c x))}{-b^2+4 a c}\right )}{\left (b^2-4 a c\right ) d (1+p)} \]

[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))

Maple [F]

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

[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)

Fricas [F]

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

[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)

Sympy [F]

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

[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

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^p}{b d+2 c d x} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{b\,d+2\,c\,d\,x} \,d x \]

[In]

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

[Out]

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