\(\int \frac {(a+b x+c x^2)^p}{(b d+2 c d x)^6} \, dx\) [1446]

Optimal result

Integrand size = 24, antiderivative size = 90 \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\frac {2 \left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b+2 c x)^2}{4 c}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,-\frac {3}{2}+p,-\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{5 \left (b^2-4 a c\right ) d^6 (b+2 c x)^5} \]

[Out]

2/5*(a-1/4*b^2/c+1/4*(2*c*x+b)^2/c)^(p+1)*hypergeom([1, -3/2+p],[-3/2],(2*c*x+b)^2/(-4*a*c+b^2))/(-4*a*c+b^2)/
d^6/(2*c*x+b)^5

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {708, 372, 371} \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=-\frac {\left (a+b x+c x^2\right )^p \left (1-\frac {(b+2 c x)^2}{b^2-4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-p,-\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{10 c d^6 (b+2 c x)^5} \]

[In]

Int[(a + b*x + c*x^2)^p/(b*d + 2*c*d*x)^6,x]

[Out]

-1/10*((a + b*x + c*x^2)^p*Hypergeometric2F1[-5/2, -p, -3/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c*d^6*(b + 2*c*x)^
5*(1 - (b + 2*c*x)^2/(b^2 - 4*a*c))^p)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 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^6} \, 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 ) \text {Subst}\left (\int \frac {\left (1+\frac {x^2}{4 \left (a-\frac {b^2}{4 c}\right ) c d^2}\right )^p}{x^6} \, 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 {5}{2},-p;-\frac {3}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{5 c d^6 (b+2 c x)^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=-\frac {2^{-1-2 p} (a+x (b+c x))^p \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-p,-\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{5 c d^6 (b+2 c x)^5} \]

[In]

Integrate[(a + b*x + c*x^2)^p/(b*d + 2*c*d*x)^6,x]

[Out]

-1/5*(2^(-1 - 2*p)*(a + x*(b + c*x))^p*Hypergeometric2F1[-5/2, -p, -3/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c*d^6*
(b + 2*c*x)^5*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^p)

Maple [F]

\[\int \frac {\left (c \,x^{2}+b x +a \right )^{p}}{\left (2 c d x +b d \right )^{6}}d x\]

[In]

int((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^6,x)

[Out]

int((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^6,x)

Fricas [F]

\[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (2 \, c d x + b d\right )}^{6}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^6,x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^p/(64*c^6*d^6*x^6 + 192*b*c^5*d^6*x^5 + 240*b^2*c^4*d^6*x^4 + 160*b^3*c^3*d^6*x^3 +
60*b^4*c^2*d^6*x^2 + 12*b^5*c*d^6*x + b^6*d^6), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\text {Timed out} \]

[In]

integrate((c*x**2+b*x+a)**p/(2*c*d*x+b*d)**6,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (2 \, c d x + b d\right )}^{6}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^6,x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^p/(2*c*d*x + b*d)^6, x)

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (2 \, c d x + b d\right )}^{6}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^6,x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^p/(2*c*d*x + b*d)^6, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (b\,d+2\,c\,d\,x\right )}^6} \,d x \]

[In]

int((a + b*x + c*x^2)^p/(b*d + 2*c*d*x)^6,x)

[Out]

int((a + b*x + c*x^2)^p/(b*d + 2*c*d*x)^6, x)