3.1445 \(\int \frac{(a+b x+c x^2)^p}{(b d+2 c d x)^5} \, dx\)

Optimal. Leaf size=63 \[ \frac{\left (a+b x+c x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;1-\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d^5 (p+1) \left (b^2-4 a c\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.105499, 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} \[ \frac{\left (a+b x+c x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;1-\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d^5 (p+1) \left (b^2-4 a c\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^p/(b*d + 2*c*d*x)^5,x]

[Out]

((a + b*x + c*x^2)^(1 + p)*Hypergeometric2F1[3, 1 + p, 2 + p, 1 - (b + 2*c*x)^2/(b^2 - 4*a*c)])/((b^2 - 4*a*c)
^3*d^5*(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)^5} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}\right )^p}{x^5} \, 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^3} \, dx,x,(b d+2 c d x)^2\right )}{4 c d}\\ &=\frac{(a+x (b+c x))^{1+p} \, _2F_1\left (3,1+p;2+p;1-\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right )^3 d^5 (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0430908, size = 64, normalized size = 1.02 \[ \frac{(a+x (b+c x))^{p+1} \, _2F_1\left (3,p+1;p+2;\frac{4 c (a+x (b+c x))}{4 a c-b^2}\right )}{d^5 (p+1) \left (b^2-4 a c\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^p/(b*d + 2*c*d*x)^5,x]

[Out]

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

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Maple [F]  time = 1.23, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{p}}{ \left ( 2\,cdx+bd \right ) ^{5}}}\, 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)^5,x)

[Out]

int((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^5,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 )}^{5}}\,{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)^5,x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^p/(2*c*d*x + b*d)^5, 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}}{32 \, c^{5} d^{5} x^{5} + 80 \, b c^{4} d^{5} x^{4} + 80 \, b^{2} c^{3} d^{5} x^{3} + 40 \, b^{3} c^{2} d^{5} x^{2} + 10 \, b^{4} c d^{5} x + b^{5} d^{5}}, 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)^5,x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^p/(32*c^5*d^5*x^5 + 80*b*c^4*d^5*x^4 + 80*b^2*c^3*d^5*x^3 + 40*b^3*c^2*d^5*x^2 + 10
*b^4*c*d^5*x + b^5*d^5), 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)**5,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 )}^{5}}\,{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)^5,x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^p/(2*c*d*x + b*d)^5, x)