3.1428 \(\int \frac{(b d+2 c d x)^m}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=70 \[ -\frac{32 c^2 (d (b+2 c x))^{m+1} \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d (m+1) \left (b^2-4 a c\right )^3} \]

[Out]

(-32*c^2*(d*(b + 2*c*x))^(1 + m)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/((b^
2 - 4*a*c)^3*d*(1 + m))

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Rubi [A]  time = 0.0516068, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {694, 364} \[ -\frac{32 c^2 (d (b+2 c x))^{m+1} \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d (m+1) \left (b^2-4 a c\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*c^2*(d*(b + 2*c*x))^(1 + m)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/((b^
2 - 4*a*c)^3*d*(1 + m))

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(b d+2 c d x)^m}{\left (a+b x+c x^2\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^m}{\left (a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}\right )^3} \, dx,x,b d+2 c d x\right )}{2 c d}\\ &=-\frac{32 c^2 (d (b+2 c x))^{1+m} \, _2F_1\left (3,\frac{1+m}{2};\frac{3+m}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right )^3 d (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0440508, size = 71, normalized size = 1.01 \[ -\frac{32 c^2 (b+2 c x) (d (b+2 c x))^m \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{(m+1) \left (b^2-4 a c\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*c^2*(b + 2*c*x)*(d*(b + 2*c*x))^m*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, (b + 2*c*x)^2/(b^2 - 4*a*c)]
)/((b^2 - 4*a*c)^3*(1 + m))

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Maple [F]  time = 1.237, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( 2\,cdx+bd \right ) ^{m}}{ \left ( c{x}^{2}+bx+a \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c d x + b d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (2 \, c d x + b d\right )}^{m}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \,{\left (b^{2} c + a c^{2}\right )} x^{4} + 3 \, a^{2} b x +{\left (b^{3} + 6 \, a b c\right )} x^{3} + a^{3} + 3 \,{\left (a b^{2} + a^{2} c\right )} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((2*c*d*x + b*d)^m/(c^3*x^6 + 3*b*c^2*x^5 + 3*(b^2*c + a*c^2)*x^4 + 3*a^2*b*x + (b^3 + 6*a*b*c)*x^3 +
a^3 + 3*(a*b^2 + a^2*c)*x^2), 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((2*c*d*x+b*d)**m/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c d x + b d\right )}^{m}}{{\left (c x^{2} + b x + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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