3.929 \(\int \frac{(d+e x)^m (a+b x+c x^2)^2}{(f+g x)^3} \, dx\)

Optimal. Leaf size=461 \[ \frac{(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right ) \left (-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (m+1))+b^2 \left (-\left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (m+2)+e^2 f^2 \left (m^2+5 m+6\right )\right )\right )+c^2 f^2 \left (12 d^2 g^2-8 d e f g (m+3)+e^2 f^2 \left (m^2+7 m+12\right )\right )\right )}{2 g^4 (m+1) (e f-d g)^3}+\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) (g (a e g (1-m)-b (4 d g-e f (m+3)))+c f (8 d g-e f (m+7)))}{2 g^4 (f+g x) (e f-d g)^2}+\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2}{2 g^4 (f+g x)^2 (e f-d g)}-\frac{c (d+e x)^{m+1} (-2 b e g+c d g+3 c e f)}{e^2 g^4 (m+1)}+\frac{c^2 (d+e x)^{m+2}}{e^2 g^3 (m+2)} \]

[Out]

-((c*(3*c*e*f + c*d*g - 2*b*e*g)*(d + e*x)^(1 + m))/(e^2*g^4*(1 + m))) + (c^2*(d + e*x)^(2 + m))/(e^2*g^3*(2 +
m)) + ((c*f^2 - b*f*g + a*g^2)^2*(d + e*x)^(1 + m))/(2*g^4*(e*f - d*g)*(f + g*x)^2) + ((c*f^2 - b*f*g + a*g^2
)*(c*f*(8*d*g - e*f*(7 + m)) + g*(a*e*g*(1 - m) - b*(4*d*g - e*f*(3 + m))))*(d + e*x)^(1 + m))/(2*g^4*(e*f - d
*g)^2*(f + g*x)) + ((c^2*f^2*(12*d^2*g^2 - 8*d*e*f*g*(3 + m) + e^2*f^2*(12 + 7*m + m^2)) - g^2*(a^2*e^2*g^2*(1
- m)*m - 2*a*b*e*g*m*(2*d*g - e*f*(1 + m)) - b^2*(2*d^2*g^2 - 4*d*e*f*g*(1 + m) + e^2*f^2*(2 + 3*m + m^2))) +
2*c*g*(a*g*(2*d^2*g^2 - 4*d*e*f*g*(1 + m) + e^2*f^2*(2 + 3*m + m^2)) - b*f*(6*d^2*g^2 - 6*d*e*f*g*(2 + m) + e
^2*f^2*(6 + 5*m + m^2))))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((g*(d + e*x))/(e*f - d*g))])/
(2*g^4*(e*f - d*g)^3*(1 + m))

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Rubi [A]  time = 1.48709, antiderivative size = 461, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {949, 1621, 951, 80, 68} \[ \frac{(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right ) \left (-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (m+1))+b^2 \left (-\left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (m+2)+e^2 f^2 \left (m^2+5 m+6\right )\right )\right )+c^2 f^2 \left (12 d^2 g^2-8 d e f g (m+3)+e^2 f^2 \left (m^2+7 m+12\right )\right )\right )}{2 g^4 (m+1) (e f-d g)^3}-\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) (g (-a e g (1-m)+4 b d g-b e f (m+3))-c f (8 d g-e f (m+7)))}{2 g^4 (f+g x) (e f-d g)^2}+\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2}{2 g^4 (f+g x)^2 (e f-d g)}-\frac{c (d+e x)^{m+1} (-2 b e g+c d g+3 c e f)}{e^2 g^4 (m+1)}+\frac{c^2 (d+e x)^{m+2}}{e^2 g^3 (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(3*c*e*f + c*d*g - 2*b*e*g)*(d + e*x)^(1 + m))/(e^2*g^4*(1 + m))) + (c^2*(d + e*x)^(2 + m))/(e^2*g^3*(2 +
m)) + ((c*f^2 - b*f*g + a*g^2)^2*(d + e*x)^(1 + m))/(2*g^4*(e*f - d*g)*(f + g*x)^2) - ((c*f^2 - b*f*g + a*g^2
)*(g*(4*b*d*g - a*e*g*(1 - m) - b*e*f*(3 + m)) - c*f*(8*d*g - e*f*(7 + m)))*(d + e*x)^(1 + m))/(2*g^4*(e*f - d
*g)^2*(f + g*x)) + ((c^2*f^2*(12*d^2*g^2 - 8*d*e*f*g*(3 + m) + e^2*f^2*(12 + 7*m + m^2)) - g^2*(a^2*e^2*g^2*(1
- m)*m - 2*a*b*e*g*m*(2*d*g - e*f*(1 + m)) - b^2*(2*d^2*g^2 - 4*d*e*f*g*(1 + m) + e^2*f^2*(2 + 3*m + m^2))) +
2*c*g*(a*g*(2*d^2*g^2 - 4*d*e*f*g*(1 + m) + e^2*f^2*(2 + 3*m + m^2)) - b*f*(6*d^2*g^2 - 6*d*e*f*g*(2 + m) + e
^2*f^2*(6 + 5*m + m^2))))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((g*(d + e*x))/(e*f - d*g))])/
(2*g^4*(e*f - d*g)^3*(1 + m))

Rule 949

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p,
d + e*x, x]}, Simp[(R*(d + e*x)^(m + 1)*(f + g*x)^(n + 1))/((m + 1)*(e*f - d*g)), x] + Dist[1/((m + 1)*(e*f -
d*g)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /;
FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&& IGtQ[p, 0] && LtQ[m, -1]

Rule 1621

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> With[{Qx = PolynomialQuotient[Px,
a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[(R*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/((m + 1)*
(b*c - a*d)), x] + Dist[1/((m + 1)*(b*c - a*d)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(b*c -
a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[m, -1] && GtQ[Expo
n[Px, x], 2]

Rule 951

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Simp[(c^p*(d + e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m +
n + 2*p + 1)), Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x + c*x^2)^p - c^p*
(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x
] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*
p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx &=\frac{\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}+\frac{\int \frac{(d+e x)^m \left (\frac{c^2 f^3 (2 d g-e f (1+m))-2 c f g (b f-a g) (2 d g-e f (1+m))+g^2 \left (a^2 e g^2 (1-m)+b^2 f (2 d g-e f (1+m))-2 a b g (2 d g-e f (1+m))\right )}{g^4}+\frac{2 (e f-d g) \left (c^2 f^2+b^2 g^2-2 c g (b f-a g)\right ) x}{g^3}-\frac{2 c (c f-2 b g) (e f-d g) x^2}{g^2}-2 c^2 \left (d-\frac{e f}{g}\right ) x^3\right )}{(f+g x)^2} \, dx}{2 (e f-d g)}\\ &=\frac{\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac{\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac{\int \frac{(d+e x)^m \left (\frac{c^2 f^2 \left (6 d^2 g^2-4 d e f g (3+2 m)+e^2 f^2 \left (6+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (4 d^2 g^2-2 d e f g (4+3 m)+e^2 f^2 \left (4+5 m+m^2\right )\right )\right )}{g^4}-\frac{4 c (c f-b g) (e f-d g)^2 x}{g^3}+\frac{2 c^2 (e f-d g)^2 x^2}{g^2}\right )}{f+g x} \, dx}{2 (e f-d g)^2}\\ &=\frac{c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac{\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac{\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac{\int \frac{(d+e x)^m \left (\frac{e (2+m) \left (c^2 f \left (10 d^2 e f g^2-2 d^3 g^3-2 d e^2 f^2 g (7+4 m)+e^3 f^3 \left (6+7 m+m^2\right )\right )-e g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c e g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (4 d^2 g^2-2 d e f g (4+3 m)+e^2 f^2 \left (4+5 m+m^2\right )\right )\right )\right )}{g^3}-\frac{2 c e (e f-d g)^2 (3 c e f+c d g-2 b e g) (2+m) x}{g^2}\right )}{f+g x} \, dx}{2 e^2 g (e f-d g)^2 (2+m)}\\ &=-\frac{c (3 c e f+c d g-2 b e g) (d+e x)^{1+m}}{e^2 g^4 (1+m)}+\frac{c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac{\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac{\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac{\left (c^2 f^2 \left (12 d^2 g^2-8 d e f g (3+m)+e^2 f^2 \left (12+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (2+m)+e^2 f^2 \left (6+5 m+m^2\right )\right )\right )\right ) \int \frac{(d+e x)^m}{f+g x} \, dx}{2 g^4 (e f-d g)^2}\\ &=-\frac{c (3 c e f+c d g-2 b e g) (d+e x)^{1+m}}{e^2 g^4 (1+m)}+\frac{c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac{\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac{\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac{\left (c^2 f^2 \left (12 d^2 g^2-8 d e f g (3+m)+e^2 f^2 \left (12+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (2+m)+e^2 f^2 \left (6+5 m+m^2\right )\right )\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{g (d+e x)}{e f-d g}\right )}{2 g^4 (e f-d g)^3 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.411693, size = 257, normalized size = 0.56 \[ \frac{(d+e x)^{m+1} \left (\frac{\left (2 c g (a g-3 b f)+b^2 g^2+6 c^2 f^2\right ) \, _2F_1\left (1,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )}{(m+1) (e f-d g)}+\frac{e^2 \left (g (a g-b f)+c f^2\right )^2 \, _2F_1\left (3,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )}{(m+1) (e f-d g)^3}-\frac{2 e (2 c f-b g) \left (g (a g-b f)+c f^2\right ) \, _2F_1\left (2,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )}{(m+1) (e f-d g)^2}-\frac{c (-2 b e g+c d g+3 c e f)}{e^2 (m+1)}+\frac{c^2 g (d+e x)}{e^2 (m+2)}\right )}{g^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*(-((c*(3*c*e*f + c*d*g - 2*b*e*g))/(e^2*(1 + m))) + (c^2*g*(d + e*x))/(e^2*(2 + m)) + ((6*c
^2*f^2 + b^2*g^2 + 2*c*g*(-3*b*f + a*g))*Hypergeometric2F1[1, 1 + m, 2 + m, (g*(d + e*x))/(-(e*f) + d*g)])/((e
*f - d*g)*(1 + m)) - (2*e*(2*c*f - b*g)*(c*f^2 + g*(-(b*f) + a*g))*Hypergeometric2F1[2, 1 + m, 2 + m, (g*(d +
e*x))/(-(e*f) + d*g)])/((e*f - d*g)^2*(1 + m)) + (e^2*(c*f^2 + g*(-(b*f) + a*g))^2*Hypergeometric2F1[3, 1 + m,
2 + m, (g*(d + e*x))/(-(e*f) + d*g)])/((e*f - d*g)^3*(1 + m))))/g^4

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Maple [F]  time = 1.459, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{2} \left ( ex+d \right ) ^{m}}{ \left ( gx+f \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x+d)^m*(c*x^2+b*x+a)^2/(g*x+f)^3,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 )}^{2}{\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*(e*x + d)^m/(g^3*x^3 + 3*f*g^2*x^2 + 3*f^2*
g*x + f^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m*(c*x**2+b*x+a)**2/(g*x+f)**3,x)

[Out]

Integral((d + e*x)**m*(a + b*x + c*x**2)**2/(f + g*x)**3, x)

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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 )}^{2}{\left (e x + d\right )}^{m}}{{\left (g x + f\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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