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

Optimal. Leaf size=287 \[ \frac{(d+e x)^{m+2} \left (2 c e g (a e g-b (2 d g+e f))+b^2 e^2 g^2+c^2 \left (3 d^2 g^2+2 d e f g+e^2 f^2\right )\right )}{e^4 g^3 (m+2)}+\frac{(d+e x)^{m+1} (b e g-c (d g+e f)) \left (e g (2 a e g-b (d g+e f))+c \left (d^2 g^2+e^2 f^2\right )\right )}{e^4 g^4 (m+1)}+\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2 \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right )}{g^4 (m+1) (e f-d g)}-\frac{c (d+e x)^{m+3} (-2 b e g+3 c d g+c e f)}{e^4 g^2 (m+3)}+\frac{c^2 (d+e x)^{m+4}}{e^4 g (m+4)} \]

[Out]

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

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Rubi [A]  time = 0.865186, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {951, 1620, 68} \[ \frac{(d+e x)^{m+2} \left (2 c e g (a e g-b (2 d g+e f))+b^2 e^2 g^2+c^2 \left (3 d^2 g^2+2 d e f g+e^2 f^2\right )\right )}{e^4 g^3 (m+2)}+\frac{(d+e x)^{m+1} (b e g-c (d g+e f)) \left (e g (2 a e g-b (d g+e f))+c \left (d^2 g^2+e^2 f^2\right )\right )}{e^4 g^4 (m+1)}+\frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2 \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right )}{g^4 (m+1) (e f-d g)}-\frac{c (d+e x)^{m+3} (-2 b e g+3 c d g+c e f)}{e^4 g^2 (m+3)}+\frac{c^2 (d+e x)^{m+4}}{e^4 g (m+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

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

Mathematica [A]  time = 0.447522, size = 265, normalized size = 0.92 \[ \frac{(d+e x)^{m+1} \left (\frac{g (d+e x) \left (2 c e g (a e g-b (2 d g+e f))+b^2 e^2 g^2+c^2 \left (3 d^2 g^2+2 d e f g+e^2 f^2\right )\right )}{e^4 (m+2)}-\frac{(-b e g+c d g+c e f) \left (e g (2 a e g-b (d g+e f))+c \left (d^2 g^2+e^2 f^2\right )\right )}{e^4 (m+1)}+\frac{\left (g (a g-b f)+c f^2\right )^2 \, _2F_1\left (1,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )}{(m+1) (e f-d g)}-\frac{c g^2 (d+e x)^2 (-2 b e g+3 c d g+c e f)}{e^4 (m+3)}+\frac{c^2 g^3 (d+e x)^3}{e^4 (m+4)}\right )}{g^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.641, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{2} \left ( ex+d \right ) ^{m}}{gx+f}}\, 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),x)

[Out]

int((e*x+d)^m*(c*x^2+b*x+a)^2/(g*x+f),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}}{g x + f}\,{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),x, algorithm="maxima")

[Out]

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

[Out]

Integral((d + e*x)**m*(a + b*x + c*x**2)**2/(f + g*x), 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}}{g x + f}\,{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),x, algorithm="giac")

[Out]

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