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

Optimal. Leaf size=245 \[ \frac{(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right ) \left (c \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )-e g m (a e g (1-m)-b (2 d g-e f (m+1)))\right )}{2 g^2 (m+1) (e f-d g)^3}+\frac{(d+e x)^{m+1} (g (a e g (1-m)-b (2 d g-e f (m+1)))+c f (4 d g-e f (m+3)))}{2 g^2 (f+g x) (e f-d g)^2}+\frac{(d+e x)^{m+1} \left (a+\frac{f (c f-b g)}{g^2}\right )}{2 (f+g x)^2 (e f-d g)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.317472, antiderivative size = 243, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {949, 78, 68} \[ \frac{(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right ) \left (e g m (-a e g (1-m)+2 b d g-b e f (m+1))+c \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )\right )}{2 g^2 (m+1) (e f-d g)^3}-\frac{(d+e x)^{m+1} (g (-a e g (1-m)+2 b d g-b e f (m+1))-c f (4 d g-e f (m+3)))}{2 g^2 (f+g x) (e f-d g)^2}+\frac{(d+e x)^{m+1} \left (a+\frac{f (c f-b g)}{g^2}\right )}{2 (f+g x)^2 (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + (f*(c*f - b*g))/g^2)*(d + e*x)^(1 + m))/(2*(e*f - d*g)*(f + g*x)^2) - ((g*(2*b*d*g - a*e*g*(1 - m) - b*e
*f*(1 + m)) - c*f*(4*d*g - e*f*(3 + m)))*(d + e*x)^(1 + m))/(2*g^2*(e*f - d*g)^2*(f + g*x)) + ((e*g*m*(2*b*d*g
- a*e*g*(1 - m) - b*e*f*(1 + m)) + c*(2*d^2*g^2 - 4*d*e*f*g*(1 + m) + e^2*f^2*(2 + 3*m + m^2)))*(d + e*x)^(1
+ m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((g*(d + e*x))/(e*f - d*g))])/(2*g^2*(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 78

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

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

Mathematica [A]  time = 0.16796, size = 157, normalized size = 0.64 \[ -\frac{(d+e x)^{m+1} \left (e \left (e \left (g (a g-b f)+c f^2\right ) \, _2F_1\left (3,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )-(2 c f-b g) (e f-d g) \, _2F_1\left (2,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )\right )+c (e f-d g)^2 \, _2F_1\left (1,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )\right )}{g^2 (m+1) (d g-e f)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.691, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) \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)/(g*x+f)^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}{\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)/(g*x+f)^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c x^{2} + b x + a\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)/(g*x+f)^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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 )}{\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)/(g*x+f)**3,x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x + a\right )}{\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)/(g*x+f)^3,x, algorithm="giac")

[Out]

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