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

Optimal. Leaf size=157 \[ \frac{(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right ) (c f (2 d g-e f (m+2))-g (a e g m+b (d g-e f (m+1))))}{g^2 (m+1) (e f-d g)^2}+\frac{(d+e x)^{m+1} \left (a+\frac{f (c f-b g)}{g^2}\right )}{(f+g x) (e f-d g)}+\frac{c (d+e x)^{m+1}}{e g^2 (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.204089, antiderivative size = 157, normalized size of antiderivative = 1., 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, 80, 68} \[ -\frac{(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right ) (g (a e g m+b d g-b e f (m+1))-c f (2 d g-e f (m+2)))}{g^2 (m+1) (e f-d g)^2}+\frac{(d+e x)^{m+1} \left (a+\frac{f (c f-b g)}{g^2}\right )}{(f+g x) (e f-d g)}+\frac{c (d+e x)^{m+1}}{e g^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(d + e*x)^(1 + m))/(e*g^2*(1 + m)) + ((a + (f*(c*f - b*g))/g^2)*(d + e*x)^(1 + m))/((e*f - d*g)*(f + g*x))
- ((g*(b*d*g + a*e*g*m - b*e*f*(1 + m)) - c*f*(2*d*g - e*f*(2 + m)))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1
+ m, 2 + m, -((g*(d + e*x))/(e*f - d*g))])/(g^2*(e*f - d*g)^2*(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 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 )}{(f+g x)^2} \, dx &=\frac{\left (a+\frac{f (c f-b g)}{g^2}\right ) (d+e x)^{1+m}}{(e f-d g) (f+g x)}+\frac{\int \frac{(d+e x)^m \left (\frac{c d f g-a e g^2 m-c e f^2 (1+m)-b g (d g-e f (1+m))}{g^2}-c \left (d-\frac{e f}{g}\right ) x\right )}{f+g x} \, dx}{e f-d g}\\ &=\frac{c (d+e x)^{1+m}}{e g^2 (1+m)}+\frac{\left (a+\frac{f (c f-b g)}{g^2}\right ) (d+e x)^{1+m}}{(e f-d g) (f+g x)}-\frac{(g (b d g+a e g m-b e f (1+m))-c f (2 d g-e f (2+m))) \int \frac{(d+e x)^m}{f+g x} \, dx}{g^2 (e f-d g)}\\ &=\frac{c (d+e x)^{1+m}}{e g^2 (1+m)}+\frac{\left (a+\frac{f (c f-b g)}{g^2}\right ) (d+e x)^{1+m}}{(e f-d g) (f+g x)}-\frac{(g (b d g+a e g m-b e f (1+m))-c f (2 d g-e f (2+m))) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{g (d+e x)}{e f-d g}\right )}{g^2 (e f-d g)^2 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.160932, size = 134, normalized size = 0.85 \[ \frac{(d+e x)^{m+1} \left (e^2 \left (g (a g-b f)+c f^2\right ) \, _2F_1\left (2,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )-e (2 c f-b g) (e f-d g) \, _2F_1\left (1,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )+c (e f-d g)^2\right )}{e g^2 (m+1) (e f-d g)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.737, 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 ) ^{2}}}\, 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)^2,x)

[Out]

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

[Out]

integrate((c*x^2 + b*x + a)*(e*x + d)^m/(g*x + f)^2, 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 )}{\left (e x + d\right )}^{m}}{g^{2} x^{2} + 2 \, f g x + f^{2}}, 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)^2,x, algorithm="fricas")

[Out]

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

[Out]

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

[Out]

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