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

Optimal. Leaf size=129 \[ \frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right )}{g^2 (m+1) (e f-d g)}-\frac{(d+e x)^{m+1} (-b e g+c d g+c e f)}{e^2 g^2 (m+1)}+\frac{c (d+e x)^{m+2}}{e^2 g (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.159453, antiderivative size = 129, 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 = {951, 80, 68} \[ \frac{(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) \, _2F_1\left (1,m+1;m+2;-\frac{g (d+e x)}{e f-d g}\right )}{g^2 (m+1) (e f-d g)}-\frac{(d+e x)^{m+1} (-b e g+c d g+c e f)}{e^2 g^2 (m+1)}+\frac{c (d+e x)^{m+2}}{e^2 g (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.146815, size = 111, normalized size = 0.86 \[ \frac{(d+e x)^{m+1} \left (\frac{\left (g (a g-b f)+c f^2\right ) \, _2F_1\left (1,m+1;m+2;\frac{g (d+e x)}{d g-e f}\right )}{(m+1) (e f-d g)}+\frac{b e g-c (d g+e f)}{e^2 (m+1)}+\frac{c g (d+e x)}{e^2 (m+2)}\right )}{g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.662, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+bx+a \right ) \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)/(g*x+f),x)

[Out]

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

[Out]

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

[Out]

integral((c*x^2 + b*x + a)*(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 )}{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)/(g*x+f),x)

[Out]

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

[Out]

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