3.809 \(\int \frac{(f+g x)^n (a+2 c d x+c e x^2)}{d+e x} \, dx\)

Optimal. Leaf size=114 \[ \frac{\left (c d^2-a e\right ) (f+g x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{e (f+g x)}{e f-d g}\right )}{e (n+1) (e f-d g)}-\frac{c (e f-d g) (f+g x)^{n+1}}{e g^2 (n+1)}+\frac{c (f+g x)^{n+2}}{g^2 (n+2)} \]

[Out]

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

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Rubi [A]  time = 0.15287, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {951, 80, 68} \[ \frac{\left (c d^2-a e\right ) (f+g x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{e (f+g x)}{e f-d g}\right )}{e (n+1) (e f-d g)}-\frac{c (e f-d g) (f+g x)^{n+1}}{e g^2 (n+1)}+\frac{c (f+g x)^{n+2}}{g^2 (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.157708, size = 93, normalized size = 0.82 \[ \frac{(f+g x)^{n+1} \left (\frac{\left (c d^2-a e\right ) \, _2F_1\left (1,n+1;n+2;\frac{e (f+g x)}{e f-d g}\right )}{e f-d g}+\frac{c (d g (n+2)-e f+e g (n+1) x)}{g^2 (n+2)}\right )}{e (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((f + g*x)^(1 + n)*((c*(-(e*f) + d*g*(2 + n) + e*g*(1 + n)*x))/(g^2*(2 + n)) + ((c*d^2 - a*e)*Hypergeometric2F
1[1, 1 + n, 2 + n, (e*(f + g*x))/(e*f - d*g)])/(e*f - d*g)))/(e*(1 + n))

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Maple [F]  time = 0.669, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( gx+f \right ) ^{n} \left ( ce{x}^{2}+2\,cdx+a \right ) }{ex+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**n*(c*e*x**2+2*c*d*x+a)/(e*x+d),x)

[Out]

Integral((f + g*x)**n*(a + 2*c*d*x + c*e*x**2)/(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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