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

Optimal. Leaf size=193 \[ \frac{(f+g x)^{n+1} \left (a e g^2 (1-n) n-c \left (d^2 g^2 \left (-n^2+n+2\right )-4 d e f g+2 e^2 f^2\right )\right ) \, _2F_1\left (1,n+1;n+2;\frac{e (f+g x)}{e f-d g}\right )}{2 e (n+1) (e f-d g)^3}-\frac{g (1-n) \left (c d^2-a e\right ) (f+g x)^{n+1}}{2 e (d+e x) (e f-d g)^2}-\frac{\left (a-\frac{c d^2}{e}\right ) (f+g x)^{n+1}}{2 (d+e x)^2 (e f-d g)} \]

[Out]

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

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Rubi [A]  time = 0.224053, antiderivative size = 193, 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 = {949, 78, 68} \[ \frac{(f+g x)^{n+1} \left (a e g^2 (1-n) n-c \left (d^2 g^2 \left (-n^2+n+2\right )-4 d e f g+2 e^2 f^2\right )\right ) \, _2F_1\left (1,n+1;n+2;\frac{e (f+g x)}{e f-d g}\right )}{2 e (n+1) (e f-d g)^3}-\frac{g (1-n) \left (c d^2-a e\right ) (f+g x)^{n+1}}{2 e (d+e x) (e f-d g)^2}-\frac{\left (a-\frac{c d^2}{e}\right ) (f+g x)^{n+1}}{2 (d+e x)^2 (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.102173, size = 106, normalized size = 0.55 \[ -\frac{(f+g x)^{n+1} \left (g^2 \left (a e-c d^2\right ) \, _2F_1\left (3,n+1;n+2;\frac{e (f+g x)}{e f-d g}\right )+c (e f-d g)^2 \, _2F_1\left (1,n+1;n+2;\frac{e (f+g x)}{e f-d g}\right )\right )}{e (n+1) (e f-d g)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.723, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( gx+f \right ) ^{n} \left ( ce{x}^{2}+2\,cdx+a \right ) }{ \left ( ex+d \right ) ^{3}}}\, 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)^3,x)

[Out]

int((g*x+f)^n*(c*e*x^2+2*c*d*x+a)/(e*x+d)^3,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}}{{\left (e x + d\right )}^{3}}\,{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)^3,x, algorithm="maxima")

[Out]

integrate((c*e*x^2 + 2*c*d*x + a)*(g*x + f)^n/(e*x + d)^3, 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^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, 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)^3,x, algorithm="fricas")

[Out]

integral((c*e*x^2 + 2*c*d*x + a)*(g*x + f)^n/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), 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 )}{\left (d + e x\right )^{3}}\, 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)**3,x)

[Out]

Integral((f + g*x)**n*(a + 2*c*d*x + c*e*x**2)/(d + e*x)**3, 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}}{{\left (e x + d\right )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

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