3.542 \(\int \frac{x^3 (e+f x)^n}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=290 \[ \frac{\left (\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b-\sqrt{b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )}+\frac{\left (-\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{(b f+c e) (e+f x)^{n+1}}{c^2 f^2 (n+1)}+\frac{(e+f x)^{n+2}}{c f^2 (n+2)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.769534, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {1628, 68} \[ \frac{\left (\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b-\sqrt{b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )}+\frac{\left (-\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{(b f+c e) (e+f x)^{n+1}}{c^2 f^2 (n+1)}+\frac{(e+f x)^{n+2}}{c f^2 (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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

Mathematica [A]  time = 0.772312, size = 261, normalized size = 0.9 \[ \frac{(e+f x)^{n+1} \left (\frac{c \left (\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e+\left (\sqrt{b^2-4 a c}-b\right ) f}\right )}{(n+1) \left (f \left (\sqrt{b^2-4 a c}-b\right )+2 c e\right )}+\frac{c \left (-\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{(n+1) \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{b f+c e}{f^2 (n+1)}+\frac{c (e+f x)}{f^2 (n+2)}\right )}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e + f*x)^(1 + n)*(-((c*e + b*f)/(f^2*(1 + n))) + (c*(e + f*x))/(f^2*(2 + n)) + (c*(a - b^2/c + (b*(b^2 - 3*a
*c))/(c*Sqrt[b^2 - 4*a*c]))*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e + (-b + Sqrt[b^2 - 4*a*c
])*f)])/((2*c*e + (-b + Sqrt[b^2 - 4*a*c])*f)*(1 + n)) + (c*(a - b^2/c - (b*(b^2 - 3*a*c))/(c*Sqrt[b^2 - 4*a*c
]))*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e - (b + Sqrt[b^2 - 4*a*c])*f)])/((2*c*e - (b + Sq
rt[b^2 - 4*a*c])*f)*(1 + n))))/c^2

________________________________________________________________________________________

Maple [F]  time = 1.278, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{n}{x}^{3}}{c{x}^{2}+bx+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{n} x^{3}}{c x^{2} + b x + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x+e)^n/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (f x + e\right )}^{n} x^{3}}{c x^{2} + b x + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x+e)^n/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(f*x+e)**n/(c*x**2+b*x+a),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{n} x^{3}}{c x^{2} + b x + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x+e)^n/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

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