\(\int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx\) [542]

Optimal result

Integrand size = 23, antiderivative size = 290 \[ \int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx=-\frac {(c e+b f) (e+f x)^{1+n}}{c^2 f^2 (1+n)}+\frac {(e+f x)^{2+n}}{c f^2 (2+n)}+\frac {\left (a-\frac {b^2}{c}+\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{c \left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}+\frac {\left (a-\frac {b^2}{c}-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{c \left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1642, 70} \[ \int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx=\frac {\left (\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{c}\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )}+\frac {\left (-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{c}\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {(b f+c e) (e+f x)^{n+1}}{c^2 f^2 (n+1)}+\frac {(e+f x)^{n+2}}{c f^2 (n+2)} \]

[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 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 1642

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.90 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.10 \[ \int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx=\frac {(e+f x)^{1+n} \left (\frac {\left (b^3-3 a b c-b^2 \sqrt {b^2-4 a c}+a c \sqrt {b^2-4 a c}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}+\frac {-\left (\left (b \left (b+\sqrt {b^2-4 a c}\right ) f-2 c \left (\sqrt {b^2-4 a c} e+2 a f\right )\right ) (b f (2+n)+c (e-f (1+n) x))\right )+\left (b^3-3 a b c+b^2 \sqrt {b^2-4 a c}-a c \sqrt {b^2-4 a c}\right ) f^2 (2+n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{f^2 \left (-2 c e+\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (2+n)}\right )}{c^2 \sqrt {b^2-4 a c} (1+n)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {x^{3} \left (f x +e \right )^{n}}{c \,x^{2}+b x +a}d x\]

[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)

Fricas [F]

\[ \int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{3}}{c x^{2} + b x + a} \,d x } \]

[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(-2)]

Exception generated. \[ \int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{3}}{c x^{2} + b x + a} \,d x } \]

[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)

Giac [F]

\[ \int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{3}}{c x^{2} + b x + a} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx=\int \frac {x^3\,{\left (e+f\,x\right )}^n}{c\,x^2+b\,x+a} \,d x \]

[In]

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

[Out]

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