\(\int \frac {x^2 (e+f x)^n}{(a+b x) (c+d x)} \, dx\) [114]

Optimal result

Integrand size = 25, antiderivative size = 158 \[ \int \frac {x^2 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\frac {(e+f x)^{1+n}}{b d f (1+n)}-\frac {a^2 (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {b (e+f x)}{b e-a f}\right )}{b (b c-a d) (b e-a f) (1+n)}+\frac {c^2 (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (e+f x)}{d e-c f}\right )}{d (b c-a d) (d e-c f) (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {186, 70} \[ \int \frac {x^2 (e+f x)^n}{(a+b x) (c+d x)} \, dx=-\frac {a^2 (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b (e+f x)}{b e-a f}\right )}{b (n+1) (b c-a d) (b e-a f)}+\frac {c^2 (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {d (e+f x)}{d e-c f}\right )}{d (n+1) (b c-a d) (d e-c f)}+\frac {(e+f x)^{n+1}}{b d f (n+1)} \]

[In]

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

[Out]

(e + f*x)^(1 + n)/(b*d*f*(1 + n)) - (a^2*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b*(e + f*x))/(b
*e - a*f)])/(b*(b*c - a*d)*(b*e - a*f)*(1 + n)) + (c^2*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (d
*(e + f*x))/(d*e - c*f)])/(d*(b*c - a*d)*(d*e - 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 186

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.97 \[ \int \frac {x^2 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\frac {(e+f x)^{1+n} \left (a^2 d f (-d e+c f) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {b (e+f x)}{b e-a f}\right )+(b e-a f) \left (-((b c-a d) (-d e+c f))+b c^2 f \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (e+f x)}{d e-c f}\right )\right )\right )}{b d (b c-a d) f (b e-a f) (d e-c f) (1+n)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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