\(\int \frac {(a+b x+c x^2)^p}{d+e x} \, dx\) [2567]

Optimal result

Integrand size = 20, antiderivative size = 184 \[ \int \frac {\left (a+b x+c x^2\right )^p}{d+e x} \, dx=\frac {2^{-1+2 p} \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e p} \]

[Out]

2^(-1+2*p)*(c*x^2+b*x+a)^p*AppellF1(-2*p,-p,-p,1-2*p,1/2*(2*d-e*(b+(-4*a*c+b^2)^(1/2))/c)/(e*x+d),1/2*(2*c*d-e
*(b-(-4*a*c+b^2)^(1/2)))/c/(e*x+d))/e/p/((e*(b+2*c*x-(-4*a*c+b^2)^(1/2))/c/(e*x+d))^p)/((e*(b+2*c*x+(-4*a*c+b^
2)^(1/2))/c/(e*x+d))^p)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {772, 138} \[ \int \frac {\left (a+b x+c x^2\right )^p}{d+e x} \, dx=\frac {2^{2 p-1} \left (a+b x+c x^2\right )^p \left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e p} \]

[In]

Int[(a + b*x + c*x^2)^p/(d + e*x),x]

[Out]

(2^(-1 + 2*p)*(a + b*x + c*x^2)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, (2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)/(2*c*(d
+ e*x)), (2*d - ((b + Sqrt[b^2 - 4*a*c])*e)/c)/(2*(d + e*x))])/(e*p*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d
+ e*x)))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^p)

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(-(1/(d + e*x))^(2*p))*((a + b*x + c*x^2)^p/(e*(e*((b - q + 2*c*x)/(2*c*(d + e*x))))^p*(e*((b + q +
2*c*x)/(2*c*(d + e*x))))^p)), Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - e*((b - q)/(2*c)))*x, x]^p*Simp[1 -
(d - e*((b + q)/(2*c)))*x, x]^p, x], x, 1/(d + e*x)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c,
0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p] && ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (2^{2 p} \left (\frac {1}{d+e x}\right )^{2 p} \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (a+b x+c x^2\right )^p\right ) \text {Subst}\left (\int x^{1-2 (1+p)} \left (1-\frac {1}{2} \left (2 d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{c}\right ) x\right )^p \left (1-\frac {1}{2} \left (2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}\right ) x\right )^p \, dx,x,\frac {1}{d+e x}\right )}{e} \\ & = \frac {2^{-1+2 p} \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (-2 p;-p,-p;1-2 p;\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e p} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x+c x^2\right )^p}{d+e x} \, dx=\frac {2^{-1+2 p} \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} (a+x (b+c x))^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 c d-b e+\sqrt {b^2-4 a c} e}{2 c d+2 c e x}\right )}{e p} \]

[In]

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

[Out]

(2^(-1 + 2*p)*(a + x*(b + c*x))^p*AppellF1[-2*p, -p, -p, 1 - 2*p, (2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)/(2*c*(d
+ e*x)), (2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)/(2*c*d + 2*c*e*x)])/(e*p*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*
(d + e*x)))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^p)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b*x + c*x^2)^p/(d + e*x),x)

[Out]

int((a + b*x + c*x^2)^p/(d + e*x), x)