### 3.2094 $$\int (d+e x) (a d e+(c d^2+a e^2) x+c d e x^2)^p \, dx$$

Optimal. Leaf size=95 $-\frac{(d+e x)^2 (a e+c d x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^p \, _2F_1\left (1,2 p+3;p+3;\frac{c d (d+e x)}{c d^2-a e^2}\right )}{(p+2) \left (c d^2-a e^2\right )}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0576793, antiderivative size = 152, normalized size of antiderivative = 1.6, number of steps used = 2, number of rules used = 2, integrand size = 33, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.061, Rules used = {640, 624} $\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{p+1}}{2 c d (p+1)}-\frac{\left (-\frac{e (a e+c d x)}{c d^2-a e^2}\right )^{-p-1} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{p+1} \, _2F_1\left (-p,p+1;p+2;\frac{c d (d+e x)}{c d^2-a e^2}\right )}{2 c d (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^p,x]

[Out]

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

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 624

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, -Simp[((a + b*x + c*
x^2)^(p + 1)*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/(2*q)])/(q*(p + 1)*((q - b - 2*c*x)/(2*q))^(p
+ 1)), x]] /; FreeQ[{a, b, c, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[4*p]

Rubi steps

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

Mathematica [A]  time = 0.0521781, size = 94, normalized size = 0.99 $\frac{\left (\frac{c d (d+e x)}{c d^2-a e^2}\right )^{-p-1} ((d+e x) (a e+c d x))^{p+1} \, _2F_1\left (-p-1,p+1;p+2;\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{c d (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.972, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) \left ( ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2} \right ) ^{p}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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