### 3.734 $$\int (d+e x) (a+c x^2)^p \, dx$$

Optimal. Leaf size=61 $\frac{d x \left (a+c x^2\right )^{p+1} \, _2F_1\left (1,p+\frac{3}{2};\frac{3}{2};-\frac{c x^2}{a}\right )}{a}+\frac{e \left (a+c x^2\right )^{p+1}}{2 c (p+1)}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0203105, antiderivative size = 70, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {641, 246, 245} $d x \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{c x^2}{a}\right )+\frac{e \left (a+c x^2\right )^{p+1}}{2 c (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)*(a + c*x^2)^p,x]

[Out]

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

Rule 641

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (d+e x) \left (a+c x^2\right )^p \, dx &=\frac{e \left (a+c x^2\right )^{1+p}}{2 c (1+p)}+d \int \left (a+c x^2\right )^p \, dx\\ &=\frac{e \left (a+c x^2\right )^{1+p}}{2 c (1+p)}+\left (d \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int \left (1+\frac{c x^2}{a}\right )^p \, dx\\ &=\frac{e \left (a+c x^2\right )^{1+p}}{2 c (1+p)}+d x \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{c x^2}{a}\right )\\ \end{align*}

Mathematica [A]  time = 0.0502764, size = 98, normalized size = 1.61 $\frac{\left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \left (2 c d (p+1) x \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{c x^2}{a}\right )+c e x^2 \left (\frac{c x^2}{a}+1\right )^p+a e \left (\left (\frac{c x^2}{a}+1\right )^p-1\right )\right )}{2 c (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)*(a + c*x^2)^p,x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.309, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) \left ( c{x}^{2}+a \right ) ^{p}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)*(c*x^2 + a)^p, x)

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)*(c*x^2 + a)^p, x)

________________________________________________________________________________________

Sympy [A]  time = 6.73627, size = 61, normalized size = 1. \begin{align*} a^{p} d x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )} + e \left (\begin{cases} \frac{a^{p} x^{2}}{2} & \text{for}\: c = 0 \\\frac{\begin{cases} \frac{\left (a + c x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a + c x^{2} \right )} & \text{otherwise} \end{cases}}{2 c} & \text{otherwise} \end{cases}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*d*x*hyper((1/2, -p), (3/2,), c*x**2*exp_polar(I*pi)/a) + e*Piecewise((a**p*x**2/2, Eq(c, 0)), (Piecewise(
((a + c*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(a + c*x**2), True))/(2*c), True))

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)*(c*x^2 + a)^p, x)