### 3.962 $$\int (d+e x) (1-\frac{e^2 x^2}{d^2})^p \, dx$$

Optimal. Leaf size=57 $-\frac{d^2 2^{p+1} \left (\frac{d-e x}{d}\right )^{p+1} \, _2F_1\left (-p-1,p+1;p+2;\frac{d-e x}{2 d}\right )}{e (p+1)}$

[Out]

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

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Rubi [A]  time = 0.0156348, antiderivative size = 56, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.095, Rules used = {641, 245} $d x \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )-\frac{d^2 \left (1-\frac{e^2 x^2}{d^2}\right )^{p+1}}{2 e (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0282315, size = 56, normalized size = 0.98 $d x \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )-\frac{d^2 \left (1-\frac{e^2 x^2}{d^2}\right )^{p+1}}{2 e (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.354, size = 47, normalized size = 0.8 \begin{align*}{\frac{e{x}^{2}}{2}{\mbox{$_2$F$_1$}(1,-p;\,2;\,{\frac{{e}^{2}{x}^{2}}{{d}^{2}}})}}+dx{\mbox{$_2$F$_1$}({\frac{1}{2}},-p;\,{\frac{3}{2}};\,{\frac{{e}^{2}{x}^{2}}{{d}^{2}}})} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(1-e^2*x^2/d^2)^p,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}{\left (-\frac{e^{2} x^{2}}{d^{2}} + 1\right )}^{p}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(1-e^2*x^2/d^2)^p,x, algorithm="maxima")

[Out]

integrate((e*x + d)*(-e^2*x^2/d^2 + 1)^p, x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(1-e^2*x^2/d^2)^p,x, algorithm="fricas")

[Out]

integral((e*x + d)*(-(e^2*x^2 - d^2)/d^2)^p, x)

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Sympy [A]  time = 3.25252, size = 78, normalized size = 1.37 \begin{align*} d x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + e \left (\begin{cases} \frac{x^{2}}{2} & \text{for}\: e^{2} = 0 \\- \frac{d^{2} \left (\begin{cases} \frac{\left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (1 - \frac{e^{2} x^{2}}{d^{2}} \right )} & \text{otherwise} \end{cases}\right )}{2 e^{2}} & \text{otherwise} \end{cases}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(1-e**2*x**2/d**2)**p,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}{\left (-\frac{e^{2} x^{2}}{d^{2}} + 1\right )}^{p}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(1-e^2*x^2/d^2)^p,x, algorithm="giac")

[Out]

integrate((e*x + d)*(-e^2*x^2/d^2 + 1)^p, x)