\(\int \frac {(1-\frac {e^2 x^2}{d^2})^p}{(d+e x)^2} \, dx\) [965]

Optimal result

Integrand size = 23, antiderivative size = 57 \[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^2} \, dx=-\frac {2^{-2+p} \left (\frac {d-e x}{d}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{d e (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {690, 71} \[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^2} \, dx=-\frac {2^{p-2} \left (\frac {d-e x}{d}\right )^{p+1} \operatorname {Hypergeometric2F1}\left (2-p,p+1,p+2,\frac {d-e x}{2 d}\right )}{d e (p+1)} \]

[In]

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

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 690

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[a^(p + 1)*d^(m - 1)*(((d - e*x)/d)^
(p + 1)/(a/d + c*(x/e))^(p + 1)), Int[(1 + e*(x/d))^(m + p)*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a, c, d, e, m}
, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) && GtQ[a, 0] &&  !(IGtQ[m, 0] &&
(IntegerQ[3*p] || IntegerQ[4*p]))

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

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.33 \[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^2} \, dx=-\frac {2^{-2+p} (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \left (1-\frac {e^2 x^2}{d^2}\right )^p \operatorname {Hypergeometric2F1}\left (2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{d^2 e (1+p)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (1-\frac {e^{2} x^{2}}{d^{2}}\right )^{p}}{\left (e x +d \right )^{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^2} \, dx=\int { \frac {{\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p}}{{\left (e x + d\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^2} \, dx=\int \frac {\left (- \left (-1 + \frac {e x}{d}\right ) \left (1 + \frac {e x}{d}\right )\right )^{p}}{\left (d + e x\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^2} \, dx=\int { \frac {{\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p}}{{\left (e x + d\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^2} \, dx=\int { \frac {{\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p}}{{\left (e x + d\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^2} \, dx=\int \frac {{\left (1-\frac {e^2\,x^2}{d^2}\right )}^p}{{\left (d+e\,x\right )}^2} \,d x \]

[In]

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

[Out]

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