3.966 \(\int \frac {(1-\frac {e^2 x^2}{d^2})^p}{(d+e x)^3} \, dx\)

Optimal. Leaf size=57 \[ -\frac {2^{p-3} \left (\frac {d-e x}{d}\right )^{p+1} \, _2F_1\left (3-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{d^2 e (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {676, 69} \[ -\frac {2^{p-3} \left (\frac {d-e x}{d}\right )^{p+1} \, _2F_1\left (3-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{d^2 e (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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 676

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*x)/e)^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*} \int \frac {\left (1-\frac {e^2 x^2}{d^2}\right )^p}{(d+e x)^3} \, dx &=\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 )^{-3+p} \, dx}{d^4}\\ &=-\frac {2^{-3+p} \left (\frac {d-e x}{d}\right )^{1+p} \, _2F_1\left (3-p,1+p;2+p;\frac {d-e x}{2 d}\right )}{d^2 e (1+p)}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 76, normalized size = 1.33 \[ -\frac {2^{p-3} (d-e x) \left (\frac {e x}{d}+1\right )^{-p} \left (1-\frac {e^2 x^2}{d^2}\right )^p \, _2F_1\left (3-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{d^3 e (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (-\frac {e^{2} x^{2} - d^{2}}{d^{2}}\right )^{p}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-e^2*x^2/d^2)^p/(e*x+d)^3,x, algorithm="fricas")

[Out]

integral((-(e^2*x^2 - d^2)/d^2)^p/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p}}{{\left (e x + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-e^2*x^2/d^2)^p/(e*x+d)^3,x, algorithm="giac")

[Out]

integrate((-e^2*x^2/d^2 + 1)^p/(e*x + d)^3, x)

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maple [F]  time = 0.83, size = 0, normalized size = 0.00 \[ \int \frac {\left (-\frac {e^{2} x^{2}}{d^{2}}+1\right )^{p}}{\left (e x +d \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1/d^2*e^2*x^2+1)^p/(e*x+d)^3,x)

[Out]

int((-1/d^2*e^2*x^2+1)^p/(e*x+d)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p}}{{\left (e x + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-e^2*x^2/d^2)^p/(e*x+d)^3,x, algorithm="maxima")

[Out]

integrate((-e^2*x^2/d^2 + 1)^p/(e*x + d)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (1-\frac {e^2\,x^2}{d^2}\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - (e^2*x^2)/d^2)^p/(d + e*x)^3,x)

[Out]

int((1 - (e^2*x^2)/d^2)^p/(d + e*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (-1 + \frac {e x}{d}\right ) \left (1 + \frac {e x}{d}\right )\right )^{p}}{\left (d + e x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-e**2*x**2/d**2)**p/(e*x+d)**3,x)

[Out]

Integral((-(-1 + e*x/d)*(1 + e*x/d))**p/(d + e*x)**3, x)

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