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

Optimal. Leaf size=108 \[ \frac{e \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},1-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{d^2 x}-\frac{e^2 \left (d^2-e^2 x^2\right )^p \, _2F_1\left (2,p;p+1;1-\frac{e^2 x^2}{d^2}\right )}{2 d^3 p} \]

[Out]

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

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Rubi [A]  time = 0.0798012, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {850, 764, 266, 65, 365, 364} \[ \frac{e \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},1-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{d^2 x}-\frac{e^2 \left (d^2-e^2 x^2\right )^p \, _2F_1\left (2,p;p+1;1-\frac{e^2 x^2}{d^2}\right )}{2 d^3 p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 850

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

Rule 764

Int[(x_)^(m_.)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[f, Int[x^m*(a + c*x^2)^p, x]
, x] + Dist[g, Int[x^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && IntegerQ[m] &&  !IntegerQ[2
*p]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 65

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

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

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

Rubi steps

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

Mathematica [B]  time = 0.615091, size = 219, normalized size = 2.03 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (\left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \left (\frac{d^3 \, _2F_1\left (1-p,-p;2-p;\frac{d^2}{e^2 x^2}\right )}{(p-1) x^2}+e^2 \left (\frac{(d-e x) \left (2-\frac{2 d^2}{e^2 x^2}\right )^p \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{d \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )}{p}\right )\right )+\frac{2 d^2 e \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x}\right )}{2 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d^2 - e^2*x^2)^p*((2*d^2*e*Hypergeometric2F1[-1/2, -p, 1/2, (e^2*x^2)/d^2])/(x*(1 - (e^2*x^2)/d^2)^p) + ((d^
3*Hypergeometric2F1[1 - p, -p, 2 - p, d^2/(e^2*x^2)])/((-1 + p)*x^2) + e^2*(((2 - (2*d^2)/(e^2*x^2))^p*(d - e*
x)*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (d*Hypergeometric2F1[-
p, -p, 1 - p, d^2/(e^2*x^2)])/p))/(1 - d^2/(e^2*x^2))^p))/(2*d^4)

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Maple [F]  time = 0.678, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{{x}^{3} \left ( ex+d \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{e x^{4} + d x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-e^2*x^2 + d^2)^p/(e*x^4 + d*x^3), x)

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Sympy [C]  time = 9.28783, size = 500, normalized size = 4.63 \begin{align*} \begin{cases} - \frac{0^{p} d^{2 p}}{2 d x^{2}} + \frac{0^{p} d^{2 p} e}{d^{2} x} + \frac{0^{p} d^{2 p} e^{2} \log{\left (\frac{e^{2} x^{2}}{d^{2}} \right )}}{2 d^{3}} - \frac{0^{p} d^{2 p} e^{2} \log{\left (-1 + \frac{e^{2} x^{2}}{d^{2}} \right )}}{2 d^{3}} - \frac{0^{p} d^{2 p} e^{2} \operatorname{acoth}{\left (\frac{e x}{d} \right )}}{d^{3}} + \frac{d e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (2 - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, 2 - p \\ 3 - p \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e^{2} x^{4} \Gamma \left (3 - p\right ) \Gamma \left (p + 1\right )} - \frac{e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (\frac{3}{2} - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, \frac{3}{2} - p \\ \frac{5}{2} - p \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e x^{3} \Gamma \left (\frac{5}{2} - p\right ) \Gamma \left (p + 1\right )} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\- \frac{0^{p} d^{2 p}}{2 d x^{2}} + \frac{0^{p} d^{2 p} e}{d^{2} x} + \frac{0^{p} d^{2 p} e^{2} \log{\left (\frac{e^{2} x^{2}}{d^{2}} \right )}}{2 d^{3}} - \frac{0^{p} d^{2 p} e^{2} \log{\left (1 - \frac{e^{2} x^{2}}{d^{2}} \right )}}{2 d^{3}} - \frac{0^{p} d^{2 p} e^{2} \operatorname{atanh}{\left (\frac{e x}{d} \right )}}{d^{3}} + \frac{d e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (2 - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, 2 - p \\ 3 - p \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e^{2} x^{4} \Gamma \left (3 - p\right ) \Gamma \left (p + 1\right )} - \frac{e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (\frac{3}{2} - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, \frac{3}{2} - p \\ \frac{5}{2} - p \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e x^{3} \Gamma \left (\frac{5}{2} - p\right ) \Gamma \left (p + 1\right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-0**p*d**(2*p)/(2*d*x**2) + 0**p*d**(2*p)*e/(d**2*x) + 0**p*d**(2*p)*e**2*log(e**2*x**2/d**2)/(2*d*
*3) - 0**p*d**(2*p)*e**2*log(-1 + e**2*x**2/d**2)/(2*d**3) - 0**p*d**(2*p)*e**2*acoth(e*x/d)/d**3 + d*e**(2*p)
*p*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(2 - p)*hyper((1 - p, 2 - p), (3 - p,), d**2/(e**2*x**2))/(2*e**2*x**4*g
amma(3 - p)*gamma(p + 1)) - e**(2*p)*p*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(3/2 - p)*hyper((1 - p, 3/2 - p), (5
/2 - p,), d**2/(e**2*x**2))/(2*e*x**3*gamma(5/2 - p)*gamma(p + 1)), Abs(e**2*x**2)/Abs(d**2) > 1), (-0**p*d**(
2*p)/(2*d*x**2) + 0**p*d**(2*p)*e/(d**2*x) + 0**p*d**(2*p)*e**2*log(e**2*x**2/d**2)/(2*d**3) - 0**p*d**(2*p)*e
**2*log(1 - e**2*x**2/d**2)/(2*d**3) - 0**p*d**(2*p)*e**2*atanh(e*x/d)/d**3 + d*e**(2*p)*p*x**(2*p)*exp(I*pi*p
)*gamma(p)*gamma(2 - p)*hyper((1 - p, 2 - p), (3 - p,), d**2/(e**2*x**2))/(2*e**2*x**4*gamma(3 - p)*gamma(p +
1)) - e**(2*p)*p*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(3/2 - p)*hyper((1 - p, 3/2 - p), (5/2 - p,), d**2/(e**2*x
**2))/(2*e*x**3*gamma(5/2 - p)*gamma(p + 1)), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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