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

Optimal. Leaf size=104 \[ -\frac{e x \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},1-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^2}-\frac{\left (d^2-e^2 x^2\right )^p \, _2F_1\left (1,p;p+1;1-\frac{e^2 x^2}{d^2}\right )}{2 d p} \]

[Out]

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

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Rubi [A]  time = 0.0679128, antiderivative size = 104, 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, 246, 245} \[ -\frac{e x \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},1-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^2}-\frac{\left (d^2-e^2 x^2\right )^p \, _2F_1\left (1,p;p+1;1-\frac{e^2 x^2}{d^2}\right )}{2 d p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*x*(d^2 - e^2*x^2)^p*Hypergeometric2F1[1/2, 1 - p, 3/2, (e^2*x^2)/d^2])/(d^2*(1 - (e^2*x^2)/d^2)^p)) - ((d
^2 - e^2*x^2)^p*Hypergeometric2F1[1, p, 1 + p, 1 - (e^2*x^2)/d^2])/(2*d*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 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 \frac{\left (d^2-e^2 x^2\right )^p}{x (d+e x)} \, dx &=\int \frac{(d-e x) \left (d^2-e^2 x^2\right )^{-1+p}}{x} \, dx\\ &=d \int \frac{\left (d^2-e^2 x^2\right )^{-1+p}}{x} \, dx-e \int \left (d^2-e^2 x^2\right )^{-1+p} \, dx\\ &=\frac{1}{2} d \operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{-1+p}}{x} \, 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 \left (1-\frac{e^2 x^2}{d^2}\right )^{-1+p} \, dx}{d^2}\\ &=-\frac{e x \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{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^2}-\frac{\left (d^2-e^2 x^2\right )^p \, _2F_1\left (1,p;1+p;1-\frac{e^2 x^2}{d^2}\right )}{2 d p}\\ \end{align*}

Mathematica [A]  time = 0.118403, size = 151, normalized size = 1.45 \[ \frac{2^{p-1} \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \left (\frac{e x}{d}+1\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (p (d-e x) \left (1-\frac{d^2}{e^2 x^2}\right )^p \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )+d (p+1) \left (\frac{e x}{2 d}+\frac{1}{2}\right )^p \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )\right )}{d^2 p (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.671, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{x \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/(e*x+d),x)

[Out]

int((-e^2*x^2+d^2)^p/x/(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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-e^2*x^2 + d^2)^p/((e*x + d)*x), 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^{2} + d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 7.5121, size = 359, normalized size = 3.45 \begin{align*} \begin{cases} - \frac{0^{p} d^{2 p} \log{\left (\frac{d^{2}}{e^{2} x^{2}} - 1 \right )}}{2 d} - \frac{0^{p} d^{2 p} \operatorname{acoth}{\left (\frac{d}{e x} \right )}}{d} + \frac{d e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (1 - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, 1 - p \\ 2 - p \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e^{2} x^{2} \Gamma \left (2 - 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{1}{2} - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, \frac{1}{2} - p \\ \frac{3}{2} - p \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e x \Gamma \left (\frac{3}{2} - p\right ) \Gamma \left (p + 1\right )} & \text{for}\: \frac{\left |{d^{2}}\right |}{\left |{e^{2}}\right | \left |{x^{2}}\right |} > 1 \\- \frac{0^{p} d^{2 p} \log{\left (- \frac{d^{2}}{e^{2} x^{2}} + 1 \right )}}{2 d} - \frac{0^{p} d^{2 p} \operatorname{atanh}{\left (\frac{d}{e x} \right )}}{d} + \frac{d e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (1 - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, 1 - p \\ 2 - p \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e^{2} x^{2} \Gamma \left (2 - 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{1}{2} - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, \frac{1}{2} - p \\ \frac{3}{2} - p \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e x \Gamma \left (\frac{3}{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/(e*x+d),x)

[Out]

Piecewise((-0**p*d**(2*p)*log(d**2/(e**2*x**2) - 1)/(2*d) - 0**p*d**(2*p)*acoth(d/(e*x))/d + d*e**(2*p)*p*x**(
2*p)*exp(I*pi*p)*gamma(p)*gamma(1 - p)*hyper((1 - p, 1 - p), (2 - p,), d**2/(e**2*x**2))/(2*e**2*x**2*gamma(2
- p)*gamma(p + 1)) - e**(2*p)*p*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(1/2 - p)*hyper((1 - p, 1/2 - p), (3/2 - p,
), d**2/(e**2*x**2))/(2*e*x*gamma(3/2 - p)*gamma(p + 1)), Abs(d**2)/(Abs(e**2)*Abs(x**2)) > 1), (-0**p*d**(2*p
)*log(-d**2/(e**2*x**2) + 1)/(2*d) - 0**p*d**(2*p)*atanh(d/(e*x))/d + d*e**(2*p)*p*x**(2*p)*exp(I*pi*p)*gamma(
p)*gamma(1 - p)*hyper((1 - p, 1 - p), (2 - p,), d**2/(e**2*x**2))/(2*e**2*x**2*gamma(2 - p)*gamma(p + 1)) - e*
*(2*p)*p*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(1/2 - p)*hyper((1 - p, 1/2 - p), (3/2 - p,), d**2/(e**2*x**2))/(2
*e*x*gamma(3/2 - 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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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