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

Optimal. Leaf size=106 \[ \frac {e \left (d^2-e^2 x^2\right )^p \, _2F_1\left (1,p;p+1;1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 p}-\frac {\left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {1}{2},1-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{d x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {850, 764, 365, 364, 266, 65} \[ \frac {e \left (d^2-e^2 x^2\right )^p \, _2F_1\left (1,p;p+1;1-\frac {e^2 x^2}{d^2}\right )}{2 d^2 p}-\frac {\left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {1}{2},1-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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])

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 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 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]))

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 167, normalized size = 1.58 \[ \frac {\left (d^2-e^2 x^2\right )^p \left (-\frac {d e \left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac {d^2}{e^2 x^2}\right )}{p}-\frac {2 d^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x}+\frac {e 2^p (e x-d) \left (\frac {e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{p+1}\right )}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d^2 - e^2*x^2)^p*((-2*d^2*Hypergeometric2F1[-1/2, -p, 1/2, (e^2*x^2)/d^2])/(x*(1 - (e^2*x^2)/d^2)^p) + (2^p*
e*(-d + e*x)*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) - (d*e*Hyperge
ometric2F1[-p, -p, 1 - p, d^2/(e^2*x^2)])/(p*(1 - d^2/(e^2*x^2))^p)))/(2*d^3)

________________________________________________________________________________________

fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{e x^{3} + d x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right ) x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^2\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 7.86, size = 450, normalized size = 4.25 \[ \begin {cases} - \frac {0^{p} d^{2 p}}{d x} - \frac {0^{p} d^{2 p} e \log {\left (\frac {e^{2} x^{2}}{d^{2}} \right )}}{2 d^{2}} + \frac {0^{p} d^{2 p} e \log {\left (-1 + \frac {e^{2} x^{2}}{d^{2}} \right )}}{2 d^{2}} + \frac {0^{p} d^{2 p} e \operatorname {acoth}{\left (\frac {e x}{d} \right )}}{d^{2}} + \frac {d e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \relax (p) \Gamma \left (\frac {3}{2} - p\right ) {{}_{2}F_{1}\left (\begin {matrix} 1 - p, \frac {3}{2} - p \\ \frac {5}{2} - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 e^{2} x^{3} \Gamma \left (\frac {5}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac {e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \relax (p) \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} 1 - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 e x^{2} \Gamma \left (2 - p\right ) \Gamma \left (p + 1\right )} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac {0^{p} d^{2 p}}{d x} - \frac {0^{p} d^{2 p} e \log {\left (\frac {e^{2} x^{2}}{d^{2}} \right )}}{2 d^{2}} + \frac {0^{p} d^{2 p} e \log {\left (1 - \frac {e^{2} x^{2}}{d^{2}} \right )}}{2 d^{2}} + \frac {0^{p} d^{2 p} e \operatorname {atanh}{\left (\frac {e x}{d} \right )}}{d^{2}} + \frac {d e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \relax (p) \Gamma \left (\frac {3}{2} - p\right ) {{}_{2}F_{1}\left (\begin {matrix} 1 - p, \frac {3}{2} - p \\ \frac {5}{2} - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 e^{2} x^{3} \Gamma \left (\frac {5}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac {e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \relax (p) \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} 1 - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 e x^{2} \Gamma \left (2 - p\right ) \Gamma \left (p + 1\right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________