\(\int \frac {(d^2-e^2 x^2)^p}{x^3 (d+e x)} \, dx\) [274]

Optimal result

Integrand size = 25, antiderivative size = 108 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)} \, dx=\frac {e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{d^2 x}-\frac {e^2 \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (2,p,1+p,1-\frac {e^2 x^2}{d^2}\right )}{2 d^3 p} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {864, 778, 272, 67, 372, 371} \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)} \, dx=\frac {e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{d^2 x}-\frac {e^2 \left (d^2-e^2 x^2\right )^p \operatorname {Hypergeometric2F1}\left (2,p,p+1,1-\frac {e^2 x^2}{d^2}\right )}{2 d^3 p} \]

[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 67

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

Rule 272

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 778

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 864

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*} \text {integral}& = \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 \text {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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(219\) vs. \(2(108)=216\).

Time = 0.59 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.03 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)} \, dx=\frac {\left (d^2-e^2 x^2\right )^p \left (\frac {2 d^2 e \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {e^2 x^2}{d^2}\right )}{x}+\left (1-\frac {d^2}{e^2 x^2}\right )^{-p} \left (\frac {d^3 \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,\frac {d^2}{e^2 x^2}\right )}{(-1+p) x^2}+e^2 \left (\frac {\left (2-\frac {2 d^2}{e^2 x^2}\right )^p (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{1+p}+\frac {d \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,\frac {d^2}{e^2 x^2}\right )}{p}\right )\right )\right )}{2 d^4} \]

[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)

Maple [F]

\[\int \frac {\left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{3} \left (e x +d \right )}d x\]

[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)

Fricas [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )} x^{3}} \,d x } \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 13.40 (sec) , antiderivative size = 484, normalized size of antiderivative = 4.48 \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)} \, dx=\begin {cases} - \frac {0^{p} d^{2} d^{2 p - 3}}{2 x^{2}} + \frac {0^{p} d d^{2 p - 3} e}{x} + \frac {0^{p} d^{2 p - 3} e^{2} \log {\left (\frac {e^{2} x^{2}}{d^{2}} \right )}}{2} - \frac {0^{p} d^{2 p - 3} e^{2} \log {\left (-1 + \frac {e^{2} x^{2}}{d^{2}} \right )}}{2} - 0^{p} d^{2 p - 3} e^{2} \operatorname {acoth}{\left (\frac {e x}{d} \right )} + \frac {d e^{2 p - 2} p x^{2 p - 4} 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 \Gamma \left (3 - p\right ) \Gamma \left (p + 1\right )} - \frac {e^{2 p - 1} p x^{2 p - 3} 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 \Gamma \left (\frac {5}{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} d^{2 p - 3}}{2 x^{2}} + \frac {0^{p} d d^{2 p - 3} e}{x} + \frac {0^{p} d^{2 p - 3} e^{2} \log {\left (\frac {e^{2} x^{2}}{d^{2}} \right )}}{2} - \frac {0^{p} d^{2 p - 3} e^{2} \log {\left (1 - \frac {e^{2} x^{2}}{d^{2}} \right )}}{2} - 0^{p} d^{2 p - 3} e^{2} \operatorname {atanh}{\left (\frac {e x}{d} \right )} + \frac {d e^{2 p - 2} p x^{2 p - 4} 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 \Gamma \left (3 - p\right ) \Gamma \left (p + 1\right )} - \frac {e^{2 p - 1} p x^{2 p - 3} 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 \Gamma \left (\frac {5}{2} - p\right ) \Gamma \left (p + 1\right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )} x^{3}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)} \, dx=\int { \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )} x^{3}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^p}{x^3\,\left (d+e\,x\right )} \,d x \]

[In]

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

[Out]

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