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

Optimal. Leaf size=110 \[ -\frac {e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x}-\frac {e^2 \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;1-\frac {e^2 x^2}{d^2}\right )}{2 d^3 (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {764, 266, 65, 365, 364} \[ -\frac {e \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x}-\frac {e^2 \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;1-\frac {e^2 x^2}{d^2}\right )}{2 d^3 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 106, normalized size = 0.96 \[ \frac {1}{2} e \left (d^2-e^2 x^2\right )^p \left (\frac {e \left (e^2 x^2-d^2\right ) \, _2F_1\left (2,p+1;p+2;1-\frac {e^2 x^2}{d^2}\right )}{d^3 (p+1)}-\frac {2 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};\frac {e^2 x^2}{d^2}\right )}{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right ) \left (-e^{2} x^{2}+d^{2}\right )^{p}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d^2-e^2\,x^2\right )}^p\,\left (d+e\,x\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 5.07, size = 85, normalized size = 0.77 \[ - \frac {d e^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (1 - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, 1 - p \\ 2 - p \end {matrix}\middle | {\frac {d^{2}}{e^{2} x^{2}}} \right )}}{2 x^{2} \Gamma \left (2 - p\right )} - \frac {d^{2 p} e {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d*e**(2*p)*x**(2*p)*exp(I*pi*p)*gamma(1 - p)*hyper((-p, 1 - p), (2 - p,), d**2/(e**2*x**2))/(2*x**2*gamma(2 -
p)) - d**(2*p)*e*hyper((-1/2, -p), (1/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/x

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