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

Optimal. Leaf size=90 \[ -\frac{e x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{3}{2},1-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^2}-\frac{d \left (d^2-e^2 x^2\right )^p}{2 e^2 p} \]

[Out]

-(d*(d^2 - e^2*x^2)^p)/(2*e^2*p) - (e*x^3*(d^2 - e^2*x^2)^p*Hypergeometric2F1[3/2, 1 - p, 5/2, (e^2*x^2)/d^2])
/(3*d^2*(1 - (e^2*x^2)/d^2)^p)

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Rubi [A]  time = 0.0581513, antiderivative size = 90, normalized size of antiderivative = 1., 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 = {785, 764, 261, 365, 364} \[ -\frac{e x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{3}{2},1-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^2}-\frac{d \left (d^2-e^2 x^2\right )^p}{2 e^2 p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(d^2 - e^2*x^2)^p)/(2*e^2*p) - (e*x^3*(d^2 - e^2*x^2)^p*Hypergeometric2F1[3/2, 1 - p, 5/2, (e^2*x^2)/d^2])
/(3*d^2*(1 - (e^2*x^2)/d^2)^p)

Rule 785

Int[(x_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^m*e^m, Int[(x*(a + c*x^2)^(m
+ p))/(a*e + c*d*x)^m, x], x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[
m, 0] && EqQ[m, -1] &&  !ILtQ[p - 1/2, 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 261

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

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{x \left (d^2-e^2 x^2\right )^p}{d+e x} \, dx &=\frac{\int x \left (d^2 e-d e^2 x\right ) \left (d^2-e^2 x^2\right )^{-1+p} \, dx}{d e}\\ &=d \int x \left (d^2-e^2 x^2\right )^{-1+p} \, dx-e \int x^2 \left (d^2-e^2 x^2\right )^{-1+p} \, dx\\ &=-\frac{d \left (d^2-e^2 x^2\right )^p}{2 e^2 p}-\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 x^2 \left (1-\frac{e^2 x^2}{d^2}\right )^{-1+p} \, dx}{d^2}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^p}{2 e^2 p}-\frac{e x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{3}{2},1-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^2}\\ \end{align*}

Mathematica [A]  time = 0.112274, size = 147, normalized size = 1.63 \[ \frac{2^{p-1} \left (\frac{e x}{d}+1\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (2 e (p+1) x \left (\frac{e x}{2 d}+\frac{1}{2}\right )^p \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )+(d-e x) \left (1-\frac{e^2 x^2}{d^2}\right )^p \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )\right )}{e^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.662, size = 0, normalized size = 0. \begin{align*} \int{\frac{x \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{ex+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 6.05309, size = 430, normalized size = 4.78 \begin{align*} \begin{cases} \frac{0^{p} d d^{2 p} \log{\left (\frac{d^{2}}{e^{2} x^{2}} \right )}}{2 e^{2}} - \frac{0^{p} d d^{2 p} \log{\left (\frac{d^{2}}{e^{2} x^{2}} - 1 \right )}}{2 e^{2}} - \frac{0^{p} d d^{2 p} \operatorname{acoth}{\left (\frac{d}{e x} \right )}}{e^{2}} + \frac{0^{p} d^{2 p} x}{e} - \frac{e^{2 p} p x x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (- p - \frac{1}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, - p - \frac{1}{2} \\ \frac{1}{2} - p \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e \Gamma \left (\frac{1}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac{d^{2 p} x^{2} \Gamma \left (p\right ) \Gamma \left (1 - p\right ){{}_{3}F_{2}\left (\begin{matrix} 2, 1, 1 - p \\ 2, 2 \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{2 d \Gamma \left (- 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 d^{2 p} \log{\left (\frac{d^{2}}{e^{2} x^{2}} \right )}}{2 e^{2}} - \frac{0^{p} d d^{2 p} \log{\left (- \frac{d^{2}}{e^{2} x^{2}} + 1 \right )}}{2 e^{2}} - \frac{0^{p} d d^{2 p} \operatorname{atanh}{\left (\frac{d}{e x} \right )}}{e^{2}} + \frac{0^{p} d^{2 p} x}{e} - \frac{e^{2 p} p x x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (- p - \frac{1}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, - p - \frac{1}{2} \\ \frac{1}{2} - p \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e \Gamma \left (\frac{1}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac{d^{2 p} x^{2} \Gamma \left (p\right ) \Gamma \left (1 - p\right ){{}_{3}F_{2}\left (\begin{matrix} 2, 1, 1 - p \\ 2, 2 \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{2 d \Gamma \left (- 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(x*(-e**2*x**2+d**2)**p/(e*x+d),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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