3.3.70 \(\int \frac {x (d^2-e^2 x^2)^p}{d+e x} \, dx\) [270]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 90, 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 = {799, 778, 267, 372, 371} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(d*(d^2 - e^2*x^2)^p)/(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 267

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 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 799

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]

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*}

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Mathematica [A]
time = 0.22, size = 147, normalized size = 1.63 \begin {gather*} \frac {2^{-1+p} \left (1+\frac {e x}{d}\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 (1+p) x \left (\frac {1}{2}+\frac {e x}{2 d}\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,1+p;2+p;\frac {d-e x}{2 d}\right )\right )}{e^2 (1+p)} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x \left (-e^{2} x^{2}+d^{2}\right )^{p}}{e x +d}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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((-x^2*e^2 + d^2)^p*x/(x*e + d), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((-x^2*e^2 + d^2)^p*x/(x*e + d), x)

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Sympy [C] Result contains complex when optimal does not.
time = 4.24, size = 427, normalized size = 4.74 \begin {gather*} \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}\: \left |{\frac {d^{2}}{e^{2} 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 {gather*}

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/(e**2*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*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*p
i)/d**2)/(2*d*gamma(-p)*gamma(p + 1)), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((-x^2*e^2 + d^2)^p*x/(x*e + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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