3.3.42 \(\int x^3 (d+e x) (d^2-e^2 x^2)^p \, dx\) [242]

Optimal. Leaf size=120 \[ -\frac {d^3 \left (d^2-e^2 x^2\right )^{1+p}}{2 e^4 (1+p)}+\frac {d \left (d^2-e^2 x^2\right )^{2+p}}{2 e^4 (2+p)}+\frac {1}{5} e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right ) \]

[Out]

-1/2*d^3*(-e^2*x^2+d^2)^(1+p)/e^4/(1+p)+1/2*d*(-e^2*x^2+d^2)^(2+p)/e^4/(2+p)+1/5*e*x^5*(-e^2*x^2+d^2)^p*hyperg
eom([5/2, -p],[7/2],e^2*x^2/d^2)/((1-e^2*x^2/d^2)^p)

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Rubi [A]
time = 0.05, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {778, 272, 45, 372, 371} \begin {gather*} \frac {1}{5} e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )+\frac {d \left (d^2-e^2 x^2\right )^{p+2}}{2 e^4 (p+2)}-\frac {d^3 \left (d^2-e^2 x^2\right )^{p+1}}{2 e^4 (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(d^3*(d^2 - e^2*x^2)^(1 + p))/(e^4*(1 + p)) + (d*(d^2 - e^2*x^2)^(2 + p))/(2*e^4*(2 + p)) + (e*x^5*(d^2 -
e^2*x^2)^p*Hypergeometric2F1[5/2, -p, 7/2, (e^2*x^2)/d^2])/(5*(1 - (e^2*x^2)/d^2)^p)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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]

Rubi steps

\begin {align*} \int x^3 (d+e x) \left (d^2-e^2 x^2\right )^p \, dx &=d \int x^3 \left (d^2-e^2 x^2\right )^p \, dx+e \int x^4 \left (d^2-e^2 x^2\right )^p \, dx\\ &=\frac {1}{2} d \text {Subst}\left (\int x \left (d^2-e^2 x\right )^p \, 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 x^4 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx\\ &=\frac {1}{5} e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )+\frac {1}{2} d \text {Subst}\left (\int \left (\frac {d^2 \left (d^2-e^2 x\right )^p}{e^2}-\frac {\left (d^2-e^2 x\right )^{1+p}}{e^2}\right ) \, dx,x,x^2\right )\\ &=-\frac {d^3 \left (d^2-e^2 x^2\right )^{1+p}}{2 e^4 (1+p)}+\frac {d \left (d^2-e^2 x^2\right )^{2+p}}{2 e^4 (2+p)}+\frac {1}{5} e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 106, normalized size = 0.88 \begin {gather*} \frac {\left (d^2-e^2 x^2\right )^p \left (-\frac {5 d \left (d^2-e^2 x^2\right ) \left (d^2+e^2 (1+p) x^2\right )}{(1+p) (2+p)}+2 e^5 x^5 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};\frac {e^2 x^2}{d^2}\right )\right )}{10 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d^2 - e^2*x^2)^p*((-5*d*(d^2 - e^2*x^2)*(d^2 + e^2*(1 + p)*x^2))/((1 + p)*(2 + p)) + (2*e^5*x^5*Hypergeometr
ic2F1[5/2, -p, 7/2, (e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^2)^p))/(10*e^4)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int x^{3} \left (e x +d \right ) \left (-e^{2} x^{2}+d^{2}\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

1/2*((p + 1)*x^4*e^4 - d^2*p*x^2*e^2 - d^4)*d*e^(p*log(-x^2*e^2 + d^2) - 4)/(p^2 + 3*p + 2) + e*integrate(x^4*
e^(p*log(x*e + d) + p*log(-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^3*(e*x+d)*(-e^2*x^2+d^2)^p,x, algorithm="fricas")

[Out]

integral((x^4*e + d*x^3)*(-x^2*e^2 + d^2)^p, x)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (97) = 194\).
time = 1.84, size = 382, normalized size = 3.18 \begin {gather*} d \left (\begin {cases} \frac {x^{4} \left (d^{2}\right )^{p}}{4} & \text {for}\: e = 0 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{2 e^{4}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{2 e^{4}} - \frac {x^{2}}{2 e^{2}} & \text {for}\: p = -1 \\- \frac {d^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} - \frac {d^{2} e^{2} p x^{2} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} p x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} & \text {otherwise} \end {cases}\right ) + \frac {d^{2 p} e x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*Piecewise((x**4*(d**2)**p/4, Eq(e, 0)), (-d**2*log(-d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2) - d**2*log(d/e + x
)/(-2*d**2*e**4 + 2*e**6*x**2) - d**2/(-2*d**2*e**4 + 2*e**6*x**2) + e**2*x**2*log(-d/e + x)/(-2*d**2*e**4 + 2
*e**6*x**2) + e**2*x**2*log(d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2), Eq(p, -2)), (-d**2*log(-d/e + x)/(2*e**4) -
d**2*log(d/e + x)/(2*e**4) - x**2/(2*e**2), Eq(p, -1)), (-d**4*(d**2 - e**2*x**2)**p/(2*e**4*p**2 + 6*e**4*p
+ 4*e**4) - d**2*e**2*p*x**2*(d**2 - e**2*x**2)**p/(2*e**4*p**2 + 6*e**4*p + 4*e**4) + e**4*p*x**4*(d**2 - e**
2*x**2)**p/(2*e**4*p**2 + 6*e**4*p + 4*e**4) + e**4*x**4*(d**2 - e**2*x**2)**p/(2*e**4*p**2 + 6*e**4*p + 4*e**
4), True)) + d**(2*p)*e*x**5*hyper((5/2, -p), (7/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/5

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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