3.4.84 \(\int x^2 (d+e x) (a+b x^2)^p \, dx\) [384]

Optimal. Leaf size=100 \[ -\frac {a e \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {e \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)}+\frac {1}{3} d x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right ) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {778, 372, 371, 272, 45} \begin {gather*} -\frac {a e \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac {e \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}+\frac {1}{3} d x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(d + e*x)*(a + b*x^2)^p,x]

[Out]

-1/2*(a*e*(a + b*x^2)^(1 + p))/(b^2*(1 + p)) + (e*(a + b*x^2)^(2 + p))/(2*b^2*(2 + p)) + (d*x^3*(a + b*x^2)^p*
Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)])/(3*(1 + (b*x^2)/a)^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^2 (d+e x) \left (a+b x^2\right )^p \, dx &=d \int x^2 \left (a+b x^2\right )^p \, dx+e \int x^3 \left (a+b x^2\right )^p \, dx\\ &=\frac {1}{2} e \text {Subst}\left (\int x (a+b x)^p \, dx,x,x^2\right )+\left (d \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int x^2 \left (1+\frac {b x^2}{a}\right )^p \, dx\\ &=\frac {1}{3} d x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )+\frac {1}{2} e \text {Subst}\left (\int \left (-\frac {a (a+b x)^p}{b}+\frac {(a+b x)^{1+p}}{b}\right ) \, dx,x,x^2\right )\\ &=-\frac {a e \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {e \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)}+\frac {1}{3} d x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 87, normalized size = 0.87 \begin {gather*} \frac {1}{6} \left (a+b x^2\right )^p \left (-\frac {3 e \left (a+b x^2\right ) \left (a-b (1+p) x^2\right )}{b^2 (1+p) (2+p)}+2 d x^3 \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(d + e*x)*(a + b*x^2)^p,x]

[Out]

((a + b*x^2)^p*((-3*e*(a + b*x^2)*(a - b*(1 + p)*x^2))/(b^2*(1 + p)*(2 + p)) + (2*d*x^3*Hypergeometric2F1[3/2,
-p, 5/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(e*x+d)*(b*x^2+a)^p,x)

[Out]

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

[Out]

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

[Out]

integral((x^3*e + d*x^2)*(b*x^2 + a)^p, x)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (82) = 164\).
time = 5.00, size = 364, normalized size = 3.64 \begin {gather*} \frac {a^{p} d x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + e \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} - \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(e*x+d)*(b*x**2+a)**p,x)

[Out]

a**p*d*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + e*Piecewise((a**p*x**4/4, Eq(b, 0)), (a*log
(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + a*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + a/(2*a*b**2 + 2*b
**3*x**2) + b*x**2*log(x - sqrt(-a/b))/(2*a*b**2 + 2*b**3*x**2) + b*x**2*log(x + sqrt(-a/b))/(2*a*b**2 + 2*b**
3*x**2), Eq(p, -2)), (-a*log(x - sqrt(-a/b))/(2*b**2) - a*log(x + sqrt(-a/b))/(2*b**2) + x**2/(2*b), Eq(p, -1)
), (-a**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + a*b*p*x**2*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2
*p + 4*b**2) + b**2*p*x**4*(a + b*x**2)**p/(2*b**2*p**2 + 6*b**2*p + 4*b**2) + b**2*x**4*(a + b*x**2)**p/(2*b*
*2*p**2 + 6*b**2*p + 4*b**2), 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^2*(e*x+d)*(b*x^2+a)^p,x, algorithm="giac")

[Out]

integrate((x*e + d)*(b*x^2 + a)^p*x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x^2)^p*(d + e*x),x)

[Out]

int(x^2*(a + b*x^2)^p*(d + e*x), x)

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