3.4.92 \(\int x^3 (d+e x)^2 (a+b x^2)^p \, dx\) [392]

Optimal. Leaf size=149 \[ -\frac {a \left (b d^2-a e^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}+\frac {\left (b d^2-2 a e^2\right ) \left (a+b x^2\right )^{2+p}}{2 b^3 (2+p)}+\frac {e^2 \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}+\frac {2}{5} d e x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right ) \]

[Out]

-1/2*a*(-a*e^2+b*d^2)*(b*x^2+a)^(1+p)/b^3/(1+p)+1/2*(-2*a*e^2+b*d^2)*(b*x^2+a)^(2+p)/b^3/(2+p)+1/2*e^2*(b*x^2+
a)^(3+p)/b^3/(3+p)+2/5*d*e*x^5*(b*x^2+a)^p*hypergeom([5/2, -p],[7/2],-b*x^2/a)/((1+b*x^2/a)^p)

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Rubi [A]
time = 0.10, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1666, 457, 78, 12, 372, 371} \begin {gather*} -\frac {a \left (b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}+\frac {\left (b d^2-2 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^3 (p+2)}+\frac {e^2 \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac {2}{5} d e x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(d + e*x)^2*(a + b*x^2)^p,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1666

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[x^m*Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2] && IGtQ[m, -2] &&  !
IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]
time = 0.19, size = 152, normalized size = 1.02 \begin {gather*} \frac {1}{10} \left (a+b x^2\right )^p \left (\frac {5 d^2 \left (a+b x^2\right ) \left (-a+b (1+p) x^2\right )}{b^2 (1+p) (2+p)}+\frac {5 e^2 \left (a+b x^2\right ) \left (2 a^2-2 a b (1+p) x^2+b^2 \left (2+3 p+p^2\right ) x^4\right )}{b^3 (1+p) (2+p) (3+p)}+4 d e x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^p*((5*d^2*(a + b*x^2)*(-a + b*(1 + p)*x^2))/(b^2*(1 + p)*(2 + p)) + (5*e^2*(a + b*x^2)*(2*a^2 - 2
*a*b*(1 + p)*x^2 + b^2*(2 + 3*p + p^2)*x^4))/(b^3*(1 + p)*(2 + p)*(3 + p)) + (4*d*e*x^5*Hypergeometric2F1[5/2,
-p, 7/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/10

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x^{3} \left (e x +d \right )^{2} \left (b \,x^{2}+a \right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

1/2*(b^2*(p + 1)*x^4 + a*b*p*x^2 - a^2)*(b*x^2 + a)^p*d^2/((p^2 + 3*p + 2)*b^2) + integrate((x^5*e^2 + 2*d*x^4
*e)*(b*x^2 + a)^p, 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)^2*(b*x^2+a)^p,x, algorithm="fricas")

[Out]

integral((x^5*e^2 + 2*d*x^4*e + d^2*x^3)*(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. 337 vs. \(2 (124) = 248\).
time = 10.34, size = 1294, normalized size = 8.68 \begin {gather*} \frac {2 a^{p} d e x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + d^{2} \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 ) + e^{2} \left (\begin {cases} \frac {a^{p} x^{6}}{6} & \text {for}\: b = 0 \\\frac {2 a^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 a^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {3 a^{2}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 b^{2} x^{4} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 b^{2} x^{4} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} & \text {for}\: p = -3 \\- \frac {2 a^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} + \frac {b^{2} x^{4}}{2 a b^{3} + 2 b^{4} x^{2}} & \text {for}\: p = -2 \\\frac {a^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{3}} + \frac {a^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{3}} - \frac {a x^{2}}{2 b^{2}} + \frac {x^{4}}{4 b} & \text {for}\: p = -1 \\\frac {2 a^{3} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} - \frac {2 a^{2} b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {a b^{2} p^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {a b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {b^{3} p^{2} x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {3 b^{3} p x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {2 b^{3} x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**p*d*e*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + d**2*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)) + e**2*Piecewise((a**p*x**6/6, Eq(b, 0)), (2*a**2*log(x - sqrt(-a/
b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*a**2*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*
b**5*x**4) + 3*a**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*log(x - sqrt(-a/b))/(4*a**2*b**3
+ 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) +
4*a*b*x**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b*
*4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4), Eq(p, -3
)), (-2*a**2*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a**2*log(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2
) - 2*a**2/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**2*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**2*log
(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) + b**2*x**4/(2*a*b**3 + 2*b**4*x**2), Eq(p, -2)), (a**2*log(x - sqrt
(-a/b))/(2*b**3) + a**2*log(x + sqrt(-a/b))/(2*b**3) - a*x**2/(2*b**2) + x**4/(4*b), Eq(p, -1)), (2*a**3*(a +
b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) - 2*a**2*b*p*x**2*(a + b*x**2)**p/(2*b**3*p**3 +
12*b**3*p**2 + 22*b**3*p + 12*b**3) + a*b**2*p**2*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*
p + 12*b**3) + a*b**2*p*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + b**3*p**2*x*
*6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + 3*b**3*p*x**6*(a + b*x**2)**p/(2*b**3*
p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + 2*b**3*x**6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3
*p + 12*b**3), 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^3*(e*x+d)^2*(b*x^2+a)^p,x, algorithm="giac")

[Out]

integrate((x*e + d)^2*(b*x^2 + a)^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 (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^2)^p*(d + e*x)^2,x)

[Out]

int(x^3*(a + b*x^2)^p*(d + e*x)^2, x)

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