3.393 \(\int x^2 (d+e x)^2 (a+b x^2)^p \, dx\)

Optimal. Leaf size=152 \[ -\frac {a d e \left (a+b x^2\right )^{p+1}}{b^2 (p+1)}+\frac {d e \left (a+b x^2\right )^{p+2}}{b^2 (p+2)}-\frac {x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a e^2-b d^2 (2 p+5)\right ) \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )}{3 b (2 p+5)}+\frac {e^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 144, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1652, 459, 365, 364, 12, 266, 43} \[ -\frac {a d e \left (a+b x^2\right )^{p+1}}{b^2 (p+1)}+\frac {d e \left (a+b x^2\right )^{p+2}}{b^2 (p+2)}+\frac {1}{3} x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (d^2-\frac {3 a e^2}{2 b p+5 b}\right ) \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )+\frac {e^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(d + e*x)^2*(a + b*x^2)^p,x]

[Out]

-((a*d*e*(a + b*x^2)^(1 + p))/(b^2*(1 + p))) + (e^2*x^3*(a + b*x^2)^(1 + p))/(b*(5 + 2*p)) + (d*e*(a + b*x^2)^
(2 + p))/(b^2*(2 + p)) + ((d^2 - (3*a*e^2)/(5*b + 2*b*p))*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 12

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

Rule 43

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 266

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 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])

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 459

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

Rule 1652

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^2 (d+e x)^2 \left (a+b x^2\right )^p \, dx &=\int 2 d e x^3 \left (a+b x^2\right )^p \, dx+\int x^2 \left (a+b x^2\right )^p \left (d^2+e^2 x^2\right ) \, dx\\ &=\frac {e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+(2 d e) \int x^3 \left (a+b x^2\right )^p \, dx-\left (-d^2+\frac {3 a e^2}{5 b+2 b p}\right ) \int x^2 \left (a+b x^2\right )^p \, dx\\ &=\frac {e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+(d e) \operatorname {Subst}\left (\int x (a+b x)^p \, dx,x,x^2\right )-\left (\left (-d^2+\frac {3 a e^2}{5 b+2 b p}\right ) \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 {e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {1}{3} \left (d^2-\frac {3 a e^2}{5 b+2 b p}\right ) 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 )+(d e) \operatorname {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 d e \left (a+b x^2\right )^{1+p}}{b^2 (1+p)}+\frac {e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {d e \left (a+b x^2\right )^{2+p}}{b^2 (2+p)}+\frac {1}{3} \left (d^2-\frac {3 a e^2}{5 b+2 b p}\right ) 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 = 0.15, size = 139, normalized size = 0.91 \[ \frac {1}{15} \left (a+b x^2\right )^p \left (\frac {3 e \left (e \left (p^2+3 p+2\right ) x^5 \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right )-\frac {5 d \left (a+b x^2\right ) \left (a-b (p+1) x^2\right )}{b^2}\right )}{(p+1) (p+2)}+5 d^2 x^3 \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^p*((5*d^2*x^3*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p + (3*e*((-5*d*(a +
b*x^2)*(a - b*(1 + p)*x^2))/b^2 + (e*(2 + 3*p + p^2)*x^5*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(1 +
(b*x^2)/a)^p))/((1 + p)*(2 + p))))/15

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{2} x^{4} + 2 \, d e x^{3} + d^{2} x^{2}\right )} {\left (b x^{2} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{2} x^{2} \left (b \,x^{2}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*(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 \[ \int {\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x^2)^p*(d + e*x)^2,x)

[Out]

int(x^2*(a + b*x^2)^p*(d + e*x)^2, x)

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sympy [C]  time = 21.18, size = 430, normalized size = 2.83 \[ \frac {a^{p} d^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {a^{p} e^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + 2 d e \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{2}} - \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*d**2*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + a**p*e**2*x**5*hyper((5/2, -p), (7/2,),
b*x**2*exp_polar(I*pi)/a)/5 + 2*d*e*Piecewise((a**p*x**4/4, Eq(b, 0)), (a*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b
**2 + 2*b**3*x**2) + a*log(I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**2 + 2*b**3*x**2) + a/(2*a*b**2 + 2*b**3*x**2) + b*
x**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**2 + 2*b**3*x**2) + b*x**2*log(I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**2 +
2*b**3*x**2), Eq(p, -2)), (-a*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*b**2) - a*log(I*sqrt(a)*sqrt(1/b) + x)/(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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