3.402 \(\int x^2 (d+e x)^3 (a+b x^2)^p \, dx\)
Optimal. Leaf size=210 \[ -\frac {a e \left (3 b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}+\frac {e \left (3 b d^2-2 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^3 (p+2)}+\frac {e^3 \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}-\frac {d x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (9 a e^2-b d^2 (2 p+5)\right ) \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )}{3 b (2 p+5)}+\frac {3 d e^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]
[Out]
-1/2*a*e*(-a*e^2+3*b*d^2)*(b*x^2+a)^(1+p)/b^3/(1+p)+3*d*e^2*x^3*(b*x^2+a)^(1+p)/b/(5+2*p)+1/2*e*(-2*a*e^2+3*b*
d^2)*(b*x^2+a)^(2+p)/b^3/(2+p)+1/2*e^3*(b*x^2+a)^(3+p)/b^3/(3+p)-1/3*d*(9*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 = 0.20, antiderivative size = 202, normalized size of antiderivative = 0.96,
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 =
{1652, 459, 365, 364, 446, 77} \[ -\frac {a e \left (3 b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}+\frac {e \left (3 b d^2-2 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^3 (p+2)}+\frac {e^3 \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac {1}{3} d x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (d^2-\frac {9 a e^2}{2 b p+5 b}\right ) \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )+\frac {3 d e^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]
Antiderivative was successfully verified.
[In]
Int[x^2*(d + e*x)^3*(a + b*x^2)^p,x]
[Out]
-(a*e*(3*b*d^2 - a*e^2)*(a + b*x^2)^(1 + p))/(2*b^3*(1 + p)) + (3*d*e^2*x^3*(a + b*x^2)^(1 + p))/(b*(5 + 2*p))
+ (e*(3*b*d^2 - 2*a*e^2)*(a + b*x^2)^(2 + p))/(2*b^3*(2 + p)) + (e^3*(a + b*x^2)^(3 + p))/(2*b^3*(3 + p)) + (
d*(d^2 - (9*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 77
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 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 446
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 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)^3 \left (a+b x^2\right )^p \, dx &=\int x^2 \left (a+b x^2\right )^p \left (d^3+3 d e^2 x^2\right ) \, dx+\int x^3 \left (a+b x^2\right )^p \left (3 d^2 e+e^3 x^2\right ) \, dx\\ &=\frac {3 d e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {1}{2} \operatorname {Subst}\left (\int x (a+b x)^p \left (3 d^2 e+e^3 x\right ) \, dx,x,x^2\right )+\left (d \left (d^2-\frac {9 a e^2}{5 b+2 b p}\right )\right ) \int x^2 \left (a+b x^2\right )^p \, dx\\ &=\frac {3 d e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {a e \left (-3 b d^2+a e^2\right ) (a+b x)^p}{b^2}+\frac {\left (3 b d^2 e-2 a e^3\right ) (a+b x)^{1+p}}{b^2}+\frac {e^3 (a+b x)^{2+p}}{b^2}\right ) \, dx,x,x^2\right )+\left (d \left (d^2-\frac {9 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 {a e \left (3 b d^2-a e^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}+\frac {3 d e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {e \left (3 b d^2-2 a e^2\right ) \left (a+b x^2\right )^{2+p}}{2 b^3 (2+p)}+\frac {e^3 \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}+\frac {1}{3} d \left (d^2-\frac {9 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 = 0.14, size = 196, normalized size = 0.93 \[ \frac {1}{30} \left (a+b x^2\right )^p \left (\frac {15 e^3 \left (a+b x^2\right ) \left (2 a^2-2 a b (p+1) x^2+b^2 \left (p^2+3 p+2\right ) x^4\right )}{b^3 (p+1) (p+2) (p+3)}+\frac {45 d^2 e \left (a+b x^2\right ) \left (b (p+1) x^2-a\right )}{b^2 (p+1) (p+2)}+10 d^3 x^3 \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )+18 d e^2 x^5 \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right )\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[x^2*(d + e*x)^3*(a + b*x^2)^p,x]
[Out]
((a + b*x^2)^p*((45*d^2*e*(a + b*x^2)*(-a + b*(1 + p)*x^2))/(b^2*(1 + p)*(2 + p)) + (15*e^3*(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)) + (10*d^3*x^3*Hypergeometric2F1
[3/2, -p, 5/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p + (18*d*e^2*x^5*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])
/(1 + (b*x^2)/a)^p))/30
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fricas [F] time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{3} x^{5} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{3} + d^{3} x^{2}\right )} {\left (b x^{2} + a\right )}^{p}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*(e*x+d)^3*(b*x^2+a)^p,x, algorithm="fricas")
[Out]
integral((e^3*x^5 + 3*d*e^2*x^4 + 3*d^2*e*x^3 + d^3*x^2)*(b*x^2 + a)^p, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p} x^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*(e*x+d)^3*(b*x^2+a)^p,x, algorithm="giac")
[Out]
integrate((e*x + d)^3*(b*x^2 + a)^p*x^2, x)
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maple [F] time = 0.06, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{3} x^{2} \left (b \,x^{2}+a \right )^{p}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*(e*x+d)^3*(b*x^2+a)^p,x)
[Out]
int(x^2*(e*x+d)^3*(b*x^2+a)^p,x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p} x^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*(e*x+d)^3*(b*x^2+a)^p,x, algorithm="maxima")
[Out]
integrate((e*x + d)^3*(b*x^2 + a)^p*x^2, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^3 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*(a + b*x^2)^p*(d + e*x)^3,x)
[Out]
int(x^2*(a + b*x^2)^p*(d + e*x)^3, x)
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sympy [C] time = 29.71, size = 1421, normalized size = 6.77 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2*(e*x+d)**3*(b*x**2+a)**p,x)
[Out]
a**p*d**3*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + 3*a**p*d*e**2*x**5*hyper((5/2, -p), (7/2
,), b*x**2*exp_polar(I*pi)/a)/5 + 3*d**2*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)) + e**3*Piecewise((a**p*x**6/6, Eq(b, 0)),
(2*a**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*a**2*log(I*sqrt(a)*sqrt(
1/b) + x)/(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(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*log(I*sqrt(a)
*sqrt(1/b) + x)/(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(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*l
og(I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4), Eq(p, -3)), (-2*a**2*log(-I*sqrt(a)*s
qrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) - 2*a**2*log(I*sqrt(a)*sqrt(1/b) + x)/(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(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**2*log
(I*sqrt(a)*sqrt(1/b) + x)/(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
(-I*sqrt(a)*sqrt(1/b) + x)/(2*b**3) + a**2*log(I*sqrt(a)*sqrt(1/b) + x)/(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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