3.392 \(\int x^3 (d+e x)^2 (a+b x^2)^p \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.138046, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1652, 446, 77, 12, 365, 364} \[ -\frac{a \left (b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}+\frac{\left (b d^2-2 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^3 (p+2)}+\frac{e^2 \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac{2}{5} d e x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*(d + e*x)^2*(a + b*x^2)^p,x]

[Out]

-(a*(b*d^2 - a*e^2)*(a + b*x^2)^(1 + p))/(2*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 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]

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 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 12

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

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 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])

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} \operatorname{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} \operatorname{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*}

Mathematica [A]  time = 0.150886, size = 152, normalized size = 1.02 \[ \frac{1}{10} \left (a+b x^2\right )^p \left (\frac{5 e^2 \left (a+b x^2\right ) \left (2 a^2-2 a b (p+1) x^2+b^2 \left (p^2+3 p+2\right ) x^4\right )}{b^3 (p+1) (p+2) (p+3)}+\frac{5 d^2 \left (a+b x^2\right ) \left (b (p+1) x^2-a\right )}{b^2 (p+1) (p+2)}+4 d e x^5 \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^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 = 0.62, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( ex+d \right ) ^{2} \left ( b{x}^{2}+a \right ) ^{p}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \frac{{\left (b^{2}{\left (p + 1\right )} x^{4} + a b p x^{2} - a^{2}\right )}{\left (b x^{2} + a\right )}^{p} d^{2}}{2 \,{\left (p^{2} + 3 \, p + 2\right )} b^{2}} + \int{\left (e^{2} x^{5} + 2 \, d e x^{4}\right )}{\left (b x^{2} + a\right )}^{p}\,{d x} \end{align*}

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((e^2*x^5 + 2*d*e*x
^4)*(b*x^2 + a)^p, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e^{2} x^{5} + 2 \, d e x^{4} + d^{2} x^{3}\right )}{\left (b x^{2} + a\right )}^{p}, x\right ) \end{align*}

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((e^2*x^5 + 2*d*e*x^4 + d^2*x^3)*(b*x^2 + a)^p, x)

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Sympy [C]  time = 44.1619, size = 1384, normalized size = 9.29 \begin{align*} \text{result too large to display} \end{align*}

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(-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*l
og(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**2*Pi
ecewise((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) + 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) + 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*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/(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)*sqrt(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 + 2
2*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., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{2}{\left (b x^{2} + a\right )}^{p} x^{3}\,{d x} \end{align*}

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((e*x + d)^2*(b*x^2 + a)^p*x^3, x)