3.391 \(\int x^4 (d+e x)^2 (a+b x^2)^p \, dx\)
Optimal. Leaf size=177 \[ \frac{a^2 d e \left (a+b x^2\right )^{p+1}}{b^3 (p+1)}-\frac{2 a d e \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac{d e \left (a+b x^2\right )^{p+3}}{b^3 (p+3)}-\frac{x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (5 a e^2-b d^2 (2 p+7)\right ) \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )}{5 b (2 p+7)}+\frac{e^2 x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)} \]
[Out]
(a^2*d*e*(a + b*x^2)^(1 + p))/(b^3*(1 + p)) + (e^2*x^5*(a + b*x^2)^(1 + p))/(b*(7 + 2*p)) - (2*a*d*e*(a + b*x^
2)^(2 + p))/(b^3*(2 + p)) + (d*e*(a + b*x^2)^(3 + p))/(b^3*(3 + p)) - ((5*a*e^2 - b*d^2*(7 + 2*p))*x^5*(a + b*
x^2)^p*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(5*b*(7 + 2*p)*(1 + (b*x^2)/a)^p)
________________________________________________________________________________________
Rubi [A] time = 0.162716, antiderivative size = 169, normalized size of antiderivative =
0.95, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules
used = {1652, 459, 365, 364, 12, 266, 43} \[ \frac{a^2 d e \left (a+b x^2\right )^{p+1}}{b^3 (p+1)}-\frac{2 a d e \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac{d e \left (a+b x^2\right )^{p+3}}{b^3 (p+3)}+\frac{1}{5} x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (d^2-\frac{5 a e^2}{2 b p+7 b}\right ) \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )+\frac{e^2 x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)} \]
Antiderivative was successfully verified.
[In]
Int[x^4*(d + e*x)^2*(a + b*x^2)^p,x]
[Out]
(a^2*d*e*(a + b*x^2)^(1 + p))/(b^3*(1 + p)) + (e^2*x^5*(a + b*x^2)^(1 + p))/(b*(7 + 2*p)) - (2*a*d*e*(a + b*x^
2)^(2 + p))/(b^3*(2 + p)) + (d*e*(a + b*x^2)^(3 + p))/(b^3*(3 + p)) + ((d^2 - (5*a*e^2)/(7*b + 2*b*p))*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 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 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])
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 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])
Rubi steps
\begin{align*} \int x^4 (d+e x)^2 \left (a+b x^2\right )^p \, dx &=\int 2 d e x^5 \left (a+b x^2\right )^p \, dx+\int x^4 \left (a+b x^2\right )^p \left (d^2+e^2 x^2\right ) \, dx\\ &=\frac{e^2 x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}+(2 d e) \int x^5 \left (a+b x^2\right )^p \, dx-\left (-d^2+\frac{5 a e^2}{7 b+2 b p}\right ) \int x^4 \left (a+b x^2\right )^p \, dx\\ &=\frac{e^2 x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}+(d e) \operatorname{Subst}\left (\int x^2 (a+b x)^p \, dx,x,x^2\right )-\left (\left (-d^2+\frac{5 a e^2}{7 b+2 b p}\right ) \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{e^2 x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}+\frac{1}{5} \left (d^2-\frac{5 a e^2}{7 b+2 b p}\right ) 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 )+(d e) \operatorname{Subst}\left (\int \left (\frac{a^2 (a+b x)^p}{b^2}-\frac{2 a (a+b x)^{1+p}}{b^2}+\frac{(a+b x)^{2+p}}{b^2}\right ) \, dx,x,x^2\right )\\ &=\frac{a^2 d e \left (a+b x^2\right )^{1+p}}{b^3 (1+p)}+\frac{e^2 x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}-\frac{2 a d e \left (a+b x^2\right )^{2+p}}{b^3 (2+p)}+\frac{d e \left (a+b x^2\right )^{3+p}}{b^3 (3+p)}+\frac{1}{5} \left (d^2-\frac{5 a e^2}{7 b+2 b p}\right ) 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 = 0.169891, size = 156, normalized size = 0.88 \[ \frac{1}{35} \left (a+b x^2\right )^p \left (\frac{35 d e \left (a+b x^2\right ) \left (2 a^2-2 a b (p+1) x^2+b^2 \left (p^2+3 p+2\right ) x^4\right )}{b^3 (p+1) (p+2) (p+3)}+7 d^2 x^5 \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )+5 e^2 x^7 \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{7}{2},-p;\frac{9}{2};-\frac{b x^2}{a}\right )\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[x^4*(d + e*x)^2*(a + b*x^2)^p,x]
[Out]
((a + b*x^2)^p*((35*d*e*(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)) + (7*d^2*x^5*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p + (5*e^2*x^7*Hypergeo
metric2F1[7/2, -p, 9/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/35
________________________________________________________________________________________
Maple [F] time = 0.598, size = 0, normalized size = 0. \begin{align*} \int{x}^{4} \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^4*(e*x+d)^2*(b*x^2+a)^p,x)
[Out]
int(x^4*(e*x+d)^2*(b*x^2+a)^p,x)
________________________________________________________________________________________
Maxima [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^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*(e*x+d)^2*(b*x^2+a)^p,x, algorithm="maxima")
[Out]
integrate((e*x + d)^2*(b*x^2 + a)^p*x^4, x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e^{2} x^{6} + 2 \, d e x^{5} + d^{2} x^{4}\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^4*(e*x+d)^2*(b*x^2+a)^p,x, algorithm="fricas")
[Out]
integral((e^2*x^6 + 2*d*e*x^5 + d^2*x^4)*(b*x^2 + a)^p, x)
________________________________________________________________________________________
Sympy [C] time = 84.8486, size = 1046, normalized size = 5.91 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**4*(e*x+d)**2*(b*x**2+a)**p,x)
[Out]
a**p*d**2*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + a**p*e**2*x**7*hyper((7/2, -p), (9/2,),
b*x**2*exp_polar(I*pi)/a)/7 + 2*d*e*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) + 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*l
og(-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))
________________________________________________________________________________________
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^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*(e*x+d)^2*(b*x^2+a)^p,x, algorithm="giac")
[Out]
integrate((e*x + d)^2*(b*x^2 + a)^p*x^4, x)