3.382 \(\int x^4 (d+e x) (a+b x^2)^p \, dx\)
Optimal. Leaf size=125 \[ \frac{a^2 e \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}-\frac{a e \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac{e \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac{1}{5} d x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right ) \]
[Out]
(a^2*e*(a + b*x^2)^(1 + p))/(2*b^3*(1 + p)) - (a*e*(a + b*x^2)^(2 + p))/(b^3*(2 + p)) + (e*(a + b*x^2)^(3 + p)
)/(2*b^3*(3 + p)) + (d*x^5*(a + b*x^2)^p*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(5*(1 + (b*x^2)/a)^p)
________________________________________________________________________________________
Rubi [A] time = 0.0848504, antiderivative size = 125, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used =
{764, 365, 364, 266, 43} \[ \frac{a^2 e \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}-\frac{a e \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac{e \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac{1}{5} d x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right ) \]
Antiderivative was successfully verified.
[In]
Int[x^4*(d + e*x)*(a + b*x^2)^p,x]
[Out]
(a^2*e*(a + b*x^2)^(1 + p))/(2*b^3*(1 + p)) - (a*e*(a + b*x^2)^(2 + p))/(b^3*(2 + p)) + (e*(a + b*x^2)^(3 + p)
)/(2*b^3*(3 + p)) + (d*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 764
Int[(x_)^(m_.)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[f, Int[x^m*(a + c*x^2)^p, x]
, x] + Dist[g, Int[x^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && IntegerQ[m] && !IntegerQ[2
*p]
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 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) \left (a+b x^2\right )^p \, dx &=d \int x^4 \left (a+b x^2\right )^p \, dx+e \int x^5 \left (a+b x^2\right )^p \, dx\\ &=\frac{1}{2} e \operatorname{Subst}\left (\int x^2 (a+b x)^p \, dx,x,x^2\right )+\left (d \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{1}{5} d 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 )+\frac{1}{2} 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 e \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}-\frac{a e \left (a+b x^2\right )^{2+p}}{b^3 (2+p)}+\frac{e \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}+\frac{1}{5} d 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 = 0.0714254, size = 112, normalized size = 0.9 \[ \frac{1}{10} \left (a+b x^2\right )^p \left (\frac{5 e \left (a+b x^2\right ) \left (2 a^2-2 a b (p+1) x^2+b^2 \left (p^2+3 p+2\right ) x^4\right )}{b^3 (p+1) (p+2) (p+3)}+2 d x^5 \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[x^4*(d + e*x)*(a + b*x^2)^p,x]
[Out]
((a + b*x^2)^p*((5*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)) + (2*d*x^5*Hypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/10
________________________________________________________________________________________
Maple [F] time = 0.418, size = 0, normalized size = 0. \begin{align*} \int{x}^{4} \left ( ex+d \right ) \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)*(b*x^2+a)^p,x)
[Out]
int(x^4*(e*x+d)*(b*x^2+a)^p,x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}{\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)*(b*x^2+a)^p,x, algorithm="maxima")
[Out]
integrate((e*x + d)*(b*x^2 + a)^p*x^4, x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e x^{5} + d 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)*(b*x^2+a)^p,x, algorithm="fricas")
[Out]
integral((e*x^5 + d*x^4)*(b*x^2 + a)^p, x)
________________________________________________________________________________________
Sympy [C] time = 37.9123, size = 1010, normalized size = 8.08 \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)*(b*x**2+a)**p,x)
[Out]
a**p*d*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + 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*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*sqr
t(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 + 1
2*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 )}{\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)*(b*x^2+a)^p,x, algorithm="giac")
[Out]
integrate((e*x + d)*(b*x^2 + a)^p*x^4, x)