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

Optimal. Leaf size=100 \[ -\frac{a d \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac{d \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}+\frac{1}{5} e x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right ) \]

[Out]

-(a*d*(a + b*x^2)^(1 + p))/(2*b^2*(1 + p)) + (d*(a + b*x^2)^(2 + p))/(2*b^2*(2 + p)) + (e*x^5*(a + b*x^2)^p*Hy
pergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])/(5*(1 + (b*x^2)/a)^p)

________________________________________________________________________________________

Rubi [A]  time = 0.064525, antiderivative size = 100, 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, 266, 43, 365, 364} \[ -\frac{a d \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac{d \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}+\frac{1}{5} e x^5 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*(d + e*x)*(a + b*x^2)^p,x]

[Out]

-(a*d*(a + b*x^2)^(1 + p))/(2*b^2*(1 + p)) + (d*(a + b*x^2)^(2 + p))/(2*b^2*(2 + p)) + (e*x^5*(a + b*x^2)^p*Hy
pergeometric2F1[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 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])

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) \left (a+b x^2\right )^p \, dx &=d \int x^3 \left (a+b x^2\right )^p \, dx+e \int x^4 \left (a+b x^2\right )^p \, dx\\ &=\frac{1}{2} d \operatorname{Subst}\left (\int x (a+b x)^p \, dx,x,x^2\right )+\left (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{1}{5} 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 )+\frac{1}{2} d \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 \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac{d \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)}+\frac{1}{5} 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 = 0.0751153, size = 87, normalized size = 0.87 \[ \frac{1}{10} \left (a+b x^2\right )^p \left (2 e x^5 \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )-\frac{5 d \left (a+b x^2\right ) \left (a-b (p+1) x^2\right )}{b^2 (p+1) (p+2)}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(d + e*x)*(a + b*x^2)^p,x]

[Out]

((a + b*x^2)^p*((-5*d*(a + b*x^2)*(a - b*(1 + p)*x^2))/(b^2*(1 + p)*(2 + p)) + (2*e*x^5*Hypergeometric2F1[5/2,
-p, 7/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/10

________________________________________________________________________________________

Maple [F]  time = 0.411, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \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^3*(e*x+d)*(b*x^2+a)^p,x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)*(b*x^2+a)^p,x, algorithm="maxima")

[Out]

e*integrate((b*x^2 + a)^p*x^4, x) + 1/2*(b^2*(p + 1)*x^4 + a*b*p*x^2 - a^2)*(b*x^2 + a)^p*d/((p^2 + 3*p + 2)*b
^2)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [C]  time = 28.6126, size = 394, normalized size = 3.94 \begin{align*} \frac{a^{p} e 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} + d \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x+d)*(b*x**2+a)**p,x)

[Out]

a**p*e*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + d*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*sq
rt(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))

________________________________________________________________________________________

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^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)*(b*x^2+a)^p,x, algorithm="giac")

[Out]

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