3.93 \(\int x^3 (a+b x^2) (a^2+2 a b x^2+b^2 x^4)^p \, dx\)

Optimal. Leaf size=86 \[ \frac{\left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (2 p+3)}-\frac{a \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (p+1)} \]

[Out]

-(a*(a + b*x^2)^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^p)/(4*b^2*(1 + p)) + ((a + b*x^2)^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^
p)/(2*b^2*(3 + 2*p))

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Rubi [A]  time = 0.0887778, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {1249, 770, 21, 43} \[ \frac{\left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (2 p+3)}-\frac{a \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

-(a*(a + b*x^2)^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^p)/(4*b^2*(1 + p)) + ((a + b*x^2)^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^
p)/(2*b^2*(3 + 2*p))

Rule 1249

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, S
ubst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && IGtQ[(m + 1)/2, 0]

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx,x,x^2\right )\\ &=\frac{1}{2} \left (\left (b \left (a+b x^2\right )\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname{Subst}\left (\int x (a+b x) \left (a b+b^2 x\right )^{2 p} \, dx,x,x^2\right )\\ &=\frac{\left (\left (b \left (a+b x^2\right )\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname{Subst}\left (\int x \left (a b+b^2 x\right )^{1+2 p} \, dx,x,x^2\right )}{2 b}\\ &=\frac{\left (\left (b \left (a+b x^2\right )\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname{Subst}\left (\int \left (-\frac{a \left (a b+b^2 x\right )^{1+2 p}}{b}+\frac{\left (a b+b^2 x\right )^{2+2 p}}{b^2}\right ) \, dx,x,x^2\right )}{2 b}\\ &=-\frac{a \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (1+p)}+\frac{\left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (3+2 p)}\\ \end{align*}

Mathematica [A]  time = 0.0248835, size = 45, normalized size = 0.52 \[ \frac{\left (\left (a+b x^2\right )^2\right )^{p+1} \left (2 b (p+1) x^2-a\right )}{4 b^2 (p+1) (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

(((a + b*x^2)^2)^(1 + p)*(-a + 2*b*(1 + p)*x^2))/(4*b^2*(1 + p)*(3 + 2*p))

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Maple [A]  time = 0.005, size = 62, normalized size = 0.7 \begin{align*} -{\frac{ \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{p} \left ( -2\,{x}^{2}pb-2\,b{x}^{2}+a \right ) \left ( b{x}^{2}+a \right ) ^{2}}{4\,{b}^{2} \left ( 2\,{p}^{2}+5\,p+3 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x)

[Out]

-1/4*(b^2*x^4+2*a*b*x^2+a^2)^p*(-2*b*p*x^2-2*b*x^2+a)*(b*x^2+a)^2/b^2/(2*p^2+5*p+3)

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Maxima [A]  time = 1.01171, size = 182, normalized size = 2.12 \begin{align*} \frac{{\left (b^{2}{\left (2 \, p + 1\right )} x^{4} + 2 \, a b p x^{2} - a^{2}\right )}{\left (b x^{2} + a\right )}^{2 \, p} a}{4 \,{\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} + \frac{{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{6} +{\left (2 \, p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + a^{3}\right )}{\left (b x^{2} + a\right )}^{2 \, p}}{2 \,{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="maxima")

[Out]

1/4*(b^2*(2*p + 1)*x^4 + 2*a*b*p*x^2 - a^2)*(b*x^2 + a)^(2*p)*a/((2*p^2 + 3*p + 1)*b^2) + 1/2*((2*p^2 + 3*p +
1)*b^3*x^6 + (2*p^2 + p)*a*b^2*x^4 - 2*a^2*b*p*x^2 + a^3)*(b*x^2 + a)^(2*p)/((4*p^3 + 12*p^2 + 11*p + 3)*b^2)

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Fricas [A]  time = 1.55942, size = 185, normalized size = 2.15 \begin{align*} \frac{{\left (2 \,{\left (b^{3} p + b^{3}\right )} x^{6} + 2 \, a^{2} b p x^{2} +{\left (4 \, a b^{2} p + 3 \, a b^{2}\right )} x^{4} - a^{3}\right )}{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \,{\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="fricas")

[Out]

1/4*(2*(b^3*p + b^3)*x^6 + 2*a^2*b*p*x^2 + (4*a*b^2*p + 3*a*b^2)*x^4 - a^3)*(b^2*x^4 + 2*a*b*x^2 + a^2)^p/(2*b
^2*p^2 + 5*b^2*p + 3*b^2)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x**2+a)*(b**2*x**4+2*a*b*x**2+a**2)**p,x)

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.14165, size = 265, normalized size = 3.08 \begin{align*} \frac{2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} p x^{6} + 2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} x^{6} + 4 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{2} p x^{4} + 3 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{2} x^{4} + 2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b p x^{2} -{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{3}}{4 \,{\left (2 \, b^{2} p^{2} + 5 \, b^{2} p + 3 \, b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="giac")

[Out]

1/4*(2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^3*p*x^6 + 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^3*x^6 + 4*(b^2*x^4 + 2*a*b*
x^2 + a^2)^p*a*b^2*p*x^4 + 3*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a*b^2*x^4 + 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^2*b*p
*x^2 - (b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^3)/(2*b^2*p^2 + 5*b^2*p + 3*b^2)