3.798 \(\int x^7 (a^2+2 a b x^2+b^2 x^4)^p \, dx\)

Optimal. Leaf size=174 \[ \frac{\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (p+2)}-\frac{3 a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (2 p+3)}+\frac{3 a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (p+1)}-\frac{a^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (2 p+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.108686, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1113, 266, 43} \[ \frac{\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (p+2)}-\frac{3 a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (2 p+3)}+\frac{3 a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (p+1)}-\frac{a^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^7*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

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

Rule 1113

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

Rubi steps

\begin{align*} \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx &=\left (\left (1+\frac{b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \int x^7 \left (1+\frac{b x^2}{a}\right )^{2 p} \, dx\\ &=\frac{1}{2} \left (\left (1+\frac{b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname{Subst}\left (\int x^3 \left (1+\frac{b x}{a}\right )^{2 p} \, dx,x,x^2\right )\\ &=\frac{1}{2} \left (\left (1+\frac{b x^2}{a}\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^3 \left (1+\frac{b x}{a}\right )^{2 p}}{b^3}+\frac{3 a^3 \left (1+\frac{b x}{a}\right )^{1+2 p}}{b^3}-\frac{3 a^3 \left (1+\frac{b x}{a}\right )^{2+2 p}}{b^3}+\frac{a^3 \left (1+\frac{b x}{a}\right )^{3+2 p}}{b^3}\right ) \, dx,x,x^2\right )\\ &=-\frac{a^3 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (1+2 p)}+\frac{3 a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (1+p)}-\frac{3 a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^4 (3+2 p)}+\frac{\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^4 (2+p)}\\ \end{align*}

Mathematica [A]  time = 0.0578408, size = 110, normalized size = 0.63 \[ \frac{\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \left (3 a^2 b (2 p+1) x^2-3 a^3-3 a b^2 \left (2 p^2+3 p+1\right ) x^4+b^3 \left (4 p^3+12 p^2+11 p+3\right ) x^6\right )}{4 b^4 (p+1) (p+2) (2 p+1) (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.05, size = 150, normalized size = 0.9 \begin{align*} -{\frac{ \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{p} \left ( -4\,{b}^{3}{p}^{3}{x}^{6}-12\,{b}^{3}{p}^{2}{x}^{6}-11\,{b}^{3}p{x}^{6}+6\,a{b}^{2}{p}^{2}{x}^{4}-3\,{b}^{3}{x}^{6}+9\,a{b}^{2}p{x}^{4}+3\,a{x}^{4}{b}^{2}-6\,{a}^{2}bp{x}^{2}-3\,{a}^{2}b{x}^{2}+3\,{a}^{3} \right ) \left ( b{x}^{2}+a \right ) }{4\,{b}^{4} \left ( 4\,{p}^{4}+20\,{p}^{3}+35\,{p}^{2}+25\,p+6 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(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*(-4*b^3*p^3*x^6-12*b^3*p^2*x^6-11*b^3*p*x^6+6*a*b^2*p^2*x^4-3*b^3*x^6+9*a*b^2*p
*x^4+3*a*b^2*x^4-6*a^2*b*p*x^2-3*a^2*b*x^2+3*a^3)*(b*x^2+a)/b^4/(4*p^4+20*p^3+35*p^2+25*p+6)

________________________________________________________________________________________

Maxima [A]  time = 0.978187, size = 155, normalized size = 0.89 \begin{align*} \frac{{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{4} x^{8} + 2 \,{\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a b^{3} x^{6} - 3 \,{\left (2 \, p^{2} + p\right )} a^{2} b^{2} x^{4} + 6 \, a^{3} b p x^{2} - 3 \, a^{4}\right )}{\left (b x^{2} + a\right )}^{2 \, p}}{4 \,{\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="maxima")

[Out]

1/4*((4*p^3 + 12*p^2 + 11*p + 3)*b^4*x^8 + 2*(2*p^3 + 3*p^2 + p)*a*b^3*x^6 - 3*(2*p^2 + p)*a^2*b^2*x^4 + 6*a^3
*b*p*x^2 - 3*a^4)*(b*x^2 + a)^(2*p)/((4*p^4 + 20*p^3 + 35*p^2 + 25*p + 6)*b^4)

________________________________________________________________________________________

Fricas [A]  time = 1.58023, size = 335, normalized size = 1.93 \begin{align*} \frac{{\left ({\left (4 \, b^{4} p^{3} + 12 \, b^{4} p^{2} + 11 \, b^{4} p + 3 \, b^{4}\right )} x^{8} + 6 \, a^{3} b p x^{2} + 2 \,{\left (2 \, a b^{3} p^{3} + 3 \, a b^{3} p^{2} + a b^{3} p\right )} x^{6} - 3 \,{\left (2 \, a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{4} - 3 \, a^{4}\right )}{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \,{\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="fricas")

[Out]

1/4*((4*b^4*p^3 + 12*b^4*p^2 + 11*b^4*p + 3*b^4)*x^8 + 6*a^3*b*p*x^2 + 2*(2*a*b^3*p^3 + 3*a*b^3*p^2 + a*b^3*p)
*x^6 - 3*(2*a^2*b^2*p^2 + a^2*b^2*p)*x^4 - 3*a^4)*(b^2*x^4 + 2*a*b*x^2 + a^2)^p/(4*b^4*p^4 + 20*b^4*p^3 + 35*b
^4*p^2 + 25*b^4*p + 6*b^4)

________________________________________________________________________________________

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**7*(b**2*x**4+2*a*b*x**2+a**2)**p,x)

[Out]

Exception raised: TypeError

________________________________________________________________________________________

Giac [B]  time = 1.24871, size = 506, normalized size = 2.91 \begin{align*} \frac{4 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} p^{3} x^{8} + 12 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} p^{2} x^{8} + 4 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} p^{3} x^{6} + 11 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} p x^{8} + 6 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} p^{2} x^{6} + 3 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{4} x^{8} + 2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{3} p x^{6} - 6 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b^{2} p^{2} x^{4} - 3 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b^{2} p x^{4} + 6 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{3} b p x^{2} - 3 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{4}}{4 \,{\left (4 \, b^{4} p^{4} + 20 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 25 \, b^{4} p + 6 \, b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="giac")

[Out]

1/4*(4*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^4*p^3*x^8 + 12*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^4*p^2*x^8 + 4*(b^2*x^4 +
2*a*b*x^2 + a^2)^p*a*b^3*p^3*x^6 + 11*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^4*p*x^8 + 6*(b^2*x^4 + 2*a*b*x^2 + a^2)
^p*a*b^3*p^2*x^6 + 3*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*b^4*x^8 + 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a*b^3*p*x^6 - 6*(
b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^2*b^2*p^2*x^4 - 3*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^2*b^2*p*x^4 + 6*(b^2*x^4 + 2*
a*b*x^2 + a^2)^p*a^3*b*p*x^2 - 3*(b^2*x^4 + 2*a*b*x^2 + a^2)^p*a^4)/(4*b^4*p^4 + 20*b^4*p^3 + 35*b^4*p^2 + 25*
b^4*p + 6*b^4)