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

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

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Rubi [A]  time = 0.11, antiderivative size = 174, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \end {gather*}

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 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 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 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]

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*}

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Mathematica [A]  time = 0.06, size = 110, normalized size = 0.63 \begin {gather*} \frac {\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \left (-3 a^3+3 a^2 b (2 p+1) x^2-3 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)} \end {gather*}

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))

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IntegrateAlgebraic [F]  time = 0.48, size = 0, normalized size = 0.00 \begin {gather*} \int x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.03, size = 163, normalized size = 0.94 \begin {gather*} \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 {gather*}

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)

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giac [B]  time = 0.19, size = 375, normalized size = 2.16 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.01, size = 150, normalized size = 0.86 \begin {gather*} -\frac {\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 \,b^{2} x^{4}-6 a^{2} b p \,x^{2}-3 a^{2} b \,x^{2}+3 a^{3}\right ) \left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{4 \left (4 p^{4}+20 p^{3}+35 p^{2}+25 p +6\right ) b^{4}} \end {gather*}

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)

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maxima [A]  time = 1.46, size = 115, normalized size = 0.66 \begin {gather*} \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 {gather*}

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)

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mupad [B]  time = 4.40, size = 206, normalized size = 1.18 \begin {gather*} {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p\,\left (\frac {x^8\,\left (p^3+3\,p^2+\frac {11\,p}{4}+\frac {3}{4}\right )}{4\,p^4+20\,p^3+35\,p^2+25\,p+6}-\frac {3\,a^4}{4\,b^4\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}+\frac {3\,a^3\,p\,x^2}{2\,b^3\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}+\frac {a\,p\,x^6\,\left (2\,p^2+3\,p+1\right )}{2\,b\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}-\frac {3\,a^2\,p\,x^4\,\left (2\,p+1\right )}{4\,b^2\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00

result too large to display

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]

Piecewise((x**8*(a**2)**p/8, Eq(b, 0)), (6*a**3*log(-I*sqrt(a)*sqrt(1/b) + x)/(12*a**3*b**4 + 36*a**2*b**5*x**
2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 6*a**3*log(I*sqrt(a)*sqrt(1/b) + x)/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 3
6*a*b**6*x**4 + 12*b**7*x**6) + 11*a**3/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 1
8*a**2*b*x**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6)
+ 18*a**2*b*x**2*log(I*sqrt(a)*sqrt(1/b) + x)/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x*
*6) + 27*a**2*b*x**2/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 18*a*b**2*x**4*log(-
I*sqrt(a)*sqrt(1/b) + x)/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 18*a*b**2*x**4*l
og(I*sqrt(a)*sqrt(1/b) + x)/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 18*a*b**2*x**
4/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 6*b**3*x**6*log(-I*sqrt(a)*sqrt(1/b) +
x)/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6) + 6*b**3*x**6*log(I*sqrt(a)*sqrt(1/b) +
x)/(12*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6), Eq(p, -2)), (Integral(x**7/((a + b*x**2
)**2)**(3/2), x), Eq(p, -3/2)), (6*a**3*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a*b**4 + 4*b**5*x**2) + 6*a**3*log(I*
sqrt(a)*sqrt(1/b) + x)/(4*a*b**4 + 4*b**5*x**2) + 6*a**3/(4*a*b**4 + 4*b**5*x**2) + 6*a**2*b*x**2*log(-I*sqrt(
a)*sqrt(1/b) + x)/(4*a*b**4 + 4*b**5*x**2) + 6*a**2*b*x**2*log(I*sqrt(a)*sqrt(1/b) + x)/(4*a*b**4 + 4*b**5*x**
2) - 3*a*b**2*x**4/(4*a*b**4 + 4*b**5*x**2) + b**3*x**6/(4*a*b**4 + 4*b**5*x**2), Eq(p, -1)), (Integral(x**7/s
qrt((a + b*x**2)**2), x), Eq(p, -1/2)), (-3*a**4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(16*b**4*p**4 + 80*b**4*p*
*3 + 140*b**4*p**2 + 100*b**4*p + 24*b**4) + 6*a**3*b*p*x**2*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(16*b**4*p**4
+ 80*b**4*p**3 + 140*b**4*p**2 + 100*b**4*p + 24*b**4) - 6*a**2*b**2*p**2*x**4*(a**2 + 2*a*b*x**2 + b**2*x**4)
**p/(16*b**4*p**4 + 80*b**4*p**3 + 140*b**4*p**2 + 100*b**4*p + 24*b**4) - 3*a**2*b**2*p*x**4*(a**2 + 2*a*b*x*
*2 + b**2*x**4)**p/(16*b**4*p**4 + 80*b**4*p**3 + 140*b**4*p**2 + 100*b**4*p + 24*b**4) + 4*a*b**3*p**3*x**6*(
a**2 + 2*a*b*x**2 + b**2*x**4)**p/(16*b**4*p**4 + 80*b**4*p**3 + 140*b**4*p**2 + 100*b**4*p + 24*b**4) + 6*a*b
**3*p**2*x**6*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(16*b**4*p**4 + 80*b**4*p**3 + 140*b**4*p**2 + 100*b**4*p + 2
4*b**4) + 2*a*b**3*p*x**6*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(16*b**4*p**4 + 80*b**4*p**3 + 140*b**4*p**2 + 10
0*b**4*p + 24*b**4) + 4*b**4*p**3*x**8*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(16*b**4*p**4 + 80*b**4*p**3 + 140*b
**4*p**2 + 100*b**4*p + 24*b**4) + 12*b**4*p**2*x**8*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(16*b**4*p**4 + 80*b**
4*p**3 + 140*b**4*p**2 + 100*b**4*p + 24*b**4) + 11*b**4*p*x**8*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(16*b**4*p*
*4 + 80*b**4*p**3 + 140*b**4*p**2 + 100*b**4*p + 24*b**4) + 3*b**4*x**8*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(16
*b**4*p**4 + 80*b**4*p**3 + 140*b**4*p**2 + 100*b**4*p + 24*b**4), True))

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