3.1039 \(\int x^7 (a+b x^2)^p \, dx\)
Optimal. Leaf size=100 \[ -\frac {a^3 \left (a+b x^2\right )^{p+1}}{2 b^4 (p+1)}+\frac {3 a^2 \left (a+b x^2\right )^{p+2}}{2 b^4 (p+2)}-\frac {3 a \left (a+b x^2\right )^{p+3}}{2 b^4 (p+3)}+\frac {\left (a+b x^2\right )^{p+4}}{2 b^4 (p+4)} \]
[Out]
-1/2*a^3*(b*x^2+a)^(1+p)/b^4/(1+p)+3/2*a^2*(b*x^2+a)^(2+p)/b^4/(2+p)-3/2*a*(b*x^2+a)^(3+p)/b^4/(3+p)+1/2*(b*x^
2+a)^(4+p)/b^4/(4+p)
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Rubi [A] time = 0.06, antiderivative size = 100, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used =
{266, 43} \[ -\frac {a^3 \left (a+b x^2\right )^{p+1}}{2 b^4 (p+1)}+\frac {3 a^2 \left (a+b x^2\right )^{p+2}}{2 b^4 (p+2)}-\frac {3 a \left (a+b x^2\right )^{p+3}}{2 b^4 (p+3)}+\frac {\left (a+b x^2\right )^{p+4}}{2 b^4 (p+4)} \]
Antiderivative was successfully verified.
[In]
Int[x^7*(a + b*x^2)^p,x]
[Out]
-(a^3*(a + b*x^2)^(1 + p))/(2*b^4*(1 + p)) + (3*a^2*(a + b*x^2)^(2 + p))/(2*b^4*(2 + p)) - (3*a*(a + b*x^2)^(3
+ p))/(2*b^4*(3 + p)) + (a + b*x^2)^(4 + p)/(2*b^4*(4 + 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]]
Rubi steps
\begin {align*} \int x^7 \left (a+b x^2\right )^p \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^3 (a+b x)^p \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {a^3 (a+b x)^p}{b^3}+\frac {3 a^2 (a+b x)^{1+p}}{b^3}-\frac {3 a (a+b x)^{2+p}}{b^3}+\frac {(a+b x)^{3+p}}{b^3}\right ) \, dx,x,x^2\right )\\ &=-\frac {a^3 \left (a+b x^2\right )^{1+p}}{2 b^4 (1+p)}+\frac {3 a^2 \left (a+b x^2\right )^{2+p}}{2 b^4 (2+p)}-\frac {3 a \left (a+b x^2\right )^{3+p}}{2 b^4 (3+p)}+\frac {\left (a+b x^2\right )^{4+p}}{2 b^4 (4+p)}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 95, normalized size = 0.95 \[ \frac {1}{2} \left (-\frac {a^3 \left (a+b x^2\right )^{p+1}}{b^4 (p+1)}+\frac {3 a^2 \left (a+b x^2\right )^{p+2}}{b^4 (p+2)}-\frac {3 a \left (a+b x^2\right )^{p+3}}{b^4 (p+3)}+\frac {\left (a+b x^2\right )^{p+4}}{b^4 (p+4)}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[x^7*(a + b*x^2)^p,x]
[Out]
(-((a^3*(a + b*x^2)^(1 + p))/(b^4*(1 + p))) + (3*a^2*(a + b*x^2)^(2 + p))/(b^4*(2 + p)) - (3*a*(a + b*x^2)^(3
+ p))/(b^4*(3 + p)) + (a + b*x^2)^(4 + p)/(b^4*(4 + p)))/2
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fricas [A] time = 0.48, size = 148, normalized size = 1.48 \[ \frac {{\left ({\left (b^{4} p^{3} + 6 \, b^{4} p^{2} + 11 \, b^{4} p + 6 \, b^{4}\right )} x^{8} + 6 \, a^{3} b p x^{2} + {\left (a b^{3} p^{3} + 3 \, a b^{3} p^{2} + 2 \, a b^{3} p\right )} x^{6} - 3 \, {\left (a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{4} - 6 \, a^{4}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^7*(b*x^2+a)^p,x, algorithm="fricas")
[Out]
1/2*((b^4*p^3 + 6*b^4*p^2 + 11*b^4*p + 6*b^4)*x^8 + 6*a^3*b*p*x^2 + (a*b^3*p^3 + 3*a*b^3*p^2 + 2*a*b^3*p)*x^6
- 3*(a^2*b^2*p^2 + a^2*b^2*p)*x^4 - 6*a^4)*(b*x^2 + a)^p/(b^4*p^4 + 10*b^4*p^3 + 35*b^4*p^2 + 50*b^4*p + 24*b^
4)
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giac [B] time = 0.59, size = 410, normalized size = 4.10 \[ \frac {{\left (b x^{2} + a\right )}^{4} {\left (b x^{2} + a\right )}^{p} p^{3} - 3 \, {\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} a p^{3} + 3 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a^{2} p^{3} - {\left (b x^{2} + a\right )} {\left (b x^{2} + a\right )}^{p} a^{3} p^{3} + 6 \, {\left (b x^{2} + a\right )}^{4} {\left (b x^{2} + a\right )}^{p} p^{2} - 21 \, {\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} a p^{2} + 24 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a^{2} p^{2} - 9 \, {\left (b x^{2} + a\right )} {\left (b x^{2} + a\right )}^{p} a^{3} p^{2} + 11 \, {\left (b x^{2} + a\right )}^{4} {\left (b x^{2} + a\right )}^{p} p - 42 \, {\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} a p + 57 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a^{2} p - 26 \, {\left (b x^{2} + a\right )} {\left (b x^{2} + a\right )}^{p} a^{3} p + 6 \, {\left (b x^{2} + a\right )}^{4} {\left (b x^{2} + a\right )}^{p} - 24 \, {\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} a + 36 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a^{2} - 24 \, {\left (b x^{2} + a\right )} {\left (b x^{2} + a\right )}^{p} a^{3}}{2 \, {\left (b^{3} p^{4} + 10 \, b^{3} p^{3} + 35 \, b^{3} p^{2} + 50 \, b^{3} p + 24 \, b^{3}\right )} b} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^7*(b*x^2+a)^p,x, algorithm="giac")
[Out]
1/2*((b*x^2 + a)^4*(b*x^2 + a)^p*p^3 - 3*(b*x^2 + a)^3*(b*x^2 + a)^p*a*p^3 + 3*(b*x^2 + a)^2*(b*x^2 + a)^p*a^2
*p^3 - (b*x^2 + a)*(b*x^2 + a)^p*a^3*p^3 + 6*(b*x^2 + a)^4*(b*x^2 + a)^p*p^2 - 21*(b*x^2 + a)^3*(b*x^2 + a)^p*
a*p^2 + 24*(b*x^2 + a)^2*(b*x^2 + a)^p*a^2*p^2 - 9*(b*x^2 + a)*(b*x^2 + a)^p*a^3*p^2 + 11*(b*x^2 + a)^4*(b*x^2
+ a)^p*p - 42*(b*x^2 + a)^3*(b*x^2 + a)^p*a*p + 57*(b*x^2 + a)^2*(b*x^2 + a)^p*a^2*p - 26*(b*x^2 + a)*(b*x^2
+ a)^p*a^3*p + 6*(b*x^2 + a)^4*(b*x^2 + a)^p - 24*(b*x^2 + a)^3*(b*x^2 + a)^p*a + 36*(b*x^2 + a)^2*(b*x^2 + a)
^p*a^2 - 24*(b*x^2 + a)*(b*x^2 + a)^p*a^3)/((b^3*p^4 + 10*b^3*p^3 + 35*b^3*p^2 + 50*b^3*p + 24*b^3)*b)
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maple [A] time = 0.01, size = 132, normalized size = 1.32 \[ -\frac {\left (-b^{3} p^{3} x^{6}-6 b^{3} p^{2} x^{6}-11 b^{3} p \,x^{6}+3 a \,b^{2} p^{2} x^{4}-6 b^{3} x^{6}+9 a \,b^{2} p \,x^{4}+6 a \,b^{2} x^{4}-6 a^{2} b p \,x^{2}-6 a^{2} b \,x^{2}+6 a^{3}\right ) \left (b \,x^{2}+a \right )^{p +1}}{2 \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right ) b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^7*(b*x^2+a)^p,x)
[Out]
-1/2*(b*x^2+a)^(p+1)*(-b^3*p^3*x^6-6*b^3*p^2*x^6-11*b^3*p*x^6+3*a*b^2*p^2*x^4-6*b^3*x^6+9*a*b^2*p*x^4+6*a*b^2*
x^4-6*a^2*b*p*x^2-6*a^2*b*x^2+6*a^3)/b^4/(p^4+10*p^3+35*p^2+50*p+24)
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maxima [A] time = 1.45, size = 106, normalized size = 1.06 \[ \frac {{\left ({\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{4} x^{8} + {\left (p^{3} + 3 \, p^{2} + 2 \, p\right )} a b^{3} x^{6} - 3 \, {\left (p^{2} + p\right )} a^{2} b^{2} x^{4} + 6 \, a^{3} b p x^{2} - 6 \, a^{4}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (p^{4} + 10 \, p^{3} + 35 \, p^{2} + 50 \, p + 24\right )} b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^7*(b*x^2+a)^p,x, algorithm="maxima")
[Out]
1/2*((p^3 + 6*p^2 + 11*p + 6)*b^4*x^8 + (p^3 + 3*p^2 + 2*p)*a*b^3*x^6 - 3*(p^2 + p)*a^2*b^2*x^4 + 6*a^3*b*p*x^
2 - 6*a^4)*(b*x^2 + a)^p/((p^4 + 10*p^3 + 35*p^2 + 50*p + 24)*b^4)
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mupad [B] time = 4.97, size = 183, normalized size = 1.83 \[ {\left (b\,x^2+a\right )}^p\,\left (\frac {x^8\,\left (p^3+6\,p^2+11\,p+6\right )}{2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {3\,a^4}{b^4\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {3\,a^3\,p\,x^2}{b^3\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}+\frac {a\,p\,x^6\,\left (p^2+3\,p+2\right )}{2\,b\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}-\frac {3\,a^2\,p\,x^4\,\left (p+1\right )}{2\,b^2\,\left (p^4+10\,p^3+35\,p^2+50\,p+24\right )}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^7*(a + b*x^2)^p,x)
[Out]
(a + b*x^2)^p*((x^8*(11*p + 6*p^2 + p^3 + 6))/(2*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) - (3*a^4)/(b^4*(50*p + 3
5*p^2 + 10*p^3 + p^4 + 24)) + (3*a^3*p*x^2)/(b^3*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) + (a*p*x^6*(3*p + p^2 +
2))/(2*b*(50*p + 35*p^2 + 10*p^3 + p^4 + 24)) - (3*a^2*p*x^4*(p + 1))/(2*b^2*(50*p + 35*p^2 + 10*p^3 + p^4 + 2
4)))
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sympy [A] time = 11.06, size = 2025, normalized size = 20.25 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**7*(b*x**2+a)**p,x)
[Out]
Piecewise((a**p*x**8/8, 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 + 3
6*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 + 36*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) + 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) + 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*sqr
t(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*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/(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)/(1
2*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)/(1
2*a**3*b**4 + 36*a**2*b**5*x**2 + 36*a*b**6*x**4 + 12*b**7*x**6), Eq(p, -4)), (-6*a**3*log(-I*sqrt(a)*sqrt(1/b
) + x)/(4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) - 6*a**3*log(I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**4 + 8*a*b*
*5*x**2 + 4*b**6*x**4) - 9*a**3/(4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) - 12*a**2*b*x**2*log(-I*sqrt(a)*sq
rt(1/b) + x)/(4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) - 12*a**2*b*x**2*log(I*sqrt(a)*sqrt(1/b) + x)/(4*a**2
*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) - 12*a**2*b*x**2/(4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) - 6*a*b**2*x
**4*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) - 6*a*b**2*x**4*log(I*sqrt(a)*sq
rt(1/b) + x)/(4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x**4) + 2*b**3*x**6/(4*a**2*b**4 + 8*a*b**5*x**2 + 4*b**6*x
**4), Eq(p, -3)), (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, -2)), (-a**3*log(-I*sqrt(a)*sqrt(1/b
) + x)/(2*b**4) - a**3*log(I*sqrt(a)*sqrt(1/b) + x)/(2*b**4) + a**2*x**2/(2*b**3) - a*x**4/(4*b**2) + x**6/(6*
b), Eq(p, -1)), (-6*a**4*(a + b*x**2)**p/(2*b**4*p**4 + 20*b**4*p**3 + 70*b**4*p**2 + 100*b**4*p + 48*b**4) +
6*a**3*b*p*x**2*(a + b*x**2)**p/(2*b**4*p**4 + 20*b**4*p**3 + 70*b**4*p**2 + 100*b**4*p + 48*b**4) - 3*a**2*b*
*2*p**2*x**4*(a + b*x**2)**p/(2*b**4*p**4 + 20*b**4*p**3 + 70*b**4*p**2 + 100*b**4*p + 48*b**4) - 3*a**2*b**2*
p*x**4*(a + b*x**2)**p/(2*b**4*p**4 + 20*b**4*p**3 + 70*b**4*p**2 + 100*b**4*p + 48*b**4) + a*b**3*p**3*x**6*(
a + b*x**2)**p/(2*b**4*p**4 + 20*b**4*p**3 + 70*b**4*p**2 + 100*b**4*p + 48*b**4) + 3*a*b**3*p**2*x**6*(a + b*
x**2)**p/(2*b**4*p**4 + 20*b**4*p**3 + 70*b**4*p**2 + 100*b**4*p + 48*b**4) + 2*a*b**3*p*x**6*(a + b*x**2)**p/
(2*b**4*p**4 + 20*b**4*p**3 + 70*b**4*p**2 + 100*b**4*p + 48*b**4) + b**4*p**3*x**8*(a + b*x**2)**p/(2*b**4*p*
*4 + 20*b**4*p**3 + 70*b**4*p**2 + 100*b**4*p + 48*b**4) + 6*b**4*p**2*x**8*(a + b*x**2)**p/(2*b**4*p**4 + 20*
b**4*p**3 + 70*b**4*p**2 + 100*b**4*p + 48*b**4) + 11*b**4*p*x**8*(a + b*x**2)**p/(2*b**4*p**4 + 20*b**4*p**3
+ 70*b**4*p**2 + 100*b**4*p + 48*b**4) + 6*b**4*x**8*(a + b*x**2)**p/(2*b**4*p**4 + 20*b**4*p**3 + 70*b**4*p**
2 + 100*b**4*p + 48*b**4), True))
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