3.689 \(\int x^7 (a+b x^2)^{4/3} \, dx\)

Optimal. Leaf size=80 \[ -\frac {3 a^3 \left (a+b x^2\right )^{7/3}}{14 b^4}+\frac {9 a^2 \left (a+b x^2\right )^{10/3}}{20 b^4}+\frac {3 \left (a+b x^2\right )^{16/3}}{32 b^4}-\frac {9 a \left (a+b x^2\right )^{13/3}}{26 b^4} \]

[Out]

-3/14*a^3*(b*x^2+a)^(7/3)/b^4+9/20*a^2*(b*x^2+a)^(10/3)/b^4-9/26*a*(b*x^2+a)^(13/3)/b^4+3/32*(b*x^2+a)^(16/3)/
b^4

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Rubi [A]  time = 0.05, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac {9 a^2 \left (a+b x^2\right )^{10/3}}{20 b^4}-\frac {3 a^3 \left (a+b x^2\right )^{7/3}}{14 b^4}+\frac {3 \left (a+b x^2\right )^{16/3}}{32 b^4}-\frac {9 a \left (a+b x^2\right )^{13/3}}{26 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^7*(a + b*x^2)^(4/3),x]

[Out]

(-3*a^3*(a + b*x^2)^(7/3))/(14*b^4) + (9*a^2*(a + b*x^2)^(10/3))/(20*b^4) - (9*a*(a + b*x^2)^(13/3))/(26*b^4)
+ (3*(a + b*x^2)^(16/3))/(32*b^4)

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 )^{4/3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^3 (a+b x)^{4/3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {a^3 (a+b x)^{4/3}}{b^3}+\frac {3 a^2 (a+b x)^{7/3}}{b^3}-\frac {3 a (a+b x)^{10/3}}{b^3}+\frac {(a+b x)^{13/3}}{b^3}\right ) \, dx,x,x^2\right )\\ &=-\frac {3 a^3 \left (a+b x^2\right )^{7/3}}{14 b^4}+\frac {9 a^2 \left (a+b x^2\right )^{10/3}}{20 b^4}-\frac {9 a \left (a+b x^2\right )^{13/3}}{26 b^4}+\frac {3 \left (a+b x^2\right )^{16/3}}{32 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 50, normalized size = 0.62 \[ \frac {3 \left (a+b x^2\right )^{7/3} \left (-81 a^3+189 a^2 b x^2-315 a b^2 x^4+455 b^3 x^6\right )}{14560 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7*(a + b*x^2)^(4/3),x]

[Out]

(3*(a + b*x^2)^(7/3)*(-81*a^3 + 189*a^2*b*x^2 - 315*a*b^2*x^4 + 455*b^3*x^6))/(14560*b^4)

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fricas [A]  time = 0.54, size = 68, normalized size = 0.85 \[ \frac {3 \, {\left (455 \, b^{5} x^{10} + 595 \, a b^{4} x^{8} + 14 \, a^{2} b^{3} x^{6} - 18 \, a^{3} b^{2} x^{4} + 27 \, a^{4} b x^{2} - 81 \, a^{5}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{14560 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b*x^2+a)^(4/3),x, algorithm="fricas")

[Out]

3/14560*(455*b^5*x^10 + 595*a*b^4*x^8 + 14*a^2*b^3*x^6 - 18*a^3*b^2*x^4 + 27*a^4*b*x^2 - 81*a^5)*(b*x^2 + a)^(
1/3)/b^4

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giac [A]  time = 0.65, size = 57, normalized size = 0.71 \[ \frac {3 \, {\left (455 \, {\left (b x^{2} + a\right )}^{\frac {16}{3}} - 1680 \, {\left (b x^{2} + a\right )}^{\frac {13}{3}} a + 2184 \, {\left (b x^{2} + a\right )}^{\frac {10}{3}} a^{2} - 1040 \, {\left (b x^{2} + a\right )}^{\frac {7}{3}} a^{3}\right )}}{14560 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b*x^2+a)^(4/3),x, algorithm="giac")

[Out]

3/14560*(455*(b*x^2 + a)^(16/3) - 1680*(b*x^2 + a)^(13/3)*a + 2184*(b*x^2 + a)^(10/3)*a^2 - 1040*(b*x^2 + a)^(
7/3)*a^3)/b^4

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maple [A]  time = 0.00, size = 47, normalized size = 0.59 \[ -\frac {3 \left (b \,x^{2}+a \right )^{\frac {7}{3}} \left (-455 b^{3} x^{6}+315 a \,b^{2} x^{4}-189 a^{2} b \,x^{2}+81 a^{3}\right )}{14560 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(b*x^2+a)^(4/3),x)

[Out]

-3/14560*(b*x^2+a)^(7/3)*(-455*b^3*x^6+315*a*b^2*x^4-189*a^2*b*x^2+81*a^3)/b^4

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maxima [A]  time = 1.38, size = 64, normalized size = 0.80 \[ \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {16}{3}}}{32 \, b^{4}} - \frac {9 \, {\left (b x^{2} + a\right )}^{\frac {13}{3}} a}{26 \, b^{4}} + \frac {9 \, {\left (b x^{2} + a\right )}^{\frac {10}{3}} a^{2}}{20 \, b^{4}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{3}} a^{3}}{14 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b*x^2+a)^(4/3),x, algorithm="maxima")

[Out]

3/32*(b*x^2 + a)^(16/3)/b^4 - 9/26*(b*x^2 + a)^(13/3)*a/b^4 + 9/20*(b*x^2 + a)^(10/3)*a^2/b^4 - 3/14*(b*x^2 +
a)^(7/3)*a^3/b^4

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mupad [B]  time = 5.19, size = 64, normalized size = 0.80 \[ {\left (b\,x^2+a\right )}^{1/3}\,\left (\frac {51\,a\,x^8}{416}+\frac {3\,b\,x^{10}}{32}-\frac {243\,a^5}{14560\,b^4}+\frac {3\,a^2\,x^6}{1040\,b}-\frac {27\,a^3\,x^4}{7280\,b^2}+\frac {81\,a^4\,x^2}{14560\,b^3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a + b*x^2)^(4/3),x)

[Out]

(a + b*x^2)^(1/3)*((51*a*x^8)/416 + (3*b*x^10)/32 - (243*a^5)/(14560*b^4) + (3*a^2*x^6)/(1040*b) - (27*a^3*x^4
)/(7280*b^2) + (81*a^4*x^2)/(14560*b^3))

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sympy [A]  time = 6.58, size = 136, normalized size = 1.70 \[ \begin {cases} - \frac {243 a^{5} \sqrt [3]{a + b x^{2}}}{14560 b^{4}} + \frac {81 a^{4} x^{2} \sqrt [3]{a + b x^{2}}}{14560 b^{3}} - \frac {27 a^{3} x^{4} \sqrt [3]{a + b x^{2}}}{7280 b^{2}} + \frac {3 a^{2} x^{6} \sqrt [3]{a + b x^{2}}}{1040 b} + \frac {51 a x^{8} \sqrt [3]{a + b x^{2}}}{416} + \frac {3 b x^{10} \sqrt [3]{a + b x^{2}}}{32} & \text {for}\: b \neq 0 \\\frac {a^{\frac {4}{3}} x^{8}}{8} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(b*x**2+a)**(4/3),x)

[Out]

Piecewise((-243*a**5*(a + b*x**2)**(1/3)/(14560*b**4) + 81*a**4*x**2*(a + b*x**2)**(1/3)/(14560*b**3) - 27*a**
3*x**4*(a + b*x**2)**(1/3)/(7280*b**2) + 3*a**2*x**6*(a + b*x**2)**(1/3)/(1040*b) + 51*a*x**8*(a + b*x**2)**(1
/3)/416 + 3*b*x**10*(a + b*x**2)**(1/3)/32, Ne(b, 0)), (a**(4/3)*x**8/8, True))

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