3.7.90 \(\int x^5 (a+b x^2)^{4/3} \, dx\) [690]

Optimal. Leaf size=59 \[ \frac {3 a^2 \left (a+b x^2\right )^{7/3}}{14 b^3}-\frac {3 a \left (a+b x^2\right )^{10/3}}{10 b^3}+\frac {3 \left (a+b x^2\right )^{13/3}}{26 b^3} \]

[Out]

3/14*a^2*(b*x^2+a)^(7/3)/b^3-3/10*a*(b*x^2+a)^(10/3)/b^3+3/26*(b*x^2+a)^(13/3)/b^3

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Rubi [A]
time = 0.03, antiderivative size = 59, 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 = {272, 45} \begin {gather*} \frac {3 a^2 \left (a+b x^2\right )^{7/3}}{14 b^3}+\frac {3 \left (a+b x^2\right )^{13/3}}{26 b^3}-\frac {3 a \left (a+b x^2\right )^{10/3}}{10 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^2)^(4/3),x]

[Out]

(3*a^2*(a + b*x^2)^(7/3))/(14*b^3) - (3*a*(a + b*x^2)^(10/3))/(10*b^3) + (3*(a + b*x^2)^(13/3))/(26*b^3)

Rule 45

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 272

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^5 \left (a+b x^2\right )^{4/3} \, dx &=\frac {1}{2} \text {Subst}\left (\int x^2 (a+b x)^{4/3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (\frac {a^2 (a+b x)^{4/3}}{b^2}-\frac {2 a (a+b x)^{7/3}}{b^2}+\frac {(a+b x)^{10/3}}{b^2}\right ) \, dx,x,x^2\right )\\ &=\frac {3 a^2 \left (a+b x^2\right )^{7/3}}{14 b^3}-\frac {3 a \left (a+b x^2\right )^{10/3}}{10 b^3}+\frac {3 \left (a+b x^2\right )^{13/3}}{26 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 39, normalized size = 0.66 \begin {gather*} \frac {3 \left (a+b x^2\right )^{7/3} \left (9 a^2-21 a b x^2+35 b^2 x^4\right )}{910 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^2)^(4/3),x]

[Out]

(3*(a + b*x^2)^(7/3)*(9*a^2 - 21*a*b*x^2 + 35*b^2*x^4))/(910*b^3)

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Maple [A]
time = 0.05, size = 36, normalized size = 0.61

method	result	size
gosper	\(\frac {3 \left (b \,x^{2}+a \right )^{\frac {7}{3}} \left (35 b^{2} x^{4}-21 a b \,x^{2}+9 a^{2}\right )}{910 b^{3}}\)	\(36\)
trager	\(\frac {3 \left (35 b^{4} x^{8}+49 a \,b^{3} x^{6}+2 a^{2} b^{2} x^{4}-3 a^{3} b \,x^{2}+9 a^{4}\right ) \left (b \,x^{2}+a \right )^{\frac {1}{3}}}{910 b^{3}}\)	\(58\)
risch	\(\frac {3 \left (35 b^{4} x^{8}+49 a \,b^{3} x^{6}+2 a^{2} b^{2} x^{4}-3 a^{3} b \,x^{2}+9 a^{4}\right ) \left (b \,x^{2}+a \right )^{\frac {1}{3}}}{910 b^{3}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^2+a)^(4/3),x,method=_RETURNVERBOSE)

[Out]

3/910*(b*x^2+a)^(7/3)*(35*b^2*x^4-21*a*b*x^2+9*a^2)/b^3

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Maxima [A]
time = 0.31, size = 47, normalized size = 0.80 \begin {gather*} \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {13}{3}}}{26 \, b^{3}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {10}{3}} a}{10 \, b^{3}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{3}} a^{2}}{14 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^(4/3),x, algorithm="maxima")

[Out]

3/26*(b*x^2 + a)^(13/3)/b^3 - 3/10*(b*x^2 + a)^(10/3)*a/b^3 + 3/14*(b*x^2 + a)^(7/3)*a^2/b^3

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Fricas [A]
time = 0.90, size = 57, normalized size = 0.97 \begin {gather*} \frac {3 \, {\left (35 \, b^{4} x^{8} + 49 \, a b^{3} x^{6} + 2 \, a^{2} b^{2} x^{4} - 3 \, a^{3} b x^{2} + 9 \, a^{4}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{910 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^(4/3),x, algorithm="fricas")

[Out]

3/910*(35*b^4*x^8 + 49*a*b^3*x^6 + 2*a^2*b^2*x^4 - 3*a^3*b*x^2 + 9*a^4)*(b*x^2 + a)^(1/3)/b^3

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (54) = 108\).
time = 0.37, size = 112, normalized size = 1.90 \begin {gather*} \begin {cases} \frac {27 a^{4} \sqrt [3]{a + b x^{2}}}{910 b^{3}} - \frac {9 a^{3} x^{2} \sqrt [3]{a + b x^{2}}}{910 b^{2}} + \frac {3 a^{2} x^{4} \sqrt [3]{a + b x^{2}}}{455 b} + \frac {21 a x^{6} \sqrt [3]{a + b x^{2}}}{130} + \frac {3 b x^{8} \sqrt [3]{a + b x^{2}}}{26} & \text {for}\: b \neq 0 \\\frac {a^{\frac {4}{3}} x^{6}}{6} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**2+a)**(4/3),x)

[Out]

Piecewise((27*a**4*(a + b*x**2)**(1/3)/(910*b**3) - 9*a**3*x**2*(a + b*x**2)**(1/3)/(910*b**2) + 3*a**2*x**4*(
a + b*x**2)**(1/3)/(455*b) + 21*a*x**6*(a + b*x**2)**(1/3)/130 + 3*b*x**8*(a + b*x**2)**(1/3)/26, Ne(b, 0)), (
a**(4/3)*x**6/6, True))

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Giac [A]
time = 1.74, size = 43, normalized size = 0.73 \begin {gather*} \frac {3 \, {\left (35 \, {\left (b x^{2} + a\right )}^{\frac {13}{3}} - 91 \, {\left (b x^{2} + a\right )}^{\frac {10}{3}} a + 65 \, {\left (b x^{2} + a\right )}^{\frac {7}{3}} a^{2}\right )}}{910 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^(4/3),x, algorithm="giac")

[Out]

3/910*(35*(b*x^2 + a)^(13/3) - 91*(b*x^2 + a)^(10/3)*a + 65*(b*x^2 + a)^(7/3)*a^2)/b^3

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Mupad [B]
time = 5.10, size = 53, normalized size = 0.90 \begin {gather*} {\left (b\,x^2+a\right )}^{1/3}\,\left (\frac {21\,a\,x^6}{130}+\frac {3\,b\,x^8}{26}+\frac {27\,a^4}{910\,b^3}+\frac {3\,a^2\,x^4}{455\,b}-\frac {9\,a^3\,x^2}{910\,b^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*x^2)^(4/3),x)

[Out]

(a + b*x^2)^(1/3)*((21*a*x^6)/130 + (3*b*x^8)/26 + (27*a^4)/(910*b^3) + (3*a^2*x^4)/(455*b) - (9*a^3*x^2)/(910
*b^2))

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