3.1.58 \(\int x^5 (a^2+2 a b x^3+b^2 x^6)^{5/2} \, dx\)

Optimal. Leaf size=78 \[ \frac {\left (a+b x^3\right )^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{21 b^2}-\frac {a \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 b^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1355, 266, 43} \begin {gather*} \frac {\left (a+b x^3\right )^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{21 b^2}-\frac {a \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

-(a*(a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(18*b^2) + ((a + b*x^3)^6*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/
(21*b^2)

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 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int x^5 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x^5 \left (a b+b^2 x^3\right )^5 \, dx}{b^4 \left (a b+b^2 x^3\right )}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \operatorname {Subst}\left (\int x \left (a b+b^2 x\right )^5 \, dx,x,x^3\right )}{3 b^4 \left (a b+b^2 x^3\right )}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \operatorname {Subst}\left (\int \left (-\frac {a \left (a b+b^2 x\right )^5}{b}+\frac {\left (a b+b^2 x\right )^6}{b^2}\right ) \, dx,x,x^3\right )}{3 b^4 \left (a b+b^2 x^3\right )}\\ &=-\frac {a \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 b^2}+\frac {\left (a+b x^3\right )^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{21 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 1.06 \begin {gather*} \frac {x^6 \sqrt {\left (a+b x^3\right )^2} \left (21 a^5+70 a^4 b x^3+105 a^3 b^2 x^6+84 a^2 b^3 x^9+35 a b^4 x^{12}+6 b^5 x^{15}\right )}{126 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(x^6*Sqrt[(a + b*x^3)^2]*(21*a^5 + 70*a^4*b*x^3 + 105*a^3*b^2*x^6 + 84*a^2*b^3*x^9 + 35*a*b^4*x^12 + 6*b^5*x^1
5))/(126*(a + b*x^3))

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IntegrateAlgebraic [A]  time = 11.58, size = 83, normalized size = 1.06 \begin {gather*} \frac {x^6 \sqrt {\left (a+b x^3\right )^2} \left (21 a^5+70 a^4 b x^3+105 a^3 b^2 x^6+84 a^2 b^3 x^9+35 a b^4 x^{12}+6 b^5 x^{15}\right )}{126 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(x^6*Sqrt[(a + b*x^3)^2]*(21*a^5 + 70*a^4*b*x^3 + 105*a^3*b^2*x^6 + 84*a^2*b^3*x^9 + 35*a*b^4*x^12 + 6*b^5*x^1
5))/(126*(a + b*x^3))

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fricas [A]  time = 1.13, size = 57, normalized size = 0.73 \begin {gather*} \frac {1}{21} \, b^{5} x^{21} + \frac {5}{18} \, a b^{4} x^{18} + \frac {2}{3} \, a^{2} b^{3} x^{15} + \frac {5}{6} \, a^{3} b^{2} x^{12} + \frac {5}{9} \, a^{4} b x^{9} + \frac {1}{6} \, a^{5} x^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/21*b^5*x^21 + 5/18*a*b^4*x^18 + 2/3*a^2*b^3*x^15 + 5/6*a^3*b^2*x^12 + 5/9*a^4*b*x^9 + 1/6*a^5*x^6

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giac [A]  time = 0.29, size = 67, normalized size = 0.86 \begin {gather*} \frac {1}{126} \, {\left (6 \, b^{5} x^{21} + 35 \, a b^{4} x^{18} + 84 \, a^{2} b^{3} x^{15} + 105 \, a^{3} b^{2} x^{12} + 70 \, a^{4} b x^{9} + 21 \, a^{5} x^{6}\right )} \mathrm {sgn}\left (b x^{3} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")

[Out]

1/126*(6*b^5*x^21 + 35*a*b^4*x^18 + 84*a^2*b^3*x^15 + 105*a^3*b^2*x^12 + 70*a^4*b*x^9 + 21*a^5*x^6)*sgn(b*x^3
+ a)

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maple [A]  time = 0.01, size = 80, normalized size = 1.03 \begin {gather*} \frac {\left (6 b^{5} x^{15}+35 a \,b^{4} x^{12}+84 a^{2} b^{3} x^{9}+105 a^{3} b^{2} x^{6}+70 a^{4} b \,x^{3}+21 a^{5}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {5}{2}} x^{6}}{126 \left (b \,x^{3}+a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)

[Out]

1/126*x^6*(6*b^5*x^15+35*a*b^4*x^12+84*a^2*b^3*x^9+105*a^3*b^2*x^6+70*a^4*b*x^3+21*a^5)*((b*x^3+a)^2)^(5/2)/(b
*x^3+a)^5

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maxima [A]  time = 0.63, size = 83, normalized size = 1.06 \begin {gather*} -\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a x^{3}}{18 \, b} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a^{2}}{18 \, b^{2}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}}}{21 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")

[Out]

-1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a*x^3/b - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a^2/b^2 + 1/21*(b^2*x
^6 + 2*a*b*x^3 + a^2)^(7/2)/b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^5\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2),x)

[Out]

int(x^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{5} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)

[Out]

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

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