3.1.51 \(\int x^{12} (a^2+2 a b x^3+b^2 x^6)^{5/2} \, dx\)

Optimal. Leaf size=255 \[ \frac {b^5 x^{28} \sqrt {a^2+2 a b x^3+b^2 x^6}}{28 \left (a+b x^3\right )}+\frac {a b^4 x^{25} \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {5 a^2 b^3 x^{22} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {a^5 x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )}+\frac {5 a^4 b x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{16 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1355, 270} \begin {gather*} \frac {b^5 x^{28} \sqrt {a^2+2 a b x^3+b^2 x^6}}{28 \left (a+b x^3\right )}+\frac {a b^4 x^{25} \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {5 a^2 b^3 x^{22} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )}+\frac {5 a^4 b x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{16 \left (a+b x^3\right )}+\frac {a^5 x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^12*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(a^5*x^13*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(13*(a + b*x^3)) + (5*a^4*b*x^16*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(
16*(a + b*x^3)) + (10*a^3*b^2*x^19*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(19*(a + b*x^3)) + (5*a^2*b^3*x^22*Sqrt[a^
2 + 2*a*b*x^3 + b^2*x^6])/(11*(a + b*x^3)) + (a*b^4*x^25*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(5*(a + b*x^3)) + (b
^5*x^28*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(28*(a + b*x^3))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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^{12} \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^{12} \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} \int \left (a^5 b^5 x^{12}+5 a^4 b^6 x^{15}+10 a^3 b^7 x^{18}+10 a^2 b^8 x^{21}+5 a b^9 x^{24}+b^{10} x^{27}\right ) \, dx}{b^4 \left (a b+b^2 x^3\right )}\\ &=\frac {a^5 x^{13} \sqrt {a^2+2 a b x^3+b^2 x^6}}{13 \left (a+b x^3\right )}+\frac {5 a^4 b x^{16} \sqrt {a^2+2 a b x^3+b^2 x^6}}{16 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{19} \sqrt {a^2+2 a b x^3+b^2 x^6}}{19 \left (a+b x^3\right )}+\frac {5 a^2 b^3 x^{22} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {a b^4 x^{25} \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {b^5 x^{28} \sqrt {a^2+2 a b x^3+b^2 x^6}}{28 \left (a+b x^3\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.33 \begin {gather*} \frac {x^{13} \sqrt {\left (a+b x^3\right )^2} \left (117040 a^5+475475 a^4 b x^3+800800 a^3 b^2 x^6+691600 a^2 b^3 x^9+304304 a b^4 x^{12}+54340 b^5 x^{15}\right )}{1521520 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^12*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(x^13*Sqrt[(a + b*x^3)^2]*(117040*a^5 + 475475*a^4*b*x^3 + 800800*a^3*b^2*x^6 + 691600*a^2*b^3*x^9 + 304304*a*
b^4*x^12 + 54340*b^5*x^15))/(1521520*(a + b*x^3))

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IntegrateAlgebraic [A]  time = 24.78, size = 83, normalized size = 0.33 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (117040 a^5 x^{13}+475475 a^4 b x^{16}+800800 a^3 b^2 x^{19}+691600 a^2 b^3 x^{22}+304304 a b^4 x^{25}+54340 b^5 x^{28}\right )}{1521520 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^12*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]

[Out]

(Sqrt[(a + b*x^3)^2]*(117040*a^5*x^13 + 475475*a^4*b*x^16 + 800800*a^3*b^2*x^19 + 691600*a^2*b^3*x^22 + 304304
*a*b^4*x^25 + 54340*b^5*x^28))/(1521520*(a + b*x^3))

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fricas [A]  time = 1.99, size = 57, normalized size = 0.22 \begin {gather*} \frac {1}{28} \, b^{5} x^{28} + \frac {1}{5} \, a b^{4} x^{25} + \frac {5}{11} \, a^{2} b^{3} x^{22} + \frac {10}{19} \, a^{3} b^{2} x^{19} + \frac {5}{16} \, a^{4} b x^{16} + \frac {1}{13} \, a^{5} x^{13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/28*b^5*x^28 + 1/5*a*b^4*x^25 + 5/11*a^2*b^3*x^22 + 10/19*a^3*b^2*x^19 + 5/16*a^4*b*x^16 + 1/13*a^5*x^13

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giac [A]  time = 0.32, size = 105, normalized size = 0.41 \begin {gather*} \frac {1}{28} \, b^{5} x^{28} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{5} \, a b^{4} x^{25} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{11} \, a^{2} b^{3} x^{22} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{19} \, a^{3} b^{2} x^{19} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{16} \, a^{4} b x^{16} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{13} \, a^{5} x^{13} \mathrm {sgn}\left (b x^{3} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")

[Out]

1/28*b^5*x^28*sgn(b*x^3 + a) + 1/5*a*b^4*x^25*sgn(b*x^3 + a) + 5/11*a^2*b^3*x^22*sgn(b*x^3 + a) + 10/19*a^3*b^
2*x^19*sgn(b*x^3 + a) + 5/16*a^4*b*x^16*sgn(b*x^3 + a) + 1/13*a^5*x^13*sgn(b*x^3 + a)

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maple [A]  time = 0.01, size = 80, normalized size = 0.31 \begin {gather*} \frac {\left (54340 b^{5} x^{15}+304304 a \,b^{4} x^{12}+691600 a^{2} b^{3} x^{9}+800800 a^{3} b^{2} x^{6}+475475 a^{4} b \,x^{3}+117040 a^{5}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {5}{2}} x^{13}}{1521520 \left (b \,x^{3}+a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)

[Out]

1/1521520*x^13*(54340*b^5*x^15+304304*a*b^4*x^12+691600*a^2*b^3*x^9+800800*a^3*b^2*x^6+475475*a^4*b*x^3+117040
*a^5)*((b*x^3+a)^2)^(5/2)/(b*x^3+a)^5

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maxima [A]  time = 0.60, size = 57, normalized size = 0.22 \begin {gather*} \frac {1}{28} \, b^{5} x^{28} + \frac {1}{5} \, a b^{4} x^{25} + \frac {5}{11} \, a^{2} b^{3} x^{22} + \frac {10}{19} \, a^{3} b^{2} x^{19} + \frac {5}{16} \, a^{4} b x^{16} + \frac {1}{13} \, a^{5} x^{13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/28*b^5*x^28 + 1/5*a*b^4*x^25 + 5/11*a^2*b^3*x^22 + 10/19*a^3*b^2*x^19 + 5/16*a^4*b*x^16 + 1/13*a^5*x^13

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^{12}\,{\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^12*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2),x)

[Out]

int(x^12*(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^{12} \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**12*(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)

[Out]

Integral(x**12*((a + b*x**3)**2)**(5/2), x)

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