3.96 \(\int x^{11} (a+b \tan ^{-1}(c x^3)) \, dx\)

Optimal. Leaf size=54 \[ \frac {1}{12} x^{12} \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac {b \tan ^{-1}\left (c x^3\right )}{12 c^4}+\frac {b x^3}{12 c^3}-\frac {b x^9}{36 c} \]

[Out]

1/12*b*x^3/c^3-1/36*b*x^9/c-1/12*b*arctan(c*x^3)/c^4+1/12*x^12*(a+b*arctan(c*x^3))

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Rubi [A]  time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5033, 275, 302, 203} \[ \frac {1}{12} x^{12} \left (a+b \tan ^{-1}\left (c x^3\right )\right )+\frac {b x^3}{12 c^3}-\frac {b \tan ^{-1}\left (c x^3\right )}{12 c^4}-\frac {b x^9}{36 c} \]

Antiderivative was successfully verified.

[In]

Int[x^11*(a + b*ArcTan[c*x^3]),x]

[Out]

(b*x^3)/(12*c^3) - (b*x^9)/(36*c) - (b*ArcTan[c*x^3])/(12*c^4) + (x^12*(a + b*ArcTan[c*x^3]))/12

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 5033

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTan
[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 + c^2*x^(2*n)), x], x]
/; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x^{11} \left (a+b \tan ^{-1}\left (c x^3\right )\right ) \, dx &=\frac {1}{12} x^{12} \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac {1}{4} (b c) \int \frac {x^{14}}{1+c^2 x^6} \, dx\\ &=\frac {1}{12} x^{12} \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac {1}{12} (b c) \operatorname {Subst}\left (\int \frac {x^4}{1+c^2 x^2} \, dx,x,x^3\right )\\ &=\frac {1}{12} x^{12} \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac {1}{12} (b c) \operatorname {Subst}\left (\int \left (-\frac {1}{c^4}+\frac {x^2}{c^2}+\frac {1}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx,x,x^3\right )\\ &=\frac {b x^3}{12 c^3}-\frac {b x^9}{36 c}+\frac {1}{12} x^{12} \left (a+b \tan ^{-1}\left (c x^3\right )\right )-\frac {b \operatorname {Subst}\left (\int \frac {1}{1+c^2 x^2} \, dx,x,x^3\right )}{12 c^3}\\ &=\frac {b x^3}{12 c^3}-\frac {b x^9}{36 c}-\frac {b \tan ^{-1}\left (c x^3\right )}{12 c^4}+\frac {1}{12} x^{12} \left (a+b \tan ^{-1}\left (c x^3\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 59, normalized size = 1.09 \[ \frac {a x^{12}}{12}-\frac {b \tan ^{-1}\left (c x^3\right )}{12 c^4}+\frac {b x^3}{12 c^3}-\frac {b x^9}{36 c}+\frac {1}{12} b x^{12} \tan ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^11*(a + b*ArcTan[c*x^3]),x]

[Out]

(b*x^3)/(12*c^3) - (b*x^9)/(36*c) + (a*x^12)/12 - (b*ArcTan[c*x^3])/(12*c^4) + (b*x^12*ArcTan[c*x^3])/12

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fricas [A]  time = 0.41, size = 51, normalized size = 0.94 \[ \frac {3 \, a c^{4} x^{12} - b c^{3} x^{9} + 3 \, b c x^{3} + 3 \, {\left (b c^{4} x^{12} - b\right )} \arctan \left (c x^{3}\right )}{36 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(a+b*arctan(c*x^3)),x, algorithm="fricas")

[Out]

1/36*(3*a*c^4*x^12 - b*c^3*x^9 + 3*b*c*x^3 + 3*(b*c^4*x^12 - b)*arctan(c*x^3))/c^4

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giac [A]  time = 0.16, size = 60, normalized size = 1.11 \[ \frac {3 \, a c x^{12} + {\left (3 \, c x^{12} \arctan \left (c x^{3}\right ) - \frac {3 \, \arctan \left (c x^{3}\right )}{c^{3}} - \frac {c^{9} x^{9} - 3 \, c^{7} x^{3}}{c^{9}}\right )} b}{36 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(a+b*arctan(c*x^3)),x, algorithm="giac")

[Out]

1/36*(3*a*c*x^12 + (3*c*x^12*arctan(c*x^3) - 3*arctan(c*x^3)/c^3 - (c^9*x^9 - 3*c^7*x^3)/c^9)*b)/c

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maple [A]  time = 0.03, size = 50, normalized size = 0.93 \[ \frac {x^{12} a}{12}+\frac {b \,x^{12} \arctan \left (c \,x^{3}\right )}{12}-\frac {b \,x^{9}}{36 c}+\frac {b \,x^{3}}{12 c^{3}}-\frac {b \arctan \left (c \,x^{3}\right )}{12 c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(a+b*arctan(c*x^3)),x)

[Out]

1/12*x^12*a+1/12*b*x^12*arctan(c*x^3)-1/36*b*x^9/c+1/12*b*x^3/c^3-1/12*b*arctan(c*x^3)/c^4

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maxima [A]  time = 0.42, size = 54, normalized size = 1.00 \[ \frac {1}{12} \, a x^{12} + \frac {1}{36} \, {\left (3 \, x^{12} \arctan \left (c x^{3}\right ) - c {\left (\frac {c^{2} x^{9} - 3 \, x^{3}}{c^{4}} + \frac {3 \, \arctan \left (c x^{3}\right )}{c^{5}}\right )}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(a+b*arctan(c*x^3)),x, algorithm="maxima")

[Out]

1/12*a*x^12 + 1/36*(3*x^12*arctan(c*x^3) - c*((c^2*x^9 - 3*x^3)/c^4 + 3*arctan(c*x^3)/c^5))*b

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mupad [B]  time = 0.45, size = 49, normalized size = 0.91 \[ \frac {a\,x^{12}}{12}+\frac {b\,x^3}{12\,c^3}-\frac {b\,x^9}{36\,c}-\frac {b\,\mathrm {atan}\left (c\,x^3\right )}{12\,c^4}+\frac {b\,x^{12}\,\mathrm {atan}\left (c\,x^3\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(a + b*atan(c*x^3)),x)

[Out]

(a*x^12)/12 + (b*x^3)/(12*c^3) - (b*x^9)/(36*c) - (b*atan(c*x^3))/(12*c^4) + (b*x^12*atan(c*x^3))/12

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11*(a+b*atan(c*x**3)),x)

[Out]

Timed out

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