\(\int x^5 (a+b \arctan (c x^2)) \, dx\) [61]

Optimal result

Integrand size = 14, antiderivative size = 47 \[ \int x^5 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=-\frac {b x^4}{12 c}+\frac {1}{6} x^6 \left (a+b \arctan \left (c x^2\right )\right )+\frac {b \log \left (1+c^2 x^4\right )}{12 c^3} \]

[Out]

-1/12*b*x^4/c+1/6*x^6*(a+b*arctan(c*x^2))+1/12*b*ln(c^2*x^4+1)/c^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4946, 272, 45} \[ \int x^5 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {1}{6} x^6 \left (a+b \arctan \left (c x^2\right )\right )+\frac {b \log \left (c^2 x^4+1\right )}{12 c^3}-\frac {b x^4}{12 c} \]

[In]

Int[x^5*(a + b*ArcTan[c*x^2]),x]

[Out]

-1/12*(b*x^4)/c + (x^6*(a + b*ArcTan[c*x^2]))/6 + (b*Log[1 + c^2*x^4])/(12*c^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]]

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} x^6 \left (a+b \arctan \left (c x^2\right )\right )-\frac {1}{3} (b c) \int \frac {x^7}{1+c^2 x^4} \, dx \\ & = \frac {1}{6} x^6 \left (a+b \arctan \left (c x^2\right )\right )-\frac {1}{12} (b c) \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^4\right ) \\ & = \frac {1}{6} x^6 \left (a+b \arctan \left (c x^2\right )\right )-\frac {1}{12} (b c) \text {Subst}\left (\int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^4\right ) \\ & = -\frac {b x^4}{12 c}+\frac {1}{6} x^6 \left (a+b \arctan \left (c x^2\right )\right )+\frac {b \log \left (1+c^2 x^4\right )}{12 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.11 \[ \int x^5 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=-\frac {b x^4}{12 c}+\frac {a x^6}{6}+\frac {1}{6} b x^6 \arctan \left (c x^2\right )+\frac {b \log \left (1+c^2 x^4\right )}{12 c^3} \]

[In]

Integrate[x^5*(a + b*ArcTan[c*x^2]),x]

[Out]

-1/12*(b*x^4)/c + (a*x^6)/6 + (b*x^6*ArcTan[c*x^2])/6 + (b*Log[1 + c^2*x^4])/(12*c^3)

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {a \,x^{6}}{6}+\frac {b \,x^{6} \arctan \left (c \,x^{2}\right )}{6}-\frac {b \,x^{4}}{12 c}+\frac {b \ln \left (c^{2} x^{4}+1\right )}{12 c^{3}}\)	\(45\)
parts	\(\frac {a \,x^{6}}{6}+\frac {b \,x^{6} \arctan \left (c \,x^{2}\right )}{6}-\frac {b \,x^{4}}{12 c}+\frac {b \ln \left (c^{2} x^{4}+1\right )}{12 c^{3}}\)	\(45\)
parallelrisch	\(\frac {2 x^{6} \arctan \left (c \,x^{2}\right ) b \,c^{3}+2 a \,c^{3} x^{6}-b \,c^{2} x^{4}+b \ln \left (c^{2} x^{4}+1\right )}{12 c^{3}}\)	\(52\)
risch	\(-\frac {i x^{6} b \ln \left (i c \,x^{2}+1\right )}{12}+\frac {i x^{6} b \ln \left (-i c \,x^{2}+1\right )}{12}+\frac {a \,x^{6}}{6}-\frac {b \,x^{4}}{12 c}+\frac {b \ln \left (-c^{2} x^{4}-1\right )}{12 c^{3}}\)	\(68\)

[In]

int(x^5*(a+b*arctan(c*x^2)),x,method=_RETURNVERBOSE)

[Out]

1/6*a*x^6+1/6*b*x^6*arctan(c*x^2)-1/12*b*x^4/c+1/12*b*ln(c^2*x^4+1)/c^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.09 \[ \int x^5 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {2 \, b c^{3} x^{6} \arctan \left (c x^{2}\right ) + 2 \, a c^{3} x^{6} - b c^{2} x^{4} + b \log \left (c^{2} x^{4} + 1\right )}{12 \, c^{3}} \]

[In]

integrate(x^5*(a+b*arctan(c*x^2)),x, algorithm="fricas")

[Out]

1/12*(2*b*c^3*x^6*arctan(c*x^2) + 2*a*c^3*x^6 - b*c^2*x^4 + b*log(c^2*x^4 + 1))/c^3

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (39) = 78\).

Time = 22.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.70 \[ \int x^5 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\begin {cases} \frac {a x^{6}}{6} + \frac {b x^{6} \operatorname {atan}{\left (c x^{2} \right )}}{6} - \frac {b x^{4}}{12 c} + \frac {b \sqrt {- \frac {1}{c^{2}}} \operatorname {atan}{\left (c x^{2} \right )}}{6 c^{2}} + \frac {b \log {\left (x^{2} + \sqrt {- \frac {1}{c^{2}}} \right )}}{6 c^{3}} & \text {for}\: c \neq 0 \\\frac {a x^{6}}{6} & \text {otherwise} \end {cases} \]

[In]

integrate(x**5*(a+b*atan(c*x**2)),x)

[Out]

Piecewise((a*x**6/6 + b*x**6*atan(c*x**2)/6 - b*x**4/(12*c) + b*sqrt(-1/c**2)*atan(c*x**2)/(6*c**2) + b*log(x*
*2 + sqrt(-1/c**2))/(6*c**3), Ne(c, 0)), (a*x**6/6, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02 \[ \int x^5 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {1}{6} \, a x^{6} + \frac {1}{12} \, {\left (2 \, x^{6} \arctan \left (c x^{2}\right ) - {\left (\frac {x^{4}}{c^{2}} - \frac {\log \left (c^{2} x^{4} + 1\right )}{c^{4}}\right )} c\right )} b \]

[In]

integrate(x^5*(a+b*arctan(c*x^2)),x, algorithm="maxima")

[Out]

1/6*a*x^6 + 1/12*(2*x^6*arctan(c*x^2) - (x^4/c^2 - log(c^2*x^4 + 1)/c^4)*c)*b

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int x^5 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {2 \, a c x^{6} + {\left (2 \, c x^{6} \arctan \left (c x^{2}\right ) - x^{4} + \frac {\log \left (c^{2} x^{4} + 1\right )}{c^{2}}\right )} b}{12 \, c} \]

[In]

integrate(x^5*(a+b*arctan(c*x^2)),x, algorithm="giac")

[Out]

1/12*(2*a*c*x^6 + (2*c*x^6*arctan(c*x^2) - x^4 + log(c^2*x^4 + 1)/c^2)*b)/c

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int x^5 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {a\,x^6}{6}+\frac {b\,\ln \left (c^2\,x^4+1\right )}{12\,c^3}-\frac {b\,x^4}{12\,c}+\frac {b\,x^6\,\mathrm {atan}\left (c\,x^2\right )}{6} \]

[In]

int(x^5*(a + b*atan(c*x^2)),x)

[Out]

(a*x^6)/6 + (b*log(c^2*x^4 + 1))/(12*c^3) - (b*x^4)/(12*c) + (b*x^6*atan(c*x^2))/6