\(\int x^3 (a+b \arctan (c x^2)) \, dx\) [62]

Optimal result

Integrand size = 14, antiderivative size = 43 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=-\frac {b x^2}{4 c}+\frac {b \arctan \left (c x^2\right )}{4 c^2}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4946, 281, 327, 209} \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )+\frac {b \arctan \left (c x^2\right )}{4 c^2}-\frac {b x^2}{4 c} \]

[In]

Int[x^3*(a + b*ArcTan[c*x^2]),x]

[Out]

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

Rule 209

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

Rule 281

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 327

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

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}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )-\frac {1}{2} (b c) \int \frac {x^5}{1+c^2 x^4} \, dx \\ & = \frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )-\frac {1}{4} (b c) \text {Subst}\left (\int \frac {x^2}{1+c^2 x^2} \, dx,x,x^2\right ) \\ & = -\frac {b x^2}{4 c}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right )+\frac {b \text {Subst}\left (\int \frac {1}{1+c^2 x^2} \, dx,x,x^2\right )}{4 c} \\ & = -\frac {b x^2}{4 c}+\frac {b \arctan \left (c x^2\right )}{4 c^2}+\frac {1}{4} x^4 \left (a+b \arctan \left (c x^2\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.12 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=-\frac {b x^2}{4 c}+\frac {a x^4}{4}+\frac {b \arctan \left (c x^2\right )}{4 c^2}+\frac {1}{4} b x^4 \arctan \left (c x^2\right ) \]

[In]

Integrate[x^3*(a + b*ArcTan[c*x^2]),x]

[Out]

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

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {a \,x^{4}}{4}+\frac {b \,x^{4} \arctan \left (c \,x^{2}\right )}{4}-\frac {b \,x^{2}}{4 c}+\frac {b \arctan \left (c \,x^{2}\right )}{4 c^{2}}\)	\(41\)
parts	\(\frac {a \,x^{4}}{4}+\frac {b \,x^{4} \arctan \left (c \,x^{2}\right )}{4}-\frac {b \,x^{2}}{4 c}+\frac {b \arctan \left (c \,x^{2}\right )}{4 c^{2}}\)	\(41\)
parallelrisch	\(\frac {\arctan \left (c \,x^{2}\right ) b \,c^{2} x^{4}+a \,c^{2} x^{4}-b c \,x^{2}+b \arctan \left (c \,x^{2}\right )}{4 c^{2}}\)	\(44\)
risch	\(-\frac {i x^{4} b \ln \left (i c \,x^{2}+1\right )}{8}+\frac {i x^{4} b \ln \left (-i c \,x^{2}+1\right )}{8}+\frac {a \,x^{4}}{4}-\frac {b \,x^{2}}{4 c}+\frac {b \arctan \left (c \,x^{2}\right )}{4 c^{2}}+\frac {b^{2}}{16 a \,c^{2}}\)	\(74\)

[In]

int(x^3*(a+b*arctan(c*x^2)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {a c^{2} x^{4} - b c x^{2} + {\left (b c^{2} x^{4} + b\right )} \arctan \left (c x^{2}\right )}{4 \, c^{2}} \]

[In]

integrate(x^3*(a+b*arctan(c*x^2)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 9.85 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.12 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\begin {cases} \frac {a x^{4}}{4} + \frac {b x^{4} \operatorname {atan}{\left (c x^{2} \right )}}{4} - \frac {b x^{2}}{4 c} + \frac {b \operatorname {atan}{\left (c x^{2} \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\\frac {a x^{4}}{4} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*(a+b*atan(c*x**2)),x)

[Out]

Piecewise((a*x**4/4 + b*x**4*atan(c*x**2)/4 - b*x**2/(4*c) + b*atan(c*x**2)/(4*c**2), Ne(c, 0)), (a*x**4/4, Tr
ue))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {1}{4} \, a x^{4} + \frac {1}{4} \, {\left (x^{4} \arctan \left (c x^{2}\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\arctan \left (c x^{2}\right )}{c^{3}}\right )}\right )} b \]

[In]

integrate(x^3*(a+b*arctan(c*x^2)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {a c x^{4} + \frac {{\left (c^{2} x^{4} \arctan \left (c x^{2}\right ) - c x^{2} + \arctan \left (c x^{2}\right )\right )} b}{c}}{4 \, c} \]

[In]

integrate(x^3*(a+b*arctan(c*x^2)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93 \[ \int x^3 \left (a+b \arctan \left (c x^2\right )\right ) \, dx=\frac {a\,x^4}{4}-\frac {b\,x^2}{4\,c}+\frac {b\,\mathrm {atan}\left (c\,x^2\right )}{4\,c^2}+\frac {b\,x^4\,\mathrm {atan}\left (c\,x^2\right )}{4} \]

[In]

int(x^3*(a + b*atan(c*x^2)),x)

[Out]

(a*x^4)/4 - (b*x^2)/(4*c) + (b*atan(c*x^2))/(4*c^2) + (b*x^4*atan(c*x^2))/4