\(\int x^5 (a+b \arctan (c x^3))^3 \, dx\) [122]

Optimal result

Integrand size = 16, antiderivative size = 147 \[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=-\frac {i b \left (a+b \arctan \left (c x^3\right )\right )^2}{2 c^2}-\frac {b x^3 \left (a+b \arctan \left (c x^3\right )\right )^2}{2 c}+\frac {\left (a+b \arctan \left (c x^3\right )\right )^3}{6 c^2}+\frac {1}{6} x^6 \left (a+b \arctan \left (c x^3\right )\right )^3-\frac {b^2 \left (a+b \arctan \left (c x^3\right )\right ) \log \left (\frac {2}{1+i c x^3}\right )}{c^2}-\frac {i b^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x^3}\right )}{2 c^2} \]

[Out]

-1/2*I*b*(a+b*arctan(c*x^3))^2/c^2-1/2*b*x^3*(a+b*arctan(c*x^3))^2/c+1/6*(a+b*arctan(c*x^3))^3/c^2+1/6*x^6*(a+
b*arctan(c*x^3))^3-b^2*(a+b*arctan(c*x^3))*ln(2/(1+I*c*x^3))/c^2-1/2*I*b^3*polylog(2,1-2/(1+I*c*x^3))/c^2

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {4948, 4946, 5036, 4930, 5040, 4964, 2449, 2352, 5004} \[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=-\frac {b^2 \log \left (\frac {2}{1+i c x^3}\right ) \left (a+b \arctan \left (c x^3\right )\right )}{c^2}+\frac {\left (a+b \arctan \left (c x^3\right )\right )^3}{6 c^2}-\frac {i b \left (a+b \arctan \left (c x^3\right )\right )^2}{2 c^2}-\frac {b x^3 \left (a+b \arctan \left (c x^3\right )\right )^2}{2 c}+\frac {1}{6} x^6 \left (a+b \arctan \left (c x^3\right )\right )^3-\frac {i b^3 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x^3+1}\right )}{2 c^2} \]

[In]

Int[x^5*(a + b*ArcTan[c*x^3])^3,x]

[Out]

((-1/2*I)*b*(a + b*ArcTan[c*x^3])^2)/c^2 - (b*x^3*(a + b*ArcTan[c*x^3])^2)/(2*c) + (a + b*ArcTan[c*x^3])^3/(6*
c^2) + (x^6*(a + b*ArcTan[c*x^3])^3)/6 - (b^2*(a + b*ArcTan[c*x^3])*Log[2/(1 + I*c*x^3)])/c^2 - ((I/2)*b^3*Pol
yLog[2, 1 - 2/(1 + I*c*x^3)])/c^2

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 4930

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

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]

Rule 4948

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m
+ 1)/n] - 1)*(a + b*ArcTan[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[Sim
plify[(m + 1)/n]]

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 + c^2*x
^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 5004

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rule 5036

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d*(f^2/e), Int[(f*x)^(m - 2)*((a + b*ArcTan[c*x])^p/
(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rule 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[c*x])^(p + 1)/(b*e*(p + 1))), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.16 \[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\frac {3 b^2 \left (a+a c^2 x^6+b \left (i-c x^3\right )\right ) \arctan \left (c x^3\right )^2+b^3 \left (1+c^2 x^6\right ) \arctan \left (c x^3\right )^3+3 b \arctan \left (c x^3\right ) \left (a \left (a-2 b c x^3+a c^2 x^6\right )-2 b^2 \log \left (1+e^{2 i \arctan \left (c x^3\right )}\right )\right )+a \left (a c x^3 \left (-3 b+a c x^3\right )+3 b^2 \log \left (1+c^2 x^6\right )\right )+3 i b^3 \operatorname {PolyLog}\left (2,-e^{2 i \arctan \left (c x^3\right )}\right )}{6 c^2} \]

[In]

Integrate[x^5*(a + b*ArcTan[c*x^3])^3,x]

[Out]

(3*b^2*(a + a*c^2*x^6 + b*(I - c*x^3))*ArcTan[c*x^3]^2 + b^3*(1 + c^2*x^6)*ArcTan[c*x^3]^3 + 3*b*ArcTan[c*x^3]
*(a*(a - 2*b*c*x^3 + a*c^2*x^6) - 2*b^2*Log[1 + E^((2*I)*ArcTan[c*x^3])]) + a*(a*c*x^3*(-3*b + a*c*x^3) + 3*b^
2*Log[1 + c^2*x^6]) + (3*I)*b^3*PolyLog[2, -E^((2*I)*ArcTan[c*x^3])])/(6*c^2)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 7.57 (sec) , antiderivative size = 935, normalized size of antiderivative = 6.36

method	result	size
risch	\(\text {Expression too large to display}\)	\(935\)
default	\(\text {Expression too large to display}\)	\(11515\)
parts	\(\text {Expression too large to display}\)	\(11515\)

[In]

int(x^5*(a+b*arctan(c*x^3))^3,x,method=_RETURNVERBOSE)

[Out]

-1/16*b^2*(I*b*c^2*x^6*ln(1-I*c*x^3)+2*a*c^2*x^6-2*b*c*x^3+I*b*ln(1-I*c*x^3)+2*I*b+2*a)/c^2*ln(1+I*c*x^3)^2+1/
2*a*b^2/c^2*ln(c^2*x^6+1)+1/2*a^2*b/c^2*arctan(c*x^3)-1/48*I*b^3*x^6*ln(1-I*c*x^3)^3+1/8*I/c^2*b^3*ln(c^2*x^6+
1)+1/6*a^3*x^6+1/8*I/c^2*b^3*ln(1-I*c*x^3)^2-1/2/c*a^2*b*x^3+1/8*b^3/c*x^3*ln(1-I*c*x^3)^2+3/4*I/c*b^2*Sum(2/3
*(ln(x-_alpha)*ln(1-I*c*x^3)+3*c*(-1/3*ln(x-_alpha)*(ln((RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootOf(c*_Z^3-I)^2,in
dex=1)-x+_alpha)/RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootOf(c*_Z^3-I)^2,index=1))+ln((RootOf(_Z^2+_Z*RootOf(c*_Z^3
-I)+RootOf(c*_Z^3-I)^2,index=2)-x+_alpha)/RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootOf(c*_Z^3-I)^2,index=2))+ln(1/2*
(2*(I/c)^(1/3)+x-_alpha)/(I/c)^(1/3)))/c-1/3*(dilog((RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootOf(c*_Z^3-I)^2,index=
1)-x+_alpha)/RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootOf(c*_Z^3-I)^2,index=1))+dilog((RootOf(_Z^2+_Z*RootOf(c*_Z^3-
I)+RootOf(c*_Z^3-I)^2,index=2)-x+_alpha)/RootOf(_Z^2+_Z*RootOf(c*_Z^3-I)+RootOf(c*_Z^3-I)^2,index=2))+dilog(1/
2*(2*(I/c)^(1/3)+x-_alpha)/(I/c)^(1/3)))/c))*b/c,_alpha=RootOf(c*_Z^3-RootOf(_Z^2+1,index=1)))+1/4*I*b*a^2*x^6
*ln(1-I*c*x^3)-1/8*a*b^2*x^6*ln(1-I*c*x^3)^2-1/8/c^2*a*b^2*ln(1-I*c*x^3)^2+1/48*I*b^3*(c^2*x^6+1)/c^2*ln(1+I*c
*x^3)^3-1/2*I/c*a*b^2*x^3*ln(1-I*c*x^3)-1/4/c^2*b^3*arctan(c*x^3)-1/48*I/c^2*b^3*ln(1-I*c*x^3)^3+(1/16*I*b^3*(
c^2*x^6+1)/c^2*ln(1-I*c*x^3)^2+1/16*b^2*(2*a*c*x^3-b)^2/c^2/a*ln(1-I*c*x^3)-1/16*b*(4*I*a^3*c^2*x^6-8*I*a^2*b*
c*x^3+4*I*ln(1-I*c*x^3)*a*b^2+4*I*a*b^2-4*ln(1-I*c*x^3)*a^2*b+ln(1-I*c*x^3)*b^3)/a/c^2)*ln(1+I*c*x^3)

Fricas [F]

\[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{3}\right ) + a\right )}^{3} x^{5} \,d x } \]

[In]

integrate(x^5*(a+b*arctan(c*x^3))^3,x, algorithm="fricas")

[Out]

integral(b^3*x^5*arctan(c*x^3)^3 + 3*a*b^2*x^5*arctan(c*x^3)^2 + 3*a^2*b*x^5*arctan(c*x^3) + a^3*x^5, x)

Sympy [F]

\[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\int x^{5} \left (a + b \operatorname {atan}{\left (c x^{3} \right )}\right )^{3}\, dx \]

[In]

integrate(x**5*(a+b*atan(c*x**3))**3,x)

[Out]

Integral(x**5*(a + b*atan(c*x**3))**3, x)

Maxima [F]

\[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{3}\right ) + a\right )}^{3} x^{5} \,d x } \]

[In]

integrate(x^5*(a+b*arctan(c*x^3))^3,x, algorithm="maxima")

[Out]

1/2*a*b^2*x^6*arctan(c*x^3)^2 + 1/6*a^3*x^6 + 1/2*(x^6*arctan(c*x^3) - c*(x^3/c^2 - arctan(c*x^3)/c^3))*a^2*b
- 1/2*(2*c*(x^3/c^2 - arctan(c*x^3)/c^3)*arctan(c*x^3) + (arctan(c*x^3)^2 - log(6*c^5*x^6 + 6*c^3))/c^2)*a*b^2
+ 1/192*(4*x^6*arctan(c*x^3)^3 - 3*x^6*arctan(c*x^3)*log(c^2*x^6 + 1)^2 + 192*integrate(1/64*(12*c^2*x^11*arc
tan(c*x^3)*log(c^2*x^6 + 1) - 12*c*x^8*arctan(c*x^3)^2 + 56*(c^2*x^11 + x^5)*arctan(c*x^3)^3 + 3*(c*x^8 + 2*(c
^2*x^11 + x^5)*arctan(c*x^3))*log(c^2*x^6 + 1)^2)/(c^2*x^6 + 1), x))*b^3

Giac [F]

\[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\int { {\left (b \arctan \left (c x^{3}\right ) + a\right )}^{3} x^{5} \,d x } \]

[In]

integrate(x^5*(a+b*arctan(c*x^3))^3,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x^3) + a)^3*x^5, x)

Mupad [F(-1)]

Timed out. \[ \int x^5 \left (a+b \arctan \left (c x^3\right )\right )^3 \, dx=\int x^5\,{\left (a+b\,\mathrm {atan}\left (c\,x^3\right )\right )}^3 \,d x \]

[In]

int(x^5*(a + b*atan(c*x^3))^3,x)

[Out]

int(x^5*(a + b*atan(c*x^3))^3, x)