3.1.77 \(\int x (a+b \text {ArcTan}(c x^2))^2 \, dx\) [77]

Optimal. Leaf size=101 \[ \frac {i \left (a+b \text {ArcTan}\left (c x^2\right )\right )^2}{2 c}+\frac {1}{2} x^2 \left (a+b \text {ArcTan}\left (c x^2\right )\right )^2+\frac {b \left (a+b \text {ArcTan}\left (c x^2\right )\right ) \log \left (\frac {2}{1+i c x^2}\right )}{c}+\frac {i b^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x^2}\right )}{2 c} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4948, 4930, 5040, 4964, 2449, 2352} \begin {gather*} \frac {1}{2} x^2 \left (a+b \text {ArcTan}\left (c x^2\right )\right )^2+\frac {i \left (a+b \text {ArcTan}\left (c x^2\right )\right )^2}{2 c}+\frac {b \log \left (\frac {2}{1+i c x^2}\right ) \left (a+b \text {ArcTan}\left (c x^2\right )\right )}{c}+\frac {i b^2 \text {Li}_2\left (1-\frac {2}{i c x^2+1}\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTan[c*x^2])^2,x]

[Out]

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

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 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 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*} \int x \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \, dx &=\int \left (\frac {1}{4} x \left (2 a+i b \log \left (1-i c x^2\right )\right )^2+\frac {1}{2} b x \left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )-\frac {1}{4} b^2 x \log ^2\left (1+i c x^2\right )\right ) \, dx\\ &=\frac {1}{4} \int x \left (2 a+i b \log \left (1-i c x^2\right )\right )^2 \, dx+\frac {1}{2} b \int x \left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right ) \, dx-\frac {1}{4} b^2 \int x \log ^2\left (1+i c x^2\right ) \, dx\\ &=\frac {1}{8} \text {Subst}\left (\int (2 a+i b \log (1-i c x))^2 \, dx,x,x^2\right )+\frac {1}{4} b \text {Subst}\left (\int (-2 i a+b \log (1-i c x)) \log (1+i c x) \, dx,x,x^2\right )-\frac {1}{8} b^2 \text {Subst}\left (\int \log ^2(1+i c x) \, dx,x,x^2\right )\\ &=-\frac {1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac {i \text {Subst}\left (\int (2 a+i b \log (x))^2 \, dx,x,1-i c x^2\right )}{8 c}+\frac {\left (i b^2\right ) \text {Subst}\left (\int \log ^2(x) \, dx,x,1+i c x^2\right )}{8 c}-\frac {1}{4} (i b c) \text {Subst}\left (\int \frac {x (-2 i a+b \log (1-i c x))}{1+i c x} \, dx,x,x^2\right )+\frac {1}{4} \left (i b^2 c\right ) \text {Subst}\left (\int \frac {x \log (1+i c x)}{1-i c x} \, dx,x,x^2\right )\\ &=\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c}-\frac {1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac {i b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c}+\frac {b \text {Subst}\left (\int (2 a+i b \log (x)) \, dx,x,1-i c x^2\right )}{4 c}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \log (x) \, dx,x,1+i c x^2\right )}{4 c}-\frac {1}{4} (i b c) \text {Subst}\left (\int \left (-\frac {i (-2 i a+b \log (1-i c x))}{c}+\frac {-2 i a+b \log (1-i c x)}{c (-i+c x)}\right ) \, dx,x,x^2\right )+\frac {1}{4} \left (i b^2 c\right ) \text {Subst}\left (\int \left (\frac {i \log (1+i c x)}{c}+\frac {\log (1+i c x)}{c (i+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{2} i a b x^2-\frac {b^2 x^2}{4}+\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c}-\frac {i b^2 \left (1+i c x^2\right ) \log \left (1+i c x^2\right )}{4 c}-\frac {1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac {i b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c}-\frac {1}{4} (i b) \text {Subst}\left (\int \frac {-2 i a+b \log (1-i c x)}{-i+c x} \, dx,x,x^2\right )-\frac {1}{4} b \text {Subst}\left (\int (-2 i a+b \log (1-i c x)) \, dx,x,x^2\right )+\frac {1}{4} \left (i b^2\right ) \text {Subst}\left (\int \frac {\log (1+i c x)}{i+c x} \, dx,x,x^2\right )-\frac {1}{4} b^2 \text {Subst}\left (\int \log (1+i c x) \, dx,x,x^2\right )+\frac {\left (i b^2\right ) \text {Subst}\left (\int \log (x) \, dx,x,1-i c x^2\right )}{4 c}\\ &=-\frac {1}{2} b^2 x^2+\frac {i b^2 \left (1-i c x^2\right ) \log \left (1-i c x^2\right )}{4 c}+\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c}+\frac {i b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+i c x^2\right )\right )}{4 c}-\frac {i b^2 \left (1+i c x^2\right ) \log \left (1+i c x^2\right )}{4 c}+\frac {i b^2 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 c}-\frac {1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac {i b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c}-\frac {1}{4} b^2 \text {Subst}\left (\int \log (1-i c x) \, dx,x,x^2\right )+\frac {1}{4} b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} i (-i+c x)\right )}{1-i c x} \, dx,x,x^2\right )+\frac {1}{4} b^2 \text {Subst}\left (\int \frac {\log \left (-\frac {1}{2} i (i+c x)\right )}{1+i c x} \, dx,x,x^2\right )+\frac {\left (i b^2\right ) \text {Subst}\left (\int \log (x) \, dx,x,1+i c x^2\right )}{4 c}\\ &=-\frac {1}{4} b^2 x^2+\frac {i b^2 \left (1-i c x^2\right ) \log \left (1-i c x^2\right )}{4 c}+\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c}+\frac {i b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+i c x^2\right )\right )}{4 c}+\frac {i b^2 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 c}-\frac {1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac {i b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c}+\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-i c x^2\right )}{4 c}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+i c x^2\right )}{4 c}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \log (x) \, dx,x,1-i c x^2\right )}{4 c}\\ &=\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c}+\frac {i b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+i c x^2\right )\right )}{4 c}+\frac {i b^2 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 c}-\frac {1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac {i b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c}-\frac {i b^2 \text {Li}_2\left (\frac {1}{2} \left (1-i c x^2\right )\right )}{4 c}+\frac {i b^2 \text {Li}_2\left (\frac {1}{2} \left (1+i c x^2\right )\right )}{4 c}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 107, normalized size = 1.06 \begin {gather*} \frac {b^2 \left (-i+c x^2\right ) \text {ArcTan}\left (c x^2\right )^2+2 b \text {ArcTan}\left (c x^2\right ) \left (a c x^2+b \log \left (1+e^{2 i \text {ArcTan}\left (c x^2\right )}\right )\right )+a \left (a c x^2-b \log \left (1+c^2 x^4\right )\right )-i b^2 \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}\left (c x^2\right )}\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(-I + c*x^2)*ArcTan[c*x^2]^2 + 2*b*ArcTan[c*x^2]*(a*c*x^2 + b*Log[1 + E^((2*I)*ArcTan[c*x^2])]) + a*(a*c*
x^2 - b*Log[1 + c^2*x^4]) - I*b^2*PolyLog[2, -E^((2*I)*ArcTan[c*x^2])])/(2*c)

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Maple [A]
time = 0.18, size = 142, normalized size = 1.41

method	result	size
derivativedivides	\(\frac {c \,x^{2} a^{2}-i \arctan \left (c \,x^{2}\right )^{2} b^{2}+\arctan \left (c \,x^{2}\right )^{2} b^{2} c \,x^{2}+2 \arctan \left (c \,x^{2}\right ) \ln \left (1+\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right ) b^{2}-i \polylog \left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right ) b^{2}+2 a b c \,x^{2} \arctan \left (c \,x^{2}\right )-a b \ln \left (c^{2} x^{4}+1\right )}{2 c}\)	\(142\)
default	\(\frac {c \,x^{2} a^{2}-i \arctan \left (c \,x^{2}\right )^{2} b^{2}+\arctan \left (c \,x^{2}\right )^{2} b^{2} c \,x^{2}+2 \arctan \left (c \,x^{2}\right ) \ln \left (1+\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right ) b^{2}-i \polylog \left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right ) b^{2}+2 a b c \,x^{2} \arctan \left (c \,x^{2}\right )-a b \ln \left (c^{2} x^{4}+1\right )}{2 c}\)	\(142\)
risch	\(-\frac {i \ln \left (-i c \,x^{2}+1\right )^{2} b^{2}}{8 c}-\frac {a b \ln \left (c^{2} x^{4}+1\right )}{2 c}+\frac {i \ln \left (-i c \,x^{2}+1\right ) x^{2} a b}{2}+\frac {b^{2} \ln \left (i c \,x^{2}+1\right ) \ln \left (-i c \,x^{2}+1\right ) x^{2}}{4}+\frac {i b^{2} \dilog \left (\frac {1}{2}-\frac {i c \,x^{2}}{2}\right )}{2 c}+\frac {i b^{2}}{2 c}+\frac {i b^{2} \ln \left (i c \,x^{2}+1\right ) \ln \left (-i c \,x^{2}+1\right )}{4 c}+\frac {i b^{2} \ln \left (i c \,x^{2}+1\right )^{2}}{8 c}-\frac {i b a \ln \left (i c \,x^{2}+1\right ) x^{2}}{2}+\frac {i a^{2}}{2 c}+\frac {i b^{2} \ln \left (\frac {1}{2}+\frac {i c \,x^{2}}{2}\right ) \ln \left (\frac {1}{2}-\frac {i c \,x^{2}}{2}\right )}{2 c}-\frac {i b^{2} \ln \left (\frac {1}{2}+\frac {i c \,x^{2}}{2}\right ) \ln \left (-i c \,x^{2}+1\right )}{2 c}-\frac {\ln \left (-i c \,x^{2}+1\right )^{2} x^{2} b^{2}}{8}-\frac {b^{2} \ln \left (i c \,x^{2}+1\right )^{2} x^{2}}{8}+\frac {a b}{c}+\frac {x^{2} a^{2}}{2}\)	\(303\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/c*(c*x^2*a^2-I*arctan(c*x^2)^2*b^2+arctan(c*x^2)^2*b^2*c*x^2+2*arctan(c*x^2)*ln(1+(1+I*c*x^2)^2/(c^2*x^4+1
))*b^2-I*polylog(2,-(1+I*c*x^2)^2/(c^2*x^4+1))*b^2+2*a*b*c*x^2*arctan(c*x^2)-a*b*ln(c^2*x^4+1))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*x^2 + 1/32*(4*x^2*arctan(c*x^2)^2 - x^2*log(c^2*x^4 + 1)^2 + 384*c^2*integrate(1/16*x^5*arctan(c*x^2)^
2/(c^2*x^4 + 1), x) + 32*c^2*integrate(1/16*x^5*log(c^2*x^4 + 1)^2/(c^2*x^4 + 1), x) + 128*c^2*integrate(1/16*
x^5*log(c^2*x^4 + 1)/(c^2*x^4 + 1), x) + 4*arctan(c*x^2)^3/c - 256*c*integrate(1/16*x^3*arctan(c*x^2)/(c^2*x^4
+ 1), x) + 32*integrate(1/16*x*log(c^2*x^4 + 1)^2/(c^2*x^4 + 1), x))*b^2 + 1/2*(2*c*x^2*arctan(c*x^2) - log(c
^2*x^4 + 1))*a*b/c

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (a + b \operatorname {atan}{\left (c x^{2} \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(a + b*atan(c*x**2))**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctan(c*x^2) + a)^2*x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\left (a+b\,\mathrm {atan}\left (c\,x^2\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atan(c*x^2))^2,x)

[Out]

int(x*(a + b*atan(c*x^2))^2, x)

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