3.261 \(\int (c+a^2 c x^2) \tan ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=128 \[ \frac {1}{3} c x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2-\frac {c \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}{3 a}+\frac {2 i c \text {Li}_2\left (1-\frac {2}{i a x+1}\right )}{3 a}+\frac {2 i c \tan ^{-1}(a x)^2}{3 a}+\frac {2}{3} c x \tan ^{-1}(a x)^2+\frac {4 c \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{3 a}+\frac {c x}{3} \]

[Out]

1/3*c*x-1/3*c*(a^2*x^2+1)*arctan(a*x)/a+2/3*I*c*arctan(a*x)^2/a+2/3*c*x*arctan(a*x)^2+1/3*c*x*(a^2*x^2+1)*arct
an(a*x)^2+4/3*c*arctan(a*x)*ln(2/(1+I*a*x))/a+2/3*I*c*polylog(2,1-2/(1+I*a*x))/a

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Rubi [A]  time = 0.10, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {4880, 4846, 4920, 4854, 2402, 2315, 8} \[ \frac {2 i c \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{3 a}+\frac {1}{3} c x \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2-\frac {c \left (a^2 x^2+1\right ) \tan ^{-1}(a x)}{3 a}+\frac {2 i c \tan ^{-1}(a x)^2}{3 a}+\frac {2}{3} c x \tan ^{-1}(a x)^2+\frac {4 c \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{3 a}+\frac {c x}{3} \]

Antiderivative was successfully verified.

[In]

Int[(c + a^2*c*x^2)*ArcTan[a*x]^2,x]

[Out]

(c*x)/3 - (c*(1 + a^2*x^2)*ArcTan[a*x])/(3*a) + (((2*I)/3)*c*ArcTan[a*x]^2)/a + (2*c*x*ArcTan[a*x]^2)/3 + (c*x
*(1 + a^2*x^2)*ArcTan[a*x]^2)/3 + (4*c*ArcTan[a*x]*Log[2/(1 + I*a*x)])/(3*a) + (((2*I)/3)*c*PolyLog[2, 1 - 2/(
1 + I*a*x)])/a

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2315

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

Rule 2402

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 4846

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

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[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 4880

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

Rule 4920

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[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 \left (c+a^2 c x^2\right ) \tan ^{-1}(a x)^2 \, dx &=-\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{3 a}+\frac {1}{3} c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2+\frac {1}{3} c \int 1 \, dx+\frac {1}{3} (2 c) \int \tan ^{-1}(a x)^2 \, dx\\ &=\frac {c x}{3}-\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{3 a}+\frac {2}{3} c x \tan ^{-1}(a x)^2+\frac {1}{3} c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2-\frac {1}{3} (4 a c) \int \frac {x \tan ^{-1}(a x)}{1+a^2 x^2} \, dx\\ &=\frac {c x}{3}-\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{3 a}+\frac {2 i c \tan ^{-1}(a x)^2}{3 a}+\frac {2}{3} c x \tan ^{-1}(a x)^2+\frac {1}{3} c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2+\frac {1}{3} (4 c) \int \frac {\tan ^{-1}(a x)}{i-a x} \, dx\\ &=\frac {c x}{3}-\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{3 a}+\frac {2 i c \tan ^{-1}(a x)^2}{3 a}+\frac {2}{3} c x \tan ^{-1}(a x)^2+\frac {1}{3} c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2+\frac {4 c \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{3 a}-\frac {1}{3} (4 c) \int \frac {\log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx\\ &=\frac {c x}{3}-\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{3 a}+\frac {2 i c \tan ^{-1}(a x)^2}{3 a}+\frac {2}{3} c x \tan ^{-1}(a x)^2+\frac {1}{3} c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2+\frac {4 c \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{3 a}+\frac {(4 i c) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i a x}\right )}{3 a}\\ &=\frac {c x}{3}-\frac {c \left (1+a^2 x^2\right ) \tan ^{-1}(a x)}{3 a}+\frac {2 i c \tan ^{-1}(a x)^2}{3 a}+\frac {2}{3} c x \tan ^{-1}(a x)^2+\frac {1}{3} c x \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^2+\frac {4 c \tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{3 a}+\frac {2 i c \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{3 a}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 82, normalized size = 0.64 \[ \frac {c \left (\left (a^3 x^3+3 a x-2 i\right ) \tan ^{-1}(a x)^2-\tan ^{-1}(a x) \left (a^2 x^2-4 \log \left (1+e^{2 i \tan ^{-1}(a x)}\right )+1\right )-2 i \text {Li}_2\left (-e^{2 i \tan ^{-1}(a x)}\right )+a x\right )}{3 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*(a*x + (-2*I + 3*a*x + a^3*x^3)*ArcTan[a*x]^2 - ArcTan[a*x]*(1 + a^2*x^2 - 4*Log[1 + E^((2*I)*ArcTan[a*x])]
) - (2*I)*PolyLog[2, -E^((2*I)*ArcTan[a*x])]))/(3*a)

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fricas [F]  time = 1.28, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.10, size = 233, normalized size = 1.82 \[ \frac {a^{2} c \arctan \left (a x \right )^{2} x^{3}}{3}+c x \arctan \left (a x \right )^{2}-\frac {a c \arctan \left (a x \right ) x^{2}}{3}-\frac {2 c \arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{3 a}+\frac {c x}{3}-\frac {c \arctan \left (a x \right )}{3 a}-\frac {i c \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{3 a}+\frac {i c \ln \left (a x -i\right )^{2}}{6 a}+\frac {i c \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{3 a}+\frac {i c \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{3 a}+\frac {i c \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{3 a}-\frac {i c \ln \left (a x +i\right )^{2}}{6 a}-\frac {i c \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{3 a}-\frac {i c \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{3 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)*arctan(a*x)^2,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 36 \, a^{4} c \int \frac {x^{4} \arctan \left (a x\right )^{2}}{48 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} + 3 \, a^{4} c \int \frac {x^{4} \log \left (a^{2} x^{2} + 1\right )^{2}}{48 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} + 4 \, a^{4} c \int \frac {x^{4} \log \left (a^{2} x^{2} + 1\right )}{48 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} - 8 \, a^{3} c \int \frac {x^{3} \arctan \left (a x\right )}{48 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} + 72 \, a^{2} c \int \frac {x^{2} \arctan \left (a x\right )^{2}}{48 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} + 6 \, a^{2} c \int \frac {x^{2} \log \left (a^{2} x^{2} + 1\right )^{2}}{48 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} + 12 \, a^{2} c \int \frac {x^{2} \log \left (a^{2} x^{2} + 1\right )}{48 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} + \frac {1}{12} \, {\left (a^{2} c x^{3} + 3 \, c x\right )} \arctan \left (a x\right )^{2} + \frac {c \arctan \left (a x\right )^{3}}{4 \, a} - 24 \, a c \int \frac {x \arctan \left (a x\right )}{48 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} - \frac {1}{48} \, {\left (a^{2} c x^{3} + 3 \, c x\right )} \log \left (a^{2} x^{2} + 1\right )^{2} + 3 \, c \int \frac {\log \left (a^{2} x^{2} + 1\right )^{2}}{48 \, {\left (a^{2} x^{2} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

36*a^4*c*integrate(1/48*x^4*arctan(a*x)^2/(a^2*x^2 + 1), x) + 3*a^4*c*integrate(1/48*x^4*log(a^2*x^2 + 1)^2/(a
^2*x^2 + 1), x) + 4*a^4*c*integrate(1/48*x^4*log(a^2*x^2 + 1)/(a^2*x^2 + 1), x) - 8*a^3*c*integrate(1/48*x^3*a
rctan(a*x)/(a^2*x^2 + 1), x) + 72*a^2*c*integrate(1/48*x^2*arctan(a*x)^2/(a^2*x^2 + 1), x) + 6*a^2*c*integrate
(1/48*x^2*log(a^2*x^2 + 1)^2/(a^2*x^2 + 1), x) + 12*a^2*c*integrate(1/48*x^2*log(a^2*x^2 + 1)/(a^2*x^2 + 1), x
) + 1/12*(a^2*c*x^3 + 3*c*x)*arctan(a*x)^2 + 1/4*c*arctan(a*x)^3/a - 24*a*c*integrate(1/48*x*arctan(a*x)/(a^2*
x^2 + 1), x) - 1/48*(a^2*c*x^3 + 3*c*x)*log(a^2*x^2 + 1)^2 + 3*c*integrate(1/48*log(a^2*x^2 + 1)^2/(a^2*x^2 +
1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a*x)^2*(c + a^2*c*x^2),x)

[Out]

int(atan(a*x)^2*(c + a^2*c*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c \left (\int a^{2} x^{2} \operatorname {atan}^{2}{\left (a x \right )}\, dx + \int \operatorname {atan}^{2}{\left (a x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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