3.1.71 \(\int (d+i c d x) (a+b \text {ArcTan}(c x))^2 \, dx\) [71]

Optimal. Leaf size=130 \[ -i a b d x-i b^2 d x \text {ArcTan}(c x)-\frac {i d (1+i c x)^2 (a+b \text {ArcTan}(c x))^2}{2 c}+\frac {2 b d (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1-i c x}\right )}{c}+\frac {i b^2 d \log \left (1+c^2 x^2\right )}{2 c}-\frac {i b^2 d \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{c} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {4974, 4930, 266, 1600, 4964, 2449, 2352} \begin {gather*} -\frac {i d (1+i c x)^2 (a+b \text {ArcTan}(c x))^2}{2 c}+\frac {2 b d \log \left (\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))}{c}-i a b d x-i b^2 d x \text {ArcTan}(c x)+\frac {i b^2 d \log \left (c^2 x^2+1\right )}{2 c}-\frac {i b^2 d \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + I*c*d*x)*(a + b*ArcTan[c*x])^2,x]

[Out]

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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 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 4974

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

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 151, normalized size = 1.16 \begin {gather*} \frac {i d \left (-2 i a^2 c x-2 a b c x+a^2 c^2 x^2+b^2 (-i+c x)^2 \text {ArcTan}(c x)^2+2 b \text {ArcTan}(c x) \left (a-2 i a c x-b c x+a c^2 x^2-2 i b \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )+2 i a b \log \left (1+c^2 x^2\right )+b^2 \log \left (1+c^2 x^2\right )-2 b^2 \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (118 ) = 236\).
time = 0.19, size = 344, normalized size = 2.65

method	result	size
derivativedivides	\(\frac {-i d \,b^{2} \arctan \left (c x \right ) c x +i d a b \arctan \left (c x \right )+b^{2} \arctan \left (c x \right )^{2} d c x -b^{2} \ln \left (c^{2} x^{2}+1\right ) \arctan \left (c x \right ) d -i d \,a^{2} \left (-\frac {1}{2} c^{2} x^{2}+i c x \right )+\frac {i d \,b^{2} \ln \left (c^{2} x^{2}+1\right )}{2}-\frac {i b^{2} d \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{2}-\frac {i b^{2} d \ln \left (c x +i\right )^{2}}{4}-i d a b c x +\frac {i b^{2} d \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\frac {i b^{2} d \ln \left (c x -i\right )^{2}}{4}-\frac {i b^{2} d \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\frac {i b^{2} d \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}+\frac {i b^{2} d \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}+\frac {i d \,b^{2} \arctan \left (c x \right )^{2}}{2}-\frac {i b^{2} d \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{2}+\frac {i d \,b^{2} \arctan \left (c x \right )^{2} c^{2} x^{2}}{2}+2 a b \arctan \left (c x \right ) d c x +i d a b \arctan \left (c x \right ) c^{2} x^{2}-a b d \ln \left (c^{2} x^{2}+1\right )}{c}\)	\(344\)
default	\(\frac {-i d \,b^{2} \arctan \left (c x \right ) c x +i d a b \arctan \left (c x \right )+b^{2} \arctan \left (c x \right )^{2} d c x -b^{2} \ln \left (c^{2} x^{2}+1\right ) \arctan \left (c x \right ) d -i d \,a^{2} \left (-\frac {1}{2} c^{2} x^{2}+i c x \right )+\frac {i d \,b^{2} \ln \left (c^{2} x^{2}+1\right )}{2}-\frac {i b^{2} d \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{2}-\frac {i b^{2} d \ln \left (c x +i\right )^{2}}{4}-i d a b c x +\frac {i b^{2} d \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\frac {i b^{2} d \ln \left (c x -i\right )^{2}}{4}-\frac {i b^{2} d \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\frac {i b^{2} d \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}+\frac {i b^{2} d \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}+\frac {i d \,b^{2} \arctan \left (c x \right )^{2}}{2}-\frac {i b^{2} d \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{2}+\frac {i d \,b^{2} \arctan \left (c x \right )^{2} c^{2} x^{2}}{2}+2 a b \arctan \left (c x \right ) d c x +i d a b \arctan \left (c x \right ) c^{2} x^{2}-a b d \ln \left (c^{2} x^{2}+1\right )}{c}\)	\(344\)
risch	\(a^{2} d x +\frac {i b^{2} \ln \left (\frac {1}{2}-\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) d}{c}+\frac {i b^{2} \dilog \left (\frac {1}{2}-\frac {i c x}{2}\right ) d}{c}-\frac {3 i \ln \left (-i c x +1\right )^{2} b^{2} d}{8 c}-\frac {a b d \ln \left (c^{2} x^{2}+1\right )}{c}-\frac {\ln \left (-i c x +1\right )^{2} x \,b^{2} d}{4}+\frac {3 \ln \left (-i c x +1\right ) x \,b^{2} d}{4}-\frac {d c a b \ln \left (-i c x +1\right ) x^{2}}{2}-\frac {i d c \,b^{2} \ln \left (-i c x +1\right )^{2} x^{2}}{8}-\frac {i d \,b^{2} \left (c^{2} x^{2}-2 i c x -1\right ) \ln \left (i c x +1\right )^{2}}{8 c}+\frac {i d a b \arctan \left (c x \right )}{c}+i \ln \left (-i c x +1\right ) x a b d -\frac {i b^{2} \ln \left (-i c x +1\right ) \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) d}{c}+\frac {3 b^{2} d \arctan \left (c x \right )}{8 c}-i a b d x +\frac {i d c \,b^{2} \ln \left (-i c x +1\right ) x^{2}}{8}+\frac {11 i b^{2} d \ln \left (c^{2} x^{2}+1\right )}{16 c}+\frac {i a^{2} c d \,x^{2}}{2}-\frac {i b^{2} \ln \left (-i c x +1\right ) \left (-i c x +1\right ) d}{2 c}+\frac {i d \,b^{2} \ln \left (-i c x +1\right ) \left (-i c x +1\right )^{2}}{8 c}+\frac {3 i d \,a^{2}}{2 c}+\left (\frac {i d \,b^{2} \left (c \,x^{2}-2 i x \right ) \ln \left (-i c x +1\right )}{4}+\frac {d b \left (2 a \,c^{2} x^{2}-4 i a c x +3 i b \ln \left (-i c x +1\right )-2 x b c \right )}{4 c}\right ) \ln \left (i c x +1\right )+\frac {d a b}{c}\)	\(462\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(-I*d*b^2*arctan(c*x)*c*x+I*d*a*b*arctan(c*x)+b^2*arctan(c*x)^2*d*c*x-b^2*ln(c^2*x^2+1)*arctan(c*x)*d-I*d*
a^2*(-1/2*c^2*x^2+I*c*x)-1/2*I*b^2*d*dilog(1/2*I*(c*x-I))+1/2*I*d*b^2*ln(c^2*x^2+1)-1/2*I*b^2*d*ln(c*x+I)*ln(1
/2*I*(c*x-I))+1/2*I*b^2*d*dilog(-1/2*I*(c*x+I))-1/4*I*b^2*d*ln(c*x+I)^2-I*d*a*b*c*x+1/2*I*b^2*d*ln(c*x+I)*ln(c
^2*x^2+1)+1/4*I*b^2*d*ln(c*x-I)^2-1/2*I*b^2*d*ln(c*x-I)*ln(c^2*x^2+1)+1/2*I*b^2*d*ln(c*x-I)*ln(-1/2*I*(c*x+I))
+1/2*I*d*b^2*arctan(c*x)^2+1/2*I*d*b^2*arctan(c*x)^2*c^2*x^2+2*a*b*arctan(c*x)*d*c*x+I*d*a*b*arctan(c*x)*c^2*x
^2-a*b*d*ln(c^2*x^2+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((d+I*c*d*x)*(a+b*arctan(c*x))^2,x, algorithm="maxima")

[Out]

4*b^2*c^3*d*integrate(1/16*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 4*b^2*c^3*d*integrate(1/16*x^3
*arctan(c*x)/(c^2*x^2 + 1), x) + 1/2*I*a^2*c*d*x^2 + 12*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^2 +
1), x) + b^2*c^2*d*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 6*b^2*c^2*d*integrate(1/16*x^2*lo
g(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + I*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*a*b*c*d + 1/4*b^2*d*arcta
n(c*x)^3/c + 4*b^2*c*d*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - 8*b^2*c*d*integrate(1
/16*x*arctan(c*x)/(c^2*x^2 + 1), x) + a^2*d*x + b^2*d*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + (2
*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b*d/c - 1/8*(-I*b^2*c*d*x^2 - 2*b^2*d*x)*arctan(c*x)^2 - 1/8*(b^2*c*d*x
^2 - 2*I*b^2*d*x)*arctan(c*x)*log(c^2*x^2 + 1) + 1/32*(-I*b^2*c*d*x^2 - 2*b^2*d*x)*log(c^2*x^2 + 1)^2 + I*inte
grate(-1/16*(12*b^2*c^2*d*x^2*arctan(c*x) - 12*(b^2*c^3*d*x^3 + b^2*c*d*x)*arctan(c*x)^2 - (b^2*c^3*d*x^3 + b^
2*c*d*x)*log(c^2*x^2 + 1)^2 - 2*(b^2*c^3*d*x^3 - 2*b^2*c*d*x - 2*(b^2*c^2*d*x^2 + b^2*d)*arctan(c*x))*log(c^2*
x^2 + 1))/(c^2*x^2 + 1), x)

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

[Out]

1/8*(-I*b^2*c*d*x^2 - 2*b^2*d*x)*log(-(c*x + I)/(c*x - I))^2 + integral(1/2*(2*I*a^2*c^3*d*x^3 + 2*a^2*c^2*d*x
^2 + 2*I*a^2*c*d*x + 2*a^2*d - (2*a*b*c^3*d*x^3 - (2*I*a*b + b^2)*c^2*d*x^2 + 2*(a*b + I*b^2)*c*d*x - 2*I*a*b*
d)*log(-(c*x + I)/(c*x - I)))/(c^2*x^2 + 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {i d \left (\int \left (- \frac {4 i a^{2}}{c^{2} x^{2} + 1}\right )\, dx + \int \left (- \frac {i b^{2} \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\right )\, dx + \int \left (- \frac {4 a b \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\right )\, dx + \int \frac {4 a^{2} c x}{c^{2} x^{2} + 1}\, dx + \int \frac {4 a^{2} c^{3} x^{3}}{c^{2} x^{2} + 1}\, dx + \int \frac {2 b^{2} c x}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {4 i a^{2} c^{2} x^{2}}{c^{2} x^{2} + 1}\right )\, dx + \int \frac {2 i b^{2} c^{2} x^{2}}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {6 a b c^{2} x^{2}}{c^{2} x^{2} + 1}\right )\, dx + \int \frac {5 b^{2} c x \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\, dx + \int \frac {4 i a b c x}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {2 i a b c^{3} x^{3}}{c^{2} x^{2} + 1}\right )\, dx + \int \frac {2 i b^{2} c^{2} x^{2} \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {4 a b c^{2} x^{2} \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\right )\, dx + \int \left (- \frac {4 i a b c x \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\right )\, dx + \int \left (- \frac {4 i a b c^{3} x^{3} \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\right )\, dx\right )}{4} + \left (- \frac {i b^{2} c d x^{2}}{8} - \frac {b^{2} d x}{4}\right ) \log {\left (i c x + 1 \right )}^{2} + \frac {\left (- i b^{2} c^{2} d x^{2} - 2 b^{2} c d x - 3 i b^{2} d\right ) \log {\left (- i c x + 1 \right )}^{2}}{8 c} + \frac {\left (- 2 a b c^{2} d x^{2} + 4 i a b c d x + i b^{2} c^{2} d x^{2} \log {\left (i c x + 1 \right )} + 2 b^{2} c d x \log {\left (i c x + 1 \right )} + 2 b^{2} c d x - i b^{2} d \log {\left (i c x + 1 \right )}\right ) \log {\left (- i c x + 1 \right )}}{4 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*d*(Integral(-4*I*a**2/(c**2*x**2 + 1), x) + Integral(-I*b**2*log(I*c*x + 1)/(c**2*x**2 + 1), x) + Integral(-
4*a*b*log(I*c*x + 1)/(c**2*x**2 + 1), x) + Integral(4*a**2*c*x/(c**2*x**2 + 1), x) + Integral(4*a**2*c**3*x**3
/(c**2*x**2 + 1), x) + Integral(2*b**2*c*x/(c**2*x**2 + 1), x) + Integral(-4*I*a**2*c**2*x**2/(c**2*x**2 + 1),
x) + Integral(2*I*b**2*c**2*x**2/(c**2*x**2 + 1), x) + Integral(-6*a*b*c**2*x**2/(c**2*x**2 + 1), x) + Integr
al(5*b**2*c*x*log(I*c*x + 1)/(c**2*x**2 + 1), x) + Integral(4*I*a*b*c*x/(c**2*x**2 + 1), x) + Integral(-2*I*a*
b*c**3*x**3/(c**2*x**2 + 1), x) + Integral(2*I*b**2*c**2*x**2*log(I*c*x + 1)/(c**2*x**2 + 1), x) + Integral(-4
*a*b*c**2*x**2*log(I*c*x + 1)/(c**2*x**2 + 1), x) + Integral(-4*I*a*b*c*x*log(I*c*x + 1)/(c**2*x**2 + 1), x) +
Integral(-4*I*a*b*c**3*x**3*log(I*c*x + 1)/(c**2*x**2 + 1), x))/4 + (-I*b**2*c*d*x**2/8 - b**2*d*x/4)*log(I*c
*x + 1)**2 + (-I*b**2*c**2*d*x**2 - 2*b**2*c*d*x - 3*I*b**2*d)*log(-I*c*x + 1)**2/(8*c) + (-2*a*b*c**2*d*x**2
+ 4*I*a*b*c*d*x + I*b**2*c**2*d*x**2*log(I*c*x + 1) + 2*b**2*c*d*x*log(I*c*x + 1) + 2*b**2*c*d*x - I*b**2*d*lo
g(I*c*x + 1))*log(-I*c*x + 1)/(4*c)

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

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^2*(d + c*d*x*1i),x)

[Out]

int((a + b*atan(c*x))^2*(d + c*d*x*1i), x)

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