3.1.69 \(\int x^2 (d+i c d x) (a+b \text {ArcTan}(c x))^2 \, dx\) [69]

Optimal. Leaf size=255 \[ \frac {i a b d x}{2 c^2}+\frac {b^2 d x}{3 c^2}+\frac {i b^2 d x^2}{12 c}-\frac {b^2 d \text {ArcTan}(c x)}{3 c^3}+\frac {i b^2 d x \text {ArcTan}(c x)}{2 c^2}-\frac {b d x^2 (a+b \text {ArcTan}(c x))}{3 c}-\frac {1}{6} i b d x^3 (a+b \text {ArcTan}(c x))-\frac {7 i d (a+b \text {ArcTan}(c x))^2}{12 c^3}+\frac {1}{3} d x^3 (a+b \text {ArcTan}(c x))^2+\frac {1}{4} i c d x^4 (a+b \text {ArcTan}(c x))^2-\frac {2 b d (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1+i c x}\right )}{3 c^3}-\frac {i b^2 d \log \left (1+c^2 x^2\right )}{3 c^3}-\frac {i b^2 d \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{3 c^3} \]

[Out]

1/2*I*a*b*d*x/c^2+1/3*b^2*d*x/c^2+1/12*I*b^2*d*x^2/c-1/3*b^2*d*arctan(c*x)/c^3+1/2*I*b^2*d*x*arctan(c*x)/c^2-1
/3*b*d*x^2*(a+b*arctan(c*x))/c-1/6*I*b*d*x^3*(a+b*arctan(c*x))-7/12*I*d*(a+b*arctan(c*x))^2/c^3+1/3*d*x^3*(a+b
*arctan(c*x))^2+1/4*I*c*d*x^4*(a+b*arctan(c*x))^2-2/3*b*d*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c^3-1/3*I*b^2*d*ln
(c^2*x^2+1)/c^3-1/3*I*b^2*d*polylog(2,1-2/(1+I*c*x))/c^3

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Rubi [A]
time = 0.36, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 14, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {4996, 4946, 5036, 327, 209, 5040, 4964, 2449, 2352, 272, 45, 4930, 266, 5004} \begin {gather*} -\frac {7 i d (a+b \text {ArcTan}(c x))^2}{12 c^3}-\frac {2 b d \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))}{3 c^3}+\frac {1}{4} i c d x^4 (a+b \text {ArcTan}(c x))^2+\frac {1}{3} d x^3 (a+b \text {ArcTan}(c x))^2-\frac {1}{6} i b d x^3 (a+b \text {ArcTan}(c x))-\frac {b d x^2 (a+b \text {ArcTan}(c x))}{3 c}+\frac {i a b d x}{2 c^2}-\frac {b^2 d \text {ArcTan}(c x)}{3 c^3}+\frac {i b^2 d x \text {ArcTan}(c x)}{2 c^2}-\frac {i b^2 d \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{3 c^3}+\frac {b^2 d x}{3 c^2}-\frac {i b^2 d \log \left (c^2 x^2+1\right )}{3 c^3}+\frac {i b^2 d x^2}{12 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*a*b*d*x)/c^2 + (b^2*d*x)/(3*c^2) + ((I/12)*b^2*d*x^2)/c - (b^2*d*ArcTan[c*x])/(3*c^3) + ((I/2)*b^2*d*x*
ArcTan[c*x])/c^2 - (b*d*x^2*(a + b*ArcTan[c*x]))/(3*c) - (I/6)*b*d*x^3*(a + b*ArcTan[c*x]) - (((7*I)/12)*d*(a
+ b*ArcTan[c*x])^2)/c^3 + (d*x^3*(a + b*ArcTan[c*x])^2)/3 + (I/4)*c*d*x^4*(a + b*ArcTan[c*x])^2 - (2*b*d*(a +
b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(3*c^3) - ((I/3)*b^2*d*Log[1 + c^2*x^2])/c^3 - ((I/3)*b^2*d*PolyLog[2, 1 -
2/(1 + I*c*x)])/c^3

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 272

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

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 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 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 4996

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

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*} \int x^2 (d+i c d x) \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\int \left (d x^2 \left (a+b \tan ^{-1}(c x)\right )^2+i c d x^3 \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d \int x^2 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx+(i c d) \int x^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=\frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} i c d x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{3} (2 b c d) \int \frac {x^3 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac {1}{2} \left (i b c^2 d\right ) \int \frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=\frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} i c d x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{2} (i b d) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx+\frac {1}{2} (i b d) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac {(2 b d) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{3 c}+\frac {(2 b d) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 c}\\ &=-\frac {b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}-\frac {1}{6} i b d x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {i d \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}+\frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} i c d x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{3} \left (b^2 d\right ) \int \frac {x^2}{1+c^2 x^2} \, dx+\frac {(i b d) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c^2}-\frac {(i b d) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{2 c^2}-\frac {(2 b d) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx}{3 c^2}+\frac {1}{6} \left (i b^2 c d\right ) \int \frac {x^3}{1+c^2 x^2} \, dx\\ &=\frac {i a b d x}{2 c^2}+\frac {b^2 d x}{3 c^2}-\frac {b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}-\frac {1}{6} i b d x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {7 i d \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^3}+\frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} i c d x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {2 b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{3 c^3}+\frac {\left (i b^2 d\right ) \int \tan ^{-1}(c x) \, dx}{2 c^2}-\frac {\left (b^2 d\right ) \int \frac {1}{1+c^2 x^2} \, dx}{3 c^2}+\frac {\left (2 b^2 d\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{3 c^2}+\frac {1}{12} \left (i b^2 c d\right ) \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )\\ &=\frac {i a b d x}{2 c^2}+\frac {b^2 d x}{3 c^2}-\frac {b^2 d \tan ^{-1}(c x)}{3 c^3}+\frac {i b^2 d x \tan ^{-1}(c x)}{2 c^2}-\frac {b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}-\frac {1}{6} i b d x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {7 i d \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^3}+\frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} i c d x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {2 b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{3 c^3}-\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 )}{3 c^3}-\frac {\left (i b^2 d\right ) \int \frac {x}{1+c^2 x^2} \, dx}{2 c}+\frac {1}{12} \left (i b^2 c d\right ) \text {Subst}\left (\int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {i a b d x}{2 c^2}+\frac {b^2 d x}{3 c^2}+\frac {i b^2 d x^2}{12 c}-\frac {b^2 d \tan ^{-1}(c x)}{3 c^3}+\frac {i b^2 d x \tan ^{-1}(c x)}{2 c^2}-\frac {b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}-\frac {1}{6} i b d x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {7 i d \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^3}+\frac {1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} i c d x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {2 b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{3 c^3}-\frac {i b^2 d \log \left (1+c^2 x^2\right )}{3 c^3}-\frac {i b^2 d \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{3 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 241, normalized size = 0.95 \begin {gather*} \frac {i d \left (b^2+6 a b c x-4 i b^2 c x+4 i a b c^2 x^2+b^2 c^2 x^2-4 i a^2 c^3 x^3-2 a b c^3 x^3+3 a^2 c^4 x^4+b^2 \left (1-4 i c^3 x^3+3 c^4 x^4\right ) \text {ArcTan}(c x)^2+2 b \text {ArcTan}(c x) \left (b \left (2 i+3 c x+2 i c^2 x^2-c^3 x^3\right )+a \left (-3-4 i c^3 x^3+3 c^4 x^4\right )+4 i b \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )-4 i a b \log \left (1+c^2 x^2\right )-4 b^2 \log \left (1+c^2 x^2\right )+4 b^2 \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )\right )}{12 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/12)*d*(b^2 + 6*a*b*c*x - (4*I)*b^2*c*x + (4*I)*a*b*c^2*x^2 + b^2*c^2*x^2 - (4*I)*a^2*c^3*x^3 - 2*a*b*c^3*x
^3 + 3*a^2*c^4*x^4 + b^2*(1 - (4*I)*c^3*x^3 + 3*c^4*x^4)*ArcTan[c*x]^2 + 2*b*ArcTan[c*x]*(b*(2*I + 3*c*x + (2*
I)*c^2*x^2 - c^3*x^3) + a*(-3 - (4*I)*c^3*x^3 + 3*c^4*x^4) + (4*I)*b*Log[1 + E^((2*I)*ArcTan[c*x])]) - (4*I)*a
*b*Log[1 + c^2*x^2] - 4*b^2*Log[1 + c^2*x^2] + 4*b^2*PolyLog[2, -E^((2*I)*ArcTan[c*x])]))/c^3

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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. 441 vs. \(2 (219 ) = 438\).
time = 0.23, size = 442, normalized size = 1.73 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(d*a^2*(1/4*I*c^4*x^4+1/3*c^3*x^3)+1/2*I*d*a*b*arctan(c*x)*c^4*x^4+2/3*d*a*b*arctan(c*x)*c^3*x^3+1/2*I*d
*a*b*c*x-1/6*I*d*a*b*c^3*x^3-1/6*I*d*b^2*arctan(c*x)*c^3*x^3+1/2*I*d*b^2*arctan(c*x)*c*x+1/4*I*d*b^2*arctan(c*
x)^2*c^4*x^4+1/3*d*b^2*c*x+1/12*I*d*b^2*ln(c*x+I)^2-1/4*I*d*b^2*arctan(c*x)^2-1/3*I*d*b^2*ln(c^2*x^2+1)-1/6*I*
d*b^2*dilog(-1/2*I*(c*x+I))-1/12*I*d*b^2*ln(c*x-I)^2+1/6*I*d*b^2*dilog(1/2*I*(c*x-I))-1/6*I*d*b^2*ln(c*x+I)*ln
(c^2*x^2+1)+1/6*I*d*b^2*ln(c*x-I)*ln(c^2*x^2+1)-1/6*I*d*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))+1/6*I*d*b^2*ln(c*x+I)
*ln(1/2*I*(c*x-I))-1/3*d*a*b*c^2*x^2+1/12*I*d*b^2*c^2*x^2+1/3*d*b^2*arctan(c*x)^2*c^3*x^3-1/3*d*b^2*arctan(c*x
)*c^2*x^2-1/2*I*d*a*b*arctan(c*x)-1/3*d*b^2*arctan(c*x)+1/3*b^2*ln(c^2*x^2+1)*arctan(c*x)*d+1/3*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(x^2*(d+I*c*d*x)*(a+b*arctan(c*x))^2,x, algorithm="maxima")

[Out]

1/4*I*a^2*c*d*x^4 + 1/3*a^2*d*x^3 + 1/6*I*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*a*
b*c*d + 1/3*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*a*b*d - 1/48*(-3*I*b^2*c*d*x^4 - 4*b^2*d*
x^3)*arctan(c*x)^2 - 1/48*(3*b^2*c*d*x^4 - 4*I*b^2*d*x^3)*arctan(c*x)*log(c^2*x^2 + 1) + 1/192*(-3*I*b^2*c*d*x
^4 - 4*b^2*d*x^3)*log(c^2*x^2 + 1)^2 + I*integrate(-1/48*(14*b^2*c^2*d*x^4*arctan(c*x) - 36*(b^2*c^3*d*x^5 + b
^2*c*d*x^3)*arctan(c*x)^2 - 3*(b^2*c^3*d*x^5 + b^2*c*d*x^3)*log(c^2*x^2 + 1)^2 - (3*b^2*c^3*d*x^5 - 4*b^2*c*d*
x^3 - 12*(b^2*c^2*d*x^4 + b^2*d*x^2)*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^2 + 1), x) + integrate(1/48*(36*(b^
2*c^2*d*x^4 + b^2*d*x^2)*arctan(c*x)^2 + 3*(b^2*c^2*d*x^4 + b^2*d*x^2)*log(c^2*x^2 + 1)^2 + 2*(3*b^2*c^3*d*x^5
- 4*b^2*c*d*x^3)*arctan(c*x) + (7*b^2*c^2*d*x^4 + 12*(b^2*c^3*d*x^5 + b^2*c*d*x^3)*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(x^2*(d+I*c*d*x)*(a+b*arctan(c*x))^2,x, algorithm="fricas")

[Out]

1/48*(-3*I*b^2*c*d*x^4 - 4*b^2*d*x^3)*log(-(c*x + I)/(c*x - I))^2 + integral(1/12*(12*I*a^2*c^3*d*x^5 + 12*a^2
*c^2*d*x^4 + 12*I*a^2*c*d*x^3 + 12*a^2*d*x^2 - (12*a*b*c^3*d*x^5 + 3*(-4*I*a*b - b^2)*c^2*d*x^4 + 4*(3*a*b + I
*b^2)*c*d*x^3 - 12*I*a*b*d*x^2)*log(-(c*x + I)/(c*x - I)))/(c^2*x^2 + 1), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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^2*(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 = -0.00 \begin {gather*} \int x^2\,{\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(x^2*(a + b*atan(c*x))^2*(d + c*d*x*1i),x)

[Out]

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

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