3.71 \(\int (d+i c d x) (a+b \tan ^{-1}(c x))^2 \, dx\)

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

[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

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Rubi [A]  time = 0.121163, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {4864, 4846, 260, 1586, 4854, 2402, 2315} \[ -\frac{i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{c}-\frac{i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac{2 b d \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-i a b d x+\frac{i b^2 d \log \left (c^2 x^2+1\right )}{2 c}-i b^2 d x \tan ^{-1}(c x) \]

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 4864

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]

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 260

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 1586

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 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 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 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]

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 ) \operatorname{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*}

Mathematica [A]  time = 0.252557, size = 151, normalized size = 1.16 \[ \frac{i d \left (-2 b^2 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+a^2 c^2 x^2-2 i a^2 c x+2 i a b \log \left (c^2 x^2+1\right )+2 b \tan ^{-1}(c x) \left (a c^2 x^2-2 i a c x+a-b c x-2 i b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )-2 a b c x+b^2 \log \left (c^2 x^2+1\right )+b^2 (c x-i)^2 \tan ^{-1}(c x)^2\right )}{2 c} \]

Warning: Unable to verify antiderivative.

[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]  time = 0.087, size = 367, normalized size = 2.8 \begin{align*} d{a}^{2}x+{\frac{{\frac{i}{2}}d{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{c}}+d{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}x+{\frac{{\frac{i}{2}}d{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{c}}-iabdx+{\frac{{\frac{i}{4}}d{b}^{2} \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{c}}-{\frac{d{b}^{2}\arctan \left ( cx \right ) \ln \left ({c}^{2}{x}^{2}+1 \right ) }{c}}-{\frac{{\frac{i}{4}}d{b}^{2} \left ( \ln \left ( cx+i \right ) \right ) ^{2}}{c}}-{\frac{{\frac{i}{2}}d{b}^{2}\ln \left ( cx+i \right ) \ln \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{c}}-{\frac{{\frac{i}{2}}d{b}^{2}{\it dilog} \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{c}}+{\frac{i}{2}}cd{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}{x}^{2}+{\frac{i}{2}}cd{a}^{2}{x}^{2}+icdab\arctan \left ( cx \right ){x}^{2}-i{b}^{2}dx\arctan \left ( cx \right ) +{\frac{{\frac{i}{2}}d{b}^{2}{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{c}}+{\frac{{\frac{i}{2}}d{b}^{2}\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{c}}+{\frac{idab\arctan \left ( cx \right ) }{c}}+{\frac{{\frac{i}{2}}d{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx+i \right ) }{c}}+2\,dab\arctan \left ( cx \right ) x-{\frac{{\frac{i}{2}}d{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx-i \right ) }{c}}-{\frac{dab\ln \left ({c}^{2}{x}^{2}+1 \right ) }{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

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 + 1/32*(4*I*b^2*c*d*x^2 + 8*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*in
tegrate(-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., size = 0, normalized size = 0. \begin{align*} \frac{1}{8} \,{\left (-i \, b^{2} c d x^{2} - 2 \, b^{2} d x\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} +{\rm integral}\left (\frac{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 -{\left (2 \, a b c^{3} d x^{3} -{\left (2 i \, a b + b^{2}\right )} c^{2} d x^{2} + 2 \,{\left (a b + i \, b^{2}\right )} c d x - 2 i \, a b d\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{2 \,{\left (c^{2} x^{2} + 1\right )}}, x\right ) \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, c d x + d\right )}{\left (b \arctan \left (c x\right ) + a\right )}^{2}\,{d x} \end{align*}

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]

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