3.23 \(\int (d+i c d x)^3 (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=100 \[ -\frac{i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 c}-\frac{b d^3 (1+i c x)^3}{12 c}-\frac{b d^3 (1+i c x)^2}{4 c}-\frac{2 b d^3 \log (1-i c x)}{c}-i b d^3 x \]

[Out]

(-I)*b*d^3*x - (b*d^3*(1 + I*c*x)^2)/(4*c) - (b*d^3*(1 + I*c*x)^3)/(12*c) - ((I/4)*d^3*(1 + I*c*x)^4*(a + b*Ar
cTan[c*x]))/c - (2*b*d^3*Log[1 - I*c*x])/c

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Rubi [A]  time = 0.0537698, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4862, 627, 43} \[ -\frac{i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 c}-\frac{b d^3 (1+i c x)^3}{12 c}-\frac{b d^3 (1+i c x)^2}{4 c}-\frac{2 b d^3 \log (1-i c x)}{c}-i b d^3 x \]

Antiderivative was successfully verified.

[In]

Int[(d + I*c*d*x)^3*(a + b*ArcTan[c*x]),x]

[Out]

(-I)*b*d^3*x - (b*d^3*(1 + I*c*x)^2)/(4*c) - (b*d^3*(1 + I*c*x)^3)/(12*c) - ((I/4)*d^3*(1 + I*c*x)^4*(a + b*Ar
cTan[c*x]))/c - (2*b*d^3*Log[1 - I*c*x])/c

Rule 4862

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 43

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])

Rubi steps

\begin{align*} \int (d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=-\frac{i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 c}+\frac{(i b) \int \frac{(d+i c d x)^4}{1+c^2 x^2} \, dx}{4 d}\\ &=-\frac{i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 c}+\frac{(i b) \int \frac{(d+i c d x)^3}{\frac{1}{d}-\frac{i c x}{d}} \, dx}{4 d}\\ &=-\frac{i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 c}+\frac{(i b) \int \left (-4 d^4+\frac{8 d^3}{\frac{1}{d}-\frac{i c x}{d}}-2 d^3 (d+i c d x)-d^2 (d+i c d x)^2\right ) \, dx}{4 d}\\ &=-i b d^3 x-\frac{b d^3 (1+i c x)^2}{4 c}-\frac{b d^3 (1+i c x)^3}{12 c}-\frac{i d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 c}-\frac{2 b d^3 \log (1-i c x)}{c}\\ \end{align*}

Mathematica [A]  time = 0.0387604, size = 77, normalized size = 0.77 \[ -\frac{i \left (3 (d+i c d x)^4 \left (a+b \tan ^{-1}(c x)\right )-b d^4 \left (c^3 x^3-6 i c^2 x^2-21 c x+24 i \log (c x+i)+4 i\right )\right )}{12 c d} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + I*c*d*x)^3*(a + b*ArcTan[c*x]),x]

[Out]

((-I/12)*(3*(d + I*c*d*x)^4*(a + b*ArcTan[c*x]) - b*d^4*(4*I - 21*c*x - (6*I)*c^2*x^2 + c^3*x^3 + (24*I)*Log[I
+ c*x])))/(c*d)

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Maple [A]  time = 0.026, size = 176, normalized size = 1.8 \begin{align*} -{\frac{i}{4}}{c}^{3}{x}^{4}a{d}^{3}-{c}^{2}{x}^{3}a{d}^{3}+{\frac{3\,i}{2}}c{x}^{2}a{d}^{3}+xa{d}^{3}-{\frac{{\frac{i}{4}}{d}^{3}a}{c}}-{\frac{i}{4}}{c}^{3}{d}^{3}b\arctan \left ( cx \right ){x}^{4}-{c}^{2}{d}^{3}b\arctan \left ( cx \right ){x}^{3}+{\frac{3\,i}{2}}c{d}^{3}b\arctan \left ( cx \right ){x}^{2}+{d}^{3}bx\arctan \left ( cx \right ) +{\frac{{\frac{7\,i}{4}}{d}^{3}b\arctan \left ( cx \right ) }{c}}-{\frac{7\,i}{4}}{d}^{3}bx+{\frac{i}{12}}{c}^{2}{d}^{3}b{x}^{3}+{\frac{c{d}^{3}b{x}^{2}}{2}}-{\frac{{d}^{3}b\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)^3*(a+b*arctan(c*x)),x)

[Out]

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

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Maxima [B]  time = 1.47576, size = 266, normalized size = 2.66 \begin{align*} -\frac{1}{4} i \, a c^{3} d^{3} x^{4} - a c^{2} d^{3} x^{3} - \frac{1}{12} i \,{\left (3 \, x^{4} \arctan \left (c x\right ) - c{\left (\frac{c^{2} x^{3} - 3 \, x}{c^{4}} + \frac{3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c^{3} d^{3} - \frac{1}{2} \,{\left (2 \, x^{3} \arctan \left (c x\right ) - c{\left (\frac{x^{2}}{c^{2}} - \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b c^{2} d^{3} + \frac{3}{2} i \, a c d^{3} x^{2} + \frac{3}{2} i \,{\left (x^{2} \arctan \left (c x\right ) - c{\left (\frac{x}{c^{2}} - \frac{\arctan \left (c x\right )}{c^{3}}\right )}\right )} b c d^{3} + a d^{3} x + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{3}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^3*(a+b*arctan(c*x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.94267, size = 367, normalized size = 3.67 \begin{align*} \frac{-6 i \, a c^{4} d^{3} x^{4} - 2 \,{\left (12 \, a - i \, b\right )} c^{3} d^{3} x^{3} +{\left (36 i \, a + 12 \, b\right )} c^{2} d^{3} x^{2} + 6 \,{\left (4 \, a - 7 i \, b\right )} c d^{3} x - 45 \, b d^{3} \log \left (\frac{c x + i}{c}\right ) - 3 \, b d^{3} \log \left (\frac{c x - i}{c}\right ) +{\left (3 \, b c^{4} d^{3} x^{4} - 12 i \, b c^{3} d^{3} x^{3} - 18 \, b c^{2} d^{3} x^{2} + 12 i \, b c d^{3} x\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{24 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^3*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

1/24*(-6*I*a*c^4*d^3*x^4 - 2*(12*a - I*b)*c^3*d^3*x^3 + (36*I*a + 12*b)*c^2*d^3*x^2 + 6*(4*a - 7*I*b)*c*d^3*x
- 45*b*d^3*log((c*x + I)/c) - 3*b*d^3*log((c*x - I)/c) + (3*b*c^4*d^3*x^4 - 12*I*b*c^3*d^3*x^3 - 18*b*c^2*d^3*
x^2 + 12*I*b*c*d^3*x)*log(-(c*x + I)/(c*x - I)))/c

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Sympy [B]  time = 3.24745, size = 228, normalized size = 2.28 \begin{align*} - \frac{i a c^{3} d^{3} x^{4}}{4} - \frac{b d^{3} \left (\frac{\log{\left (x - \frac{i}{c} \right )}}{8} + \frac{15 \log{\left (x + \frac{i}{c} \right )}}{8}\right )}{c} - x^{3} \left (a c^{2} d^{3} - \frac{i b c^{2} d^{3}}{12}\right ) - x^{2} \left (- \frac{3 i a c d^{3}}{2} - \frac{b c d^{3}}{2}\right ) - x \left (- a d^{3} + \frac{7 i b d^{3}}{4}\right ) + \left (- \frac{b c^{3} d^{3} x^{4}}{8} + \frac{i b c^{2} d^{3} x^{3}}{2} + \frac{3 b c d^{3} x^{2}}{4} - \frac{i b d^{3} x}{2}\right ) \log{\left (i c x + 1 \right )} + \left (\frac{b c^{3} d^{3} x^{4}}{8} - \frac{i b c^{2} d^{3} x^{3}}{2} - \frac{3 b c d^{3} x^{2}}{4} + \frac{i b d^{3} x}{2}\right ) \log{\left (- i c x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)**3*(a+b*atan(c*x)),x)

[Out]

-I*a*c**3*d**3*x**4/4 - b*d**3*(log(x - I/c)/8 + 15*log(x + I/c)/8)/c - x**3*(a*c**2*d**3 - I*b*c**2*d**3/12)
- x**2*(-3*I*a*c*d**3/2 - b*c*d**3/2) - x*(-a*d**3 + 7*I*b*d**3/4) + (-b*c**3*d**3*x**4/8 + I*b*c**2*d**3*x**3
/2 + 3*b*c*d**3*x**2/4 - I*b*d**3*x/2)*log(I*c*x + 1) + (b*c**3*d**3*x**4/8 - I*b*c**2*d**3*x**3/2 - 3*b*c*d**
3*x**2/4 + I*b*d**3*x/2)*log(-I*c*x + 1)

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Giac [B]  time = 1.15569, size = 235, normalized size = 2.35 \begin{align*} -\frac{6 \, b c^{4} d^{3} i x^{4} \arctan \left (c x\right ) + 6 \, a c^{4} d^{3} i x^{4} - 2 \, b c^{3} d^{3} i x^{3} + 24 \, b c^{3} d^{3} x^{3} \arctan \left (c x\right ) + 24 \, a c^{3} d^{3} x^{3} - 36 \, b c^{2} d^{3} i x^{2} \arctan \left (c x\right ) - 36 \, a c^{2} d^{3} i x^{2} - 12 \, b c^{2} d^{3} x^{2} + 42 \, b c d^{3} i x - 24 \, b c d^{3} x \arctan \left (c x\right ) - 24 \, a c d^{3} x + 45 \, b d^{3} \log \left (c x + i\right ) + 3 \, b d^{3} \log \left (c x - i\right )}{24 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^3*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

-1/24*(6*b*c^4*d^3*i*x^4*arctan(c*x) + 6*a*c^4*d^3*i*x^4 - 2*b*c^3*d^3*i*x^3 + 24*b*c^3*d^3*x^3*arctan(c*x) +
24*a*c^3*d^3*x^3 - 36*b*c^2*d^3*i*x^2*arctan(c*x) - 36*a*c^2*d^3*i*x^2 - 12*b*c^2*d^3*x^2 + 42*b*c*d^3*i*x - 2
4*b*c*d^3*x*arctan(c*x) - 24*a*c*d^3*x + 45*b*d^3*log(c*x + i) + 3*b*d^3*log(c*x - i))/c