3.34 \(\int (d+i c d x)^4 (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=125 \[ -\frac{i d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c}-\frac{b d^4 (1+i c x)^4}{20 c}-\frac{2 b d^4 (1+i c x)^3}{15 c}-\frac{2 b d^4 (1+i c x)^2}{5 c}-\frac{16 b d^4 \log (1-i c x)}{5 c}-\frac{8}{5} i b d^4 x \]

[Out]

((-8*I)/5)*b*d^4*x - (2*b*d^4*(1 + I*c*x)^2)/(5*c) - (2*b*d^4*(1 + I*c*x)^3)/(15*c) - (b*d^4*(1 + I*c*x)^4)/(2
0*c) - ((I/5)*d^4*(1 + I*c*x)^5*(a + b*ArcTan[c*x]))/c - (16*b*d^4*Log[1 - I*c*x])/(5*c)

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Rubi [A]  time = 0.063127, antiderivative size = 125, 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} \[ -\frac{i d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c}-\frac{b d^4 (1+i c x)^4}{20 c}-\frac{2 b d^4 (1+i c x)^3}{15 c}-\frac{2 b d^4 (1+i c x)^2}{5 c}-\frac{16 b d^4 \log (1-i c x)}{5 c}-\frac{8}{5} i b d^4 x \]

Antiderivative was successfully verified.

[In]

Int[(d + I*c*d*x)^4*(a + b*ArcTan[c*x]),x]

[Out]

((-8*I)/5)*b*d^4*x - (2*b*d^4*(1 + I*c*x)^2)/(5*c) - (2*b*d^4*(1 + I*c*x)^3)/(15*c) - (b*d^4*(1 + I*c*x)^4)/(2
0*c) - ((I/5)*d^4*(1 + I*c*x)^5*(a + b*ArcTan[c*x]))/c - (16*b*d^4*Log[1 - I*c*x])/(5*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)^4 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=-\frac{i d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c}+\frac{(i b) \int \frac{(d+i c d x)^5}{1+c^2 x^2} \, dx}{5 d}\\ &=-\frac{i d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c}+\frac{(i b) \int \frac{(d+i c d x)^4}{\frac{1}{d}-\frac{i c x}{d}} \, dx}{5 d}\\ &=-\frac{i d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c}+\frac{(i b) \int \left (-8 d^5+\frac{16 d^4}{\frac{1}{d}-\frac{i c x}{d}}-4 d^4 (d+i c d x)-2 d^3 (d+i c d x)^2-d^2 (d+i c d x)^3\right ) \, dx}{5 d}\\ &=-\frac{8}{5} i b d^4 x-\frac{2 b d^4 (1+i c x)^2}{5 c}-\frac{2 b d^4 (1+i c x)^3}{15 c}-\frac{b d^4 (1+i c x)^4}{20 c}-\frac{i d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c}-\frac{16 b d^4 \log (1-i c x)}{5 c}\\ \end{align*}

Mathematica [A]  time = 0.0298645, size = 77, normalized size = 0.62 \[ \frac{d^4 \left (12 (c x-i)^5 \left (a+b \tan ^{-1}(c x)\right )-b \left (3 c^4 x^4-20 i c^3 x^3-66 c^2 x^2+180 i c x+192 \log (c x+i)+35\right )\right )}{60 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + I*c*d*x)^4*(a + b*ArcTan[c*x]),x]

[Out]

(d^4*(12*(-I + c*x)^5*(a + b*ArcTan[c*x]) - b*(35 + (180*I)*c*x - 66*c^2*x^2 - (20*I)*c^3*x^3 + 3*c^4*x^4 + 19
2*Log[I + c*x])))/(60*c)

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Maple [B]  time = 0.029, size = 216, normalized size = 1.7 \begin{align*}{\frac{{c}^{4}{x}^{5}a{d}^{4}}{5}}+2\,ic{d}^{4}b\arctan \left ( cx \right ){x}^{2}-2\,{c}^{2}{x}^{3}a{d}^{4}-{\frac{{\frac{i}{5}}{d}^{4}a}{c}}+xa{d}^{4}-i{c}^{3}{d}^{4}b\arctan \left ( cx \right ){x}^{4}+{\frac{{c}^{4}{d}^{4}b\arctan \left ( cx \right ){x}^{5}}{5}}+2\,ic{x}^{2}a{d}^{4}-2\,{c}^{2}{d}^{4}b\arctan \left ( cx \right ){x}^{3}-3\,i{d}^{4}bx+{d}^{4}bx\arctan \left ( cx \right ) -i{c}^{3}{x}^{4}a{d}^{4}+{\frac{i}{3}}{c}^{2}{d}^{4}b{x}^{3}-{\frac{{c}^{3}{d}^{4}b{x}^{4}}{20}}+{\frac{3\,i{d}^{4}b\arctan \left ( cx \right ) }{c}}+{\frac{11\,c{d}^{4}b{x}^{2}}{10}}-{\frac{8\,{d}^{4}b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{5\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^4*(a+b*arctan(c*x)),x)

[Out]

1/5*c^4*x^5*a*d^4+2*I*c*d^4*b*arctan(c*x)*x^2-2*c^2*x^3*a*d^4-1/5*I/c*d^4*a+x*a*d^4-I*c^3*d^4*b*arctan(c*x)*x^
4+1/5*c^4*d^4*b*arctan(c*x)*x^5+2*I*c*x^2*a*d^4-2*c^2*d^4*b*arctan(c*x)*x^3-3*I*d^4*b*x+d^4*b*x*arctan(c*x)-I*
c^3*x^4*a*d^4+1/3*I*c^2*d^4*b*x^3-1/20*c^3*d^4*b*x^4+3*I/c*d^4*b*arctan(c*x)+11/10*c*d^4*b*x^2-8/5/c*d^4*b*ln(
c^2*x^2+1)

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Maxima [B]  time = 1.49166, size = 356, normalized size = 2.85 \begin{align*} \frac{1}{5} \, a c^{4} d^{4} x^{5} - i \, a c^{3} d^{4} x^{4} + \frac{1}{20} \,{\left (4 \, x^{5} \arctan \left (c x\right ) - c{\left (\frac{c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c^{4} d^{4} - 2 \, a c^{2} d^{4} x^{3} - \frac{1}{3} 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^{4} -{\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^{4} + 2 i \, a c d^{4} x^{2} + 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^{4} + a d^{4} x + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{4}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x)),x, algorithm="maxima")

[Out]

1/5*a*c^4*d^4*x^5 - I*a*c^3*d^4*x^4 + 1/20*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/
c^6))*b*c^4*d^4 - 2*a*c^2*d^4*x^3 - 1/3*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^4 - (2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*c^2*d^4 + 2*I*a*c*d^4*x^2 + 2*I*(x^2*arct
an(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*b*c*d^4 + a*d^4*x + 1/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*d^4/c

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Fricas [A]  time = 2.53474, size = 435, normalized size = 3.48 \begin{align*} \frac{12 \, a c^{5} d^{4} x^{5} +{\left (-60 i \, a - 3 \, b\right )} c^{4} d^{4} x^{4} - 20 \,{\left (6 \, a - i \, b\right )} c^{3} d^{4} x^{3} +{\left (120 i \, a + 66 \, b\right )} c^{2} d^{4} x^{2} + 60 \,{\left (a - 3 i \, b\right )} c d^{4} x - 186 \, b d^{4} \log \left (\frac{c x + i}{c}\right ) - 6 \, b d^{4} \log \left (\frac{c x - i}{c}\right ) +{\left (6 i \, b c^{5} d^{4} x^{5} + 30 \, b c^{4} d^{4} x^{4} - 60 i \, b c^{3} d^{4} x^{3} - 60 \, b c^{2} d^{4} x^{2} + 30 i \, b c d^{4} x\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{60 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

1/60*(12*a*c^5*d^4*x^5 + (-60*I*a - 3*b)*c^4*d^4*x^4 - 20*(6*a - I*b)*c^3*d^4*x^3 + (120*I*a + 66*b)*c^2*d^4*x
^2 + 60*(a - 3*I*b)*c*d^4*x - 186*b*d^4*log((c*x + I)/c) - 6*b*d^4*log((c*x - I)/c) + (6*I*b*c^5*d^4*x^5 + 30*
b*c^4*d^4*x^4 - 60*I*b*c^3*d^4*x^3 - 60*b*c^2*d^4*x^2 + 30*I*b*c*d^4*x)*log(-(c*x + I)/(c*x - I)))/c

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Sympy [B]  time = 3.96163, size = 272, normalized size = 2.18 \begin{align*} \frac{a c^{4} d^{4} x^{5}}{5} + \frac{b d^{4} \left (- \frac{\log{\left (x - \frac{i}{c} \right )}}{10} - \frac{31 \log{\left (x + \frac{i}{c} \right )}}{10}\right )}{c} + x^{4} \left (- i a c^{3} d^{4} - \frac{b c^{3} d^{4}}{20}\right ) + x^{3} \left (- 2 a c^{2} d^{4} + \frac{i b c^{2} d^{4}}{3}\right ) + x^{2} \left (2 i a c d^{4} + \frac{11 b c d^{4}}{10}\right ) + x \left (a d^{4} - 3 i b d^{4}\right ) + \left (- \frac{i b c^{4} d^{4} x^{5}}{10} - \frac{b c^{3} d^{4} x^{4}}{2} + i b c^{2} d^{4} x^{3} + b c d^{4} x^{2} - \frac{i b d^{4} x}{2}\right ) \log{\left (i c x + 1 \right )} + \left (\frac{i b c^{4} d^{4} x^{5}}{10} + \frac{b c^{3} d^{4} x^{4}}{2} - i b c^{2} d^{4} x^{3} - b c d^{4} x^{2} + \frac{i b d^{4} 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)**4*(a+b*atan(c*x)),x)

[Out]

a*c**4*d**4*x**5/5 + b*d**4*(-log(x - I/c)/10 - 31*log(x + I/c)/10)/c + x**4*(-I*a*c**3*d**4 - b*c**3*d**4/20)
+ x**3*(-2*a*c**2*d**4 + I*b*c**2*d**4/3) + x**2*(2*I*a*c*d**4 + 11*b*c*d**4/10) + x*(a*d**4 - 3*I*b*d**4) +
(-I*b*c**4*d**4*x**5/10 - b*c**3*d**4*x**4/2 + I*b*c**2*d**4*x**3 + b*c*d**4*x**2 - I*b*d**4*x/2)*log(I*c*x +
1) + (I*b*c**4*d**4*x**5/10 + b*c**3*d**4*x**4/2 - I*b*c**2*d**4*x**3 - b*c*d**4*x**2 + I*b*d**4*x/2)*log(-I*c
*x + 1)

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Giac [B]  time = 1.20288, size = 289, normalized size = 2.31 \begin{align*} \frac{12 \, b c^{5} d^{4} x^{5} \arctan \left (c x\right ) + 12 \, a c^{5} d^{4} x^{5} - 60 \, b c^{4} d^{4} i x^{4} \arctan \left (c x\right ) - 60 \, a c^{4} d^{4} i x^{4} - 3 \, b c^{4} d^{4} x^{4} + 20 \, b c^{3} d^{4} i x^{3} - 120 \, b c^{3} d^{4} x^{3} \arctan \left (c x\right ) - 120 \, a c^{3} d^{4} x^{3} + 120 \, b c^{2} d^{4} i x^{2} \arctan \left (c x\right ) + 120 \, a c^{2} d^{4} i x^{2} + 66 \, b c^{2} d^{4} x^{2} - 180 \, b c d^{4} i x + 60 \, b c d^{4} x \arctan \left (c x\right ) + 60 \, a c d^{4} x - 186 \, b d^{4} \log \left (c x + i\right ) - 6 \, b d^{4} \log \left (c x - i\right )}{60 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

1/60*(12*b*c^5*d^4*x^5*arctan(c*x) + 12*a*c^5*d^4*x^5 - 60*b*c^4*d^4*i*x^4*arctan(c*x) - 60*a*c^4*d^4*i*x^4 -
3*b*c^4*d^4*x^4 + 20*b*c^3*d^4*i*x^3 - 120*b*c^3*d^4*x^3*arctan(c*x) - 120*a*c^3*d^4*x^3 + 120*b*c^2*d^4*i*x^2
*arctan(c*x) + 120*a*c^2*d^4*i*x^2 + 66*b*c^2*d^4*x^2 - 180*b*c*d^4*i*x + 60*b*c*d^4*x*arctan(c*x) + 60*a*c*d^
4*x - 186*b*d^4*log(c*x + i) - 6*b*d^4*log(c*x - i))/c