3.1.26 \(\int x^3 (a+b \text {ArcTan}(c x))^3 \, dx\) [26]

Optimal. Leaf size=194 \[ -\frac {b^3 x}{4 c^3}+\frac {b^3 \text {ArcTan}(c x)}{4 c^4}+\frac {b^2 x^2 (a+b \text {ArcTan}(c x))}{4 c^2}+\frac {i b (a+b \text {ArcTan}(c x))^2}{c^4}+\frac {3 b x (a+b \text {ArcTan}(c x))^2}{4 c^3}-\frac {b x^3 (a+b \text {ArcTan}(c x))^2}{4 c}-\frac {(a+b \text {ArcTan}(c x))^3}{4 c^4}+\frac {1}{4} x^4 (a+b \text {ArcTan}(c x))^3+\frac {2 b^2 (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1+i c x}\right )}{c^4}+\frac {i b^3 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4} \]

[Out]

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

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Rubi [A]
time = 0.38, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {4946, 5036, 327, 209, 5040, 4964, 2449, 2352, 4930, 5004} \begin {gather*} \frac {2 b^2 \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))}{c^4}+\frac {b^2 x^2 (a+b \text {ArcTan}(c x))}{4 c^2}-\frac {(a+b \text {ArcTan}(c x))^3}{4 c^4}+\frac {i b (a+b \text {ArcTan}(c x))^2}{c^4}+\frac {3 b x (a+b \text {ArcTan}(c x))^2}{4 c^3}+\frac {1}{4} x^4 (a+b \text {ArcTan}(c x))^3-\frac {b x^3 (a+b \text {ArcTan}(c x))^2}{4 c}+\frac {b^3 \text {ArcTan}(c x)}{4 c^4}+\frac {i b^3 \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{c^4}-\frac {b^3 x}{4 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTan[c*x])^3,x]

[Out]

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

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

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Mathematica [A]
time = 0.41, size = 225, normalized size = 1.16 \begin {gather*} \frac {a b^2+3 a^2 b c x-b^3 c x+a b^2 c^2 x^2-a^2 b c^3 x^3+a^3 c^4 x^4-b^2 \left (b \left (4 i-3 c x+c^3 x^3\right )+a \left (3-3 c^4 x^4\right )\right ) \text {ArcTan}(c x)^2+b^3 \left (-1+c^4 x^4\right ) \text {ArcTan}(c x)^3+b \text {ArcTan}(c x) \left (-2 a b c x \left (-3+c^2 x^2\right )+b^2 \left (1+c^2 x^2\right )+3 a^2 \left (-1+c^4 x^4\right )+8 b^2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )-4 a b^2 \log \left (1+c^2 x^2\right )-4 i b^3 \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )}{4 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b^2 + 3*a^2*b*c*x - b^3*c*x + a*b^2*c^2*x^2 - a^2*b*c^3*x^3 + a^3*c^4*x^4 - b^2*(b*(4*I - 3*c*x + c^3*x^3)
+ a*(3 - 3*c^4*x^4))*ArcTan[c*x]^2 + b^3*(-1 + c^4*x^4)*ArcTan[c*x]^3 + b*ArcTan[c*x]*(-2*a*b*c*x*(-3 + c^2*x^
2) + b^2*(1 + c^2*x^2) + 3*a^2*(-1 + c^4*x^4) + 8*b^2*Log[1 + E^((2*I)*ArcTan[c*x])]) - 4*a*b^2*Log[1 + c^2*x^
2] - (4*I)*b^3*PolyLog[2, -E^((2*I)*ArcTan[c*x])])/(4*c^4)

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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. 410 vs. \(2 (176 ) = 352\).
time = 0.54, size = 411, normalized size = 2.12 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctan(c*x))^3,x,method=_RETURNVERBOSE)

[Out]

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

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

[Out]

3/4*a*b^2*x^4*arctan(c*x)^2 + 1/4*a^3*x^4 + 1/4*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^
5))*a^2*b - 1/4*(2*c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5)*arctan(c*x) - (c^2*x^2 + 3*arctan(c*x)^2 - 4*lo
g(c^2*x^2 + 1))/c^4)*a*b^2 + 1/64*(4*(512*c^5*integrate(1/64*x^5*arctan(c*x)^3/(c^5*x^2 + c^3), x) - 192*c^4*i
ntegrate(1/64*x^4*arctan(c*x)^2/(c^5*x^2 + c^3), x) - 48*c^4*integrate(1/64*x^4*log(c^2*x^2 + 1)^2/(c^5*x^2 +
c^3), x) - 64*c^4*integrate(1/64*x^4*log(c^2*x^2 + 1)/(c^5*x^2 + c^3), x) + 512*c^3*integrate(1/64*x^3*arctan(
c*x)^3/(c^5*x^2 + c^3), x) + 128*c^3*integrate(1/64*x^3*arctan(c*x)/(c^5*x^2 + c^3), x) + 192*c^2*integrate(1/
64*x^2*log(c^2*x^2 + 1)/(c^5*x^2 + c^3), x) - 384*c*integrate(1/64*x*arctan(c*x)/(c^5*x^2 + c^3), x) + arctan(
c*x)^3/c^4 + 48*integrate(1/64*log(c^2*x^2 + 1)^2/(c^5*x^2 + c^3), x))*c^4 + 8*(c^4*x^4 - 1)*arctan(c*x)^3 - 4
*(c^3*x^3 - 3*c*x)*arctan(c*x)^2 + (c^3*x^3 - 3*c*x)*log(c^2*x^2 + 1)^2)*b^3/c^4

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

[Out]

integral(b^3*x^3*arctan(c*x)^3 + 3*a*b^2*x^3*arctan(c*x)^2 + 3*a^2*b*x^3*arctan(c*x) + a^3*x^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (a + b \operatorname {atan}{\left (c x \right )}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(a + b*atan(c*x))**3, x)

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atan(c*x))^3,x)

[Out]

int(x^3*(a + b*atan(c*x))^3, x)

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