3.14 \(\int x^4 (a+b \tan ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=170 \[ \frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac {2 b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^5}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {b x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac {i b^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{5 c^5}+\frac {3 b^2 \tan ^{-1}(c x)}{10 c^5}-\frac {3 b^2 x}{10 c^4}+\frac {b^2 x^3}{30 c^2} \]

[Out]

-3/10*b^2*x/c^4+1/30*b^2*x^3/c^2+3/10*b^2*arctan(c*x)/c^5+1/5*b*x^2*(a+b*arctan(c*x))/c^3-1/10*b*x^4*(a+b*arct
an(c*x))/c+1/5*I*(a+b*arctan(c*x))^2/c^5+1/5*x^5*(a+b*arctan(c*x))^2+2/5*b*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c
^5+1/5*I*b^2*polylog(2,1-2/(1+I*c*x))/c^5

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Rubi [A]  time = 0.29, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {4852, 4916, 302, 203, 321, 4920, 4854, 2402, 2315} \[ \frac {i b^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{5 c^5}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac {2 b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {b x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac {b^2 x^3}{30 c^2}-\frac {3 b^2 x}{10 c^4}+\frac {3 b^2 \tan ^{-1}(c x)}{10 c^5} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTan[c*x])^2,x]

[Out]

(-3*b^2*x)/(10*c^4) + (b^2*x^3)/(30*c^2) + (3*b^2*ArcTan[c*x])/(10*c^5) + (b*x^2*(a + b*ArcTan[c*x]))/(5*c^3)
- (b*x^4*(a + b*ArcTan[c*x]))/(10*c) + ((I/5)*(a + b*ArcTan[c*x])^2)/c^5 + (x^5*(a + b*ArcTan[c*x])^2)/5 + (2*
b*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(5*c^5) + ((I/5)*b^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/c^5

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 321

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 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]

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 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

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 4916

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 4920

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[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^4 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{5} (2 b c) \int \frac {x^5 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {(2 b) \int x^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{5 c}+\frac {(2 b) \int \frac {x^3 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{5 c}\\ &=-\frac {b x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{10} b^2 \int \frac {x^4}{1+c^2 x^2} \, dx+\frac {(2 b) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{5 c^3}-\frac {(2 b) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{5 c^3}\\ &=\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}-\frac {b x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{10} b^2 \int \left (-\frac {1}{c^4}+\frac {x^2}{c^2}+\frac {1}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx+\frac {(2 b) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx}{5 c^4}-\frac {b^2 \int \frac {x^2}{1+c^2 x^2} \, dx}{5 c^2}\\ &=-\frac {3 b^2 x}{10 c^4}+\frac {b^2 x^3}{30 c^2}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}-\frac {b x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{5 c^5}+\frac {b^2 \int \frac {1}{1+c^2 x^2} \, dx}{10 c^4}+\frac {b^2 \int \frac {1}{1+c^2 x^2} \, dx}{5 c^4}-\frac {\left (2 b^2\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{5 c^4}\\ &=-\frac {3 b^2 x}{10 c^4}+\frac {b^2 x^3}{30 c^2}+\frac {3 b^2 \tan ^{-1}(c x)}{10 c^5}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}-\frac {b x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{5 c^5}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{5 c^5}\\ &=-\frac {3 b^2 x}{10 c^4}+\frac {b^2 x^3}{30 c^2}+\frac {3 b^2 \tan ^{-1}(c x)}{10 c^5}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}-\frac {b x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{5 c^5}+\frac {i b^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{5 c^5}\\ \end {align*}

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Mathematica [A]  time = 0.51, size = 169, normalized size = 0.99 \[ \frac {6 a^2 c^5 x^5-3 a b c^4 x^4+6 a b c^2 x^2-6 a b \log \left (c^2 x^2+1\right )-3 b \tan ^{-1}(c x) \left (-4 a c^5 x^5+b \left (c^4 x^4-2 c^2 x^2-3\right )-4 b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )+9 a b+6 b^2 \left (c^5 x^5-i\right ) \tan ^{-1}(c x)^2+b^2 c^3 x^3-6 i b^2 \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )-9 b^2 c x}{30 c^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4*(a + b*ArcTan[c*x])^2,x]

[Out]

(9*a*b - 9*b^2*c*x + 6*a*b*c^2*x^2 + b^2*c^3*x^3 - 3*a*b*c^4*x^4 + 6*a^2*c^5*x^5 + 6*b^2*(-I + c^5*x^5)*ArcTan
[c*x]^2 - 3*b*ArcTan[c*x]*(-4*a*c^5*x^5 + b*(-3 - 2*c^2*x^2 + c^4*x^4) - 4*b*Log[1 + E^((2*I)*ArcTan[c*x])]) -
6*a*b*Log[1 + c^2*x^2] - (6*I)*b^2*PolyLog[2, -E^((2*I)*ArcTan[c*x])])/(30*c^5)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x^{4} \arctan \left (c x\right )^{2} + 2 \, a b x^{4} \arctan \left (c x\right ) + a^{2} x^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^4*arctan(c*x)^2 + 2*a*b*x^4*arctan(c*x) + a^2*x^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))^2,x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.10, size = 334, normalized size = 1.96 \[ \frac {x^{5} a^{2}}{5}+\frac {x^{5} b^{2} \arctan \left (c x \right )^{2}}{5}-\frac {b^{2} \arctan \left (c x \right ) x^{4}}{10 c}+\frac {b^{2} \arctan \left (c x \right ) x^{2}}{5 c^{3}}-\frac {b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{5 c^{5}}+\frac {b^{2} x^{3}}{30 c^{2}}-\frac {3 b^{2} x}{10 c^{4}}+\frac {3 b^{2} \arctan \left (c x \right )}{10 c^{5}}+\frac {i b^{2} \ln \left (c x -i\right )^{2}}{20 c^{5}}-\frac {i b^{2} \ln \left (c x +i\right )^{2}}{20 c^{5}}+\frac {i b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{10 c^{5}}-\frac {i b^{2} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{10 c^{5}}+\frac {i b^{2} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{10 c^{5}}-\frac {i b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{10 c^{5}}+\frac {i b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{10 c^{5}}-\frac {i b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{10 c^{5}}+\frac {2 x^{5} a b \arctan \left (c x \right )}{5}-\frac {x^{4} a b}{10 c}+\frac {a b \,x^{2}}{5 c^{3}}-\frac {a b \ln \left (c^{2} x^{2}+1\right )}{5 c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctan(c*x))^2,x)

[Out]

1/5*x^5*a^2+1/5*x^5*b^2*arctan(c*x)^2-1/10/c*b^2*arctan(c*x)*x^4+1/5/c^3*b^2*arctan(c*x)*x^2-1/5/c^5*b^2*arcta
n(c*x)*ln(c^2*x^2+1)+1/30*b^2*x^3/c^2-3/10*b^2*x/c^4+3/10*b^2*arctan(c*x)/c^5+1/20*I/c^5*b^2*ln(c*x-I)^2-1/10*
I/c^5*b^2*dilog(1/2*I*(c*x-I))-1/20*I/c^5*b^2*ln(I+c*x)^2+1/10*I/c^5*b^2*dilog(-1/2*I*(I+c*x))+1/10*I/c^5*b^2*
ln(c*x-I)*ln(-1/2*I*(I+c*x))-1/10*I/c^5*b^2*ln(c*x-I)*ln(c^2*x^2+1)+1/10*I/c^5*b^2*ln(I+c*x)*ln(c^2*x^2+1)-1/1
0*I/c^5*b^2*ln(I+c*x)*ln(1/2*I*(c*x-I))+2/5*x^5*a*b*arctan(c*x)-1/10/c*x^4*a*b+1/5*a*b*x^2/c^3-1/5/c^5*a*b*ln(
c^2*x^2+1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{5} \, a^{2} x^{5} + \frac {1}{10} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} a b + \frac {1}{80} \, {\left (4 \, x^{5} \arctan \left (c x\right )^{2} - x^{5} \log \left (c^{2} x^{2} + 1\right )^{2} + 80 \, \int \frac {4 \, c^{2} x^{6} \log \left (c^{2} x^{2} + 1\right ) - 8 \, c x^{5} \arctan \left (c x\right ) + 60 \, {\left (c^{2} x^{6} + x^{4}\right )} \arctan \left (c x\right )^{2} + 5 \, {\left (c^{2} x^{6} + x^{4}\right )} \log \left (c^{2} x^{2} + 1\right )^{2}}{80 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x}\right )} b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))^2,x, algorithm="maxima")

[Out]

1/5*a^2*x^5 + 1/10*(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))*a*b + 1/80*(4*x^5*
arctan(c*x)^2 - x^5*log(c^2*x^2 + 1)^2 + 80*integrate(1/80*(4*c^2*x^6*log(c^2*x^2 + 1) - 8*c*x^5*arctan(c*x) +
60*(c^2*x^6 + x^4)*arctan(c*x)^2 + 5*(c^2*x^6 + x^4)*log(c^2*x^2 + 1)^2)/(c^2*x^2 + 1), x))*b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*atan(c*x))^2,x)

[Out]

int(x^4*(a + b*atan(c*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*atan(c*x))**2,x)

[Out]

Integral(x**4*(a + b*atan(c*x))**2, x)

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