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

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

[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

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Rubi [A]  time = 0.288737, antiderivative size = 170, normalized size of antiderivative = 1., 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 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 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 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 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 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 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]

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 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*}

Mathematica [A]  time = 0.484662, size = 169, normalized size = 0.99 \[ \frac{-6 i b^2 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+6 a^2 c^5 x^5-3 a b c^4 x^4+6 a b c^2 x^2-6 a b \log \left (c^2 x^2+1\right )-3 b \tan ^{-1}(c x) \left (-4 a c^5 x^5+b \left (c^4 x^4-2 c^2 x^2-3\right )-4 b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )+9 a b+b^2 c^3 x^3+6 b^2 \left (c^5 x^5-i\right ) \tan ^{-1}(c x)^2-9 b^2 c x}{30 c^5} \]

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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Maple [B]  time = 0.044, size = 334, normalized size = 2. \begin{align*}{\frac{{x}^{5}{a}^{2}}{5}}+{\frac{{x}^{5}{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{5}}-{\frac{{b}^{2}\arctan \left ( cx \right ){x}^{4}}{10\,c}}+{\frac{{b}^{2}\arctan \left ( cx \right ){x}^{2}}{5\,{c}^{3}}}-{\frac{{b}^{2}\arctan \left ( cx \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 ( cx \right ) }{10\,{c}^{5}}}+{\frac{{\frac{i}{10}}{b}^{2}{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{5}}}-{\frac{{\frac{i}{10}}{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx-i \right ) }{{c}^{5}}}+{\frac{{\frac{i}{20}}{b}^{2} \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{{c}^{5}}}+{\frac{{\frac{i}{10}}{b}^{2}\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{5}}}-{\frac{{\frac{i}{10}}{b}^{2}\ln \left ( cx+i \right ) \ln \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{{c}^{5}}}-{\frac{{\frac{i}{10}}{b}^{2}{\it dilog} \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{{c}^{5}}}+{\frac{{\frac{i}{10}}{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx+i \right ) }{{c}^{5}}}-{\frac{{\frac{i}{20}}{b}^{2} \left ( \ln \left ( cx+i \right ) \right ) ^{2}}{{c}^{5}}}+{\frac{2\,{x}^{5}ab\arctan \left ( cx \right ) }{5}}-{\frac{{x}^{4}ab}{10\,c}}+{\frac{ab{x}^{2}}{5\,{c}^{3}}}-{\frac{ab\ln \left ({c}^{2}{x}^{2}+1 \right ) }{5\,{c}^{5}}} \end{align*}

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+1/10*I/c^5*b^2*dilog(-1/2*I*(c*x
+I))-1/10*I/c^5*b^2*ln(c^2*x^2+1)*ln(c*x-I)+1/20*I/c^5*b^2*ln(c*x-I)^2+1/10*I/c^5*b^2*ln(c*x-I)*ln(-1/2*I*(c*x
+I))-1/10*I/c^5*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))-1/10*I/c^5*b^2*dilog(1/2*I*(c*x-I))+1/10*I/c^5*b^2*ln(c^2*x^2+
1)*ln(c*x+I)-1/20*I/c^5*b^2*ln(c*x+I)^2+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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \left (a + b \operatorname{atan}{\left (c x \right )}\right )^{2}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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