3.11 \(\int (d+e x) (a+b \tan ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=171 \[ -\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac {2 b d \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac {a b e x}{c}+\frac {b^2 e \log \left (c^2 x^2+1\right )}{2 c^2}+\frac {i b^2 d \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{c}-\frac {b^2 e x \tan ^{-1}(c x)}{c} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {4864, 4846, 260, 4984, 4884, 4920, 4854, 2402, 2315} \[ \frac {i b^2 d \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac {2 b d \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac {a b e x}{c}+\frac {b^2 e \log \left (c^2 x^2+1\right )}{2 c^2}-\frac {b^2 e x \tan ^{-1}(c x)}{c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcTan[c*x])^2,x]

[Out]

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && 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 4864

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

Rule 4884

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 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 4984

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> I
nt[ExpandIntegrand[(a + b*ArcTan[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& IGtQ[p, 0] && EqQ[e, c^2*d] && IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 172, normalized size = 1.01 \[ \frac {2 a^2 c^2 d x+a^2 c^2 e x^2+2 b \tan ^{-1}(c x) \left (a \left (2 c^2 d x+c^2 e x^2+e\right )+2 b c d \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-b c e x\right )-2 a b c d \log \left (c^2 x^2+1\right )-2 a b c e x+b^2 e \log \left (c^2 x^2+1\right )+b^2 (c x-i) \tan ^{-1}(c x)^2 (2 c d+c e x+i e)-2 i b^2 c d \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )}{2 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 1.35, size = 0, normalized size = 0.00 \[ {\rm integral}\left (a^{2} e x + a^{2} d + {\left (b^{2} e x + b^{2} d\right )} \arctan \left (c x\right )^{2} + 2 \, {\left (a b e x + a b d\right )} \arctan \left (c x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(a^2*e*x + a^2*d + (b^2*e*x + b^2*d)*arctan(c*x)^2 + 2*(a*b*e*x + a*b*d)*arctan(c*x), 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((e*x+d)*(a+b*arctan(c*x))^2,x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.10, size = 360, normalized size = 2.11 \[ \frac {a^{2} x^{2} e}{2}+a^{2} d x +\frac {b^{2} \arctan \left (c x \right )^{2} x^{2} e}{2}+b^{2} \arctan \left (c x \right )^{2} x d -\frac {b^{2} \ln \left (c^{2} x^{2}+1\right ) \arctan \left (c x \right ) d}{c}+\frac {b^{2} \arctan \left (c x \right )^{2} e}{2 c^{2}}-\frac {b^{2} e x \arctan \left (c x \right )}{c}+\frac {b^{2} e \ln \left (c^{2} x^{2}+1\right )}{2 c^{2}}+\frac {i b^{2} d \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 c}-\frac {i b^{2} d \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{2 c}+\frac {i b^{2} d \ln \left (c x -i\right )^{2}}{4 c}+\frac {i b^{2} d \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{2 c}-\frac {i b^{2} d \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{2 c}-\frac {i b^{2} d \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{2 c}-\frac {i b^{2} d \ln \left (c x +i\right )^{2}}{4 c}+\frac {i b^{2} d \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 c}+a b \arctan \left (c x \right ) x^{2} e +2 a b \arctan \left (c x \right ) x d -\frac {a b e x}{c}-\frac {a b d \ln \left (c^{2} x^{2}+1\right )}{c}+\frac {a b e \arctan \left (c x \right )}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 12 \, b^{2} c^{2} e \int \frac {x^{3} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + b^{2} c^{2} e \int \frac {x^{3} \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 12 \, b^{2} c^{2} d \int \frac {x^{2} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 2 \, b^{2} c^{2} e \int \frac {x^{3} \log \left (c^{2} x^{2} + 1\right )}{16 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + b^{2} c^{2} d \int \frac {x^{2} \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + 4 \, b^{2} c^{2} d \int \frac {x^{2} \log \left (c^{2} x^{2} + 1\right )}{16 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + \frac {1}{2} \, a^{2} e x^{2} + \frac {b^{2} d \arctan \left (c x\right )^{3}}{4 \, c} - 4 \, b^{2} c e \int \frac {x^{2} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} - 8 \, b^{2} c d \int \frac {x \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} a b e + a^{2} d x + 12 \, b^{2} e \int \frac {x \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + b^{2} e \int \frac {x \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + b^{2} d \int \frac {\log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} a b d}{c} + \frac {1}{8} \, {\left (b^{2} e x^{2} + 2 \, b^{2} d x\right )} \arctan \left (c x\right )^{2} - \frac {1}{32} \, {\left (b^{2} e x^{2} + 2 \, b^{2} d x\right )} \log \left (c^{2} x^{2} + 1\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*b^2*c^2*e*integrate(1/16*x^3*arctan(c*x)^2/(c^2*x^2 + 1), x) + b^2*c^2*e*integrate(1/16*x^3*log(c^2*x^2 + 1
)^2/(c^2*x^2 + 1), x) + 12*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 2*b^2*c^2*e*integrat
e(1/16*x^3*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + b^2*c^2*d*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1),
x) + 4*b^2*c^2*d*integrate(1/16*x^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 1/2*a^2*e*x^2 + 1/4*b^2*d*arctan(c*x
)^3/c - 4*b^2*c*e*integrate(1/16*x^2*arctan(c*x)/(c^2*x^2 + 1), x) - 8*b^2*c*d*integrate(1/16*x*arctan(c*x)/(c
^2*x^2 + 1), x) + (x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*a*b*e + a^2*d*x + 12*b^2*e*integrate(1/16*x*
arctan(c*x)^2/(c^2*x^2 + 1), x) + b^2*e*integrate(1/16*x*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + b^2*d*integrat
e(1/16*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + (2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b*d/c + 1/8*(b^2*e*x^2
+ 2*b^2*d*x)*arctan(c*x)^2 - 1/32*(b^2*e*x^2 + 2*b^2*d*x)*log(c^2*x^2 + 1)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^2*(d + e*x),x)

[Out]

int((a + b*atan(c*x))^2*(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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