3.121 \(\int \frac {\tan ^{-1}(1+2 x)}{(4+3 x)^3} \, dx\)

Optimal. Leaf size=64 \[ -\frac {5}{578} \log \left (2 x^2+2 x+1\right )-\frac {1}{34 (3 x+4)}+\frac {5}{289} \log (3 x+4)-\frac {\tan ^{-1}(2 x+1)}{6 (3 x+4)^2}+\frac {8}{867} \tan ^{-1}(2 x+1) \]

[Out]

-1/34/(4+3*x)+8/867*arctan(1+2*x)-1/6*arctan(1+2*x)/(4+3*x)^2+5/289*ln(4+3*x)-5/578*ln(2*x^2+2*x+1)

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Rubi [A]  time = 0.07, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5045, 1982, 709, 800, 634, 617, 204, 628} \[ -\frac {5}{578} \log \left (2 x^2+2 x+1\right )-\frac {1}{34 (3 x+4)}+\frac {5}{289} \log (3 x+4)-\frac {\tan ^{-1}(2 x+1)}{6 (3 x+4)^2}+\frac {8}{867} \tan ^{-1}(2 x+1) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[1 + 2*x]/(4 + 3*x)^3,x]

[Out]

-1/(34*(4 + 3*x)) + (8*ArcTan[1 + 2*x])/867 - ArcTan[1 + 2*x]/(6*(4 + 3*x)^2) + (5*Log[4 + 3*x])/289 - (5*Log[
1 + 2*x + 2*x^2])/578

Rule 204

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 709

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d + e*x)^(m + 1)*Simp[c*d - b*e - c
*e*x, x])/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[m, -1]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 1982

Int[(u_)^(m_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^p, x] /; FreeQ[{m, p}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rule 5045

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*(a + b*ArcTan[c + d*x])^p)/(f*(m + 1)), x] - Dist[(b*d*p)/(f*(m + 1)), Int[((e + f*x)^(m + 1)*(a + b*Arc
Tan[c + d*x])^(p - 1))/(1 + (c + d*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(1+2 x)}{(4+3 x)^3} \, dx &=-\frac {\tan ^{-1}(1+2 x)}{6 (4+3 x)^2}+\frac {1}{3} \int \frac {1}{(4+3 x)^2 \left (1+(1+2 x)^2\right )} \, dx\\ &=-\frac {\tan ^{-1}(1+2 x)}{6 (4+3 x)^2}+\frac {1}{3} \int \frac {1}{(4+3 x)^2 \left (2+4 x+4 x^2\right )} \, dx\\ &=-\frac {1}{34 (4+3 x)}-\frac {\tan ^{-1}(1+2 x)}{6 (4+3 x)^2}+\frac {1}{102} \int \frac {4-12 x}{(4+3 x) \left (2+4 x+4 x^2\right )} \, dx\\ &=-\frac {1}{34 (4+3 x)}-\frac {\tan ^{-1}(1+2 x)}{6 (4+3 x)^2}+\frac {1}{102} \int \left (\frac {90}{17 (4+3 x)}-\frac {2 (7+30 x)}{17 \left (1+2 x+2 x^2\right )}\right ) \, dx\\ &=-\frac {1}{34 (4+3 x)}-\frac {\tan ^{-1}(1+2 x)}{6 (4+3 x)^2}+\frac {5}{289} \log (4+3 x)-\frac {1}{867} \int \frac {7+30 x}{1+2 x+2 x^2} \, dx\\ &=-\frac {1}{34 (4+3 x)}-\frac {\tan ^{-1}(1+2 x)}{6 (4+3 x)^2}+\frac {5}{289} \log (4+3 x)-\frac {5}{578} \int \frac {2+4 x}{1+2 x+2 x^2} \, dx+\frac {8}{867} \int \frac {1}{1+2 x+2 x^2} \, dx\\ &=-\frac {1}{34 (4+3 x)}-\frac {\tan ^{-1}(1+2 x)}{6 (4+3 x)^2}+\frac {5}{289} \log (4+3 x)-\frac {5}{578} \log \left (1+2 x+2 x^2\right )-\frac {8}{867} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+2 x\right )\\ &=-\frac {1}{34 (4+3 x)}+\frac {8}{867} \tan ^{-1}(1+2 x)-\frac {\tan ^{-1}(1+2 x)}{6 (4+3 x)^2}+\frac {5}{289} \log (4+3 x)-\frac {5}{578} \log \left (1+2 x+2 x^2\right )\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 81, normalized size = 1.27 \[ \frac {-289 \tan ^{-1}(2 x+1)+(3 x+4) ((-15+8 i) (3 x+4) \log ((1+i) x+i)-(15+8 i) (3 x+4) \log (1+(1+i) x)+90 x \log (3 x+4)+120 \log (3 x+4)-51)}{1734 (3 x+4)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[1 + 2*x]/(4 + 3*x)^3,x]

[Out]

(-289*ArcTan[1 + 2*x] + (4 + 3*x)*(-51 - (15 - 8*I)*(4 + 3*x)*Log[I + (1 + I)*x] - (15 + 8*I)*(4 + 3*x)*Log[1
+ (1 + I)*x] + 120*Log[4 + 3*x] + 90*x*Log[4 + 3*x]))/(1734*(4 + 3*x)^2)

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fricas [A]  time = 1.13, size = 77, normalized size = 1.20 \[ \frac {{\left (48 \, x^{2} + 128 \, x - 11\right )} \arctan \left (2 \, x + 1\right ) - 5 \, {\left (9 \, x^{2} + 24 \, x + 16\right )} \log \left (2 \, x^{2} + 2 \, x + 1\right ) + 10 \, {\left (9 \, x^{2} + 24 \, x + 16\right )} \log \left (3 \, x + 4\right ) - 51 \, x - 68}{578 \, {\left (9 \, x^{2} + 24 \, x + 16\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(1+2*x)/(4+3*x)^3,x, algorithm="fricas")

[Out]

1/578*((48*x^2 + 128*x - 11)*arctan(2*x + 1) - 5*(9*x^2 + 24*x + 16)*log(2*x^2 + 2*x + 1) + 10*(9*x^2 + 24*x +
16)*log(3*x + 4) - 51*x - 68)/(9*x^2 + 24*x + 16)

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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(arctan(1+2*x)/(4+3*x)^3,x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.05, size = 54, normalized size = 0.84 \[ -\frac {2 \arctan \left (1+2 x \right )}{3 \left (8+6 x \right )^{2}}-\frac {1}{17 \left (8+6 x \right )}+\frac {5 \ln \left (8+6 x \right )}{289}-\frac {5 \ln \left (\left (1+2 x \right )^{2}+1\right )}{578}+\frac {8 \arctan \left (1+2 x \right )}{867} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(1+2*x)/(4+3*x)^3,x)

[Out]

-2/3/(8+6*x)^2*arctan(1+2*x)-1/17/(8+6*x)+5/289*ln(8+6*x)-5/578*ln((1+2*x)^2+1)+8/867*arctan(1+2*x)

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maxima [A]  time = 0.40, size = 54, normalized size = 0.84 \[ -\frac {1}{34 \, {\left (3 \, x + 4\right )}} - \frac {\arctan \left (2 \, x + 1\right )}{6 \, {\left (3 \, x + 4\right )}^{2}} + \frac {8}{867} \, \arctan \left (2 \, x + 1\right ) - \frac {5}{578} \, \log \left (2 \, x^{2} + 2 \, x + 1\right ) + \frac {5}{289} \, \log \left (3 \, x + 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(1+2*x)/(4+3*x)^3,x, algorithm="maxima")

[Out]

-1/34/(3*x + 4) - 1/6*arctan(2*x + 1)/(3*x + 4)^2 + 8/867*arctan(2*x + 1) - 5/578*log(2*x^2 + 2*x + 1) + 5/289
*log(3*x + 4)

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mupad [B]  time = 0.70, size = 46, normalized size = 0.72 \[ \frac {5\,\ln \left (x+\frac {4}{3}\right )}{289}-\frac {5\,\ln \left (x^2+x+\frac {1}{2}\right )}{578}+\frac {8\,\mathrm {atan}\left (2\,x+1\right )}{867}-\frac {\frac {3\,x}{34}+\frac {\mathrm {atan}\left (2\,x+1\right )}{6}+\frac {2}{17}}{{\left (3\,x+4\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(2*x + 1)/(3*x + 4)^3,x)

[Out]

(5*log(x + 4/3))/289 - (5*log(x + x^2 + 1/2))/578 + (8*atan(2*x + 1))/867 - ((3*x)/34 + atan(2*x + 1)/6 + 2/17
)/(3*x + 4)^2

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sympy [B]  time = 0.60, size = 223, normalized size = 3.48 \[ \frac {90 x^{2} \log {\left (x + \frac {4}{3} \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {45 x^{2} \log {\left (2 x^{2} + 2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} + \frac {48 x^{2} \operatorname {atan}{\left (2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} + \frac {240 x \log {\left (x + \frac {4}{3} \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {120 x \log {\left (2 x^{2} + 2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} + \frac {128 x \operatorname {atan}{\left (2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {51 x}{5202 x^{2} + 13872 x + 9248} + \frac {160 \log {\left (x + \frac {4}{3} \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {80 \log {\left (2 x^{2} + 2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {11 \operatorname {atan}{\left (2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {68}{5202 x^{2} + 13872 x + 9248} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(1+2*x)/(4+3*x)**3,x)

[Out]

90*x**2*log(x + 4/3)/(5202*x**2 + 13872*x + 9248) - 45*x**2*log(2*x**2 + 2*x + 1)/(5202*x**2 + 13872*x + 9248)
+ 48*x**2*atan(2*x + 1)/(5202*x**2 + 13872*x + 9248) + 240*x*log(x + 4/3)/(5202*x**2 + 13872*x + 9248) - 120*
x*log(2*x**2 + 2*x + 1)/(5202*x**2 + 13872*x + 9248) + 128*x*atan(2*x + 1)/(5202*x**2 + 13872*x + 9248) - 51*x
/(5202*x**2 + 13872*x + 9248) + 160*log(x + 4/3)/(5202*x**2 + 13872*x + 9248) - 80*log(2*x**2 + 2*x + 1)/(5202
*x**2 + 13872*x + 9248) - 11*atan(2*x + 1)/(5202*x**2 + 13872*x + 9248) - 68/(5202*x**2 + 13872*x + 9248)

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