Integrand size = 14, antiderivative size = 64 \[ \int \frac {\arctan (1+2 x)}{(4+3 x)^3} \, dx=-\frac {1}{34 (4+3 x)}+\frac {8}{867} \arctan (1+2 x)-\frac {\arctan (1+2 x)}{6 (4+3 x)^2}+\frac {5}{289} \log (4+3 x)-\frac {5}{578} \log \left (1+2 x+2 x^2\right ) \]
-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)
Result contains complex when optimal does not.
Time = 0.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.27 \[ \int \frac {\arctan (1+2 x)}{(4+3 x)^3} \, dx=\frac {-289 \arctan (1+2 x)+(4+3 x) (-51-(15-8 i) (4+3 x) \log (i+(1+i) x)-(15+8 i) (4+3 x) \log (1+(1+i) x)+120 \log (4+3 x)+90 x \log (4+3 x))}{1734 (4+3 x)^2} \]
(-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)
Time = 0.32 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5568, 2081, 1145, 27, 1200, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\arctan (2 x+1)}{(3 x+4)^3} \, dx\) |
\(\Big \downarrow \) 5568 |
\(\displaystyle \frac {1}{3} \int \frac {1}{(3 x+4)^2 \left ((2 x+1)^2+1\right )}dx-\frac {\arctan (2 x+1)}{6 (3 x+4)^2}\) |
\(\Big \downarrow \) 2081 |
\(\displaystyle \frac {1}{3} \int \frac {1}{(3 x+4)^2 \left (4 x^2+4 x+2\right )}dx-\frac {\arctan (2 x+1)}{6 (3 x+4)^2}\) |
\(\Big \downarrow \) 1145 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{34} \int \frac {2 (1-3 x)}{(3 x+4) \left (2 x^2+2 x+1\right )}dx-\frac {3}{34 (3 x+4)}\right )-\frac {\arctan (2 x+1)}{6 (3 x+4)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{17} \int \frac {1-3 x}{(3 x+4) \left (2 x^2+2 x+1\right )}dx-\frac {3}{34 (3 x+4)}\right )-\frac {\arctan (2 x+1)}{6 (3 x+4)^2}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{17} \int \left (\frac {-30 x-7}{17 \left (2 x^2+2 x+1\right )}+\frac {45}{17 (3 x+4)}\right )dx-\frac {3}{34 (3 x+4)}\right )-\frac {\arctan (2 x+1)}{6 (3 x+4)^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{17} \left (\frac {8}{17} \arctan (2 x+1)-\frac {15}{34} \log \left (2 x^2+2 x+1\right )+\frac {15}{17} \log (3 x+4)\right )-\frac {3}{34 (3 x+4)}\right )-\frac {\arctan (2 x+1)}{6 (3 x+4)^2}\) |
-1/6*ArcTan[1 + 2*x]/(4 + 3*x)^2 + (-3/(34*(4 + 3*x)) + ((8*ArcTan[1 + 2*x ])/17 + (15*Log[4 + 3*x])/17 - (15*Log[1 + 2*x + 2*x^2])/34)/17)/3
3.2.21.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] + Simp [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}, x] && ILtQ[m, -1]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
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])
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] - Simp[b*d*(p/(f*(m + 1))) Int[(e + f*x)^(m + 1)*((a + b*ArcTan[ 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]
Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(-\frac {2 \arctan \left (1+2 x \right )}{3 \left (8+6 x \right )^{2}}-\frac {1}{17 \left (8+6 x \right )}+\frac {5 \ln \left (8+6 x \right )}{289}-\frac {5 \ln \left (\left (1+2 x \right )^{2}+1\right )}{578}+\frac {8 \arctan \left (1+2 x \right )}{867}\) | \(54\) |
default | \(-\frac {2 \arctan \left (1+2 x \right )}{3 \left (8+6 x \right )^{2}}-\frac {1}{17 \left (8+6 x \right )}+\frac {5 \ln \left (8+6 x \right )}{289}-\frac {5 \ln \left (\left (1+2 x \right )^{2}+1\right )}{578}+\frac {8 \arctan \left (1+2 x \right )}{867}\) | \(54\) |
parts | \(-\frac {1}{34 \left (4+3 x \right )}+\frac {8 \arctan \left (1+2 x \right )}{867}-\frac {\arctan \left (1+2 x \right )}{6 \left (4+3 x \right )^{2}}+\frac {5 \ln \left (4+3 x \right )}{289}-\frac {5 \ln \left (2 x^{2}+2 x +1\right )}{578}\) | \(55\) |
parallelrisch | \(\frac {810 \ln \left (\frac {4}{3}+x \right ) x^{2}-405 \ln \left (x^{2}+x +\frac {1}{2}\right ) x^{2}+432 \arctan \left (1+2 x \right ) x^{2}-612+2160 \ln \left (\frac {4}{3}+x \right ) x -1080 \ln \left (x^{2}+x +\frac {1}{2}\right ) x +1152 \arctan \left (1+2 x \right ) x +1440 \ln \left (\frac {4}{3}+x \right )-720 \ln \left (x^{2}+x +\frac {1}{2}\right )-459 x -99 \arctan \left (1+2 x \right )}{5202 \left (4+3 x \right )^{2}}\) | \(96\) |
risch | \(\frac {i \ln \left (1+i \left (1+2 x \right )\right )}{12 \left (4+3 x \right )^{2}}-\frac {i \left (-270 i \ln \left (2 x +1-i\right ) x^{2}-720 i \ln \left (2 x +1-i\right ) x -306 i x +960 i \ln \left (4+3 x \right )+1440 i \ln \left (4+3 x \right ) x -720 i \ln \left (2 x +1+i\right ) x -144 \ln \left (2 x +1+i\right ) x^{2}+144 \ln \left (2 x +1-i\right ) x^{2}+540 i \ln \left (4+3 x \right ) x^{2}-270 i \ln \left (2 x +1+i\right ) x^{2}-480 i \ln \left (2 x +1+i\right )-480 i \ln \left (2 x +1-i\right )-384 \ln \left (2 x +1+i\right ) x +384 \ln \left (2 x +1-i\right ) x -408 i-256 \ln \left (2 x +1+i\right )+256 \ln \left (2 x +1-i\right )+289 \ln \left (1-i \left (1+2 x \right )\right )\right )}{3468 \left (4+3 x \right )^{2}}\) | \(226\) |
-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)
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.20 \[ \int \frac {\arctan (1+2 x)}{(4+3 x)^3} \, dx=\frac {{\left (48 \, x^{2} + 128 \, x - 11\right )} \arctan \left (2 \, x + 1\right ) - 5 \, {\left (9 \, x^{2} + 24 \, x + 16\right )} \log \left (2 \, x^{2} + 2 \, x + 1\right ) + 10 \, {\left (9 \, x^{2} + 24 \, x + 16\right )} \log \left (3 \, x + 4\right ) - 51 \, x - 68}{578 \, {\left (9 \, x^{2} + 24 \, x + 16\right )}} \]
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)
Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (56) = 112\).
Time = 0.25 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.48 \[ \int \frac {\arctan (1+2 x)}{(4+3 x)^3} \, dx=\frac {90 x^{2} \log {\left (3 x + 4 \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {45 x^{2} \log {\left (2 x^{2} + 2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} + \frac {48 x^{2} \operatorname {atan}{\left (2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} + \frac {240 x \log {\left (3 x + 4 \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {120 x \log {\left (2 x^{2} + 2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} + \frac {128 x \operatorname {atan}{\left (2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {51 x}{5202 x^{2} + 13872 x + 9248} + \frac {160 \log {\left (3 x + 4 \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {80 \log {\left (2 x^{2} + 2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {11 \operatorname {atan}{\left (2 x + 1 \right )}}{5202 x^{2} + 13872 x + 9248} - \frac {68}{5202 x^{2} + 13872 x + 9248} \]
90*x**2*log(3*x + 4)/(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(3*x + 4)/(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( 3*x + 4)/(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)
Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.84 \[ \int \frac {\arctan (1+2 x)}{(4+3 x)^3} \, dx=-\frac {1}{34 \, {\left (3 \, x + 4\right )}} - \frac {\arctan \left (2 \, x + 1\right )}{6 \, {\left (3 \, x + 4\right )}^{2}} + \frac {8}{867} \, \arctan \left (2 \, x + 1\right ) - \frac {5}{578} \, \log \left (2 \, x^{2} + 2 \, x + 1\right ) + \frac {5}{289} \, \log \left (3 \, x + 4\right ) \]
-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)
\[ \int \frac {\arctan (1+2 x)}{(4+3 x)^3} \, dx=\int { \frac {\arctan \left (2 \, x + 1\right )}{{\left (3 \, x + 4\right )}^{3}} \,d x } \]
Time = 0.87 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.72 \[ \int \frac {\arctan (1+2 x)}{(4+3 x)^3} \, dx=\frac {5\,\ln \left (x+\frac {4}{3}\right )}{289}-\frac {5\,\ln \left (x^2+x+\frac {1}{2}\right )}{578}+\frac {8\,\mathrm {atan}\left (2\,x+1\right )}{867}-\frac {\frac {3\,x}{34}+\frac {\mathrm {atan}\left (2\,x+1\right )}{6}+\frac {2}{17}}{{\left (3\,x+4\right )}^2} \]