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]
________________________________________________________________________________________
Rubi [A] time = 0.067086, antiderivative size = 64, normalized size of antiderivative = 1., 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]
[Out]
Rule 5045
Rule 1982
Rule 709
Rule 800
Rule 634
Rule 617
Rule 204
Rule 628
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*}
Mathematica [C] time = 0.0439864, 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]
[Out]
________________________________________________________________________________________
Maple [A] time = 0.044, size = 54, normalized size = 0.8 \begin{align*} -{\frac{2\,\arctan \left ( 1+2\,x \right ) }{3\, \left ( 8+6\,x \right ) ^{2}}}-{\frac{5\,\ln \left ( \left ( 1+2\,x \right ) ^{2}+1 \right ) }{578}}+{\frac{8\,\arctan \left ( 1+2\,x \right ) }{867}}-{\frac{1}{136+102\,x}}+{\frac{5\,\ln \left ( 8+6\,x \right ) }{289}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [A] time = 1.44811, size = 73, normalized size = 1.14 \begin{align*} -\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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 2.42057, size = 219, normalized size = 3.42 \begin{align*} \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 )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [B] time = 0.779718, size = 223, normalized size = 3.48 \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.10174, size = 74, normalized size = 1.16 \begin{align*} -\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 ({\left | 3 \, x + 4 \right |}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]