Optimal. Leaf size=96 \[ -\frac {b}{2 \left (1+a^2\right ) x}-\frac {\left (1-a^2\right ) b^2 \text {ArcTan}(a+b x)}{2 \left (1+a^2\right )^2}-\frac {\text {ArcTan}(a+b x)}{2 x^2}-\frac {a b^2 \log (x)}{\left (1+a^2\right )^2}+\frac {a b^2 \log \left (1+(a+b x)^2\right )}{2 \left (1+a^2\right )^2} \]
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Rubi [A]
time = 0.07, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5153, 378, 724,
815, 649, 209, 266} \begin {gather*} -\frac {\left (1-a^2\right ) b^2 \text {ArcTan}(a+b x)}{2 \left (a^2+1\right )^2}-\frac {a b^2 \log (x)}{\left (a^2+1\right )^2}+\frac {a b^2 \log \left ((a+b x)^2+1\right )}{2 \left (a^2+1\right )^2}-\frac {b}{2 \left (a^2+1\right ) x}-\frac {\text {ArcTan}(a+b x)}{2 x^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 266
Rule 378
Rule 649
Rule 724
Rule 815
Rule 5153
Rubi steps
\begin {align*} \int \frac {\tan ^{-1}(a+b x)}{x^3} \, dx &=-\frac {\tan ^{-1}(a+b x)}{2 x^2}+\frac {1}{2} b \int \frac {1}{x^2 \left (1+(a+b x)^2\right )} \, dx\\ &=-\frac {\tan ^{-1}(a+b x)}{2 x^2}+\frac {1}{2} b^2 \text {Subst}\left (\int \frac {1}{(-a+x)^2 \left (1+x^2\right )} \, dx,x,a+b x\right )\\ &=-\frac {b}{2 \left (1+a^2\right ) x}-\frac {\tan ^{-1}(a+b x)}{2 x^2}+\frac {b^2 \text {Subst}\left (\int \frac {-a-x}{(-a+x) \left (1+x^2\right )} \, dx,x,a+b x\right )}{2 \left (1+a^2\right )}\\ &=-\frac {b}{2 \left (1+a^2\right ) x}-\frac {\tan ^{-1}(a+b x)}{2 x^2}+\frac {b^2 \text {Subst}\left (\int \left (\frac {2 a}{\left (1+a^2\right ) (a-x)}+\frac {-1+a^2+2 a x}{\left (1+a^2\right ) \left (1+x^2\right )}\right ) \, dx,x,a+b x\right )}{2 \left (1+a^2\right )}\\ &=-\frac {b}{2 \left (1+a^2\right ) x}-\frac {\tan ^{-1}(a+b x)}{2 x^2}-\frac {a b^2 \log (x)}{\left (1+a^2\right )^2}+\frac {b^2 \text {Subst}\left (\int \frac {-1+a^2+2 a x}{1+x^2} \, dx,x,a+b x\right )}{2 \left (1+a^2\right )^2}\\ &=-\frac {b}{2 \left (1+a^2\right ) x}-\frac {\tan ^{-1}(a+b x)}{2 x^2}-\frac {a b^2 \log (x)}{\left (1+a^2\right )^2}+\frac {\left (a b^2\right ) \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,a+b x\right )}{\left (1+a^2\right )^2}-\frac {\left (\left (1-a^2\right ) b^2\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,a+b x\right )}{2 \left (1+a^2\right )^2}\\ &=-\frac {b}{2 \left (1+a^2\right ) x}-\frac {\left (1-a^2\right ) b^2 \tan ^{-1}(a+b x)}{2 \left (1+a^2\right )^2}-\frac {\tan ^{-1}(a+b x)}{2 x^2}-\frac {a b^2 \log (x)}{\left (1+a^2\right )^2}+\frac {a b^2 \log \left (1+(a+b x)^2\right )}{2 \left (1+a^2\right )^2}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 0.08, size = 92, normalized size = 0.96 \begin {gather*} \frac {-2 \text {ArcTan}(a+b x)+\frac {b x \left (-4 a b x \log (x)-i (i+a)^2 b x \log (i-a-b x)+(-i+a) (-2 (i+a)+(1+i a) b x \log (i+a+b x))\right )}{\left (1+a^2\right )^2}}{4 x^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.09, size = 84, normalized size = 0.88
method | result | size |
derivativedivides | \(b^{2} \left (-\frac {\arctan \left (b x +a \right )}{2 b^{2} x^{2}}+\frac {a \ln \left (1+\left (b x +a \right )^{2}\right )+\left (a^{2}-1\right ) \arctan \left (b x +a \right )}{2 \left (a^{2}+1\right )^{2}}-\frac {1}{2 \left (a^{2}+1\right ) b x}-\frac {a \ln \left (-b x \right )}{\left (a^{2}+1\right )^{2}}\right )\) | \(84\) |
default | \(b^{2} \left (-\frac {\arctan \left (b x +a \right )}{2 b^{2} x^{2}}+\frac {a \ln \left (1+\left (b x +a \right )^{2}\right )+\left (a^{2}-1\right ) \arctan \left (b x +a \right )}{2 \left (a^{2}+1\right )^{2}}-\frac {1}{2 \left (a^{2}+1\right ) b x}-\frac {a \ln \left (-b x \right )}{\left (a^{2}+1\right )^{2}}\right )\) | \(84\) |
risch | \(\frac {i \ln \left (1+i \left (b x +a \right )\right )}{4 x^{2}}-\frac {i \left (a^{4} \ln \left (1-i \left (b x +a \right )\right )+2 a^{2} \ln \left (1-i \left (b x +a \right )\right )+\ln \left (1-i \left (b x +a \right )\right )-\ln \left (\left (a^{6} b -4 i a^{5} b +9 a^{4} b +8 i a^{3} b -9 a^{2} b -4 i a b -b \right ) x +13 a^{5}-3 i a^{6}+a^{7}-17 a^{3}+17 i a^{4}+3 a -13 i a^{2}-i\right ) a^{2} b^{2} x^{2}+\ln \left (\left (a^{6} b +4 i a^{5} b +9 a^{4} b -8 i a^{3} b -9 a^{2} b +4 i a b -b \right ) x +13 a^{5}+3 i a^{6}+a^{7}-17 a^{3}-17 i a^{4}+3 a +13 i a^{2}+i\right ) a^{2} b^{2} x^{2}+\ln \left (\left (a^{6} b -4 i a^{5} b +9 a^{4} b +8 i a^{3} b -9 a^{2} b -4 i a b -b \right ) x +13 a^{5}-3 i a^{6}+a^{7}-17 a^{3}+17 i a^{4}+3 a -13 i a^{2}-i\right ) b^{2} x^{2}+2 i \ln \left (\left (a^{6} b -4 i a^{5} b +9 a^{4} b +8 i a^{3} b -9 a^{2} b -4 i a b -b \right ) x +13 a^{5}-3 i a^{6}+a^{7}-17 a^{3}+17 i a^{4}+3 a -13 i a^{2}-i\right ) a \,b^{2} x^{2}-\ln \left (\left (a^{6} b +4 i a^{5} b +9 a^{4} b -8 i a^{3} b -9 a^{2} b +4 i a b -b \right ) x +13 a^{5}+3 i a^{6}+a^{7}-17 a^{3}-17 i a^{4}+3 a +13 i a^{2}+i\right ) b^{2} x^{2}+2 i \ln \left (\left (a^{6} b +4 i a^{5} b +9 a^{4} b -8 i a^{3} b -9 a^{2} b +4 i a b -b \right ) x +13 a^{5}+3 i a^{6}+a^{7}-17 a^{3}-17 i a^{4}+3 a +13 i a^{2}+i\right ) a \,b^{2} x^{2}-4 i \ln \left (\left (a^{4} b +34 a^{2} b +b \right ) x \right ) a \,b^{2} x^{2}-2 i x \,a^{2} b -2 i b x \right )}{4 x^{2} \left (a^{2}+2 i a -1\right ) \left (a^{2}-2 i a -1\right )}\) | \(665\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.47, size = 112, normalized size = 1.17 \begin {gather*} \frac {1}{2} \, {\left (\frac {{\left (a^{2} - 1\right )} b \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{4} + 2 \, a^{2} + 1} + \frac {a b \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{4} + 2 \, a^{2} + 1} - \frac {2 \, a b \log \left (x\right )}{a^{4} + 2 \, a^{2} + 1} - \frac {1}{{\left (a^{2} + 1\right )} x}\right )} b - \frac {\arctan \left (b x + a\right )}{2 \, x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.30, size = 95, normalized size = 0.99 \begin {gather*} \frac {a b^{2} x^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, a b^{2} x^{2} \log \left (x\right ) - {\left (a^{2} + 1\right )} b x + {\left ({\left (a^{2} - 1\right )} b^{2} x^{2} - a^{4} - 2 \, a^{2} - 1\right )} \arctan \left (b x + a\right )}{2 \, {\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [C] Result contains complex when optimal does not.
time = 1.02, size = 382, normalized size = 3.98 \begin {gather*} \begin {cases} - \frac {b^{2} \operatorname {atan}{\left (b x - i \right )}}{8} - \frac {b}{8 x} - \frac {\operatorname {atan}{\left (b x - i \right )}}{2 x^{2}} - \frac {i}{8 x^{2}} & \text {for}\: a = - i \\- \frac {b^{2} \operatorname {atan}{\left (b x + i \right )}}{8} - \frac {b}{8 x} - \frac {\operatorname {atan}{\left (b x + i \right )}}{2 x^{2}} + \frac {i}{8 x^{2}} & \text {for}\: a = i \\- \frac {a^{4} \operatorname {atan}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} + \frac {a^{2} b^{2} x^{2} \operatorname {atan}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac {a^{2} b x}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac {2 a^{2} \operatorname {atan}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac {2 a b^{2} x^{2} \log {\left (x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} + \frac {a b^{2} x^{2} \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac {b^{2} x^{2} \operatorname {atan}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac {b x}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} - \frac {\operatorname {atan}{\left (a + b x \right )}}{2 a^{4} x^{2} + 4 a^{2} x^{2} + 2 x^{2}} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.22, size = 232, normalized size = 2.42 \begin {gather*} \frac {a\,b^2\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2\,{\left (a^2+1\right )}^2}-\frac {\frac {b\,x}{2}+\mathrm {atan}\left (a+b\,x\right )\,\left (\frac {a^2}{2}+\frac {1}{2}\right )+\frac {b^2\,x^2\,\mathrm {atan}\left (a+b\,x\right )}{2}+\frac {x^3\,\left (b^3-3\,a^2\,b^3\right )}{2\,\left (a^4+2\,a^2+1\right )}-\frac {a\,b^4\,x^4}{{\left (a^2+1\right )}^2}+a\,b\,x\,\mathrm {atan}\left (a+b\,x\right )}{a^2\,x^2+2\,a\,b\,x^3+b^2\,x^4+x^2}-\frac {\mathrm {atan}\left (\frac {2\,x\,b^2+2\,a\,b}{2\,\sqrt {b^2\,\left (a^2+1\right )-a^2\,b^2}}\right )\,\left (b^3-a^2\,b^3\right )}{\sqrt {b^2}\,\left (2\,a^4+4\,a^2+2\right )}-\frac {a\,b^2\,\ln \left (x\right )}{{\left (a^2+1\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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