Optimal. Leaf size=62 \[ -\frac {\cot ^{-1}(a+b x)}{x}+\frac {a b \text {ArcTan}(a+b x)}{1+a^2}-\frac {b \log (x)}{1+a^2}+\frac {b \log \left (1+(a+b x)^2\right )}{2 \left (1+a^2\right )} \]
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Rubi [A]
time = 0.03, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5154, 378, 720,
31, 649, 209, 266} \begin {gather*} \frac {a b \text {ArcTan}(a+b x)}{a^2+1}-\frac {b \log (x)}{a^2+1}+\frac {b \log \left ((a+b x)^2+1\right )}{2 \left (a^2+1\right )}-\frac {\cot ^{-1}(a+b x)}{x} \end {gather*}
Antiderivative was successfully verified.
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Rule 31
Rule 209
Rule 266
Rule 378
Rule 649
Rule 720
Rule 5154
Rubi steps
\begin {align*} \int \frac {\cot ^{-1}(a+b x)}{x^2} \, dx &=-\frac {\cot ^{-1}(a+b x)}{x}-b \int \frac {1}{x \left (1+(a+b x)^2\right )} \, dx\\ &=-\frac {\cot ^{-1}(a+b x)}{x}-b \text {Subst}\left (\int \frac {1}{(-a+x) \left (1+x^2\right )} \, dx,x,a+b x\right )\\ &=-\frac {\cot ^{-1}(a+b x)}{x}-\frac {b \text {Subst}\left (\int \frac {1}{-a+x} \, dx,x,a+b x\right )}{1+a^2}-\frac {b \text {Subst}\left (\int \frac {-a-x}{1+x^2} \, dx,x,a+b x\right )}{1+a^2}\\ &=-\frac {\cot ^{-1}(a+b x)}{x}-\frac {b \log (x)}{1+a^2}+\frac {b \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,a+b x\right )}{1+a^2}+\frac {(a b) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,a+b x\right )}{1+a^2}\\ &=-\frac {\cot ^{-1}(a+b x)}{x}+\frac {a b \tan ^{-1}(a+b x)}{1+a^2}-\frac {b \log (x)}{1+a^2}+\frac {b \log \left (1+(a+b x)^2\right )}{2 \left (1+a^2\right )}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 0.04, size = 66, normalized size = 1.06 \begin {gather*} -\frac {\cot ^{-1}(a+b x)}{x}+\frac {b (-2 \log (x)+(1-i a) \log (i-a-b x)+(1+i a) \log (i+a+b x))}{2 \left (1+a^2\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.08, size = 61, normalized size = 0.98
method | result | size |
derivativedivides | \(b \left (-\frac {\mathrm {arccot}\left (b x +a \right )}{b x}-\frac {\ln \left (-b x \right )}{a^{2}+1}+\frac {\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}+\arctan \left (b x +a \right ) a}{a^{2}+1}\right )\) | \(61\) |
default | \(b \left (-\frac {\mathrm {arccot}\left (b x +a \right )}{b x}-\frac {\ln \left (-b x \right )}{a^{2}+1}+\frac {\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{2}+\arctan \left (b x +a \right ) a}{a^{2}+1}\right )\) | \(61\) |
risch | \(-\frac {i \ln \left (1+i \left (b x +a \right )\right )}{2 x}-\frac {-i a^{2} \ln \left (1-i \left (b x +a \right )\right )-i \ln \left (1-i \left (b x +a \right )\right )+\pi \,a^{2}+\pi +2 \ln \left (x \right ) x b -x b \ln \left (\left (i a b -3 b \right ) x +i a^{2}+3 i-2 a \right )+i x b \ln \left (\left (i a b -3 b \right ) x +i a^{2}+3 i-2 a \right ) a -x b \ln \left (\left (i a b +3 b \right ) x +i a^{2}+3 i+2 a \right )-i x b \ln \left (\left (i a b +3 b \right ) x +i a^{2}+3 i+2 a \right ) a}{2 x \left (a -i\right ) \left (i+a \right )}\) | \(196\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.48, size = 77, normalized size = 1.24 \begin {gather*} \frac {1}{2} \, b {\left (\frac {2 \, a \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{2} + 1} + \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{2} + 1} - \frac {2 \, \log \left (x\right )}{a^{2} + 1}\right )} - \frac {\operatorname {arccot}\left (b x + a\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 3.32, size = 64, normalized size = 1.03 \begin {gather*} \frac {2 \, a b x \arctan \left (b x + a\right ) + b x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \, b x \log \left (x\right ) - 2 \, {\left (a^{2} + 1\right )} \operatorname {arccot}\left (b x + a\right )}{2 \, {\left (a^{2} + 1\right )} x} \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 = 0.80, size = 167, normalized size = 2.69 \begin {gather*} \begin {cases} - \frac {i b \operatorname {acot}{\left (b x - i \right )}}{2} - \frac {\operatorname {acot}{\left (b x - i \right )}}{x} + \frac {i}{2 x} & \text {for}\: a = - i \\\frac {i b \operatorname {acot}{\left (b x + i \right )}}{2} - \frac {\operatorname {acot}{\left (b x + i \right )}}{x} - \frac {i}{2 x} & \text {for}\: a = i \\- \frac {2 a^{2} \operatorname {acot}{\left (a + b x \right )}}{2 a^{2} x + 2 x} - \frac {2 a b x \operatorname {acot}{\left (a + b x \right )}}{2 a^{2} x + 2 x} - \frac {2 b x \log {\left (x \right )}}{2 a^{2} x + 2 x} + \frac {b x \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 a^{2} x + 2 x} - \frac {2 \operatorname {acot}{\left (a + b x \right )}}{2 a^{2} x + 2 x} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 498 vs.
\(2 (60) = 120\).
time = 0.54, size = 498, normalized size = 8.03 \begin {gather*} -\frac {{\left (2 \, a \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 2 \, a \log \left (\frac {4 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 4 \, a \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} + \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} - 4 \, a \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1\right )}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) + \log \left (\frac {4 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 4 \, a \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} + \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} - 4 \, a \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1\right )}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 2 \, a \arctan \left (\frac {1}{b x + a}\right ) - 4 \, \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) - \log \left (\frac {4 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 4 \, a \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} + \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} - 4 \, a \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1\right )}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right )\right )} b}{2 \, {\left (2 \, a^{3} \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) + a^{2} \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - a^{2} + 2 \, a \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right ) + \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.22, size = 62, normalized size = 1.00 \begin {gather*} -\frac {\mathrm {acot}\left (a+b\,x\right )}{x}-\frac {b\,x\,\ln \left (x\right )-\frac {b\,x\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2}+a\,b\,x\,\mathrm {acot}\left (a+b\,x\right )}{x\,\left (a^2+1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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