Integrand size = 33, antiderivative size = 216 \[ \int \frac {\arctan (a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=-\frac {2 i \sqrt {1+(a+b x)^2} \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}+\frac {i \sqrt {1+(a+b x)^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}-\frac {i \sqrt {1+(a+b x)^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}} \]
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Time = 0.12 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {5163, 5010, 5006} \[ \int \frac {\arctan (a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=-\frac {2 i \sqrt {(a+b x)^2+1} \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c (a+b x)^2+c}}+\frac {i \sqrt {(a+b x)^2+1} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c (a+b x)^2+c}}-\frac {i \sqrt {(a+b x)^2+1} \operatorname {PolyLog}\left (2,\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c (a+b x)^2+c}} \]
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Rule 5006
Rule 5010
Rule 5163
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\arctan (x)}{\sqrt {c+c x^2}} \, dx,x,a+b x\right )}{b} \\ & = \frac {\sqrt {1+(a+b x)^2} \text {Subst}\left (\int \frac {\arctan (x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b \sqrt {c+c (a+b x)^2}} \\ & = -\frac {2 i \sqrt {1+(a+b x)^2} \arctan (a+b x) \arctan \left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}+\frac {i \sqrt {1+(a+b x)^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}}-\frac {i \sqrt {1+(a+b x)^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b \sqrt {c+c (a+b x)^2}} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.44 \[ \int \frac {\arctan (a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=-\frac {i \sqrt {1+(a+b x)^2} \left (2 \arctan \left (e^{i \arctan (a+b x)}\right ) \arctan (a+b x)-\operatorname {PolyLog}\left (2,-i e^{i \arctan (a+b x)}\right )+\operatorname {PolyLog}\left (2,i e^{i \arctan (a+b x)}\right )\right )}{b \sqrt {c \left (1+(a+b x)^2\right )}} \]
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Time = 0.51 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.81
method | result | size |
default | \(-\frac {\left (\arctan \left (b x +a \right ) \ln \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-\arctan \left (b x +a \right ) \ln \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )\right ) \sqrt {c \left (b x +a -i\right ) \left (b x +a +i\right )}}{\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b c}\) | \(176\) |
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\[ \int \frac {\arctan (a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\int { \frac {\arctan \left (b x + a\right )}{\sqrt {b^{2} c x^{2} + 2 \, a b c x + {\left (a^{2} + 1\right )} c}} \,d x } \]
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Timed out. \[ \int \frac {\arctan (a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\text {Timed out} \]
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\[ \int \frac {\arctan (a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\int { \frac {\arctan \left (b x + a\right )}{\sqrt {b^{2} c x^{2} + 2 \, a b c x + {\left (a^{2} + 1\right )} c}} \,d x } \]
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\[ \int \frac {\arctan (a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\int { \frac {\arctan \left (b x + a\right )}{\sqrt {b^{2} c x^{2} + 2 \, a b c x + {\left (a^{2} + 1\right )} c}} \,d x } \]
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Timed out. \[ \int \frac {\arctan (a+b x)}{\sqrt {\left (1+a^2\right ) c+2 a b c x+b^2 c x^2}} \, dx=\int \frac {\mathrm {atan}\left (a+b\,x\right )}{\sqrt {c\,b^2\,x^2+2\,a\,c\,b\,x+c\,\left (a^2+1\right )}} \,d x \]
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