Optimal. Leaf size=152 \[ -\frac {i \text {Li}_2\left (1-\frac {2 b (c+d x)}{(b c-a d+i d) (1-i (a+b x))}\right )}{2 d}+\frac {\tan ^{-1}(a+b x) \log \left (\frac {2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{d}+\frac {i \text {Li}_2\left (1-\frac {2}{1-i (a+b x)}\right )}{2 d}-\frac {\log \left (\frac {2}{1-i (a+b x)}\right ) \tan ^{-1}(a+b x)}{d} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.14, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5047, 4856, 2402, 2315, 2447} \[ -\frac {i \text {PolyLog}\left (2,1-\frac {2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{2 d}+\frac {i \text {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )}{2 d}+\frac {\tan ^{-1}(a+b x) \log \left (\frac {2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{d}-\frac {\log \left (\frac {2}{1-i (a+b x)}\right ) \tan ^{-1}(a+b x)}{d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 2315
Rule 2402
Rule 2447
Rule 4856
Rule 5047
Rubi steps
\begin {align*} \int \frac {\tan ^{-1}(a+b x)}{c+d x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\tan ^{-1}(x)}{\frac {b c-a d}{b}+\frac {d x}{b}} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\tan ^{-1}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}+\frac {\tan ^{-1}(a+b x) \log \left (\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{d}+\frac {\operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,a+b x\right )}{d}-\frac {\operatorname {Subst}\left (\int \frac {\log \left (\frac {2 \left (\frac {b c-a d}{b}+\frac {d x}{b}\right )}{\left (\frac {i d}{b}+\frac {b c-a d}{b}\right ) (1-i x)}\right )}{1+x^2} \, dx,x,a+b x\right )}{d}\\ &=-\frac {\tan ^{-1}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}+\frac {\tan ^{-1}(a+b x) \log \left (\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{d}-\frac {i \text {Li}_2\left (1-\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{2 d}+\frac {i \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i (a+b x)}\right )}{d}\\ &=-\frac {\tan ^{-1}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}+\frac {\tan ^{-1}(a+b x) \log \left (\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{d}+\frac {i \text {Li}_2\left (1-\frac {2}{1-i (a+b x)}\right )}{2 d}-\frac {i \text {Li}_2\left (1-\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{2 d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.02, size = 231, normalized size = 1.52 \[ \frac {i \text {Li}_2\left (-\frac {i d (1-i (a+b x))}{b c-a d-i d}\right )}{2 d}-\frac {i \text {Li}_2\left (\frac {i d (i (a+b x)+1)}{b c-a d+i d}\right )}{2 d}+\frac {i \log (1-i (a+b x)) \log \left (-\frac {i \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{-\frac {d}{b}-\frac {i (b c-a d)}{b}}\right )}{2 d}-\frac {i \log (1+i (a+b x)) \log \left (\frac {i \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{-\frac {d}{b}+\frac {i (b c-a d)}{b}}\right )}{2 d} \]
Warning: Unable to verify antiderivative.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (b x + a\right )}{d x + c}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.06, size = 198, normalized size = 1.30 \[ \frac {\ln \left (d \left (b x +a \right )-a d +b c \right ) \arctan \left (b x +a \right )}{d}+\frac {i \ln \left (d \left (b x +a \right )-a d +b c \right ) \ln \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )}{2 d}-\frac {i \ln \left (d \left (b x +a \right )-a d +b c \right ) \ln \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )}{2 d}+\frac {i \dilog \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )}{2 d}-\frac {i \dilog \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 0.53, size = 284, normalized size = 1.87 \[ \frac {\arctan \left (b x + a\right ) \log \left (d x + c\right )}{d} - \frac {\arctan \left (\frac {b^{2} x + a b}{b}\right ) \log \left (d x + c\right )}{d} - \frac {\arctan \left (\frac {b d^{2} x + b c d}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}, \frac {b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - \arctan \left (b x + a\right ) \log \left (\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}\right ) + i \, {\rm Li}_2\left (\frac {i \, b d x + {\left (i \, a + 1\right )} d}{-i \, b c + {\left (i \, a + 1\right )} d}\right ) - i \, {\rm Li}_2\left (\frac {i \, b d x + {\left (i \, a - 1\right )} d}{-i \, b c + {\left (i \, a - 1\right )} d}\right )}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atan}\left (a+b\,x\right )}{c+d\,x} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atan}{\left (a + b x \right )}}{c + d x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________