Optimal. Leaf size=126 \[ -\frac {b (c+d x) \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{f \left (a^2+b^2\right )}+\frac {i b d \text {Li}_2\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f^2 \left (a^2+b^2\right )}+\frac {(c+d x)^2}{2 d (a-i b)} \]
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Rubi [A] time = 0.16, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3731, 2190, 2279, 2391} \[ \frac {i b d \text {PolyLog}\left (2,\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f^2 \left (a^2+b^2\right )}-\frac {b (c+d x) \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{f \left (a^2+b^2\right )}+\frac {(c+d x)^2}{2 d (a-i b)} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2391
Rule 3731
Rubi steps
\begin {align*} \int \frac {c+d x}{a+b \cot (e+f x)} \, dx &=\frac {(c+d x)^2}{2 (a-i b) d}+(2 i b) \int \frac {e^{2 i (e+f x)} (c+d x)}{(a-i b)^2+\left (-a^2-b^2\right ) e^{2 i (e+f x)}} \, dx\\ &=\frac {(c+d x)^2}{2 (a-i b) d}-\frac {b (c+d x) \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{\left (a^2+b^2\right ) f}+\frac {(b d) \int \log \left (1+\frac {\left (-a^2-b^2\right ) e^{2 i (e+f x)}}{(a-i b)^2}\right ) \, dx}{\left (a^2+b^2\right ) f}\\ &=\frac {(c+d x)^2}{2 (a-i b) d}-\frac {b (c+d x) \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{\left (a^2+b^2\right ) f}-\frac {(i b d) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {\left (-a^2-b^2\right ) x}{(a-i b)^2}\right )}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 \left (a^2+b^2\right ) f^2}\\ &=\frac {(c+d x)^2}{2 (a-i b) d}-\frac {b (c+d x) \log \left (1-\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{\left (a^2+b^2\right ) f}+\frac {i b d \text {Li}_2\left (\frac {(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 \left (a^2+b^2\right ) f^2}\\ \end {align*}
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Mathematica [A] time = 1.86, size = 182, normalized size = 1.44 \[ \frac {x \sin (e) (2 c+d x)}{2 (a \sin (e)+b \cos (e))}+\frac {1}{2} b \left (-\frac {2 (c+d x) \log \left (1+\frac {(-a+i b) e^{-2 i (e+f x)}}{a+i b}\right )}{f \left (a^2+b^2\right )}-\frac {i d \text {Li}_2\left (\frac {(a-i b) e^{-2 i (e+f x)}}{a+i b}\right )}{f^2 \left (a^2+b^2\right )}+\frac {2 i (c+d x)^2}{d (a+i b) \left (a \left (-1+e^{2 i e}\right )+i b \left (1+e^{2 i e}\right )\right )}\right ) \]
Antiderivative was successfully verified.
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fricas [B] time = 0.66, size = 475, normalized size = 3.77 \[ \frac {2 \, a d f^{2} x^{2} + 4 \, a c f^{2} x + i \, b d {\rm Li}_2\left (-\frac {a^{2} + b^{2} - {\left (a^{2} + 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (-i \, a^{2} + 2 \, a b + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}} + 1\right ) - i \, b d {\rm Li}_2\left (-\frac {a^{2} + b^{2} - {\left (a^{2} - 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (i \, a^{2} + 2 \, a b - i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}} + 1\right ) + 2 \, {\left (b d e - b c f\right )} \log \left (\frac {1}{2} \, a^{2} + i \, a b - \frac {1}{2} \, b^{2} - \frac {1}{2} \, {\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + \frac {1}{2} \, {\left (i \, a^{2} + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )\right ) + 2 \, {\left (b d e - b c f\right )} \log \left (-\frac {1}{2} \, a^{2} + i \, a b + \frac {1}{2} \, b^{2} + \frac {1}{2} \, {\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + \frac {1}{2} \, {\left (i \, a^{2} + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )\right ) - 2 \, {\left (b d f x + b d e\right )} \log \left (\frac {a^{2} + b^{2} - {\left (a^{2} + 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (-i \, a^{2} + 2 \, a b + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - 2 \, {\left (b d f x + b d e\right )} \log \left (\frac {a^{2} + b^{2} - {\left (a^{2} - 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (i \, a^{2} + 2 \, a b - i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right )}{4 \, {\left (a^{2} + b^{2}\right )} f^{2}} \]
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 \[ \int \frac {d x + c}{b \cot \left (f x + e\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 2.25, size = 445, normalized size = 3.53 \[ \frac {d \,x^{2}}{2 i b +2 a}+\frac {c x}{i b +a}-\frac {2 b c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (i b +a \right ) \left (i b -a \right )}+\frac {b c \ln \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+i {\mathrm e}^{2 i \left (f x +e \right )} b -a +i b \right )}{f \left (i b +a \right ) \left (i b -a \right )}-\frac {b d \ln \left (1-\frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right ) x}{f \left (i b +a \right ) \left (-i b +a \right )}-\frac {b d \ln \left (1-\frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right ) e}{f^{2} \left (i b +a \right ) \left (-i b +a \right )}+\frac {i b d \,x^{2}}{\left (i b +a \right ) \left (-i b +a \right )}+\frac {2 i b d e x}{f \left (i b +a \right ) \left (-i b +a \right )}+\frac {i b d \,e^{2}}{f^{2} \left (i b +a \right ) \left (-i b +a \right )}+\frac {i b d \polylog \left (2, \frac {\left (i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b +a}\right )}{2 f^{2} \left (i b +a \right ) \left (-i b +a \right )}+\frac {2 b d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2} \left (i b +a \right ) \left (i b -a \right )}-\frac {b d e \ln \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+i {\mathrm e}^{2 i \left (f x +e \right )} b -a +i b \right )}{f^{2} \left (i b +a \right ) \left (i b -a \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.67, size = 406, normalized size = 3.22 \[ \frac {{\left (a + i \, b\right )} d f^{2} x^{2} + 2 \, {\left (a + i \, b\right )} c f^{2} x - 2 i \, b d f x \arctan \left (-\frac {2 \, a b \cos \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} - b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, f x + 2 \, e\right ) + a^{2} + b^{2} - {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - b d f x \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} - 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - 2 i \, b c f \arctan \left (b \cos \left (2 \, f x + 2 \, e\right ) + a \sin \left (2 \, f x + 2 \, e\right ) + b, a \cos \left (2 \, f x + 2 \, e\right ) - b \sin \left (2 \, f x + 2 \, e\right ) - a\right ) - b c f \log \left ({\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} - 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right ) + i \, b d {\rm Li}_2\left (\frac {{\left (i \, a - b\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{i \, a + b}\right )}{2 \, {\left (a^{2} + b^{2}\right )} f^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {c+d\,x}{a+b\,\mathrm {cot}\left (e+f\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {c + d x}{a + b \cot {\left (e + f x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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