Optimal. Leaf size=250 \[ \frac {i \text {Li}_3\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d^2 \log ^2(f)}-\frac {i \text {Li}_3\left (-\frac {b f^{c+d x}}{a+i}\right )}{2 d^2 \log ^2(f)}-\frac {i x \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x \text {Li}_2\left (-\frac {b f^{c+d x}}{a+i}\right )}{2 d \log (f)}-\frac {1}{4} i x^2 \log \left (1-\frac {b f^{c+d x}}{-a+i}\right )+\frac {1}{4} i x^2 \log \left (1+\frac {b f^{c+d x}}{a+i}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right ) \]
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Rubi [A] time = 2.65, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5144, 2551, 12, 6742, 2190, 2531, 2282, 6589} \[ \frac {i \text {PolyLog}\left (3,\frac {b f^{c+d x}}{-a+i}\right )}{2 d^2 \log ^2(f)}-\frac {i \text {PolyLog}\left (3,-\frac {b f^{c+d x}}{a+i}\right )}{2 d^2 \log ^2(f)}-\frac {i x \text {PolyLog}\left (2,\frac {b f^{c+d x}}{-a+i}\right )}{2 d \log (f)}+\frac {i x \text {PolyLog}\left (2,-\frac {b f^{c+d x}}{a+i}\right )}{2 d \log (f)}-\frac {1}{4} i x^2 \log \left (1-\frac {b f^{c+d x}}{-a+i}\right )+\frac {1}{4} i x^2 \log \left (1+\frac {b f^{c+d x}}{a+i}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right ) \]
Antiderivative was successfully verified.
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Rule 12
Rule 2190
Rule 2282
Rule 2531
Rule 2551
Rule 5144
Rule 6589
Rule 6742
Rubi steps
\begin {align*} \int x \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx &=\frac {1}{2} i \int x \log \left (1-\frac {i}{a+b f^{c+d x}}\right ) \, dx-\frac {1}{2} i \int x \log \left (1+\frac {i}{a+b f^{c+d x}}\right ) \, dx\\ &=\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )+\frac {1}{4} \int \frac {b d f^{c+d x} x^2 \log (f)}{\left (i (1-i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx+\frac {1}{4} \int \frac {b d f^{c+d x} x^2 \log (f)}{\left (-i (1+i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx\\ &=\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )+\frac {1}{4} (b d \log (f)) \int \frac {f^{c+d x} x^2}{\left (i (1-i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx+\frac {1}{4} (b d \log (f)) \int \frac {f^{c+d x} x^2}{\left (-i (1+i a)+b f^{c+d x}\right ) \left (a+b f^{c+d x}\right )} \, dx\\ &=\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )+\frac {1}{4} (b d \log (f)) \int \left (\frac {i f^{c+d x} x^2}{a+b f^{c+d x}}-\frac {i f^{c+d x} x^2}{-i+a+b f^{c+d x}}\right ) \, dx+\frac {1}{4} (b d \log (f)) \int \left (-\frac {i f^{c+d x} x^2}{a+b f^{c+d x}}+\frac {i f^{c+d x} x^2}{i+a+b f^{c+d x}}\right ) \, dx\\ &=\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} (i b d \log (f)) \int \frac {f^{c+d x} x^2}{-i+a+b f^{c+d x}} \, dx+\frac {1}{4} (i b d \log (f)) \int \frac {f^{c+d x} x^2}{i+a+b f^{c+d x}} \, dx\\ &=-\frac {1}{4} i x^2 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{4} i x^2 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )+\frac {1}{2} i \int x \log \left (1+\frac {b f^{c+d x}}{-i+a}\right ) \, dx-\frac {1}{2} i \int x \log \left (1+\frac {b f^{c+d x}}{i+a}\right ) \, dx\\ &=-\frac {1}{4} i x^2 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{4} i x^2 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {i x \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x \text {Li}_2\left (-\frac {b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac {i \int \text {Li}_2\left (-\frac {b f^{c+d x}}{-i+a}\right ) \, dx}{2 d \log (f)}-\frac {i \int \text {Li}_2\left (-\frac {b f^{c+d x}}{i+a}\right ) \, dx}{2 d \log (f)}\\ &=-\frac {1}{4} i x^2 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{4} i x^2 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {i x \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x \text {Li}_2\left (-\frac {b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {b x}{i-a}\right )}{x} \, dx,x,f^{c+d x}\right )}{2 d^2 \log ^2(f)}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {b x}{i+a}\right )}{x} \, dx,x,f^{c+d x}\right )}{2 d^2 \log ^2(f)}\\ &=-\frac {1}{4} i x^2 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{4} i x^2 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {i x \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x \text {Li}_2\left (-\frac {b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac {i \text {Li}_3\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d^2 \log ^2(f)}-\frac {i \text {Li}_3\left (-\frac {b f^{c+d x}}{i+a}\right )}{2 d^2 \log ^2(f)}\\ \end {align*}
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Mathematica [A] time = 0.31, size = 250, normalized size = 1.00 \[ \frac {i \text {Li}_3\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d^2 \log ^2(f)}-\frac {i \text {Li}_3\left (-\frac {b f^{c+d x}}{a+i}\right )}{2 d^2 \log ^2(f)}-\frac {i x \text {Li}_2\left (\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x \text {Li}_2\left (-\frac {b f^{c+d x}}{a+i}\right )}{2 d \log (f)}-\frac {1}{4} i x^2 \log \left (1-\frac {b f^{c+d x}}{-a+i}\right )+\frac {1}{4} i x^2 \log \left (1+\frac {b f^{c+d x}}{a+i}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right ) \]
Antiderivative was successfully verified.
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fricas [C] time = 0.70, size = 304, normalized size = 1.22 \[ \frac {2 \, d^{2} x^{2} \operatorname {arccot}\left (b f^{d x + c} + a\right ) \log \relax (f)^{2} + i \, c^{2} \log \left (b f^{d x + c} + a + i\right ) \log \relax (f)^{2} - i \, c^{2} \log \left (b f^{d x + c} + a - i\right ) \log \relax (f)^{2} - 2 i \, d x {\rm Li}_2\left (-\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \relax (f) + 2 i \, d x {\rm Li}_2\left (-\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \relax (f) + {\left (-i \, d^{2} x^{2} + i \, c^{2}\right )} \log \relax (f)^{2} \log \left (\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + {\left (i \, d^{2} x^{2} - i \, c^{2}\right )} \log \relax (f)^{2} \log \left (\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + 2 i \, {\rm polylog}\left (3, -\frac {{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) - 2 i \, {\rm polylog}\left (3, -\frac {{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right )}{4 \, d^{2} \log \relax (f)^{2}} \]
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 \[ \int x \operatorname {arccot}\left (b f^{d x + c} + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.48, size = 678, normalized size = 2.71 \[ \frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c^{2}}{4 d^{2}}+\frac {\pi \,x^{2}}{4}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x c}{2 d}+\frac {i \polylog \left (3, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right )}{2 d^{2} \ln \relax (f )^{2}}-\frac {i c^{2} \ln \left (i f^{d x} f^{c} b +i a +1\right )}{4 d^{2}}-\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x}{2 d \ln \relax (f )}+\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c}{2 d^{2} \ln \relax (f )}+\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x}{2 d \ln \relax (f )}-\frac {i c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+i+a}{i+a}\right )}{2 d^{2}}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x^{2}}{4}-\frac {i \polylog \left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c}{2 d^{2} \ln \relax (f )}+\frac {i c \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{-i+a}\right ) x}{2 d}+\frac {i c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{-i+a}\right )}{2 d^{2}}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x^{2}}{4}-\frac {i \polylog \left (3, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right )}{2 d^{2} \ln \relax (f )^{2}}+\frac {i c \dilog \left (\frac {b \,f^{d x} f^{c}+a -i}{-i+a}\right )}{2 d^{2} \ln \relax (f )}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x c}{2 d}+\frac {i c^{2} \ln \left (1-i a -i f^{d x} f^{c} b \right )}{4 d^{2}}+\frac {i x^{2} \ln \left (1+i \left (a +b \,f^{d x +c}\right )\right )}{4}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c^{2}}{4 d^{2}}-\frac {i c \ln \left (\frac {b \,f^{d x} f^{c}+i+a}{i+a}\right ) x}{2 d}-\frac {i c \dilog \left (\frac {b \,f^{d x} f^{c}+i+a}{i+a}\right )}{2 d^{2} \ln \relax (f )}-\frac {i x^{2} \ln \left (1-i \left (a +b \,f^{d x +c}\right )\right )}{4} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ b d f^{c} \int \frac {f^{d x} x^{2}}{2 \, {\left (b^{2} f^{2 \, d x} f^{2 \, c} + 2 \, a b f^{d x} f^{c} + a^{2} + 1\right )}}\,{d x} \log \relax (f) + \frac {1}{2} \, x^{2} \arctan \left (\frac {1}{b f^{d x} f^{c} + a}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,\mathrm {acot}\left (a+b\,f^{c+d\,x}\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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