Optimal. Leaf size=72 \[ \frac {a^2 d (d \cot (e+f x))^{n-1}}{f (1-n)}-\frac {2 a^2 d (d \cot (e+f x))^{n-1} \, _2F_1(1,n-1;n;-i \cot (e+f x))}{f (1-n)} \]
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Rubi [A] time = 0.18, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3673, 3543, 3537, 12, 64} \[ \frac {a^2 d (d \cot (e+f x))^{n-1}}{f (1-n)}-\frac {2 a^2 d (d \cot (e+f x))^{n-1} \, _2F_1(1,n-1;n;-i \cot (e+f x))}{f (1-n)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 64
Rule 3537
Rule 3543
Rule 3673
Rubi steps
\begin {align*} \int (d \cot (e+f x))^n (a+i a \tan (e+f x))^2 \, dx &=d^2 \int (d \cot (e+f x))^{-2+n} (i a+a \cot (e+f x))^2 \, dx\\ &=\frac {a^2 d (d \cot (e+f x))^{-1+n}}{f (1-n)}+d^2 \int (d \cot (e+f x))^{-2+n} \left (-2 a^2+2 i a^2 \cot (e+f x)\right ) \, dx\\ &=\frac {a^2 d (d \cot (e+f x))^{-1+n}}{f (1-n)}+\frac {\left (4 i a^4 d^2\right ) \operatorname {Subst}\left (\int \frac {2^{2-n} \left (-\frac {i d x}{a^2}\right )^{-2+n}}{-4 a^4-2 a^2 x} \, dx,x,2 i a^2 \cot (e+f x)\right )}{f}\\ &=\frac {a^2 d (d \cot (e+f x))^{-1+n}}{f (1-n)}+\frac {\left (i 2^{4-n} a^4 d^2\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {i d x}{a^2}\right )^{-2+n}}{-4 a^4-2 a^2 x} \, dx,x,2 i a^2 \cot (e+f x)\right )}{f}\\ &=\frac {a^2 d (d \cot (e+f x))^{-1+n}}{f (1-n)}-\frac {2 a^2 d (d \cot (e+f x))^{-1+n} \, _2F_1(1,-1+n;n;-i \cot (e+f x))}{f (1-n)}\\ \end {align*}
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Mathematica [B] time = 2.04, size = 198, normalized size = 2.75 \[ -\frac {e^{-2 i e} \left (1+e^{2 i (e+f x)}\right )^{-n} \left (\frac {i \left (1+e^{2 i (e+f x)}\right )}{-1+e^{2 i (e+f x)}}\right )^{n-1} \cos ^2(e+f x) (a+i a \tan (e+f x))^2 \left (2^n \left (1+e^{2 i (e+f x)}\right ) \, _2F_1\left (1-n,1-n;2-n;\frac {1}{2} \left (1-e^{2 i (e+f x)}\right )\right )-\left (1+e^{2 i (e+f x)}\right )^n\right ) \cot ^{-n}(e+f x) (d \cot (e+f x))^n}{f (n-1) (\cos (f x)+i \sin (f x))^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 2.10, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {4 \, a^{2} \left (\frac {i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} - 1}\right )^{n} e^{\left (4 i \, f x + 4 i \, e\right )}}{e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1}, x\right ) \]
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 \[ \int {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \left (d \cot \left (f x + e\right )\right )^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.84, size = 0, normalized size = 0.00 \[ \int \left (d \cot \left (f x +e \right )\right )^{n} \left (a +i a \tan \left (f x +e \right )\right )^{2}\, dx \]
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 \[ \int {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \left (d \cot \left (f x + e\right )\right )^{n}\,{d x} \]
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 \[ \int {\left (d\,\mathrm {cot}\left (e+f\,x\right )\right )}^n\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \]
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 \[ - a^{2} \left (\int \left (- \left (d \cot {\left (e + f x \right )}\right )^{n}\right )\, dx + \int \left (d \cot {\left (e + f x \right )}\right )^{n} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- 2 i \left (d \cot {\left (e + f x \right )}\right )^{n} \tan {\left (e + f x \right )}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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