Optimal. Leaf size=189 \[ -\frac {2 a^4 \left (16+11 n+2 n^2\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n) (3+n)}+\frac {8 a^4 \, _2F_1(1,1+n;2+n;i \tan (e+f x)) (d \tan (e+f x))^{1+n}}{d f (1+n)}-\frac {(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}-\frac {2 (4+n) (d \tan (e+f x))^{1+n} \left (a^4+i a^4 \tan (e+f x)\right )}{d f (2+n) (3+n)} \]
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Rubi [A]
time = 0.37, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3637, 3675,
3673, 3618, 12, 66} \begin {gather*} \frac {8 a^4 (d \tan (e+f x))^{n+1} \, _2F_1(1,n+1;n+2;i \tan (e+f x))}{d f (n+1)}-\frac {2 a^4 \left (2 n^2+11 n+16\right ) (d \tan (e+f x))^{n+1}}{d f (n+1) (n+2) (n+3)}-\frac {2 (n+4) \left (a^4+i a^4 \tan (e+f x)\right ) (d \tan (e+f x))^{n+1}}{d f (n+2) (n+3)}-\frac {\left (a^2+i a^2 \tan (e+f x)\right )^2 (d \tan (e+f x))^{n+1}}{d f (n+3)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 66
Rule 3618
Rule 3637
Rule 3673
Rule 3675
Rubi steps
\begin {align*} \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^4 \, dx &=-\frac {(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}+\frac {a \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^2 (2 a d (2+n)+2 i a d (4+n) \tan (e+f x)) \, dx}{d (3+n)}\\ &=-\frac {(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}-\frac {2 (4+n) (d \tan (e+f x))^{1+n} \left (a^4+i a^4 \tan (e+f x)\right )}{d f (2+n) (3+n)}+\frac {a \int (d \tan (e+f x))^n (a+i a \tan (e+f x)) \left (2 a^2 d^2 \left (8+9 n+2 n^2\right )+2 i a^2 d^2 \left (16+11 n+2 n^2\right ) \tan (e+f x)\right ) \, dx}{d^2 (2+n) (3+n)}\\ &=-\frac {2 a^4 \left (16+11 n+2 n^2\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n) (3+n)}-\frac {(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}-\frac {2 (4+n) (d \tan (e+f x))^{1+n} \left (a^4+i a^4 \tan (e+f x)\right )}{d f (2+n) (3+n)}+\frac {a \int (d \tan (e+f x))^n \left (8 a^3 d^2 (2+n) (3+n)+8 i a^3 d^2 (2+n) (3+n) \tan (e+f x)\right ) \, dx}{d^2 (2+n) (3+n)}\\ &=-\frac {2 a^4 \left (16+11 n+2 n^2\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n) (3+n)}-\frac {(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}-\frac {2 (4+n) (d \tan (e+f x))^{1+n} \left (a^4+i a^4 \tan (e+f x)\right )}{d f (2+n) (3+n)}+\frac {\left (64 i a^7 d^2 (2+n) (3+n)\right ) \text {Subst}\left (\int \frac {8^{-n} \left (-\frac {i x}{a^3 d (2+n) (3+n)}\right )^n}{-64 a^6 d^4 (2+n)^2 (3+n)^2+8 a^3 d^2 (2+n) (3+n) x} \, dx,x,8 i a^3 d^2 (2+n) (3+n) \tan (e+f x)\right )}{f}\\ &=-\frac {2 a^4 \left (16+11 n+2 n^2\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n) (3+n)}-\frac {(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}-\frac {2 (4+n) (d \tan (e+f x))^{1+n} \left (a^4+i a^4 \tan (e+f x)\right )}{d f (2+n) (3+n)}+\frac {\left (i 8^{2-n} a^7 d^2 (2+n) (3+n)\right ) \text {Subst}\left (\int \frac {\left (-\frac {i x}{a^3 d (2+n) (3+n)}\right )^n}{-64 a^6 d^4 (2+n)^2 (3+n)^2+8 a^3 d^2 (2+n) (3+n) x} \, dx,x,8 i a^3 d^2 (2+n) (3+n) \tan (e+f x)\right )}{f}\\ &=-\frac {2 a^4 \left (16+11 n+2 n^2\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n) (3+n)}+\frac {8 a^4 \, _2F_1(1,1+n;2+n;i \tan (e+f x)) (d \tan (e+f x))^{1+n}}{d f (1+n)}-\frac {(d \tan (e+f x))^{1+n} \left (a^2+i a^2 \tan (e+f x)\right )^2}{d f (3+n)}-\frac {2 (4+n) (d \tan (e+f x))^{1+n} \left (a^4+i a^4 \tan (e+f x)\right )}{d f (2+n) (3+n)}\\ \end {align*}
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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice
the leaf count of optimal. \(1065\) vs. \(2(189)=378\).
time = 8.95, size = 1065, normalized size = 5.63 \begin {gather*} \frac {\cos ^4(e+f x) \left (\frac {\sec ^2(e+f x) (-4 i \cos (4 e)-4 \sin (4 e))}{2+n}+\frac {(-3-2 n+\cos (2 e)) \sec ^2(e) (-2 i \cos (4 e)-2 \sin (4 e))}{(1+n) (2+n)}+\frac {(-\cos (e-f x)+\cos (e+f x)) \sec ^2(e) \sec (e+f x) (-2 i \cos (4 e)-2 \sin (4 e))}{1+n}\right ) (d \tan (e+f x))^n (a+i a \tan (e+f x))^4}{f (\cos (f x)+i \sin (f x))^4}+\frac {\cos ^4(e+f x) \left (\frac {\sec ^2(e) (-1+\cos (2 e)+2 i \sin (2 e)) (2 i \cos (4 e)+2 \sin (4 e))}{1+n}+\frac {\sec ^2(e) \sec (e+f x) (2 i \cos (4 e)+2 \sin (4 e)) (-\cos (e-f x)+\cos (e+f x)-2 i \sin (e-f x)+2 i \sin (e+f x))}{1+n}\right ) (d \tan (e+f x))^n (a+i a \tan (e+f x))^4}{f (\cos (f x)+i \sin (f x))^4}+\frac {\cos ^4(e+f x) \left (\frac {\sec (e) \sec ^3(e+f x) (\cos (4 e)-i \sin (4 e)) \sin (f x)}{3+n}+\frac {\sec (e) \sec (e+f x) (2 \cos (4 e)-2 i \sin (4 e)) \sin (f x)}{(1+n) (3+n)}+\frac {\sec ^2(e+f x) (\cos (4 e)-i \sin (4 e)) \tan (e)}{3+n}+\frac {(2 \cos (4 e)-2 i \sin (4 e)) \tan (e)}{(1+n) (3+n)}\right ) (d \tan (e+f x))^n (a+i a \tan (e+f x))^4}{f (\cos (f x)+i \sin (f x))^4}+\frac {i 2^{3-n} \left (-\frac {i \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )^n \cos ^4(e+f x) \left (2^n \, _2F_1\left (1,n;1+n;-\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}\right )-\left (1+e^{2 i (e+f x)}\right )^n \, _2F_1\left (n,n;1+n;\frac {1}{2} \left (1-e^{2 i (e+f x)}\right )\right )\right ) \tan ^{-n}(e+f x) (d \tan (e+f x))^n (a+i a \tan (e+f x))^4}{\left (e^{2 i e}+e^{4 i e}\right ) f n (\cos (f x)+i \sin (f x))^4}-\frac {8 i e^{-4 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 \left (\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{-n} \cos ^4(e+f x) \left (-\frac {\left (1+e^{2 i (e+f x)}\right )^{-n} \, _2F_1\left (1,n;1+n;\frac {1-e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}\right )}{n}-\frac {\left (1+e^{2 i e}\right ) \left (-1+e^{2 i (e+f x)}\right ) \left (1+e^{2 i (e+f x)}\right )^{-1-n} \, _2F_1\left (1,1+n;2+n;\frac {1-e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}\right )}{1+n}+\frac {2^{-n} \, _2F_1\left (n,n;1+n;\frac {1}{2} \left (1-e^{2 i (e+f x)}\right )\right )}{n}\right ) \tan ^{-n}(e+f x) (d \tan (e+f x))^n (a+i a \tan (e+f x))^4}{\left (1+e^{2 i e}\right ) f (\cos (f x)+i \sin (f x))^4} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 1.12, size = 0, normalized size = 0.00 \[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +i a \tan \left (f x +e \right )\right )^{4}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} a^{4} \left (\int \left (d \tan {\left (e + f x \right )}\right )^{n}\, dx + \int \left (- 6 \left (d \tan {\left (e + f x \right )}\right )^{n} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{n} \tan ^{4}{\left (e + f x \right )}\, dx + \int 4 i \left (d \tan {\left (e + f x \right )}\right )^{n} \tan {\left (e + f x \right )}\, dx + \int \left (- 4 i \left (d \tan {\left (e + f x \right )}\right )^{n} \tan ^{3}{\left (e + f x \right )}\right )\, dx\right ) \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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