Optimal. Leaf size=157 \[ \frac {a^3 (i c-d (5+2 n)) (c+d \tan (e+f x))^{1+n}}{d^2 f (1+n) (2+n)}+\frac {4 a^3 \, _2F_1\left (1,1+n;2+n;\frac {c+d \tan (e+f x)}{c-i d}\right ) (c+d \tan (e+f x))^{1+n}}{(i c+d) f (1+n)}-\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{1+n}}{d f (2+n)} \]
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Rubi [A]
time = 0.24, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3637, 3673,
3618, 70} \begin {gather*} \frac {a^3 (-d (2 n+5)+i c) (c+d \tan (e+f x))^{n+1}}{d^2 f (n+1) (n+2)}+\frac {4 a^3 (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {c+d \tan (e+f x)}{c-i d}\right )}{f (n+1) (d+i c)}-\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{n+1}}{d f (n+2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 70
Rule 3618
Rule 3637
Rule 3673
Rubi steps
\begin {align*} \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^n \, dx &=-\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{1+n}}{d f (2+n)}+\frac {a \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^n (a (i c+d (3+2 n))+a (c+i d (5+2 n)) \tan (e+f x)) \, dx}{d (2+n)}\\ &=\frac {a^3 (i c-d (5+2 n)) (c+d \tan (e+f x))^{1+n}}{d^2 f (1+n) (2+n)}-\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{1+n}}{d f (2+n)}+\frac {a \int (c+d \tan (e+f x))^n \left (4 a^2 d (2+n)+4 i a^2 d (2+n) \tan (e+f x)\right ) \, dx}{d (2+n)}\\ &=\frac {a^3 (i c-d (5+2 n)) (c+d \tan (e+f x))^{1+n}}{d^2 f (1+n) (2+n)}-\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{1+n}}{d f (2+n)}+\frac {\left (16 i a^5 d (2+n)\right ) \text {Subst}\left (\int \frac {\left (c-\frac {i x}{4 a^2 (2+n)}\right )^n}{-16 a^4 d^2 (2+n)^2+4 a^2 d (2+n) x} \, dx,x,4 i a^2 d (2+n) \tan (e+f x)\right )}{f}\\ &=\frac {a^3 (i c-d (5+2 n)) (c+d \tan (e+f x))^{1+n}}{d^2 f (1+n) (2+n)}+\frac {4 a^3 \, _2F_1\left (1,1+n;2+n;\frac {c+d \tan (e+f x)}{c-i d}\right ) (c+d \tan (e+f x))^{1+n}}{(i c+d) f (1+n)}-\frac {\left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{1+n}}{d f (2+n)}\\ \end {align*}
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Mathematica [F]
time = 16.17, size = 0, normalized size = 0.00 \begin {gather*} \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^n \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 2.93, size = 0, normalized size = 0.00 \[\int \left (a +i a \tan \left (f x +e \right )\right )^{3} \left (c +d \tan \left (f x +e \right )\right )^{n}\, dx\]
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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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 (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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