Optimal. Leaf size=376 \[ -\frac {(B+i (A-C)) F_1\left (1+m;-n,1;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n}}{2 (a-i b) f (1+m)}-\frac {(A+i B-C) F_1\left (1+m;-n,1;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n}}{2 (i a-b) f (1+m)}+\frac {C \, _2F_1\left (1+m,-n;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n}}{b f (1+m)} \]
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Rubi [A]
time = 0.62, antiderivative size = 376, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3736, 6857, 72,
71, 142, 141} \begin {gather*} -\frac {(B+i (A-C)) (a+b \tan (e+f x))^{m+1} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (a-i b)}-\frac {(A+i B-C) (a+b \tan (e+f x))^{m+1} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a)}+\frac {C (a+b \tan (e+f x))^{m+1} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right )}{b f (m+1)} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 141
Rule 142
Rule 3736
Rule 6857
Rubi steps
\begin {align*} \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^n \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac {\text {Subst}\left (\int \frac {(a+b x)^m (c+d x)^n \left (A+B x+C x^2\right )}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \left (C (a+b x)^m (c+d x)^n+\frac {(-B+i (A-C)) (a+b x)^m (c+d x)^n}{2 (i-x)}+\frac {(B+i (A-C)) (a+b x)^m (c+d x)^n}{2 (i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(-B+i (A-C)) \text {Subst}\left (\int \frac {(a+b x)^m (c+d x)^n}{i-x} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {(B+i (A-C)) \text {Subst}\left (\int \frac {(a+b x)^m (c+d x)^n}{i+x} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {C \text {Subst}\left (\int (a+b x)^m (c+d x)^n \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\left ((-B+i (A-C)) (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n}\right ) \text {Subst}\left (\int \frac {(a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n}{i-x} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {\left ((B+i (A-C)) (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n}\right ) \text {Subst}\left (\int \frac {(a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n}{i+x} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac {\left (C (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n}\right ) \text {Subst}\left (\int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(B+i (A-C)) F_1\left (1+m;-n,1;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n}}{2 (a-i b) f (1+m)}-\frac {(A+i B-C) F_1\left (1+m;-n,1;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n}}{2 (i a-b) f (1+m)}+\frac {C \, _2F_1\left (1+m,-n;2+m;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n}}{b f (1+m)}\\ \end {align*}
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Mathematica [F]
time = 16.61, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^n \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.37, size = 0, normalized size = 0.00 \[\int \left (a +b \tan \left (f x +e \right )\right )^{m} \left (c +d \tan \left (f x +e \right )\right )^{n} \left (A +B \tan \left (f x +e \right )+C \left (\tan ^{2}\left (f x +e \right )\right )\right )\, dx\]
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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \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 = -0.00 \begin {gather*} \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,\left (C\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,\mathrm {tan}\left (e+f\,x\right )+A\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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