Optimal. Leaf size=376 \[ -\frac {(B+i (A-C)) (a+b \tan (e+f x))^{m+1} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (a-i b)}-\frac {(A+i B-C) (a+b \tan (e+f x))^{m+1} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a)}+\frac {C (a+b \tan (e+f x))^{m+1} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right )}{b f (m+1)} \]
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Rubi [A] time = 0.90, 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 = {3655, 6725, 70, 69, 137, 136} \[ -\frac {(B+i (A-C)) (a+b \tan (e+f x))^{m+1} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (a-i b)}-\frac {(A+i B-C) (a+b \tan (e+f x))^{m+1} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a)}+\frac {C (a+b \tan (e+f x))^{m+1} (c+d \tan (e+f x))^n \left (\frac {b (c+d \tan (e+f x))}{b c-a d}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;-\frac {d (a+b \tan (e+f x))}{b c-a d}\right )}{b f (m+1)} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 136
Rule 137
Rule 3655
Rule 6725
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 {\operatorname {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 {\operatorname {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)) \operatorname {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)) \operatorname {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 \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 25.50, size = 0, normalized size = 0.00 \[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^n \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 1.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m} {\left (d \tan \left (f x + e\right ) + c\right )}^{n}, x\right ) \]
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 \[ \int {\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m} {\left (d \tan \left (f x + e\right ) + c\right )}^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 4.34, 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 \[ \int {\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m} {\left (d \tan \left (f x + e\right ) + c\right )}^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,\left (C\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,\mathrm {tan}\left (e+f\,x\right )+A\right ) \,d x \]
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 \[ \text {Exception raised: HeuristicGCDFailed} \]
Verification of antiderivative is not currently implemented for this CAS.
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