Optimal. Leaf size=353 \[ \frac {b \tan (e+f x) \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{f}-\frac {\log (\cos (e+f x)) \left (a^3 (d (A-C)+B c)+3 a^2 b (A c-B d-c C)-3 a b^2 (d (A-C)+B c)-b^3 (A c-B d-c C)\right )}{f}+x \left (a^3 (A c-B d-c C)-3 a^2 b (d (A-C)+B c)-3 a b^2 (A c-B d-c C)+b^3 (d (A-C)+B c)\right )+\frac {(d (A-C)+B c) (a+b \tan (e+f x))^3}{3 f}+\frac {(a+b \tan (e+f x))^2 (a A d+a B c-a C d+A b c-b B d-b c C)}{2 f}-\frac {(a C d-5 b (B d+c C)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac {C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f} \]
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Rubi [A] time = 0.79, antiderivative size = 353, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3637, 3630, 3528, 3525, 3475} \[ \frac {b \tan (e+f x) \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{f}-\frac {\log (\cos (e+f x)) \left (3 a^2 b (A c-B d-c C)+a^3 (d (A-C)+B c)-3 a b^2 (d (A-C)+B c)-b^3 (A c-B d-c C)\right )}{f}+x \left (-3 a^2 b (d (A-C)+B c)+a^3 (A c-B d-c C)-3 a b^2 (A c-B d-c C)+b^3 (d (A-C)+B c)\right )+\frac {(d (A-C)+B c) (a+b \tan (e+f x))^3}{3 f}+\frac {(a+b \tan (e+f x))^2 (a A d+a B c-a C d+A b c-b B d-b c C)}{2 f}-\frac {(a C d-5 b (B d+c C)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac {C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f} \]
Antiderivative was successfully verified.
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Rule 3475
Rule 3525
Rule 3528
Rule 3630
Rule 3637
Rubi steps
\begin {align*} \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac {C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\frac {\int (a+b \tan (e+f x))^3 \left (-5 A b c+a C d-5 b (B c+(A-C) d) \tan (e+f x)+(a C d-5 b (c C+B d)) \tan ^2(e+f x)\right ) \, dx}{5 b}\\ &=-\frac {(a C d-5 b (c C+B d)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac {C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\frac {\int (a+b \tan (e+f x))^3 (-5 b (A c-c C-B d)-5 b (B c+(A-C) d) \tan (e+f x)) \, dx}{5 b}\\ &=\frac {(B c+(A-C) d) (a+b \tan (e+f x))^3}{3 f}-\frac {(a C d-5 b (c C+B d)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac {C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\frac {\int (a+b \tan (e+f x))^2 (5 b (b B c+b (A-C) d-a (A c-c C-B d))-5 b (A b c+a B c-b c C+a A d-b B d-a C d) \tan (e+f x)) \, dx}{5 b}\\ &=\frac {(A b c+a B c-b c C+a A d-b B d-a C d) (a+b \tan (e+f x))^2}{2 f}+\frac {(B c+(A-C) d) (a+b \tan (e+f x))^3}{3 f}-\frac {(a C d-5 b (c C+B d)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac {C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\frac {\int (a+b \tan (e+f x)) \left (-5 b \left (a^2 (A c-c C-B d)-b^2 (A c-c C-B d)-2 a b (B c+(A-C) d)\right )-5 b \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \tan (e+f x)\right ) \, dx}{5 b}\\ &=\left (a^3 (A c-c C-B d)-3 a b^2 (A c-c C-B d)-3 a^2 b (B c+(A-C) d)+b^3 (B c+(A-C) d)\right ) x+\frac {b \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \tan (e+f x)}{f}+\frac {(A b c+a B c-b c C+a A d-b B d-a C d) (a+b \tan (e+f x))^2}{2 f}+\frac {(B c+(A-C) d) (a+b \tan (e+f x))^3}{3 f}-\frac {(a C d-5 b (c C+B d)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac {C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\left (-3 a^2 b (A c-c C-B d)+b^3 (A c-c C-B d)-a^3 (B c+(A-C) d)+3 a b^2 (B c+(A-C) d)\right ) \int \tan (e+f x) \, dx\\ &=\left (a^3 (A c-c C-B d)-3 a b^2 (A c-c C-B d)-3 a^2 b (B c+(A-C) d)+b^3 (B c+(A-C) d)\right ) x-\frac {\left (3 a^2 b (A c-c C-B d)-b^3 (A c-c C-B d)+a^3 (B c+(A-C) d)-3 a b^2 (B c+(A-C) d)\right ) \log (\cos (e+f x))}{f}+\frac {b \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \tan (e+f x)}{f}+\frac {(A b c+a B c-b c C+a A d-b B d-a C d) (a+b \tan (e+f x))^2}{2 f}+\frac {(B c+(A-C) d) (a+b \tan (e+f x))^3}{3 f}-\frac {(a C d-5 b (c C+B d)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac {C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}\\ \end {align*}
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Mathematica [C] time = 6.38, size = 300, normalized size = 0.85 \[ \frac {C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\frac {\frac {(a C d-5 b (B d+c C)) (a+b \tan (e+f x))^4}{4 b f}-\frac {5 \left (3 (-a A d-a B c+a C d+A b c-b B d-b c C) \left (6 a b^2 \tan (e+f x)+(-b+i a)^3 \log (-\tan (e+f x)+i)-(b+i a)^3 \log (\tan (e+f x)+i)+b^3 \tan ^2(e+f x)\right )-(d (A-C)+B c) \left (-6 b^2 \left (6 a^2-b^2\right ) \tan (e+f x)-12 a b^3 \tan ^2(e+f x)-3 i (a-i b)^4 \log (\tan (e+f x)+i)+3 i (a+i b)^4 \log (-\tan (e+f x)+i)-2 b^4 \tan ^3(e+f x)\right )\right )}{6 f}}{5 b} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.04, size = 415, normalized size = 1.18 \[ \frac {12 \, C b^{3} d \tan \left (f x + e\right )^{5} + 15 \, {\left (C b^{3} c + {\left (3 \, C a b^{2} + B b^{3}\right )} d\right )} \tan \left (f x + e\right )^{4} + 20 \, {\left ({\left (3 \, C a b^{2} + B b^{3}\right )} c + {\left (3 \, C a^{2} b + 3 \, B a b^{2} + {\left (A - C\right )} b^{3}\right )} d\right )} \tan \left (f x + e\right )^{3} + 60 \, {\left ({\left ({\left (A - C\right )} a^{3} - 3 \, B a^{2} b - 3 \, {\left (A - C\right )} a b^{2} + B b^{3}\right )} c - {\left (B a^{3} + 3 \, {\left (A - C\right )} a^{2} b - 3 \, B a b^{2} - {\left (A - C\right )} b^{3}\right )} d\right )} f x + 30 \, {\left ({\left (3 \, C a^{2} b + 3 \, B a b^{2} + {\left (A - C\right )} b^{3}\right )} c + {\left (C a^{3} + 3 \, B a^{2} b + 3 \, {\left (A - C\right )} a b^{2} - B b^{3}\right )} d\right )} \tan \left (f x + e\right )^{2} - 30 \, {\left ({\left (B a^{3} + 3 \, {\left (A - C\right )} a^{2} b - 3 \, B a b^{2} - {\left (A - C\right )} b^{3}\right )} c + {\left ({\left (A - C\right )} a^{3} - 3 \, B a^{2} b - 3 \, {\left (A - C\right )} a b^{2} + B b^{3}\right )} d\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 60 \, {\left ({\left (C a^{3} + 3 \, B a^{2} b + 3 \, {\left (A - C\right )} a b^{2} - B b^{3}\right )} c + {\left (B a^{3} + 3 \, {\left (A - C\right )} a^{2} b - 3 \, B a b^{2} - {\left (A - C\right )} b^{3}\right )} d\right )} \tan \left (f x + e\right )}{60 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.03, size = 994, normalized size = 2.82 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 416, normalized size = 1.18 \[ \frac {12 \, C b^{3} d \tan \left (f x + e\right )^{5} + 15 \, {\left (C b^{3} c + {\left (3 \, C a b^{2} + B b^{3}\right )} d\right )} \tan \left (f x + e\right )^{4} + 20 \, {\left ({\left (3 \, C a b^{2} + B b^{3}\right )} c + {\left (3 \, C a^{2} b + 3 \, B a b^{2} + {\left (A - C\right )} b^{3}\right )} d\right )} \tan \left (f x + e\right )^{3} + 30 \, {\left ({\left (3 \, C a^{2} b + 3 \, B a b^{2} + {\left (A - C\right )} b^{3}\right )} c + {\left (C a^{3} + 3 \, B a^{2} b + 3 \, {\left (A - C\right )} a b^{2} - B b^{3}\right )} d\right )} \tan \left (f x + e\right )^{2} + 60 \, {\left ({\left ({\left (A - C\right )} a^{3} - 3 \, B a^{2} b - 3 \, {\left (A - C\right )} a b^{2} + B b^{3}\right )} c - {\left (B a^{3} + 3 \, {\left (A - C\right )} a^{2} b - 3 \, B a b^{2} - {\left (A - C\right )} b^{3}\right )} d\right )} {\left (f x + e\right )} + 30 \, {\left ({\left (B a^{3} + 3 \, {\left (A - C\right )} a^{2} b - 3 \, B a b^{2} - {\left (A - C\right )} b^{3}\right )} c + {\left ({\left (A - C\right )} a^{3} - 3 \, B a^{2} b - 3 \, {\left (A - C\right )} a b^{2} + B b^{3}\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + 60 \, {\left ({\left (C a^{3} + 3 \, B a^{2} b + 3 \, {\left (A - C\right )} a b^{2} - B b^{3}\right )} c + {\left (B a^{3} + 3 \, {\left (A - C\right )} a^{2} b - 3 \, B a b^{2} - {\left (A - C\right )} b^{3}\right )} d\right )} \tan \left (f x + e\right )}{60 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.00, size = 477, normalized size = 1.35 \[ x\,\left (A\,a^3\,c+A\,b^3\,d-B\,a^3\,d+B\,b^3\,c-C\,a^3\,c-C\,b^3\,d-3\,A\,a\,b^2\,c-3\,A\,a^2\,b\,d-3\,B\,a^2\,b\,c+3\,B\,a\,b^2\,d+3\,C\,a\,b^2\,c+3\,C\,a^2\,b\,d\right )+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (\frac {B\,b^3\,d}{4}+\frac {C\,b^3\,c}{4}+\frac {3\,C\,a\,b^2\,d}{4}\right )}{f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {A\,b^3\,d}{3}+\frac {B\,b^3\,c}{3}-\frac {C\,b^3\,d}{3}+B\,a\,b^2\,d+C\,a\,b^2\,c+C\,a^2\,b\,d\right )}{f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {A\,b^3\,c}{2}-\frac {B\,b^3\,d}{2}+\frac {C\,a^3\,d}{2}-\frac {C\,b^3\,c}{2}+\frac {3\,A\,a\,b^2\,d}{2}+\frac {3\,B\,a\,b^2\,c}{2}+\frac {3\,B\,a^2\,b\,d}{2}+\frac {3\,C\,a^2\,b\,c}{2}-\frac {3\,C\,a\,b^2\,d}{2}\right )}{f}+\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {A\,a^3\,d}{2}-\frac {A\,b^3\,c}{2}+\frac {B\,a^3\,c}{2}+\frac {B\,b^3\,d}{2}-\frac {C\,a^3\,d}{2}+\frac {C\,b^3\,c}{2}+\frac {3\,A\,a^2\,b\,c}{2}-\frac {3\,A\,a\,b^2\,d}{2}-\frac {3\,B\,a\,b^2\,c}{2}-\frac {3\,B\,a^2\,b\,d}{2}-\frac {3\,C\,a^2\,b\,c}{2}+\frac {3\,C\,a\,b^2\,d}{2}\right )}{f}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (B\,a^3\,d-A\,b^3\,d-B\,b^3\,c+C\,a^3\,c+C\,b^3\,d+3\,A\,a\,b^2\,c+3\,A\,a^2\,b\,d+3\,B\,a^2\,b\,c-3\,B\,a\,b^2\,d-3\,C\,a\,b^2\,c-3\,C\,a^2\,b\,d\right )}{f}+\frac {C\,b^3\,d\,{\mathrm {tan}\left (e+f\,x\right )}^5}{5\,f} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.65, size = 1001, normalized size = 2.84 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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