Optimal. Leaf size=138 \[ -\frac {x}{a^3}+\frac {(3 a+4 b) \tan (e+f x)}{8 a^2 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\left (3 a^2+12 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 \sqrt {b} f (a+b)^{3/2}}+\frac {\tan (e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]
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Rubi [A] time = 0.23, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4141, 1975, 471, 527, 522, 203, 205} \[ \frac {\left (3 a^2+12 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 \sqrt {b} f (a+b)^{3/2}}+\frac {(3 a+4 b) \tan (e+f x)}{8 a^2 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {x}{a^3}+\frac {\tan (e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]
Antiderivative was successfully verified.
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Rule 203
Rule 205
Rule 471
Rule 522
Rule 527
Rule 1975
Rule 4141
Rubi steps
\begin {align*} \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {\operatorname {Subst}\left (\int \frac {1-3 x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=\frac {\tan (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(3 a+4 b) \tan (e+f x)}{8 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {5 a+4 b+(-3 a-4 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a+b) f}\\ &=\frac {\tan (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(3 a+4 b) \tan (e+f x)}{8 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a^3 f}+\frac {\left (3 a^2+12 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a+b) f}\\ &=-\frac {x}{a^3}+\frac {\left (3 a^2+12 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 \sqrt {b} (a+b)^{3/2} f}+\frac {\tan (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(3 a+4 b) \tan (e+f x)}{8 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}
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Mathematica [C] time = 10.60, size = 1334, normalized size = 9.67 \[ \frac {(\cos (2 (e+f x)) a+a+2 b)^3 \sec ^6(e+f x) \left (-\frac {6 a (a+2 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}+\frac {4 \left (3 a^2+8 b a+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}-\frac {4 a \sqrt {b} \left (3 a^2+16 b a+3 (a+2 b) \cos (2 (e+f x)) a+16 b^2\right ) \sin (2 (e+f x))}{(a+b)^2 (\cos (2 (e+f x)) a+a+2 b)^2}+\frac {2 \sqrt {b} \left (3 a^3+14 b a^2+24 b^2 a+\left (3 a^2+4 b a+4 b^2\right ) \cos (2 (e+f x)) a+16 b^3\right ) \sin (2 (e+f x))}{(a+b)^2 (\cos (2 (e+f x)) a+a+2 b)^2}-\frac {\sqrt {b} \left (\frac {2 \left (3 a^5-10 b a^4+80 b^2 a^3+480 b^3 a^2+640 b^4 a+256 b^5\right ) \tan ^{-1}\left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {\sec (2 e) \left (-9 \sin (2 e) a^6+9 \sin (2 f x) a^6+3 \sin (2 (e+2 f x)) a^6-3 \sin (4 e+2 f x) a^6+12 b \sin (2 e) a^5-14 b \sin (2 f x) a^5-12 b \sin (2 (e+2 f x)) a^5+10 b \sin (4 e+2 f x) a^5+128 b^2 f x \cos (2 (e+2 f x)) a^4+512 b^2 f x \cos (4 e+2 f x) a^4+128 b^2 f x \cos (6 e+4 f x) a^4+684 b^2 \sin (2 e) a^4-608 b^2 \sin (2 f x) a^4-204 b^2 \sin (2 (e+2 f x)) a^4+304 b^2 \sin (4 e+2 f x) a^4+256 b^3 f x \cos (2 (e+2 f x)) a^3+2048 b^3 f x \cos (4 e+2 f x) a^3+256 b^3 f x \cos (6 e+4 f x) a^3+2880 b^3 \sin (2 e) a^3-2112 b^3 \sin (2 f x) a^3-384 b^3 \sin (2 (e+2 f x)) a^3+1056 b^3 \sin (4 e+2 f x) a^3+128 b^4 f x \cos (2 (e+2 f x)) a^2+2560 b^4 f x \cos (4 e+2 f x) a^2+128 b^4 f x \cos (6 e+4 f x) a^2+5280 b^4 \sin (2 e) a^2-2560 b^4 \sin (2 f x) a^2-192 b^4 \sin (2 (e+2 f x)) a^2+1280 b^4 \sin (4 e+2 f x) a^2+512 b^2 (a+b)^2 (a+2 b) f x \cos (2 f x) a+1024 b^5 f x \cos (4 e+2 f x) a+4608 b^5 \sin (2 e) a-1024 b^5 \sin (2 f x) a+512 b^5 \sin (4 e+2 f x) a+256 b^2 (a+b)^2 \left (3 a^2+8 b a+8 b^2\right ) f x \cos (2 e)+1536 b^6 \sin (2 e)\right )}{(\cos (2 (e+f x)) a+a+2 b)^2}\right )}{a^3 (a+b)^2}-\frac {2 \sqrt {b} \left (\frac {a \sec (2 e) \left (\left (-9 a^4-16 b a^3+48 b^2 a^2+128 b^3 a+64 b^4\right ) \sin (2 f x)+a \left (-3 a^3+2 b a^2+24 b^2 a+16 b^3\right ) \sin (2 (e+2 f x))+\left (3 a^4-64 b^2 a^2-128 b^3 a-64 b^4\right ) \sin (4 e+2 f x)\right )+\left (9 a^5+18 b a^4-64 b^2 a^3-256 b^3 a^2-320 b^4 a-128 b^5\right ) \tan (2 e)}{a^2 (\cos (2 (e+f x)) a+a+2 b)^2}-\frac {6 a^2 \tan ^{-1}\left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{(a+b)^2}\right )}{4096 b^{5/2} f \left (b \sec ^2(e+f x)+a\right )^3} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.55, size = 860, normalized size = 6.23 \[ \left [-\frac {32 \, {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f x \cos \left (f x + e\right )^{4} + 64 \, {\left (a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4}\right )} f x \cos \left (f x + e\right )^{2} + 32 \, {\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} f x + {\left ({\left (3 \, a^{4} + 12 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} + 12 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (3 \, a^{3} b + 12 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 4 \, {\left ({\left (5 \, a^{4} b + 11 \, a^{3} b^{2} + 6 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{3} b^{2} + 7 \, a^{2} b^{3} + 4 \, a b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\left ({\left (a^{7} b + 2 \, a^{6} b^{2} + a^{5} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b^{2} + 2 \, a^{5} b^{3} + a^{4} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{3} + 2 \, a^{4} b^{4} + a^{3} b^{5}\right )} f\right )}}, -\frac {16 \, {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f x \cos \left (f x + e\right )^{4} + 32 \, {\left (a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4}\right )} f x \cos \left (f x + e\right )^{2} + 16 \, {\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} f x + {\left ({\left (3 \, a^{4} + 12 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} + 12 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (3 \, a^{3} b + 12 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - 2 \, {\left ({\left (5 \, a^{4} b + 11 \, a^{3} b^{2} + 6 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{3} b^{2} + 7 \, a^{2} b^{3} + 4 \, a b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{7} b + 2 \, a^{6} b^{2} + a^{5} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b^{2} + 2 \, a^{5} b^{3} + a^{4} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{3} + 2 \, a^{4} b^{4} + a^{3} b^{5}\right )} f\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.31, size = 181, normalized size = 1.31 \[ \frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )}}{{\left (a^{4} + a^{3} b\right )} \sqrt {a b + b^{2}}} + \frac {3 \, a b \tan \left (f x + e\right )^{3} + 4 \, b^{2} \tan \left (f x + e\right )^{3} + 5 \, a^{2} \tan \left (f x + e\right ) + 9 \, a b \tan \left (f x + e\right ) + 4 \, b^{2} \tan \left (f x + e\right )}{{\left (a^{3} + a^{2} b\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} - \frac {8 \, {\left (f x + e\right )}}{a^{3}}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.72, size = 263, normalized size = 1.91 \[ \frac {3 b \left (\tan ^{3}\left (f x +e \right )\right )}{8 f a \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} \left (a +b \right )}+\frac {b^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{2 f \,a^{2} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} \left (a +b \right )}+\frac {5 \tan \left (f x +e \right )}{8 a f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {b \tan \left (f x +e \right )}{2 a^{2} f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {3 \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{8 f a \left (a +b \right ) \sqrt {\left (a +b \right ) b}}+\frac {3 \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) b}{2 f \,a^{2} \left (a +b \right ) \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) b^{2}}{f \,a^{3} \left (a +b \right ) \sqrt {\left (a +b \right ) b}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 191, normalized size = 1.38 \[ \frac {\frac {{\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} b}} + \frac {{\left (3 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + {\left (5 \, a^{2} + 9 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )}{a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3} + {\left (a^{3} b^{2} + a^{2} b^{3}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} \tan \left (f x + e\right )^{2}} - \frac {8 \, {\left (f x + e\right )}}{a^{3}}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.44, size = 2405, normalized size = 17.43 \[ \text {result too large to display} \]
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 \[ \int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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