Optimal. Leaf size=189 \[ -\frac {b (9 a-5 b) \cot (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}+\frac {b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{7/2} f (a-b)^3}-\frac {\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 f (a-b)^2}-\frac {b \cot (e+f x)}{4 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {x}{(a-b)^3} \]
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Rubi [A] time = 0.29, antiderivative size = 189, 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 = {3670, 472, 579, 583, 522, 203, 205} \[ \frac {b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{7/2} f (a-b)^3}-\frac {\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 f (a-b)^2}-\frac {b (9 a-5 b) \cot (e+f x)}{8 a^2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {b \cot (e+f x)}{4 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {x}{(a-b)^3} \]
Antiderivative was successfully verified.
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Rule 203
Rule 205
Rule 472
Rule 522
Rule 579
Rule 583
Rule 3670
Rubi steps
\begin {align*} \int \frac {\cot ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {4 a-5 b-5 b x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a (a-b) f}\\ &=-\frac {b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {8 a^2-27 a b+15 b^2-3 (9 a-5 b) b x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 (a-b)^2 f}\\ &=-\frac {\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac {b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {8 a^3+8 a^2 b-27 a b^2+15 b^3+b \left (8 a^2-27 a b+15 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a-b)^2 f}\\ &=-\frac {\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac {b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+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-b)^3 f}+\frac {\left (b^2 \left (35 a^2-42 a b+15 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 (a-b)^3 f}\\ &=-\frac {x}{(a-b)^3}+\frac {b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{7/2} (a-b)^3 f}-\frac {\left (8 a^2-27 a b+15 b^2\right ) \cot (e+f x)}{8 a^3 (a-b)^2 f}-\frac {b \cot (e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(9 a-5 b) b \cot (e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}
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Mathematica [A] time = 2.40, size = 174, normalized size = 0.92 \[ \frac {\frac {b^2 (13 a-7 b) \sin (2 (e+f x))}{a^3 (a-b)^2 ((a-b) \cos (2 (e+f x))+a+b)}-\frac {8 \cot (e+f x)}{a^3}-\frac {4 b^3 \sin (2 (e+f x))}{a^2 (a-b)^2 ((a-b) \cos (2 (e+f x))+a+b)^2}+\frac {b^{3/2} \left (35 a^2-42 a b+15 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{7/2} (a-b)^3}+\frac {8 (e+f x)}{(b-a)^3}}{8 f} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.51, size = 881, normalized size = 4.66 \[ \left [-\frac {32 \, a^{3} b^{2} f x \tan \left (f x + e\right )^{5} + 64 \, a^{4} b f x \tan \left (f x + e\right )^{3} + 32 \, a^{5} f x \tan \left (f x + e\right ) + 32 \, a^{5} - 96 \, a^{4} b + 96 \, a^{3} b^{2} - 32 \, a^{2} b^{3} + 4 \, {\left (8 \, a^{3} b^{2} - 35 \, a^{2} b^{3} + 42 \, a b^{4} - 15 \, b^{5}\right )} \tan \left (f x + e\right )^{4} + 4 \, {\left (16 \, a^{4} b - 61 \, a^{3} b^{2} + 70 \, a^{2} b^{3} - 25 \, a b^{4}\right )} \tan \left (f x + e\right )^{2} + {\left ({\left (35 \, a^{2} b^{3} - 42 \, a b^{4} + 15 \, b^{5}\right )} \tan \left (f x + e\right )^{5} + 2 \, {\left (35 \, a^{3} b^{2} - 42 \, a^{2} b^{3} + 15 \, a b^{4}\right )} \tan \left (f x + e\right )^{3} + {\left (35 \, a^{4} b - 42 \, a^{3} b^{2} + 15 \, a^{2} b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{4} - 6 \, a b \tan \left (f x + e\right )^{2} + a^{2} - 4 \, {\left (a b \tan \left (f x + e\right )^{3} - a^{2} \tan \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}}}{b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}}\right )}{32 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} - a^{3} b^{5}\right )} f \tan \left (f x + e\right )^{5} + 2 \, {\left (a^{7} b - 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} - a^{4} b^{4}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{8} - 3 \, a^{7} b + 3 \, a^{6} b^{2} - a^{5} b^{3}\right )} f \tan \left (f x + e\right )\right )}}, -\frac {16 \, a^{3} b^{2} f x \tan \left (f x + e\right )^{5} + 32 \, a^{4} b f x \tan \left (f x + e\right )^{3} + 16 \, a^{5} f x \tan \left (f x + e\right ) + 16 \, a^{5} - 48 \, a^{4} b + 48 \, a^{3} b^{2} - 16 \, a^{2} b^{3} + 2 \, {\left (8 \, a^{3} b^{2} - 35 \, a^{2} b^{3} + 42 \, a b^{4} - 15 \, b^{5}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (16 \, a^{4} b - 61 \, a^{3} b^{2} + 70 \, a^{2} b^{3} - 25 \, a b^{4}\right )} \tan \left (f x + e\right )^{2} - {\left ({\left (35 \, a^{2} b^{3} - 42 \, a b^{4} + 15 \, b^{5}\right )} \tan \left (f x + e\right )^{5} + 2 \, {\left (35 \, a^{3} b^{2} - 42 \, a^{2} b^{3} + 15 \, a b^{4}\right )} \tan \left (f x + e\right )^{3} + {\left (35 \, a^{4} b - 42 \, a^{3} b^{2} + 15 \, a^{2} b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt {\frac {b}{a}}}{2 \, b \tan \left (f x + e\right )}\right )}{16 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} - a^{3} b^{5}\right )} f \tan \left (f x + e\right )^{5} + 2 \, {\left (a^{7} b - 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} - a^{4} b^{4}\right )} f \tan \left (f x + e\right )^{3} + {\left (a^{8} - 3 \, a^{7} b + 3 \, a^{6} b^{2} - a^{5} b^{3}\right )} f \tan \left (f x + e\right )\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 5.48, size = 231, normalized size = 1.22 \[ \frac {\frac {{\left (35 \, a^{2} b^{2} - 42 \, a b^{3} + 15 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )}}{{\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} \sqrt {a b}} - \frac {8 \, {\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {11 \, a b^{3} \tan \left (f x + e\right )^{3} - 7 \, b^{4} \tan \left (f x + e\right )^{3} + 13 \, a^{2} b^{2} \tan \left (f x + e\right ) - 9 \, a b^{3} \tan \left (f x + e\right )}{{\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2}} - \frac {8}{a^{3} \tan \left (f x + e\right )}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.98, size = 379, normalized size = 2.01 \[ \frac {11 b^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} a}-\frac {9 b^{4} \left (\tan ^{3}\left (f x +e \right )\right )}{4 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} a^{2}}+\frac {7 b^{5} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \,a^{3} \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {13 \tan \left (f x +e \right ) b^{2}}{8 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {11 b^{3} \tan \left (f x +e \right )}{4 f \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} a}+\frac {9 b^{4} \tan \left (f x +e \right )}{8 f \,a^{2} \left (a -b \right )^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {35 \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right ) b^{2}}{8 f \left (a -b \right )^{3} a \sqrt {a b}}-\frac {21 b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{4 f \left (a -b \right )^{3} a^{2} \sqrt {a b}}+\frac {15 b^{4} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{8 f \,a^{3} \left (a -b \right )^{3} \sqrt {a b}}-\frac {1}{f \,a^{3} \tan \left (f x +e \right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \left (a -b \right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.82, size = 275, normalized size = 1.46 \[ \frac {\frac {{\left (35 \, a^{2} b^{2} - 42 \, a b^{3} + 15 \, b^{4}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} \sqrt {a b}} - \frac {{\left (8 \, a^{2} b^{2} - 27 \, a b^{3} + 15 \, b^{4}\right )} \tan \left (f x + e\right )^{4} + 8 \, a^{4} - 16 \, a^{3} b + 8 \, a^{2} b^{2} + {\left (16 \, a^{3} b - 45 \, a^{2} b^{2} + 25 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}}{{\left (a^{5} b^{2} - 2 \, a^{4} b^{3} + a^{3} b^{4}\right )} \tan \left (f x + e\right )^{5} + 2 \, {\left (a^{6} b - 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (a^{7} - 2 \, a^{6} b + a^{5} b^{2}\right )} \tan \left (f x + e\right )} - \frac {8 \, {\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 15.05, size = 915, normalized size = 4.84 \[ -\frac {2\,\mathrm {atan}\left (\frac {2\,\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {262144\,a^{28}\,b^2-2883584\,a^{27}\,b^3+14155776\,a^{26}\,b^4-40370176\,a^{25}\,b^5+72089600\,a^{24}\,b^6-77856768\,a^{23}\,b^7+34603008\,a^{22}\,b^8+34603008\,a^{21}\,b^9-77856768\,a^{20}\,b^{10}+72089600\,a^{19}\,b^{11}-40370176\,a^{18}\,b^{12}+14155776\,a^{17}\,b^{13}-2883584\,a^{16}\,b^{14}+262144\,a^{15}\,b^{15}}{{\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )}^2}-230400\,a^9\,b^{15}+2672640\,a^{10}\,b^{14}-14078976\,a^{11}\,b^{13}+44261376\,a^{12}\,b^{12}-91801600\,a^{13}\,b^{11}+131051520\,a^{14}\,b^{10}-130287616\,a^{15}\,b^9+89219072\,a^{16}\,b^8-40743936\,a^{17}\,b^7+11847680\,a^{18}\,b^6-2237440\,a^{19}\,b^5+393216\,a^{20}\,b^4-65536\,a^{21}\,b^3\right )}{\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )\,\left (\frac {2\,\left (131072\,a^{25}\,b^2-1179648\,a^{24}\,b^3+4145152\,a^{23}\,b^4-5160960\,a^{22}\,b^5-10567680\,a^{21}\,b^6+58638336\,a^{20}\,b^7-127893504\,a^{19}\,b^8+174882816\,a^{18}\,b^9-164659200\,a^{17}\,b^{10}+109281280\,a^{16}\,b^{11}-50577408\,a^{15}\,b^{12}+15613952\,a^{14}\,b^{13}-2899968\,a^{13}\,b^{14}+245760\,a^{12}\,b^{15}\right )}{{\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )}^2}+230400\,a^9\,b^{12}-1981440\,a^{10}\,b^{11}+7443456\,a^{11}\,b^{10}-15879168\,a^{12}\,b^9+20933632\,a^{13}\,b^8-17363968\,a^{14}\,b^7+8788992\,a^{15}\,b^6-2458624\,a^{16}\,b^5+286720\,a^{17}\,b^4\right )}\right )}{f\,\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )}-\frac {\frac {1}{a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (8\,a^2\,b^2-27\,a\,b^3+15\,b^4\right )}{8\,a^3\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (16\,a^2\,b-45\,a\,b^2+25\,b^3\right )}{8\,a^2\,\left (a^2-2\,a\,b+b^2\right )}}{f\,\left (a^2\,\mathrm {tan}\left (e+f\,x\right )+2\,a\,b\,{\mathrm {tan}\left (e+f\,x\right )}^3+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^5\right )}+\frac {\mathrm {atan}\left (\frac {b^5\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^7\,b^3\right )}^{3/2}\,225{}\mathrm {i}-a\,b^4\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^7\,b^3\right )}^{3/2}\,1260{}\mathrm {i}+a^4\,b\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^7\,b^3\right )}^{3/2}\,1225{}\mathrm {i}+a^{14}\,b\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^7\,b^3}\,64{}\mathrm {i}+a^2\,b^3\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^7\,b^3\right )}^{3/2}\,2814{}\mathrm {i}-a^3\,b^2\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^7\,b^3\right )}^{3/2}\,2940{}\mathrm {i}}{-64\,a^{18}\,b^2+1225\,a^{15}\,b^5-2940\,a^{14}\,b^6+2814\,a^{13}\,b^7-1260\,a^{12}\,b^8+225\,a^{11}\,b^9}\right )\,\sqrt {-a^7\,b^3}\,\left (35\,a^2-42\,a\,b+15\,b^2\right )\,1{}\mathrm {i}}{8\,f\,\left (-a^{10}+3\,a^9\,b-3\,a^8\,b^2+a^7\,b^3\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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