Integrand size = 23, antiderivative size = 240 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {x}{(a-b)^3}-\frac {b^{5/2} \left (63 a^2-90 a b+35 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{9/2} (a-b)^3 f}+\frac {\left (8 a^3+8 a^2 b-55 a b^2+35 b^3\right ) \cot (e+f x)}{8 a^4 (a-b)^2 f}-\frac {\left (8 a^2-55 a b+35 b^2\right ) \cot ^3(e+f x)}{24 a^3 (a-b)^2 f}-\frac {b \cot ^3(e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(11 a-7 b) b \cot ^3(e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )} \]
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Time = 0.44 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3751, 483, 593, 597, 536, 209, 211} \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {b (11 a-7 b) \cot ^3(e+f x)}{8 a^2 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {b^{5/2} \left (63 a^2-90 a b+35 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{9/2} f (a-b)^3}-\frac {\left (8 a^2-55 a b+35 b^2\right ) \cot ^3(e+f x)}{24 a^3 f (a-b)^2}+\frac {\left (8 a^3+8 a^2 b-55 a b^2+35 b^3\right ) \cot (e+f x)}{8 a^4 f (a-b)^2}-\frac {b \cot ^3(e+f x)}{4 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^2}+\frac {x}{(a-b)^3} \]
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Rule 209
Rule 211
Rule 483
Rule 536
Rule 593
Rule 597
Rule 3751
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x^4 \left (1+x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {b \cot ^3(e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\text {Subst}\left (\int \frac {4 a-7 b-7 b x^2}{x^4 \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 ^3(e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(11 a-7 b) b \cot ^3(e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {8 a^2-55 a b+35 b^2-5 (11 a-7 b) b x^2}{x^4 \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^2-55 a b+35 b^2\right ) \cot ^3(e+f x)}{24 a^3 (a-b)^2 f}-\frac {b \cot ^3(e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(11 a-7 b) b \cot ^3(e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {3 \left (8 a^3+8 a^2 b-55 a b^2+35 b^3\right )+3 b \left (8 a^2-55 a b+35 b^2\right ) x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{24 a^3 (a-b)^2 f} \\ & = \frac {\left (8 a^3+8 a^2 b-55 a b^2+35 b^3\right ) \cot (e+f x)}{8 a^4 (a-b)^2 f}-\frac {\left (8 a^2-55 a b+35 b^2\right ) \cot ^3(e+f x)}{24 a^3 (a-b)^2 f}-\frac {b \cot ^3(e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(11 a-7 b) b \cot ^3(e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {3 \left (8 a^4+8 a^3 b+8 a^2 b^2-55 a b^3+35 b^4\right )+3 b \left (8 a^3+8 a^2 b-55 a b^2+35 b^3\right ) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{24 a^4 (a-b)^2 f} \\ & = \frac {\left (8 a^3+8 a^2 b-55 a b^2+35 b^3\right ) \cot (e+f x)}{8 a^4 (a-b)^2 f}-\frac {\left (8 a^2-55 a b+35 b^2\right ) \cot ^3(e+f x)}{24 a^3 (a-b)^2 f}-\frac {b \cot ^3(e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(11 a-7 b) b \cot ^3(e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b)^3 f}-\frac {\left (b^3 \left (63 a^2-90 a b+35 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^4 (a-b)^3 f} \\ & = \frac {x}{(a-b)^3}-\frac {b^{5/2} \left (63 a^2-90 a b+35 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{9/2} (a-b)^3 f}+\frac {\left (8 a^3+8 a^2 b-55 a b^2+35 b^3\right ) \cot (e+f x)}{8 a^4 (a-b)^2 f}-\frac {\left (8 a^2-55 a b+35 b^2\right ) \cot ^3(e+f x)}{24 a^3 (a-b)^2 f}-\frac {b \cot ^3(e+f x)}{4 a (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {(11 a-7 b) b \cot ^3(e+f x)}{8 a^2 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )} \\ \end{align*}
Time = 4.81 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.77 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {-\frac {3 b^{5/2} \left (63 a^2-90 a b+35 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{9/2} (a-b)^3}-\frac {8 \cot (e+f x) \left (-4 a-9 b+a \csc ^2(e+f x)\right )}{a^4}+\frac {3 \left (8 (e+f x)-\frac {(a-b) b^3 \left (17 a^2+2 a b-11 b^2+\left (17 a^2-28 a b+11 b^2\right ) \cos (2 (e+f x))\right ) \sin (2 (e+f x))}{a^4 (a+b+(a-b) \cos (2 (e+f x)))^2}\right )}{(a-b)^3}}{24 f} \]
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Time = 0.38 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.72
method | result | size |
derivativedivides | \(\frac {-\frac {1}{3 a^{3} \tan \left (f x +e \right )^{3}}-\frac {-3 b -a}{a^{4} \tan \left (f x +e \right )}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{\left (a -b \right )^{3}}-\frac {b^{3} \left (\frac {\left (\frac {15}{8} a^{2} b -\frac {13}{4} a \,b^{2}+\frac {11}{8} b^{3}\right ) \tan \left (f x +e \right )^{3}+\frac {a \left (17 a^{2}-30 a b +13 b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (63 a^{2}-90 a b +35 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{4} \left (a -b \right )^{3}}}{f}\) | \(173\) |
default | \(\frac {-\frac {1}{3 a^{3} \tan \left (f x +e \right )^{3}}-\frac {-3 b -a}{a^{4} \tan \left (f x +e \right )}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{\left (a -b \right )^{3}}-\frac {b^{3} \left (\frac {\left (\frac {15}{8} a^{2} b -\frac {13}{4} a \,b^{2}+\frac {11}{8} b^{3}\right ) \tan \left (f x +e \right )^{3}+\frac {a \left (17 a^{2}-30 a b +13 b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (63 a^{2}-90 a b +35 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{4} \left (a -b \right )^{3}}}{f}\) | \(173\) |
risch | \(\text {Expression too large to display}\) | \(1124\) |
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Leaf count of result is larger than twice the leaf count of optimal. 473 vs. \(2 (222) = 444\).
Time = 0.35 (sec) , antiderivative size = 1006, normalized size of antiderivative = 4.19 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
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Time = 0.41 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.38 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {\frac {3 \, {\left (63 \, a^{2} b^{3} - 90 \, a b^{4} + 35 \, b^{5}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} \sqrt {a b}} - \frac {3 \, {\left (8 \, a^{3} b^{2} + 8 \, a^{2} b^{3} - 55 \, a b^{4} + 35 \, b^{5}\right )} \tan \left (f x + e\right )^{6} - 8 \, a^{5} + 16 \, a^{4} b - 8 \, a^{3} b^{2} + {\left (48 \, a^{4} b + 40 \, a^{3} b^{2} - 275 \, a^{2} b^{3} + 175 \, a b^{4}\right )} \tan \left (f x + e\right )^{4} + 8 \, {\left (3 \, a^{5} + a^{4} b - 11 \, a^{3} b^{2} + 7 \, a^{2} b^{3}\right )} \tan \left (f x + e\right )^{2}}{{\left (a^{6} b^{2} - 2 \, a^{5} b^{3} + a^{4} b^{4}\right )} \tan \left (f x + e\right )^{7} + 2 \, {\left (a^{7} b - 2 \, a^{6} b^{2} + a^{5} b^{3}\right )} \tan \left (f x + e\right )^{5} + {\left (a^{8} - 2 \, a^{7} b + a^{6} b^{2}\right )} \tan \left (f x + e\right )^{3}} - \frac {24 \, {\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}}}{24 \, f} \]
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Time = 1.16 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.04 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {\frac {3 \, {\left (63 \, a^{2} b^{3} - 90 \, a b^{4} + 35 \, b^{5}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )}}{{\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} \sqrt {a b}} - \frac {24 \, {\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {3 \, {\left (15 \, a b^{4} \tan \left (f x + e\right )^{3} - 11 \, b^{5} \tan \left (f x + e\right )^{3} + 17 \, a^{2} b^{3} \tan \left (f x + e\right ) - 13 \, a b^{4} \tan \left (f x + e\right )\right )}}{{\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2}} - \frac {8 \, {\left (3 \, a \tan \left (f x + e\right )^{2} + 9 \, b \tan \left (f x + e\right )^{2} - a\right )}}{a^{4} \tan \left (f x + e\right )^{3}}}{24 \, f} \]
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Time = 14.76 (sec) , antiderivative size = 986, normalized size of antiderivative = 4.11 \[ \int \frac {\cot ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {2\,\mathrm {atan}\left (\frac {2\,\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {262144\,a^{33}\,b^2-2883584\,a^{32}\,b^3+14155776\,a^{31}\,b^4-40370176\,a^{30}\,b^5+72089600\,a^{29}\,b^6-77856768\,a^{28}\,b^7+34603008\,a^{27}\,b^8+34603008\,a^{26}\,b^9-77856768\,a^{25}\,b^{10}+72089600\,a^{24}\,b^{11}-40370176\,a^{23}\,b^{12}+14155776\,a^{22}\,b^{13}-2883584\,a^{21}\,b^{14}+262144\,a^{20}\,b^{15}}{{\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )}^2}-1254400\,a^{12}\,b^{17}+13977600\,a^{13}\,b^{16}-70333440\,a^{14}\,b^{15}+210329600\,a^{15}\,b^{14}-413730816\,a^{16}\,b^{13}+559067136\,a^{17}\,b^{12}-525322240\,a^{18}\,b^{11}+338780160\,a^{19}\,b^{10}-143512576\,a^{20}\,b^9+36390912\,a^{21}\,b^8-5047296\,a^{22}\,b^7+1310720\,a^{23}\,b^6-983040\,a^{24}\,b^5+393216\,a^{25}\,b^4-65536\,a^{26}\,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^{30}\,b^2-1179648\,a^{29}\,b^3+4718592\,a^{28}\,b^4-12042240\,a^{27}\,b^5+27279360\,a^{26}\,b^6-67518464\,a^{25}\,b^7+155959296\,a^{24}\,b^8-279281664\,a^{23}\,b^9+365199360\,a^{22}\,b^{10}-344883200\,a^{21}\,b^{11}+233275392\,a^{20}\,b^{12}-110542848\,a^{19}\,b^{13}+34947072\,a^{18}\,b^{14}-6635520\,a^{17}\,b^{15}+573440\,a^{16}\,b^{16}\right )}{{\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )}^2}+1254400\,a^{12}\,b^{14}-10214400\,a^{13}\,b^{13}+35927040\,a^{14}\,b^{12}-70650880\,a^{15}\,b^{11}+83495936\,a^{16}\,b^{10}-58242048\,a^{17}\,b^9+20216832\,a^{18}\,b^8-17408\,a^{19}\,b^7-2285568\,a^{20}\,b^6+516096\,a^{21}\,b^5\right )}\right )}{f\,\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )}+\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (3\,a+7\,b\right )}{3\,a^2}-\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (8\,a^3\,b^2+8\,a^2\,b^3-55\,a\,b^4+35\,b^5\right )}{8\,a^4\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (48\,a^3\,b+40\,a^2\,b^2-275\,a\,b^3+175\,b^4\right )}{24\,a^3\,\left (a^2-2\,a\,b+b^2\right )}}{f\,\left (a^2\,{\mathrm {tan}\left (e+f\,x\right )}^3+2\,a\,b\,{\mathrm {tan}\left (e+f\,x\right )}^5+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^7\right )}-\frac {\mathrm {atan}\left (\frac {b^5\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^9\,b^5\right )}^{3/2}\,1225{}\mathrm {i}-a\,b^4\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^9\,b^5\right )}^{3/2}\,6300{}\mathrm {i}+a^4\,b\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^9\,b^5\right )}^{3/2}\,3969{}\mathrm {i}+a^{18}\,b\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^9\,b^5}\,64{}\mathrm {i}+a^2\,b^3\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^9\,b^5\right )}^{3/2}\,12510{}\mathrm {i}-a^3\,b^2\,\mathrm {tan}\left (e+f\,x\right )\,{\left (-a^9\,b^5\right )}^{3/2}\,11340{}\mathrm {i}}{-64\,a^{23}\,b^3+3969\,a^{18}\,b^8-11340\,a^{17}\,b^9+12510\,a^{16}\,b^{10}-6300\,a^{15}\,b^{11}+1225\,a^{14}\,b^{12}}\right )\,\sqrt {-a^9\,b^5}\,\left (63\,a^2-90\,a\,b+35\,b^2\right )\,1{}\mathrm {i}}{8\,f\,\left (-a^{12}+3\,a^{11}\,b-3\,a^{10}\,b^2+a^9\,b^3\right )} \]
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