Integrand size = 23, antiderivative size = 117 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\left (3 a^2-4 a b+8 b^2\right ) x}{8 a^3}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b} f}+\frac {(3 a-4 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f} \]
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Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4231, 425, 541, 536, 209, 211} \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^3 f \sqrt {a+b}}+\frac {(3 a-4 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f}+\frac {x \left (3 a^2-4 a b+8 b^2\right )}{8 a^3}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 a f} \]
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Rule 209
Rule 211
Rule 425
Rule 536
Rule 541
Rule 4231
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f}-\frac {\text {Subst}\left (\int \frac {-3 a+b-3 b x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{4 a f} \\ & = \frac {(3 a-4 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f}+\frac {\text {Subst}\left (\int \frac {3 a^2-a b+4 b^2+(3 a-4 b) 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 f} \\ & = \frac {(3 a-4 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f}-\frac {b^3 \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{a^3 f}+\frac {\left (3 a^2-4 a b+8 b^2\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 f} \\ & = \frac {\left (3 a^2-4 a b+8 b^2\right ) x}{8 a^3}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b} f}+\frac {(3 a-4 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f} \\ \end{align*}
Time = 3.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {4 \left (3 a^2-4 a b+8 b^2\right ) (e+f x)-\frac {32 b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+8 a (a-b) \sin (2 (e+f x))+a^2 \sin (4 (e+f x))}{32 a^3 f} \]
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Time = 1.37 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(\frac {-\frac {b^{3} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}+\frac {\frac {\left (\frac {3}{8} a^{2}-\frac {1}{2} a b \right ) \tan \left (f x +e \right )^{3}+\left (-\frac {1}{2} a b +\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{3}}}{f}\) | \(116\) |
default | \(\frac {-\frac {b^{3} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}+\frac {\frac {\left (\frac {3}{8} a^{2}-\frac {1}{2} a b \right ) \tan \left (f x +e \right )^{3}+\left (-\frac {1}{2} a b +\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{3}}}{f}\) | \(116\) |
risch | \(\frac {3 x}{8 a}-\frac {x b}{2 a^{2}}+\frac {x \,b^{2}}{a^{3}}-\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{8 a^{2} f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{8 a^{2} f}+\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right ) f \,a^{3}}-\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right ) f \,a^{3}}+\frac {\sin \left (4 f x +4 e \right )}{32 a f}\) | \(227\) |
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Time = 0.29 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.93 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {2 \, b^{2} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} f x + {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, a^{3} f}, \frac {4 \, b^{2} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} f x + {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, a^{3} f}\right ] \]
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\[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\cos ^{4}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]
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Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {8 \, b^{3} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3}} - \frac {{\left (3 \, a - 4 \, b\right )} \tan \left (f x + e\right )^{3} + {\left (5 \, a - 4 \, b\right )} \tan \left (f x + e\right )}{a^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{2} \tan \left (f x + e\right )^{2} + a^{2}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{3}}}{8 \, f} \]
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Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.21 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {8 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b^{3}}{\sqrt {a b + b^{2}} a^{3}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{3}} - \frac {3 \, a \tan \left (f x + e\right )^{3} - 4 \, b \tan \left (f x + e\right )^{3} + 5 \, a \tan \left (f x + e\right ) - 4 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{2}}}{8 \, f} \]
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Time = 20.26 (sec) , antiderivative size = 1114, normalized size of antiderivative = 9.52 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\text {Too large to display} \]
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