Optimal. Leaf size=117 \[ -\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a^3 f \sqrt{a+b}}+\frac{x \left (3 a^2-4 a b+8 b^2\right )}{8 a^3}+\frac{(3 a-4 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a f} \]
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Rubi [A] time = 0.156664, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4146, 414, 527, 522, 203, 205} \[ -\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a^3 f \sqrt{a+b}}+\frac{x \left (3 a^2-4 a b+8 b^2\right )}{8 a^3}+\frac{(3 a-4 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a f} \]
Antiderivative was successfully verified.
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Rule 4146
Rule 414
Rule 527
Rule 522
Rule 203
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{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{\operatorname{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{\operatorname{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 \operatorname{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 ) \operatorname{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} \tan ^{-1}\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*}
Mathematica [A] time = 0.463789, size = 95, normalized size = 0.81 \[ \frac{4 \left (3 a^2-4 a b+8 b^2\right ) (e+f x)+a^2 \sin (4 (e+f x))-\frac{32 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{\sqrt{a+b}}+8 a (a-b) \sin (2 (e+f x))}{32 a^3 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.097, size = 194, normalized size = 1.7 \begin{align*}{\frac{3\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,fa \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}b}{2\,f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{\tan \left ( fx+e \right ) b}{2\,f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{5\,\tan \left ( fx+e \right ) }{8\,fa \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}}{f{a}^{3}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ) }{8\,fa}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) b}{2\,f{a}^{2}}}-{\frac{{b}^{3}}{f{a}^{3}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.59508, size = 811, normalized size = 6.93 \begin{align*} \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 ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.2972, size = 201, normalized size = 1.72 \begin{align*} -\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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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