Optimal. Leaf size=203 \[ -\frac{b^{5/2} (7 a+6 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^4 f (a+b)^{3/2}}+\frac{x \left (3 a^2-8 a b+24 b^2\right )}{8 a^4}+\frac{b (a-3 b) (3 a+4 b) \tan (e+f x)}{8 a^3 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}+\frac{3 (a-2 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )} \]
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Rubi [A] time = 0.289948, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{b^{5/2} (7 a+6 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^4 f (a+b)^{3/2}}+\frac{x \left (3 a^2-8 a b+24 b^2\right )}{8 a^4}+\frac{b (a-3 b) (3 a+4 b) \tan (e+f x)}{8 a^3 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}+\frac{3 (a-2 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )} \]
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)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-3 a+b-5 b x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=\frac{3 (a-2 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2+a b+6 b^2+9 (a-2 b) b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{8 a^2 f}\\ &=\frac{3 (a-2 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{(a-3 b) b (3 a+4 b) \tan (e+f x)}{8 a^3 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 \left (3 a^3-2 a^2 b+11 a b^2+12 b^3\right )+2 (a-3 b) 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 )}{16 a^3 (a+b) f}\\ &=\frac{3 (a-2 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{(a-3 b) b (3 a+4 b) \tan (e+f x)}{8 a^3 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\left (b^3 (7 a+6 b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^4 (a+b) f}+\frac{\left (3 a^2-8 a b+24 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a^4 f}\\ &=\frac{\left (3 a^2-8 a b+24 b^2\right ) x}{8 a^4}-\frac{b^{5/2} (7 a+6 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^4 (a+b)^{3/2} f}+\frac{3 (a-2 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{(a-3 b) b (3 a+4 b) \tan (e+f x)}{8 a^3 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 1.58873, size = 138, normalized size = 0.68 \[ \frac{4 \left (3 a^2-8 a b+24 b^2\right ) (e+f x)+a^2 \sin (4 (e+f x))-\frac{16 b^{5/2} (7 a+6 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{(a+b)^{3/2}}-\frac{16 a b^3 \sin (2 (e+f x))}{(a+b) (a \cos (2 (e+f x))+a+2 b)}+8 a (a-2 b) \sin (2 (e+f x))}{32 a^4 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.106, size = 276, normalized size = 1.4 \begin{align*}{\frac{3\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}b}{f{a}^{3} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{\tan \left ( fx+e \right ) b}{f{a}^{3} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{5\,\tan \left ( fx+e \right ) }{8\,f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}+3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}}{f{a}^{4}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ) }{8\,f{a}^{2}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) b}{f{a}^{3}}}-{\frac{{b}^{3}\tan \left ( fx+e \right ) }{2\,f{a}^{3} \left ( a+b \right ) \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{7\,{b}^{3}}{2\,f{a}^{3} \left ( a+b \right ) }\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}}}}-3\,{\frac{{b}^{4}}{f{a}^{4} \left ( a+b \right ) \sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{\tan \left ( fx+e \right ) b}{\sqrt{ \left ( a+b \right ) b}}} \right ) } \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 = 0.710846, size = 1492, normalized size = 7.35 \begin{align*} \text{result too large to display} \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 = 1.25679, size = 277, normalized size = 1.36 \begin{align*} -\frac{\frac{4 \, b^{3} \tan \left (f x + e\right )}{{\left (a^{4} + a^{3} b\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac{4 \,{\left (7 \, a b^{3} + 6 \, b^{4}\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 + b^{2}}}\right )\right )}}{{\left (a^{5} + a^{4} b\right )} \sqrt{a b + b^{2}}} - \frac{{\left (3 \, a^{2} - 8 \, a b + 24 \, b^{2}\right )}{\left (f x + e\right )}}{a^{4}} - \frac{3 \, a \tan \left (f x + e\right )^{3} - 8 \, b \tan \left (f x + e\right )^{3} + 5 \, a \tan \left (f x + e\right ) - 8 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{3}}}{8 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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