Optimal. Leaf size=122 \[ -\frac{\left (a^2-10 a b+13 b^2\right ) \tan (e+f x)}{4 f}+\frac{1}{8} x \left (3 a^2-30 a b+35 b^2\right )+\frac{(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}-\frac{(a-9 b) (a-b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{b^2 \tan ^3(e+f x)}{3 f} \]
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Rubi [A] time = 0.131028, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3663, 463, 455, 1153, 203} \[ -\frac{\left (a^2-10 a b+13 b^2\right ) \tan (e+f x)}{4 f}+\frac{1}{8} x \left (3 a^2-30 a b+35 b^2\right )+\frac{(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}-\frac{(a-9 b) (a-b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{b^2 \tan ^3(e+f x)}{3 f} \]
Antiderivative was successfully verified.
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Rule 3663
Rule 463
Rule 455
Rule 1153
Rule 203
Rubi steps
\begin{align*} \int \sin ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+b x^2\right )^2}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}-\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a^2-10 a b+5 b^2-4 b^2 x^2\right )}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=-\frac{(a-9 b) (a-b) \cos (e+f x) \sin (e+f x)}{8 f}+\frac{(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac{\operatorname{Subst}\left (\int \frac{(a-9 b) (a-b)-2 (a-9 b) (a-b) x^2+8 b^2 x^4}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac{(a-9 b) (a-b) \cos (e+f x) \sin (e+f x)}{8 f}+\frac{(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac{\operatorname{Subst}\left (\int \left (-2 \left (a^2-10 a b+13 b^2\right )+8 b^2 x^2+\frac{3 a^2-30 a b+35 b^2}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac{(a-9 b) (a-b) \cos (e+f x) \sin (e+f x)}{8 f}-\frac{\left (a^2-10 a b+13 b^2\right ) \tan (e+f x)}{4 f}+\frac{(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac{b^2 \tan ^3(e+f x)}{3 f}+\frac{\left (3 a^2-30 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac{1}{8} \left (3 a^2-30 a b+35 b^2\right ) x-\frac{(a-9 b) (a-b) \cos (e+f x) \sin (e+f x)}{8 f}-\frac{\left (a^2-10 a b+13 b^2\right ) \tan (e+f x)}{4 f}+\frac{(a-b)^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac{b^2 \tan ^3(e+f x)}{3 f}\\ \end{align*}
Mathematica [A] time = 1.4757, size = 96, normalized size = 0.79 \[ \frac{12 \left (3 a^2-30 a b+35 b^2\right ) (e+f x)-24 \left (a^2-4 a b+3 b^2\right ) \sin (2 (e+f x))+3 (a-b)^2 \sin (4 (e+f x))+32 b \tan (e+f x) \left (6 a+b \sec ^2(e+f x)-10 b\right )}{96 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.047, size = 199, normalized size = 1.6 \begin{align*}{\frac{1}{f} \left ({a}^{2} \left ( -{\frac{\cos \left ( fx+e \right ) }{4} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) +2\,ab \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{7}}{\cos \left ( fx+e \right ) }}+ \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+5/4\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) \cos \left ( fx+e \right ) -{\frac{15\,fx}{8}}-{\frac{15\,e}{8}} \right ) +{b}^{2} \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{9}}{3\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}}-2\,{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{9}}{\cos \left ( fx+e \right ) }}-2\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{7}+7/6\, \left ( \sin \left ( fx+e \right ) \right ) ^{5}+{\frac{35\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{24}}+{\frac{35\,\sin \left ( fx+e \right ) }{16}} \right ) \cos \left ( fx+e \right ) +{\frac{35\,fx}{8}}+{\frac{35\,e}{8}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.49387, size = 176, normalized size = 1.44 \begin{align*} \frac{8 \, b^{2} \tan \left (f x + e\right )^{3} + 3 \,{\left (3 \, a^{2} - 30 \, a b + 35 \, b^{2}\right )}{\left (f x + e\right )} + 24 \,{\left (2 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right ) - \frac{3 \,{\left ({\left (5 \, a^{2} - 18 \, a b + 13 \, b^{2}\right )} \tan \left (f x + e\right )^{3} +{\left (3 \, a^{2} - 14 \, a b + 11 \, b^{2}\right )} \tan \left (f x + e\right )\right )}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}}{24 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.06931, size = 293, normalized size = 2.4 \begin{align*} \frac{3 \,{\left (3 \, a^{2} - 30 \, a b + 35 \, b^{2}\right )} f x \cos \left (f x + e\right )^{3} +{\left (6 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{6} - 3 \,{\left (5 \, a^{2} - 18 \, a b + 13 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 16 \,{\left (3 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, b^{2}\right )} \sin \left (f x + e\right )}{24 \, f \cos \left (f x + e\right )^{3}} \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 [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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