Optimal. Leaf size=258 \[ \frac{5 d^2 (2 c-d) \left (2 c^2-3 c d+2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{2 a^2 f}-\frac{d \left (c^2+10 c d-12 d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{3 a^2 f}-\frac{d \tan (e+f x) \left (d \left (20 c^2 d+2 c^3-57 c d^2+30 d^3\right ) \sec (e+f x)+4 \left (-44 c^2 d^2+10 c^3 d+c^4+40 c d^3-12 d^4\right )\right )}{6 a^2 f}+\frac{(c-d) (c+10 d) \tan (e+f x) (c+d \sec (e+f x))^3}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{(c-d) \tan (e+f x) (c+d \sec (e+f x))^4}{3 f (a \sec (e+f x)+a)^2} \]
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Rubi [A] time = 0.437776, antiderivative size = 315, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {3987, 98, 150, 153, 147, 63, 217, 203} \[ -\frac{d \left (c^2+10 c d-12 d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{3 a^2 f}-\frac{d \tan (e+f x) \left (d \left (20 c^2 d+2 c^3-57 c d^2+30 d^3\right ) \sec (e+f x)+4 \left (-44 c^2 d^2+10 c^3 d+c^4+40 c d^3-12 d^4\right )\right )}{6 a^2 f}+\frac{(c-d) (c+10 d) \tan (e+f x) (c+d \sec (e+f x))^3}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{5 d^2 (2 c-d) \left (2 c^2-3 c d+2 d^2\right ) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{a f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{(c-d) \tan (e+f x) (c+d \sec (e+f x))^4}{3 f (a \sec (e+f x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 3987
Rule 98
Rule 150
Rule 153
Rule 147
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \frac{\sec (e+f x) (c+d \sec (e+f x))^5}{(a+a \sec (e+f x))^2} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^5}{\sqrt{a-a x} (a+a x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x)^3 \left (-a^2 \left (c^2+6 c d-4 d^2\right )+3 a^2 (c-2 d) d x\right )}{\sqrt{a-a x} (a+a x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{3 a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) (c+10 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x)^2 \left (-3 a^4 (11 c-10 d) d^2+3 a^4 d \left (c^2+10 c d-12 d^2\right ) x\right )}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{3 a^4 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d \left (c^2+10 c d-12 d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 a^2 f}+\frac{(c-d) (c+10 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x) \left (3 a^6 d^2 \left (31 c^2-50 c d+24 d^2\right )-3 a^6 d \left (2 c^3+20 c^2 d-57 c d^2+30 d^3\right ) x\right )}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{9 a^6 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d \left (c^2+10 c d-12 d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 a^2 f}+\frac{(c-d) (c+10 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{d \left (4 \left (c^4+10 c^3 d-44 c^2 d^2+40 c d^3-12 d^4\right )+d \left (2 c^3+20 c^2 d-57 c d^2+30 d^3\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a^2 f}-\frac{\left (5 (2 c-d) d^2 \left (2 c^2-3 c d+2 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d \left (c^2+10 c d-12 d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 a^2 f}+\frac{(c-d) (c+10 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{d \left (4 \left (c^4+10 c^3 d-44 c^2 d^2+40 c d^3-12 d^4\right )+d \left (2 c^3+20 c^2 d-57 c d^2+30 d^3\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a^2 f}+\frac{\left (5 (2 c-d) d^2 \left (2 c^2-3 c d+2 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 a-x^2}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{d \left (c^2+10 c d-12 d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 a^2 f}+\frac{(c-d) (c+10 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{d \left (4 \left (c^4+10 c^3 d-44 c^2 d^2+40 c d^3-12 d^4\right )+d \left (2 c^3+20 c^2 d-57 c d^2+30 d^3\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a^2 f}+\frac{\left (5 (2 c-d) d^2 \left (2 c^2-3 c d+2 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right )}{a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{5 (2 c-d) d^2 \left (2 c^2-3 c d+2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right ) \tan (e+f x)}{a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}-\frac{d \left (c^2+10 c d-12 d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{3 a^2 f}+\frac{(c-d) (c+10 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{d \left (4 \left (c^4+10 c^3 d-44 c^2 d^2+40 c d^3-12 d^4\right )+d \left (2 c^3+20 c^2 d-57 c d^2+30 d^3\right ) \sec (e+f x)\right ) \tan (e+f x)}{6 a^2 f}\\ \end{align*}
Mathematica [A] time = 4.15357, size = 446, normalized size = 1.73 \[ \frac{240 d^2 \left (8 c^2 d-4 c^3-7 c d^2+2 d^3\right ) \cos ^4\left (\frac{1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )+2 \sin \left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x) \left (-100 c^3 d^2 \cos (3 (e+f x))-40 c^3 d^2 \cos (4 (e+f x))+280 c^2 d^3 \cos (3 (e+f x))+100 c^2 d^3 \cos (4 (e+f x))+\left (-300 c^3 d^2+840 c^2 d^3+60 c^4 d+6 c^5-585 c d^4+190 d^5\right ) \cos (e+f x)+4 \left (-40 c^3 d^2+130 c^2 d^3+5 c^4 d+2 c^5-95 c d^4+30 d^5\right ) \cos (2 (e+f x))-120 c^3 d^2+420 c^2 d^3+20 c^4 d \cos (3 (e+f x))+5 c^4 d \cos (4 (e+f x))+15 c^4 d+2 c^5 \cos (3 (e+f x))+2 c^5 \cos (4 (e+f x))+6 c^5-215 c d^4 \cos (3 (e+f x))-80 c d^4 \cos (4 (e+f x))-300 c d^4+66 d^5 \cos (3 (e+f x))+24 d^5 \cos (4 (e+f x))+104 d^5\right )}{24 a^2 f (\cos (e+f x)+1)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.099, size = 766, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.04235, size = 1042, normalized size = 4.04 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.539205, size = 1037, normalized size = 4.02 \begin{align*} \frac{15 \,{\left ({\left (4 \, c^{3} d^{2} - 8 \, c^{2} d^{3} + 7 \, c d^{4} - 2 \, d^{5}\right )} \cos \left (f x + e\right )^{5} + 2 \,{\left (4 \, c^{3} d^{2} - 8 \, c^{2} d^{3} + 7 \, c d^{4} - 2 \, d^{5}\right )} \cos \left (f x + e\right )^{4} +{\left (4 \, c^{3} d^{2} - 8 \, c^{2} d^{3} + 7 \, c d^{4} - 2 \, d^{5}\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \,{\left ({\left (4 \, c^{3} d^{2} - 8 \, c^{2} d^{3} + 7 \, c d^{4} - 2 \, d^{5}\right )} \cos \left (f x + e\right )^{5} + 2 \,{\left (4 \, c^{3} d^{2} - 8 \, c^{2} d^{3} + 7 \, c d^{4} - 2 \, d^{5}\right )} \cos \left (f x + e\right )^{4} +{\left (4 \, c^{3} d^{2} - 8 \, c^{2} d^{3} + 7 \, c d^{4} - 2 \, d^{5}\right )} \cos \left (f x + e\right )^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (2 \, d^{5} + 2 \,{\left (2 \, c^{5} + 5 \, c^{4} d - 40 \, c^{3} d^{2} + 100 \, c^{2} d^{3} - 80 \, c d^{4} + 24 \, d^{5}\right )} \cos \left (f x + e\right )^{4} +{\left (2 \, c^{5} + 20 \, c^{4} d - 100 \, c^{3} d^{2} + 280 \, c^{2} d^{3} - 215 \, c d^{4} + 66 \, d^{5}\right )} \cos \left (f x + e\right )^{3} + 6 \,{\left (10 \, c^{2} d^{3} - 5 \, c d^{4} + 2 \, d^{5}\right )} \cos \left (f x + e\right )^{2} +{\left (15 \, c d^{4} - 2 \, d^{5}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{12 \,{\left (a^{2} f \cos \left (f x + e\right )^{5} + 2 \, a^{2} f \cos \left (f x + e\right )^{4} + a^{2} f \cos \left (f x + e\right )^{3}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c^{5} \sec{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{5} \sec ^{6}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{5 c d^{4} \sec ^{5}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{10 c^{2} d^{3} \sec ^{4}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{10 c^{3} d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{5 c^{4} d \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx}{a^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.29513, size = 716, normalized size = 2.78 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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