Optimal. Leaf size=202 \[ \frac{2 (7 A-B) \tan (c+d x) \sec ^2(c+d x)}{35 d \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} (A-B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (7 A-31 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 a d}+\frac{4 (49 A-37 B) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 B \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt{a \sec (c+d x)+a}} \]
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Rubi [A] time = 0.605622, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4021, 4010, 4001, 3795, 203} \[ \frac{2 (7 A-B) \tan (c+d x) \sec ^2(c+d x)}{35 d \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} (A-B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (7 A-31 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 a d}+\frac{4 (49 A-37 B) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 B \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt{a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 4021
Rule 4010
Rule 4001
Rule 3795
Rule 203
Rubi steps
\begin{align*} \int \frac{\sec ^4(c+d x) (A+B \sec (c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{2 B \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 \int \frac{\sec ^3(c+d x) \left (3 a B+\frac{1}{2} a (7 A-B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{7 a}\\ &=\frac{2 (7 A-B) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 B \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{4 \int \frac{\sec ^2(c+d x) \left (a^2 (7 A-B)-\frac{1}{4} a^2 (7 A-31 B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{35 a^2}\\ &=\frac{2 (7 A-B) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 B \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}-\frac{2 (7 A-31 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+\frac{8 \int \frac{\sec (c+d x) \left (-\frac{1}{8} a^3 (7 A-31 B)+\frac{1}{4} a^3 (49 A-37 B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{105 a^3}\\ &=\frac{4 (49 A-37 B) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (7 A-B) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 B \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}-\frac{2 (7 A-31 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+(-A+B) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{4 (49 A-37 B) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (7 A-B) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 B \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}-\frac{2 (7 A-31 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+\frac{(2 (A-B)) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac{\sqrt{2} (A-B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{4 (49 A-37 B) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (7 A-B) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 B \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}-\frac{2 (7 A-31 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 a d}\\ \end{align*}
Mathematica [A] time = 0.52584, size = 140, normalized size = 0.69 \[ \frac{\tan (c+d x) \left (2 \sqrt{1-\sec (c+d x)} \left (3 (7 A-B) \sec ^2(c+d x)+(31 B-7 A) \sec (c+d x)+91 A+15 B \sec ^3(c+d x)-43 B\right )-105 \sqrt{2} (A-B) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{105 d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.34, size = 785, normalized size = 3.9 \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 [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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Fricas [A] time = 0.598308, size = 1116, normalized size = 5.52 \begin{align*} \left [-\frac{105 \, \sqrt{2}{\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{4} +{\left (A - B\right )} a \cos \left (d x + c\right )^{3}\right )} \sqrt{-\frac{1}{a}} \log \left (-\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left ({\left (91 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} -{\left (7 \, A - 31 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (7 \, A - B\right )} \cos \left (d x + c\right ) + 15 \, B\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{210 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}, \frac{2 \,{\left ({\left (91 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} -{\left (7 \, A - 31 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (7 \, A - B\right )} \cos \left (d x + c\right ) + 15 \, B\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac{105 \, \sqrt{2}{\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{4} +{\left (A - B\right )} a \cos \left (d x + c\right )^{3}\right )} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right )}{\sqrt{a}}}{105 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}\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*} \int \frac{\left (A + B \sec{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 9.47519, size = 387, normalized size = 1.92 \begin{align*} -\frac{\frac{105 \, \sqrt{2}{\left (A - B\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{2 \,{\left (\frac{105 \, \sqrt{2} A a^{3}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} -{\left ({\left (\frac{\sqrt{2}{\left (119 \, A a^{3} - 92 \, B a^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{7 \, \sqrt{2}{\left (37 \, A a^{3} - 16 \, B a^{3}\right )}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{35 \, \sqrt{2}{\left (7 \, A a^{3} - 4 \, B a^{3}\right )}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{105 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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