Optimal. Leaf size=159 \[ \frac{\sqrt{2} (B-C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (5 B-C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 a d}-\frac{4 (5 B-7 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.514366, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4072, 4021, 4010, 4001, 3795, 203} \[ \frac{\sqrt{2} (B-C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (5 B-C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 a d}-\frac{4 (5 B-7 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 4072
Rule 4021
Rule 4010
Rule 4001
Rule 3795
Rule 203
Rubi steps
\begin{align*} \int \frac{\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\int \frac{\sec ^3(c+d x) (B+C \sec (c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 \int \frac{\sec ^2(c+d x) \left (2 a C+\frac{1}{2} a (5 B-C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{5 a}\\ &=\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+\frac{4 \int \frac{\sec (c+d x) \left (\frac{1}{4} a^2 (5 B-C)-\frac{1}{2} a^2 (5 B-7 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{15 a^2}\\ &=-\frac{4 (5 B-7 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+(B-C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=-\frac{4 (5 B-7 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 a d}-\frac{(2 (B-C)) \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} (B-C) \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 (5 B-7 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 a d}\\ \end{align*}
Mathematica [A] time = 0.376557, size = 123, normalized size = 0.77 \[ \frac{\tan (c+d x) \left (2 \sqrt{1-\sec (c+d x)} \left ((5 B-C) \sec (c+d x)-5 B+3 C \sec ^2(c+d x)+13 C\right )+15 \sqrt{2} (B-C) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{15 d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.332, size = 595, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{2}}{\sqrt{a \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 0.652144, size = 1019, normalized size = 6.41 \begin{align*} \left [-\frac{15 \, \sqrt{2}{\left ({\left (B - C\right )} a \cos \left (d x + c\right )^{3} +{\left (B - C\right )} a \cos \left (d x + c\right )^{2}\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 (5 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{2} -{\left (5 \, B - C\right )} \cos \left (d x + c\right ) - 3 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{30 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}, -\frac{2 \,{\left ({\left (5 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{2} -{\left (5 \, B - C\right )} \cos \left (d x + c\right ) - 3 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac{15 \, \sqrt{2}{\left ({\left (B - C\right )} a \cos \left (d x + c\right )^{3} +{\left (B - C\right )} a \cos \left (d x + c\right )^{2}\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}}}{15 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (B + C \sec{\left (c + d x \right )}\right ) \sec ^{3}{\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.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 8.94, size = 366, normalized size = 2.3 \begin{align*} \frac{\frac{15 \,{\left (\sqrt{2} B - \sqrt{2} C\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 ({\left (10 \, \sqrt{2} B a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) - 20 \, \sqrt{2} C a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) -{\left (10 \, \sqrt{2} B a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) - 17 \, \sqrt{2} C a^{2} \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 )^{2} + \frac{15 \, \sqrt{2} C a^{2}}{\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 )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{15 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]