Optimal. Leaf size=216 \[ -\frac{(75 B-163 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(39 B-95 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{48 a^3 d}+\frac{(93 B-197 C) \tan (c+d x)}{24 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(B-C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac{(9 B-17 C) \tan (c+d x) \sec ^2(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]
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Rubi [A] time = 0.747935, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4072, 4019, 4010, 4001, 3795, 203} \[ -\frac{(75 B-163 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(39 B-95 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{48 a^3 d}+\frac{(93 B-197 C) \tan (c+d x)}{24 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(B-C) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac{(9 B-17 C) \tan (c+d x) \sec ^2(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 4019
Rule 4010
Rule 4001
Rule 3795
Rule 203
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=\int \frac{\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{\int \frac{\sec ^3(c+d x) \left (3 a (B-C)-\frac{1}{2} a (3 B-11 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(9 B-17 C) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\sec ^2(c+d x) \left (a^2 (9 B-17 C)-\frac{1}{4} a^2 (39 B-95 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(9 B-17 C) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(39 B-95 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}+\frac{\int \frac{\sec (c+d x) \left (-\frac{1}{8} a^3 (39 B-95 C)+\frac{1}{4} a^3 (93 B-197 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{12 a^5}\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(9 B-17 C) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(93 B-197 C) \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(39 B-95 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}-\frac{(75 B-163 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(9 B-17 C) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(93 B-197 C) \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(39 B-95 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}+\frac{(75 B-163 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 )}{16 a^2 d}\\ &=-\frac{(75 B-163 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(B-C) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(9 B-17 C) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(93 B-197 C) \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(39 B-95 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}\\ \end{align*}
Mathematica [A] time = 2.59787, size = 161, normalized size = 0.75 \[ \frac{\tan (c+d x) \left (\sqrt{1-\sec (c+d x)} \left (32 (3 B-5 C) \sec ^2(c+d x)+(255 B-503 C) \sec (c+d x)+147 B+32 C \sec ^3(c+d x)-299 C\right )-6 \sqrt{2} (75 B-163 C) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{48 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.306, size = 795, normalized size = 3.7 \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 = 0.652811, size = 1474, normalized size = 6.82 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (B + C \sec{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 9.78534, size = 420, normalized size = 1.94 \begin{align*} \frac{\frac{{\left ({\left (3 \,{\left (\frac{2 \, \sqrt{2}{\left (B a^{5} - C a^{5}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{\sqrt{2}{\left (15 \, B a^{5} - 23 \, C a^{5}\right )}}{a^{6} \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{4 \, \sqrt{2}{\left (75 \, B a^{5} - 167 \, C a^{5}\right )}}{a^{6} \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{3 \, \sqrt{2}{\left (83 \, B a^{5} - 155 \, C a^{5}\right )}}{a^{6} \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 )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}} - \frac{3 \, \sqrt{2}{\left (75 \, B - 163 \, 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} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{96 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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