Optimal. Leaf size=164 \[ -\frac{(43 A-3 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(11 A-3 B) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.253715, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3922, 3920, 3774, 203, 3795} \[ -\frac{(43 A-3 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(11 A-3 B) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3922
Rule 3920
Rule 3774
Rule 203
Rule 3795
Rubi steps
\begin{align*} \int \frac{A+B \sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac{(A-B) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{\int \frac{-4 a A+\frac{3}{2} a (A-B) \sec (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(11 A-3 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{8 a^2 A-\frac{1}{4} a^2 (11 A-3 B) \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(11 A-3 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{A \int \sqrt{a+a \sec (c+d x)} \, dx}{a^3}-\frac{(43 A-3 B) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A-B) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(11 A-3 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(2 A) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^2 d}+\frac{(43 A-3 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 )}{16 a^2 d}\\ &=\frac{2 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac{(43 A-3 B) \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{(A-B) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(11 A-3 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 26.6766, size = 10177, normalized size = 62.05 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.204, size = 824, normalized size = 5. \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{B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [B] time = 16.1611, size = 1754, normalized size = 10.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Giac [B] time = 11.3761, size = 471, normalized size = 2.87 \begin{align*} -\frac{2 \, \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}{\left (\frac{2 \, \sqrt{2}{\left (A a^{5} - B a^{5}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{8} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{\sqrt{2}{\left (13 \, A a^{5} - 5 \, B a^{5}\right )}}{a^{8} \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 ) + \frac{\sqrt{2}{\left (43 \, A - 3 \, 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 )}^{2}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{64 \, A \log \left ({\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 )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{64 \, A \log \left ({\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 )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]