Optimal. Leaf size=169 \[ \frac {(19 A-75 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-9 B) \tan (c+d x)}{4 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {(5 A-13 B) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]
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Rubi [A] time = 0.45, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4019, 4008, 4001, 3795, 203} \[ \frac {(19 A-75 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-9 B) \tan (c+d x)}{4 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {(5 A-13 B) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 203
Rule 3795
Rule 4001
Rule 4008
Rule 4019
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {\int \frac {\sec ^2(c+d x) \left (2 a (A-B)-\frac {1}{2} a (A-9 B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {\int \frac {\sec (c+d x) \left (-\frac {3}{4} a^2 (5 A-13 B)+a^2 (A-9 B) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(A-9 B) \tan (c+d x)}{4 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(19 A-75 B) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(A-9 B) \tan (c+d x)}{4 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(19 A-75 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 {(19 A-75 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) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 A-13 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(A-9 B) \tan (c+d x)}{4 a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 1.58, size = 144, normalized size = 0.85 \[ \frac {\tan (c+d x) \left (\sqrt {1-\sec (c+d x)} \left ((85 B-13 A) \sec (c+d x)-9 A+32 B \sec ^2(c+d x)+49 B\right )+2 \sqrt {2} (19 A-75 B) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )\right )}{16 d \sqrt {1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 484, normalized size = 2.86 \[ \left [\frac {\sqrt {2} {\left ({\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right ) + 19 \, A - 75 \, B\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (9 \, A - 49 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (13 \, A - 85 \, B\right )} \cos \left (d x + c\right ) - 32 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {\sqrt {2} {\left ({\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right ) + 19 \, A - 75 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (9 \, A - 49 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (13 \, A - 85 \, B\right )} \cos \left (d x + c\right ) - 32 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 6.39, size = 289, normalized size = 1.71 \[ -\frac {\frac {{\left ({\left (\frac {2 \, {\left (\sqrt {2} A a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - \sqrt {2} B a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8}} + \frac {9 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 17 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {11 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 83 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {{\left (19 \, \sqrt {2} A - 75 \, \sqrt {2} B\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{32 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.63, size = 597, normalized size = 3.53 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (19 A \sin \left (d x +c \right ) \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )-75 B \sin \left (d x +c \right ) \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+38 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-150 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+19 A \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+18 A \left (\cos ^{3}\left (d x +c \right )\right )-75 B \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-98 B \left (\cos ^{3}\left (d x +c \right )\right )+8 A \left (\cos ^{2}\left (d x +c \right )\right )-72 B \left (\cos ^{2}\left (d x +c \right )\right )-26 A \cos \left (d x +c \right )+106 B \cos \left (d x +c \right )+64 B \right )}{32 d \sin \left (d x +c \right )^{5} a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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