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 = 0.61, antiderivative size = 202, normalized size of antiderivative = 1.00, 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 203
Rule 3795
Rule 4001
Rule 4010
Rule 4021
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*}
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Mathematica [A] time = 0.54, 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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fricas [A] time = 0.51, size = 432, normalized size = 2.14 \[ \left [-\frac {105 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{4} + {\left (A - B\right )} a \cos \left (d x + c\right )^{3}\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (91 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} - {\left (7 \, A - 31 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, A - B\right )} \cos \left (d x + c\right ) + 15 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{210 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}, \frac {2 \, {\left ({\left (91 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} - {\left (7 \, A - 31 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, A - B\right )} \cos \left (d x + c\right ) + 15 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac {105 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{4} + {\left (A - B\right )} a \cos \left (d x + c\right )^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{105 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.35, size = 287, normalized size = 1.42 \[ -\frac {\frac {105 \, \sqrt {2} {\left (A - B\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {2 \, {\left (\frac {105 \, \sqrt {2} A a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - {\left ({\left (\frac {\sqrt {2} {\left (119 \, A a^{3} - 92 \, B a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {7 \, \sqrt {2} {\left (37 \, A a^{3} - 16 \, B a^{3}\right )}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {35 \, \sqrt {2} {\left (7 \, A a^{3} - 4 \, B a^{3}\right )}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{105 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.88, size = 785, normalized size = 3.89 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{4}}{\sqrt {a \sec \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]
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 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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