Optimal. Leaf size=194 \[ \frac {31 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac {11 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}} \]
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Rubi [A] time = 0.53, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4020, 4022, 3920, 3774, 203, 3795} \[ \frac {31 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac {11 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}-\frac {A \sin (c+d x) \cos (c+d x)}{d (a-a \sec (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 203
Rule 3774
Rule 3795
Rule 3920
Rule 4020
Rule 4022
Rubi steps
\begin {align*} \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx &=-\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {\int \frac {\cos ^2(c+d x) (6 a A+5 a A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}-\frac {\int \frac {\cos (c+d x) \left (-13 a^2 A-9 a^2 A \sec (c+d x)\right )}{\sqrt {a-a \sec (c+d x)}} \, dx}{4 a^3}\\ &=-\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}+\frac {\int \frac {\frac {31 a^3 A}{2}+\frac {13}{2} a^3 A \sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{4 a^4}\\ &=-\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}+\frac {(31 A) \int \sqrt {a-a \sec (c+d x)} \, dx}{8 a^2}+\frac {(11 A) \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a}\\ &=-\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}+\frac {(31 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 )}{4 a d}-\frac {(11 A) \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 )}{a d}\\ &=\frac {31 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac {11 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}-\frac {A \cos (c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {13 A \sin (c+d x)}{4 a d \sqrt {a-a \sec (c+d x)}}+\frac {3 A \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a-a \sec (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 6.77, size = 408, normalized size = 2.10 \[ A \left (\frac {\sin ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x) \left (\frac {3 \sin \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )}{2 d}+\frac {5 \sin \left (\frac {3 c}{2}\right ) \sin \left (\frac {3 d x}{2}\right )}{d}+\frac {\sin \left (\frac {5 c}{2}\right ) \sin \left (\frac {5 d x}{2}\right )}{2 d}-\frac {3 \cos \left (\frac {c}{2}\right ) \cos \left (\frac {d x}{2}\right )}{2 d}-\frac {5 \cos \left (\frac {3 c}{2}\right ) \cos \left (\frac {3 d x}{2}\right )}{d}-\frac {\cos \left (\frac {5 c}{2}\right ) \cos \left (\frac {5 d x}{2}\right )}{2 d}-\frac {2 \cot \left (\frac {c}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}+\frac {2 \csc \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \csc ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )}{(a-a \sec (c+d x))^{3/2}}-\frac {e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \sin ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) \left (31 \sinh ^{-1}\left (e^{i (c+d x)}\right )-44 \sqrt {2} \tanh ^{-1}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+31 \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{2 \sqrt {2} d (a-a \sec (c+d x))^{3/2}}\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 550, normalized size = 2.84 \[ \left [-\frac {22 \, \sqrt {2} {\left (A \cos \left (d x + c\right ) - A\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} + {\left (3 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 31 \, {\left (A \cos \left (d x + c\right ) - A\right )} \sqrt {-a} \log \left (\frac {2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} - {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 2 \, {\left (2 \, A \cos \left (d x + c\right )^{4} + 9 \, A \cos \left (d x + c\right )^{3} - 6 \, A \cos \left (d x + c\right )^{2} - 13 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{8 \, {\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}, \frac {22 \, \sqrt {2} {\left (A \cos \left (d x + c\right ) - A\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 ) \sin \left (d x + c\right ) - 31 \, {\left (A \cos \left (d x + c\right ) - A\right )} \sqrt {a} \arctan \left (\frac {\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 ) \sin \left (d x + c\right ) - {\left (2 \, A \cos \left (d x + c\right )^{4} + 9 \, A \cos \left (d x + c\right )^{3} - 6 \, A \cos \left (d x + c\right )^{2} - 13 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.45, size = 295, normalized size = 1.52 \[ -\frac {\frac {22 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {31 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (7 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A + 18 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{2} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.93, size = 883, normalized size = 4.55 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (A \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {A}{\cos \left (c+d\,x\right )}\right )}{{\left (a-\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
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 \[ A \left (\int \frac {\cos ^{2}{\left (c + d x \right )}}{- a \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx + \int \frac {\cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{- a \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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