Optimal. Leaf size=236 \[ \frac {85 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{8 a^{3/2} d}-\frac {15 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}+\frac {35 A \sin (c+d x)}{8 a d \sqrt {a-a \sec (c+d x)}}+\frac {4 A \sin (c+d x) \cos ^2(c+d x)}{3 a d \sqrt {a-a \sec (c+d x)}}-\frac {A \sin (c+d x) \cos ^2(c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {25 A \sin (c+d x) \cos (c+d x)}{12 a d \sqrt {a-a \sec (c+d x)}} \]
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Rubi [A] time = 0.70, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {85 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{8 a^{3/2} d}-\frac {15 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}+\frac {35 A \sin (c+d x)}{8 a d \sqrt {a-a \sec (c+d x)}}+\frac {4 A \sin (c+d x) \cos ^2(c+d x)}{3 a d \sqrt {a-a \sec (c+d x)}}-\frac {A \sin (c+d x) \cos ^2(c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {25 A \sin (c+d x) \cos (c+d x)}{12 a d \sqrt {a-a \sec (c+d x)}} \]
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 ^3(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx &=-\frac {A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {\int \frac {\cos ^3(c+d x) (8 a A+7 a A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a-a \sec (c+d x)}}-\frac {\int \frac {\cos ^2(c+d x) \left (-25 a^2 A-20 a^2 A \sec (c+d x)\right )}{\sqrt {a-a \sec (c+d x)}} \, dx}{6 a^3}\\ &=-\frac {A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {25 A \cos (c+d x) \sin (c+d x)}{12 a d \sqrt {a-a \sec (c+d x)}}+\frac {4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a-a \sec (c+d x)}}+\frac {\int \frac {\cos (c+d x) \left (\frac {105 a^3 A}{2}+\frac {75}{2} a^3 A \sec (c+d x)\right )}{\sqrt {a-a \sec (c+d x)}} \, dx}{12 a^4}\\ &=-\frac {A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {35 A \sin (c+d x)}{8 a d \sqrt {a-a \sec (c+d x)}}+\frac {25 A \cos (c+d x) \sin (c+d x)}{12 a d \sqrt {a-a \sec (c+d x)}}+\frac {4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a-a \sec (c+d x)}}-\frac {\int \frac {-\frac {255 a^4 A}{4}-\frac {105}{4} a^4 A \sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{12 a^5}\\ &=-\frac {A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {35 A \sin (c+d x)}{8 a d \sqrt {a-a \sec (c+d x)}}+\frac {25 A \cos (c+d x) \sin (c+d x)}{12 a d \sqrt {a-a \sec (c+d x)}}+\frac {4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a-a \sec (c+d x)}}+\frac {(85 A) \int \sqrt {a-a \sec (c+d x)} \, dx}{16 a^2}+\frac {(15 A) \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{2 a}\\ &=-\frac {A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {35 A \sin (c+d x)}{8 a d \sqrt {a-a \sec (c+d x)}}+\frac {25 A \cos (c+d x) \sin (c+d x)}{12 a d \sqrt {a-a \sec (c+d x)}}+\frac {4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a-a \sec (c+d x)}}+\frac {(85 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 )}{8 a d}-\frac {(15 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 {85 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{8 a^{3/2} d}-\frac {15 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 ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac {35 A \sin (c+d x)}{8 a d \sqrt {a-a \sec (c+d x)}}+\frac {25 A \cos (c+d x) \sin (c+d x)}{12 a d \sqrt {a-a \sec (c+d x)}}+\frac {4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt {a-a \sec (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 6.73, size = 452, normalized size = 1.92 \[ A \left (\frac {\sin ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x) \left (\frac {65 \sin \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )}{12 d}+\frac {25 \sin \left (\frac {3 c}{2}\right ) \sin \left (\frac {3 d x}{2}\right )}{3 d}+\frac {5 \sin \left (\frac {5 c}{2}\right ) \sin \left (\frac {5 d x}{2}\right )}{4 d}+\frac {\sin \left (\frac {7 c}{2}\right ) \sin \left (\frac {7 d x}{2}\right )}{6 d}-\frac {65 \cos \left (\frac {c}{2}\right ) \cos \left (\frac {d x}{2}\right )}{12 d}-\frac {25 \cos \left (\frac {3 c}{2}\right ) \cos \left (\frac {3 d x}{2}\right )}{3 d}-\frac {5 \cos \left (\frac {5 c}{2}\right ) \cos \left (\frac {5 d x}{2}\right )}{4 d}-\frac {\cos \left (\frac {7 c}{2}\right ) \cos \left (\frac {7 d x}{2}\right )}{6 d}-\frac {2 \cot \left (\frac {c}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}+\frac {2 \csc \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \csc ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )}{(a-a \sec (c+d x))^{3/2}}-\frac {5 e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \sin ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) \left (17 \sinh ^{-1}\left (e^{i (c+d x)}\right )-24 \sqrt {2} \tanh ^{-1}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+17 \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{4 \sqrt {2} d (a-a \sec (c+d x))^{3/2}}\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 572, normalized size = 2.42 \[ \left [-\frac {180 \, \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 ) + 255 \, {\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 (8 \, A \cos \left (d x + c\right )^{5} + 26 \, A \cos \left (d x + c\right )^{4} + 73 \, A \cos \left (d x + c\right )^{3} - 50 \, A \cos \left (d x + c\right )^{2} - 105 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{48 \, {\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}, \frac {180 \, \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 ) - 255 \, {\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 (8 \, A \cos \left (d x + c\right )^{5} + 26 \, A \cos \left (d x + c\right )^{4} + 73 \, A \cos \left (d x + c\right )^{3} - 50 \, A \cos \left (d x + c\right )^{2} - 105 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{24 \, {\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 = 1.53, size = 320, normalized size = 1.36 \[ -\frac {\frac {180 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {255 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {12 \, \sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - \frac {\sqrt {2} {\left (63 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {5}{2}} A + 272 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A a + 324 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{3} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.86, size = 1104, normalized size = 4.68 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (A \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^3\,\left (A+\frac {A}{\cos \left (c+d\,x\right )}\right )}{{\left (a-\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
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 \[ A \left (\int \frac {\cos ^{3}{\left (c + d x \right )}}{- a \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx + \int \frac {\cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{- a \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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