Optimal. Leaf size=184 \[ \frac {a^{5/2} (5 A+2 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^3 (15 A+70 B+64 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (15 A-10 B-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}-\frac {a (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d} \]
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Rubi [A] time = 0.40, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4086, 3917, 3915, 3774, 203, 3792} \[ \frac {a^3 (15 A+70 B+64 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (15 A-10 B-16 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 d}+\frac {a^{5/2} (5 A+2 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {a (5 A-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{d} \]
Antiderivative was successfully verified.
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Rule 203
Rule 3774
Rule 3792
Rule 3915
Rule 3917
Rule 4086
Rubi steps
\begin {align*} \int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}+\frac {\int (a+a \sec (c+d x))^{5/2} \left (\frac {1}{2} a (5 A+2 B)-\frac {1}{2} a (5 A-2 C) \sec (c+d x)\right ) \, dx}{a}\\ &=\frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {2 \int (a+a \sec (c+d x))^{3/2} \left (\frac {5}{4} a^2 (5 A+2 B)-\frac {1}{4} a^2 (15 A-10 B-16 C) \sec (c+d x)\right ) \, dx}{5 a}\\ &=\frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {a^2 (15 A-10 B-16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4 \int \sqrt {a+a \sec (c+d x)} \left (\frac {15}{8} a^3 (5 A+2 B)+\frac {1}{8} a^3 (15 A+70 B+64 C) \sec (c+d x)\right ) \, dx}{15 a}\\ &=\frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {a^2 (15 A-10 B-16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{2} \left (a^2 (5 A+2 B)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx+\frac {1}{30} \left (a^2 (15 A+70 B+64 C)\right ) \int \sec (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}+\frac {a^3 (15 A+70 B+64 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (15 A-10 B-16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}-\frac {\left (a^3 (5 A+2 B)\right ) \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 )}{d}\\ &=\frac {a^{5/2} (5 A+2 B) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{d}+\frac {a^3 (15 A+70 B+64 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (15 A-10 B-16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end {align*}
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Mathematica [A] time = 2.01, size = 209, normalized size = 1.14 \[ \frac {\cos (c+d x) (a (\sec (c+d x)+1))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {\cos (c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) ((45 A+40 B+112 C) \cos (c+d x)+4 (15 A+40 B+43 C) \cos (2 (c+d x))+15 A \cos (3 (c+d x))+60 A+160 B+196 C)}{(\cos (c+d x)+1)^2}+\frac {60 (5 A+2 B) \sin (c+d x) \tan ^{-1}\left (\sqrt {\sec (c+d x)-1}\right )}{\sqrt {\sec (c+d x)-1} (\sec (c+d x)+1)^3}\right )}{30 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 440, normalized size = 2.39 \[ \left [\frac {15 \, {\left ({\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (15 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (15 \, A + 40 \, B + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, B + 14 \, C\right )} a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{30 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, -\frac {15 \, {\left ({\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (15 \, A a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (15 \, A + 40 \, B + 43 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, B + 14 \, C\right )} a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 2.41, size = 559, normalized size = 3.04 \[ -\frac {15 \, {\left (5 \, A \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, B \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \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 ) - 15 \, {\left (5 \, A \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 2 \, B \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \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 ) + \frac {60 \, {\left (3 \, \sqrt {2} {\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 \sqrt {-a} a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} A \sqrt {-a} a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{{\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 )}^{4} - 6 \, {\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 + a^{2}} - \frac {4 \, {\left ({\left (\sqrt {2} {\left (15 \, A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 35 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 32 \, C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, \sqrt {2} {\left (3 \, A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 8 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 8 \, C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \sqrt {2} {\left (A a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, B a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 4 \, C a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\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 )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{30 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.73, size = 604, normalized size = 3.28 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (75 A \sqrt {2}\, \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+30 B \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}\, \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )+150 A \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+60 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )+75 A \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )+30 B \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \sin \left (d x +c \right )+120 A \left (\cos ^{4}\left (d x +c \right )\right )+120 A \left (\cos ^{3}\left (d x +c \right )\right )+640 B \left (\cos ^{3}\left (d x +c \right )\right )+688 C \left (\cos ^{3}\left (d x +c \right )\right )-240 A \left (\cos ^{2}\left (d x +c \right )\right )-560 B \left (\cos ^{2}\left (d x +c \right )\right )-464 C \left (\cos ^{2}\left (d x +c \right )\right )-80 B \cos \left (d x +c \right )-176 C \cos \left (d x +c \right )-48 C \right ) a^{2}}{120 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.83, size = 2780, normalized size = 15.11 \[ \text {result too large to display} \]
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 \[ \int \cos \left (c+d\,x\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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