Optimal. Leaf size=270 \[ \frac {(163 A-75 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(95 A-39 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{48 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {(17 A-9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}} \]
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Rubi [A] time = 0.91, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2955, 4020, 4022, 4013, 3808, 206} \[ \frac {(95 A-39 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{48 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}+\frac {(163 A-75 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(17 A-9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 206
Rule 2955
Rule 3808
Rule 4013
Rule 4020
Rule 4022
Rubi steps
\begin {align*} \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (11 A-3 B)-3 a (A-B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(17 A-9 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} a^2 (95 A-39 B)-a^2 (17 A-9 B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(17 A-9 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(95 A-39 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{8} a^3 (299 A-147 B)+\frac {1}{4} a^3 (95 A-39 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{12 a^5}\\ &=-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(17 A-9 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(95 A-39 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {\left ((163 A-75 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(17 A-9 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(95 A-39 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {\left ((163 A-75 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=\frac {(163 A-75 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(17 A-9 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(299 A-147 B) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(95 A-39 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 1.70, size = 183, normalized size = 0.68 \[ \frac {2 \tan (c+d x) \sqrt {1-\sec (c+d x)} \sec (c+d x) ((255 B-479 A) \cos (c+d x)+(48 B-80 A) \cos (2 (c+d x))+8 A \cos (3 (c+d x))-379 A+195 B)-12 \sqrt {2} (163 A-75 B) \sin (c+d x) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )}{96 d \sqrt {\cos (c+d x)-1} (a (\sec (c+d x)+1))^{5/2}} \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.47, size = 542, normalized size = 2.01 \[ \left [-\frac {3 \, \sqrt {2} {\left ({\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right ) + 163 \, A - 75 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (32 \, A \cos \left (d x + c\right )^{3} - 32 \, {\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (503 \, A - 255 \, B\right )} \cos \left (d x + c\right ) - 299 \, A + 147 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{192 \, {\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 {3 \, \sqrt {2} {\left ({\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (163 \, A - 75 \, B\right )} \cos \left (d x + c\right ) + 163 \, A - 75 \, B\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) - 2 \, {\left (32 \, A \cos \left (d x + c\right )^{3} - 32 \, {\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (503 \, A - 255 \, B\right )} \cos \left (d x + c\right ) - 299 \, A + 147 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{96 \, {\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 [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 2.29, size = 439, normalized size = 1.63 \[ -\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (489 A \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-225 B \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+978 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+64 A \left (\cos ^{4}\left (d x +c \right )\right )-450 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+489 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, A \sin \left (d x +c \right )-384 A \left (\cos ^{3}\left (d x +c \right )\right )-225 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, B \sin \left (d x +c \right )+192 B \left (\cos ^{3}\left (d x +c \right )\right )-686 A \left (\cos ^{2}\left (d x +c \right )\right )+318 B \left (\cos ^{2}\left (d x +c \right )\right )+408 A \cos \left (d x +c \right )-216 B \cos \left (d x +c \right )+598 A -294 B \right )}{96 d \,a^{3} \sin \left (d x +c \right )^{5}} \]
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 = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-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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