Optimal. Leaf size=387 \[ \frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)}+\frac {\sin (c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x)}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}+\frac {\sin (c+d x) \left (-5 a^3 C+3 a^2 b B-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}}-\frac {\left (-5 a^4 C+3 a^3 b B-a^2 b^2 (A-7 C)-5 a b^3 B+3 A b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2} \]
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Rubi [A] time = 1.50, antiderivative size = 387, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4112, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}-\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 b B-5 a^3 C-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right )}-\frac {\left (-a^2 b^2 (A-7 C)+3 a^3 b B-5 a^4 C-5 a b^3 B+3 A b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2}-\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)}+\frac {\sin (c+d x) \left (5 a^2 C-3 a b B+3 A b^2-2 b^2 C\right )}{3 b^2 d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x)}+\frac {\sin (c+d x) \left (3 a^2 b B-5 a^3 C-a b^2 (A-4 C)-2 b^3 B\right )}{b^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2805
Rule 3002
Rule 3055
Rule 3059
Rule 4112
Rubi steps
\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx &=\int \frac {C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))^2} \, dx\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {\int \frac {\frac {1}{2} \left (-3 A b^2+3 a b B-5 a^2 C+2 b^2 C\right )+b (b B-a (A+C)) \cos (c+d x)+\frac {3}{2} \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {2 \int \frac {-\frac {3}{4} \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right )+\frac {1}{2} b \left (3 A b^2-3 a b B+2 a^2 C+b^2 C\right ) \cos (c+d x)-\frac {1}{4} a \left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {4 \int \frac {\frac {1}{8} \left (9 a^3 b B-12 a b^3 B-a^2 b^2 (3 A-16 C)-15 a^4 C+2 b^4 (3 A+C)\right )+\frac {1}{4} b \left (6 a^2 b B-3 b^3 B-a b^2 (3 A-7 C)-10 a^3 C\right ) \cos (c+d x)+\frac {3}{8} a \left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 b^3 \left (a^2-b^2\right )}\\ &=\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}+\frac {4 \int \frac {-\frac {1}{8} a \left (9 a^3 b B-12 a b^3 B-a^2 b^2 (3 A-16 C)-15 a^4 C+2 b^4 (3 A+C)\right )+\frac {1}{8} a^2 b \left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a b^3 \left (a^2-b^2\right )}-\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 b^3 \left (a^2-b^2\right )}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^2 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^4+3 a^3 b B-5 a b^3 B-a^2 b^2 (A-7 C)-5 a^4 C\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{(a-b) b^3 (a+b)^2 d}+\frac {\left (3 A b^2-3 a b B+5 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (3 a^2 b B-2 b^3 B-a b^2 (A-4 C)-5 a^3 C\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}\\ \end {align*}
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Mathematica [A] time = 6.77, size = 423, normalized size = 1.09 \[ \frac {\frac {8 b \left (b^2-a^2\right ) (3 b B-5 a C) \sin (c+d x)+8 b^2 C \left (b^2-a^2\right ) \tan (c+d x)+6 a \sin (2 (c+d x)) \left (5 a^3 C-3 a^2 b B+a b^2 (A-4 C)+2 b^3 B\right )}{\left (b^2-a^2\right ) \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)}+\frac {\frac {8 b \left (10 a^3 C-6 a^2 b B+a b^2 (3 A-7 C)+3 b^3 B\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a (a+b)}+\frac {6 \sin (c+d x) \left (5 a^3 C-3 a^2 b B+a b^2 (A-4 C)+2 b^3 B\right ) \left (\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt {\sin ^2(c+d x)}}+\frac {2 \left (45 a^4 C-27 a^3 b B+a^2 b^2 (9 A-44 C)+30 a b^3 B-4 b^4 (3 A+C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b) (a+b)}}{12 b^3 d} \]
Antiderivative was successfully verified.
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fricas [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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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 24.03, size = 1031, normalized size = 2.66 \[ \text {result too large to display} \]
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 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
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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