Optimal. Leaf size=237 \[ \frac {(4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {5 (A-2 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a^2 d}+\frac {(4 A-7 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {5 (A-2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {(4 A-7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.37, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4019, 3787, 3768, 3771, 2639, 2641} \[ \frac {(4 A-7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {5 (A-2 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a^2 d}+\frac {(4 A-7 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {5 (A-2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {(4 A-7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 3768
Rule 3771
Rule 3787
Rule 4019
Rubi steps
\begin {align*} \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx &=\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (\frac {5}{2} a (A-B)-\frac {3}{2} a (A-3 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac {(4 A-7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {\int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {3}{2} a^2 (4 A-7 B)-\frac {15}{2} a^2 (A-2 B) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac {(4 A-7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(4 A-7 B) \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{2 a^2}-\frac {(5 (A-2 B)) \int \sec ^{\frac {5}{2}}(c+d x) \, dx}{2 a^2}\\ &=\frac {(4 A-7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac {5 (A-2 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {(4 A-7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(4 A-7 B) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a^2}-\frac {(5 (A-2 B)) \int \sqrt {\sec (c+d x)} \, dx}{6 a^2}\\ &=\frac {(4 A-7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac {5 (A-2 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {(4 A-7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\left ((4 A-7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2}-\frac {\left (5 (A-2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2}\\ &=-\frac {(4 A-7 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {5 (A-2 B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {(4 A-7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac {5 (A-2 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {(4 A-7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}
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Mathematica [C] time = 8.03, size = 865, normalized size = 3.65 \[ \frac {4 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc \left (\frac {c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt {1+e^{2 i (c+d x)}}\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}-\frac {7 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc \left (\frac {c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt {1+e^{2 i (c+d x)}}\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}-\frac {10 A \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sec \left (\frac {c}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac {20 B \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sec \left (\frac {c}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)) \left (\frac {2 \sec \left (\frac {c}{2}\right ) \left (B \sin \left (\frac {d x}{2}\right )-A \sin \left (\frac {d x}{2}\right )\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d}+\frac {2 (B-A) \tan \left (\frac {c}{2}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d}+\frac {4 \sec \left (\frac {c}{2}\right ) \left (8 B \sin \left (\frac {d x}{2}\right )-5 A \sin \left (\frac {d x}{2}\right )\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d}-\frac {2 (7 B-4 A) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{d}+\frac {8 B \sec (c) \sec (c+d x) \sin (d x)}{3 d}+\frac {4 (10 \cos (c) B+2 B-5 A \cos (c)) \sec (c) \tan \left (\frac {c}{2}\right )}{3 d}\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{(B+A \cos (c+d x)) (\sec (c+d x) a+a)^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right )^{4} + A \sec \left (d x + c\right )^{3}\right )} \sqrt {\sec \left (d x + c\right )}}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, x\right ) \]
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 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 15.80, size = 750, normalized size = 3.16 \[ \text {result too large to display} \]
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 \[ \int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
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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