Optimal. Leaf size=408 \[ \frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (-5 a^4 C+a^2 b^2 (A+9 C)+3 A b^4\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^4 d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {2 \left (8 a^3 C+6 a^2 b C-a b^2 (A+9 C)+3 b^3 (A-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^3 d \sqrt {a+b} \left (a^2-b^2\right )} \]
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Rubi [A] time = 0.82, antiderivative size = 408, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4091, 4080, 4005, 3832, 4004} \[ \frac {2 \left (a^2 b^2 (A+9 C)-5 a^4 C+3 A b^4\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (6 a^2 b C+8 a^3 C-a b^2 (A+9 C)+3 b^3 (A-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^3 d \sqrt {a+b} \left (a^2-b^2\right )}+\frac {2 \left (a^2 b^2 (A+15 C)-8 a^4 C+3 b^4 (A-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^4 d \sqrt {a+b} \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
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Rule 3832
Rule 4004
Rule 4005
Rule 4080
Rule 4091
Rubi steps
\begin {align*} \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx &=\frac {2 a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \int \frac {\sec (c+d x) \left (-\frac {3}{2} b \left (A b^2+a^2 C\right )+\frac {1}{2} a \left (A b^2-2 a^2 C+3 b^2 C\right ) \sec (c+d x)+\frac {3}{2} b \left (a^2-b^2\right ) C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac {2 a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {4 \int \frac {\sec (c+d x) \left (-\frac {1}{2} a b^2 \left (a^2 C-b^2 (2 A+3 C)\right )+\frac {1}{4} b \left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=\frac {2 a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {\left (3 b^3 (A-C)+8 a^3 C+6 a^2 b C-a b^2 (A+9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 (a-b) b^2 (a+b)^2}-\frac {\left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )^2}\\ &=\frac {2 \left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^4 (a+b)^{3/2} d}-\frac {2 \left (3 b^3 (A-C)+8 a^3 C+6 a^2 b C-a b^2 (A+9 C)\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^3 (a+b)^{3/2} d}+\frac {2 a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 29.56, size = 702, normalized size = 1.72 \[ \frac {\sec (c+d x) (a \cos (c+d x)+b)^3 \left (A+C \sec ^2(c+d x)\right ) \left (\frac {4 \left (a^2 C \sin (c+d x)+A b^2 \sin (c+d x)\right )}{3 b \left (b^2-a^2\right ) (a \cos (c+d x)+b)^2}+\frac {8 \left (-2 a^4 C \sin (c+d x)+a^2 A b^2 \sin (c+d x)+4 a^2 b^2 C \sin (c+d x)+A b^4 \sin (c+d x)\right )}{3 b^2 \left (b^2-a^2\right )^2 (a \cos (c+d x)+b)}-\frac {4 \left (-8 a^4 C+a^2 A b^2+15 a^2 b^2 C+3 A b^4-3 b^4 C\right ) \sin (c+d x)}{3 b^3 \left (b^2-a^2\right )^2}\right )}{d (a+b \sec (c+d x))^{5/2} (A \cos (2 c+2 d x)+A+2 C)}+\frac {4 \sqrt {2} \sqrt {\frac {\cos (c+d x)}{(\cos (c+d x)+1)^2}} \sqrt {\sec (c+d x)} \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} (a \cos (c+d x)+b)^2 \left (A+C \sec ^2(c+d x)\right ) \left (\left (-8 a^4 C+a^2 b^2 (A+15 C)+3 b^4 (A-C)\right ) \cos (c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)-(a+b) \sec (c+d x) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \left (\left (8 a^4 C-a^2 b^2 (A+15 C)+3 b^4 (C-A)\right ) E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+b \left (-8 a^3 C+6 a^2 b C+a b^2 (A+9 C)+3 b^3 (A-C)\right ) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )\right )\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {\frac {1}{\cos (c+d x)+1}} \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a+b \sec (c+d x))^{5/2} (A \cos (2 c+2 d x)+A+2 C)} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{4} + A \sec \left (d x + c\right )^{2}\right )} \sqrt {b \sec \left (d x + c\right ) + a}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, x\right ) \]
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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 2.46, size = 6135, normalized size = 15.04 \[ \text {output too large to display} \]
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 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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