Optimal. Leaf size=368 \[ \frac {2 b (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (-5 a^3 B+8 a^2 A b+a b^2 B-4 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (-3 a^3 B+9 a^2 A b+2 a b^2 B-8 A b^3\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^4 A+6 a^3 b B-15 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]
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Rubi [A] time = 0.94, antiderivative size = 368, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4030, 4100, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac {2 b \left (8 a^2 A b-5 a^3 B+a b^2 B-4 A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {2 b (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (9 a^2 A b-3 a^3 B+2 a b^2 B-8 A b^3\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-15 a^2 A b^2+3 a^4 A+6 a^3 b B-2 a b^3 B+8 A b^4\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]
Antiderivative was successfully verified.
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Rule 2653
Rule 2655
Rule 2661
Rule 2663
Rule 3856
Rule 3858
Rule 4030
Rule 4035
Rule 4100
Rubi steps
\begin {align*} \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{5/2}} \, dx &=\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \int \frac {\frac {1}{2} \left (-3 a^2 A+4 A b^2-a b B\right )+\frac {3}{2} a (A b-a B) \sec (c+d x)-b (A b-a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 \int \frac {\frac {1}{4} \left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right )-\frac {1}{4} a \left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {\left (9 a^2 A b-8 A b^3-3 a^3 B+2 a b^2 B\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )}+\frac {\left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (9 a^2 A b-8 A b^3-3 a^3 B+2 a b^2 B\right ) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{3 a^3 \left (a^2-b^2\right )^2 \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (9 a^2 A b-8 A b^3-3 a^3 B+2 a b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{3 a^3 \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{3 a^3 \left (a^2-b^2\right )^2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ &=-\frac {2 \left (9 a^2 A b-8 A b^3-3 a^3 B+2 a b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{3 a^3 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^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 = 2.69, size = 297, normalized size = 0.81 \[ \frac {2 \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+b) \left (-\frac {a b \sin (c+d x) \left (a \left (6 a^3 B-9 a^2 A b-2 a b^2 B+5 A b^3\right ) \cos (c+d x)+b \left (5 a^3 B-8 a^2 A b-a b^2 B+4 A b^3\right )\right )}{\left (a^2-b^2\right )^2}-\frac {\left (\frac {a \cos (c+d x)+b}{a+b}\right )^{3/2} \left (-\left (a^2 \left (3 a^3 B-6 a^2 A b+a b^2 B+2 A b^3\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )\right )-\left (3 a^4 A+6 a^3 b B-15 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )-b F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )\right )\right )}{(a-b)^2 (a+b)}\right )}{3 a^3 d (a+b \sec (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}}{b^{3} \sec \left (d x + c\right )^{4} + 3 \, a b^{2} \sec \left (d x + c\right )^{3} + 3 \, a^{2} b \sec \left (d x + c\right )^{2} + a^{3} \sec \left (d x + c\right )}, x\right ) \]
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 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\sec \left (d x + c\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 2.44, size = 5169, normalized size = 14.05 \[ \text {output too large to display} \]
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 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]
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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