Optimal. Leaf size=388 \[ -\frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac {a \left (5 a^2-11 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right )^2}+\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 b^3 d \left (a^2-b^2\right )^2}-\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac {a \left (15 a^4-38 a^2 b^2+35 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d (a-b)^2 (a+b)^3} \]
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Rubi [A] time = 0.96, antiderivative size = 388, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3845, 4098, 4102, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac {a^2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac {\left (-29 a^2 b^2+15 a^4+8 b^4\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 b^3 d \left (a^2-b^2\right )^2}-\frac {a \left (5 a^2-11 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right )^2}-\frac {\left (-29 a^2 b^2+15 a^4+8 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac {a \left (-38 a^2 b^2+15 a^4+35 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d (a-b)^2 (a+b)^3} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 2641
Rule 2805
Rule 3771
Rule 3787
Rule 3845
Rule 3849
Rule 4098
Rule 4102
Rule 4106
Rubi steps
\begin {align*} \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a+b \sec (c+d x))^3} \, dx &=-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3 a^2}{2}-2 a b \sec (c+d x)-\frac {1}{2} \left (5 a^2-4 b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\int \frac {\sqrt {\sec (c+d x)} \left (-\frac {1}{4} a^2 \left (5 a^2-11 b^2\right )+a b \left (a^2-4 b^2\right ) \sec (c+d x)+\frac {1}{4} \left (15 a^4-29 a^2 b^2+8 b^4\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\int \frac {-\frac {1}{8} a \left (15 a^4-29 a^2 b^2+8 b^4\right )-\frac {1}{2} b \left (5 a^4-10 a^2 b^2+2 b^4\right ) \sec (c+d x)-\frac {3}{8} a \left (5 a^4-11 a^2 b^2+8 b^4\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{b^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\int \frac {-\frac {1}{8} a^2 \left (15 a^4-29 a^2 b^2+8 b^4\right )-\left (\frac {1}{2} a b \left (5 a^4-10 a^2 b^2+2 b^4\right )-\frac {1}{8} a b \left (15 a^4-29 a^2 b^2+8 b^4\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{a^2 b^3 \left (a^2-b^2\right )^2}-\frac {\left (a \left (15 a^4-38 a^2 b^2+35 b^4\right )\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\left (a \left (5 a^2-11 b^2\right )\right ) \int \sqrt {\sec (c+d x)} \, dx}{8 b^2 \left (a^2-b^2\right )^2}-\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{8 b^3 \left (a^2-b^2\right )^2}-\frac {\left (a \left (15 a^4-38 a^2 b^2+35 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {a \left (15 a^4-38 a^2 b^2+35 b^4\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 (a-b)^2 b^3 (a+b)^3 d}+\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\left (a \left (5 a^2-11 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 b^2 \left (a^2-b^2\right )^2}-\frac {\left (\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {a \left (5 a^2-11 b^2\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 b^2 \left (a^2-b^2\right )^2 d}-\frac {a \left (15 a^4-38 a^2 b^2+35 b^4\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 (a-b)^2 b^3 (a+b)^3 d}+\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {a^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a^2 \left (5 a^2-11 b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [A] time = 6.88, size = 721, normalized size = 1.86 \[ \frac {\sqrt {\sec (c+d x)} \left (\frac {a^2 \sin (c+d x)}{2 b \left (b^2-a^2\right ) (a \cos (c+d x)+b)^2}+\frac {11 a^2 b^2 \sin (c+d x)-5 a^4 \sin (c+d x)}{4 b^2 \left (b^2-a^2\right )^2 (a \cos (c+d x)+b)}+\frac {\left (15 a^4-29 a^2 b^2+8 b^4\right ) \sin (c+d x)}{4 b^3 \left (b^2-a^2\right )^2}\right )}{d}-\frac {\frac {2 \left (45 a^5-95 a^3 b^2+56 a b^4\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \left (F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-\Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )\right )}{b \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}+\frac {2 \left (40 a^4 b-80 a^2 b^3+16 b^5\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )}{a \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}+\frac {\left (15 a^5-29 a^3 b^2+8 a b^4\right ) \sin (c+d x) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (2 a^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 b^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)-2 a (a-2 b) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a^2 b \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a \cos (c+d x)+b)}}{16 b^3 d (a-b)^2 (a+b)^2} \]
Warning: Unable to verify antiderivative.
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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 \[ \int \frac {\sec \left (d x + c\right )^{\frac {9}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 16.75, size = 2014, normalized size = 5.19 \[ \text {Expression too large to display} \]
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 \[ \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{9/2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]
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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