Optimal. Leaf size=257 \[ -\frac{(A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{a d \left (a^2-b^2\right )}+\frac{a (A b-a B) \sin (c+d x) \sqrt{\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{(A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}+\frac{\left (a^2 A b+a^3 B-3 a b^2 B+A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d (a-b) (a+b)^2} \]
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Rubi [A] time = 0.527313, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4029, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{a (A b-a B) \sin (c+d x) \sqrt{\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{(A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d \left (a^2-b^2\right )}-\frac{(A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}+\frac{\left (a^2 A b+a^3 B-3 a b^2 B+A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d (a-b) (a+b)^2} \]
Antiderivative was successfully verified.
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Rule 4029
Rule 4106
Rule 3849
Rule 2805
Rule 3787
Rule 3771
Rule 2639
Rule 2641
Rubi steps
\begin{align*} \int \frac{\sec ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx &=\frac{a (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{-\frac{1}{2} a (A b-a B)-b (A b-a B) \sec (c+d x)+\frac{1}{2} \left (a A b+a^2 B-2 b^2 B\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{a (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{-\frac{1}{2} a^2 (A b-a B)-\frac{1}{2} a b (A b-a B) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{a^2 b \left (a^2-b^2\right )}+\frac{\left (a^2 A b+A b^3+a^3 B-3 a b^2 B\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 a b \left (a^2-b^2\right )}\\ &=\frac{a (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{(A b-a B) \int \sqrt{\sec (c+d x)} \, dx}{2 a \left (a^2-b^2\right )}-\frac{(A b-a B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 b \left (a^2-b^2\right )}+\frac{\left (\left (a^2 A b+A b^3+a^3 B-3 a b^2 B\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}{2 a b \left (a^2-b^2\right )}\\ &=\frac{\left (a^2 A b+A b^3+a^3 B-3 a b^2 B\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)}}{a (a-b) b (a+b)^2 d}+\frac{a (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left ((A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 a \left (a^2-b^2\right )}-\frac{\left ((A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{(A b-a B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b \left (a^2-b^2\right ) d}-\frac{(A b-a B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a \left (a^2-b^2\right ) d}+\frac{\left (a^2 A b+A b^3+a^3 B-3 a b^2 B\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)}}{a (a-b) b (a+b)^2 d}+\frac{a (A b-a B) \sqrt{\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [B] time = 6.88949, size = 643, normalized size = 2.5 \[ \frac{\frac{2 \left (-3 a^2 B-a A b+4 b^2 B\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \left (\text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right )+\Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )}{b \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}-\frac{2 \left (a A b-a^2 B\right ) \sin (c+d x) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (a (a-2 b) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right )+a^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 b^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 a b \sec ^2(c+d x)+2 a b \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+2 a b\right )}{a^2 b \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a \cos (c+d x)+b)}-\frac{2 \left (4 A b^2-4 a b B\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a+b \sec (c+d x)) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )}{a \left (1-\cos ^2(c+d x)\right ) (a \cos (c+d x)+b)}}{4 b d (b-a) (a+b)}+\frac{\sqrt{\sec (c+d x)} \left (\frac{A b \sin (c+d x)-a B \sin (c+d x)}{\left (b^2-a^2\right ) (a \cos (c+d x)+b)}-\frac{(A b-a B) \sin (c+d x)}{b \left (b^2-a^2\right )}\right )}{d} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 5.18, size = 715, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )}\right ) \sec ^{\frac{3}{2}}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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