Optimal. Leaf size=223 \[ \frac {\left (2 a^2 A-3 A b^2+a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {b \left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 (a-b) (a+b)^2 d}+\frac {b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))} \]
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Rubi [A]
time = 0.45, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3033, 3068,
3138, 2719, 3081, 2720, 2884} \begin {gather*} \frac {\left (2 a^2 A+a b B-3 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {\left (-2 a^3 B+4 a^2 A b+a b^2 B-3 A b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 B+5 a^2 A b+a b^2 B-3 A b^3\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 2720
Rule 2884
Rule 3033
Rule 3068
Rule 3081
Rule 3138
Rubi steps
\begin {align*} \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {\cos ^{\frac {3}{2}}(c+d x) (B+A \cos (c+d x))}{(b+a \cos (c+d x))^2} \, dx\\ &=\frac {b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{2} b (A b-a B)-a (A b-a B) \cos (c+d x)+\frac {1}{2} \left (2 a^2 A-3 A b^2+a b B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}-\frac {\int \frac {-\frac {1}{2} a b (A b-a B)+\frac {1}{2} \left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a^2 \left (a^2-b^2\right )}+\frac {\left (2 a^2 A-3 A b^2+a b B\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2 A-3 A b^2+a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\left (b \left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )}-\frac {\left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2 A-3 A b^2+a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {b \left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 (a-b) (a+b)^2 d}+\frac {b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 12.89, size = 281, normalized size = 1.26 \begin {gather*} \frac {\frac {4 b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (b+a \cos (c+d x))}+\frac {\frac {2 \left (2 a^2 A-A b^2-a b B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {8 (-A b+a B) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac {2 \left (2 a^2 A-3 A b^2+a b B\right ) \left (-2 a b E\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{4 a d} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(842\) vs.
\(2(299)=598\).
time = 5.76, size = 843, normalized size = 3.78
method | result | size |
default | \(\text {Expression too large to display}\) | \(843\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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