Optimal. Leaf size=346 \[ \frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (15 A b^4+12 a^3 b B-9 a b^3 B-a^2 b^2 (16 A-3 C)-2 a^4 (A+3 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^4 \left (a^2-b^2\right ) d}+\frac {b \left (5 A b^4+5 a^3 b B-3 a b^3 B-a^2 b^2 (7 A-C)-3 a^4 C\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 (a-b) (a+b)^2 d}-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))} \]
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Rubi [A]
time = 0.84, antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {4197, 3126,
3128, 3138, 2719, 3081, 2720, 2884} \begin {gather*} \frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (-\left (a^2 (2 A-3 C)\right )-3 a b B+5 A b^2\right )}{3 a^2 d \left (a^2-b^2\right )}+\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (2 a^3 B-a^2 b (4 A-C)-3 a b^2 B+5 A b^3\right )}{a^3 d \left (a^2-b^2\right )}-\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-2 a^4 (A+3 C)+12 a^3 b B-a^2 b^2 (16 A-3 C)-9 a b^3 B+15 A b^4\right )}{3 a^4 d \left (a^2-b^2\right )}+\frac {b \left (-3 a^4 C+5 a^3 b B-a^2 b^2 (7 A-C)-3 a b^3 B+5 A b^4\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 d (a-b) (a+b)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 2720
Rule 2884
Rule 3081
Rule 3126
Rule 3128
Rule 3138
Rule 4197
Rubi steps
\begin {align*} \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{(b+a \cos (c+d x))^2} \, dx\\ &=\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3}{2} \left (A b^2-a (b B-a C)\right )-a (A b-a B+b C) \cos (c+d x)-\frac {1}{2} \left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \cos ^2(c+d x)\right )}{b+a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {2 \int \frac {-\frac {1}{4} b \left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right )+\frac {1}{2} a \left (2 A b^2-3 a b B+a^2 (A+3 C)\right ) \cos (c+d x)+\frac {3}{4} \left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}-\frac {2 \int \frac {\frac {1}{4} a b \left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right )+\frac {1}{4} \left (15 A b^4+12 a^3 b B-9 a b^3 B-a^2 b^2 (16 A-3 C)-2 a^4 (A+3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a^3 \left (a^2-b^2\right )}+\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(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 b^4+5 a^3 b B-3 a b^3 B-a^2 b^2 (7 A-C)-3 a^4 C\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^4 \left (a^2-b^2\right )}-\frac {\left (15 A b^4+12 a^3 b B-9 a b^3 B-a^2 b^2 (16 A-3 C)-2 a^4 (A+3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=\frac {\left (5 A b^3+2 a^3 B-3 a b^2 B-a^2 b (4 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (15 A b^4+12 a^3 b B-9 a b^3 B-a^2 b^2 (16 A-3 C)-2 a^4 (A+3 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^4 \left (a^2-b^2\right ) d}+\frac {b \left (5 A b^4+5 a^3 b B-3 a b^3 B-a^2 b^2 (7 A-C)-3 a^4 C\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 (a-b) (a+b)^2 d}-\frac {\left (5 A b^2-3 a b B-a^2 (2 A-3 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2-a (b B-a C)\right ) \cos ^{\frac {3}{2}}(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 = 13.93, size = 337, normalized size = 0.97 \begin {gather*} \frac {4 \sqrt {\cos (c+d x)} \left (2 A+\frac {3 b \left (A b^2+a (-b B+a C)\right )}{\left (-a^2+b^2\right ) (b+a \cos (c+d x))}\right ) \sin (c+d x)-\frac {\frac {2 \left (5 A b^3+6 a^3 B-3 a b^2 B-a^2 b (8 A+3 C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {8 \left (2 A b^2-3 a b B+a^2 (A+3 C)\right ) \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 {6 \left (5 A b^3+2 a^3 B-3 a b^2 B+a^2 b (-4 A+C)\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)}}{12 a^2 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. \(1122\) vs.
\(2(416)=832\).
time = 0.39, size = 1123, normalized size = 3.25
method | result | size |
default | \(\text {Expression too large to display}\) | \(1123\) |
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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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 {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\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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