Optimal. Leaf size=496 \[ \frac {\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}+\frac {\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (3 A b^6-3 a^5 b B+6 a^3 b^3 B-15 a b^5 B+15 a^6 C+5 a^2 b^4 (2 A+7 C)-a^4 b^2 (A+38 C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a (a-b)^2 b^3 (a+b)^3 d}-\frac {\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \]
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Rubi [A]
time = 1.30, antiderivative size = 496, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4197, 3134,
3138, 2719, 3081, 2720, 2884} \begin {gather*} -\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)^2}+\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^4 C+a^3 b B+a^2 b^2 (3 A+11 C)-7 a b^3 B+3 A b^4\right )}{4 a b^2 d \left (a^2-b^2\right )^2}+\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-15 a^4 C+3 a^3 b B+a^2 b^2 (A+29 C)-9 a b^3 B+b^4 (5 A-8 C)\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac {\sin (c+d x) \left (-15 a^4 C+3 a^3 b B+a^2 b^2 (A+29 C)-9 a b^3 B+b^4 (5 A-8 C)\right )}{4 b^3 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)}}+\frac {\sin (c+d x) \left (-5 a^4 C+a^3 b B+a^2 b^2 (3 A+11 C)-7 a b^3 B+3 A b^4\right )}{4 b^2 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)}-\frac {\left (15 a^6 C-3 a^5 b B-a^4 b^2 (A+38 C)+6 a^3 b^3 B+5 a^2 b^4 (2 A+7 C)-15 a b^5 B+3 A b^6\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b^3 d (a-b)^2 (a+b)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 2720
Rule 2884
Rule 3081
Rule 3134
Rule 3138
Rule 4197
Rubi steps
\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx &=\int \frac {C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^3} \, dx\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}-\frac {\int \frac {\frac {1}{2} \left (-A b^2+a b B-5 a^2 C+4 b^2 C\right )+2 b (b B-a (A+C)) \cos (c+d x)+\frac {3}{2} \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (-3 a^3 b B+9 a b^3 B-b^4 (5 A-8 C)+15 a^4 C-a^2 b^2 (A+29 C)\right )+b \left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right ) \cos (c+d x)+\frac {1}{4} \left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{8} \left (3 a^4 b B-5 a^2 b^3 B+8 b^5 B-15 a^5 C-a b^4 (7 A+24 C)+a^3 b^2 (A+33 C)\right )+\frac {1}{2} b \left (a^3 b B-4 a b^3 B+2 b^4 (A-C)-5 a^4 C+a^2 b^2 (A+10 C)\right ) \cos (c+d x)+\frac {1}{8} a \left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}-\frac {\int \frac {-\frac {1}{8} a \left (3 a^4 b B-5 a^2 b^3 B+8 b^5 B-15 a^5 C-a b^4 (7 A+24 C)+a^3 b^2 (A+33 C)\right )-\frac {1}{8} a b \left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a b^3 \left (a^2-b^2\right )^2}+\frac {\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}+\frac {\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a b^2 \left (a^2-b^2\right )^2}-\frac {\left (3 A b^6-3 a^5 b B+6 a^3 b^3 B-15 a b^5 B+15 a^6 C+5 a^2 b^4 (2 A+7 C)-a^4 b^2 (A+38 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a b^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}+\frac {\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (3 A b^6-3 a^5 b B+6 a^3 b^3 B-15 a b^5 B+15 a^6 C+5 a^2 b^4 (2 A+7 C)-a^4 b^2 (A+38 C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a (a-b)^2 b^3 (a+b)^3 d}-\frac {\left (3 a^3 b B-9 a b^3 B+b^4 (5 A-8 C)-15 a^4 C+a^2 b^2 (A+29 C)\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+a^3 b B-7 a b^3 B-5 a^4 C+a^2 b^2 (3 A+11 C)\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 17.44, size = 592, normalized size = 1.19 \begin {gather*} -\frac {\frac {2 \left (-3 a^3 A b^2+9 a A b^4-9 a^4 b B+19 a^2 b^3 B-16 b^5 B+45 a^5 C-95 a^3 b^2 C+56 a b^4 C\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {\left (-8 a^2 A b^3-16 A b^5-8 a^3 b^2 B+32 a b^4 B+40 a^4 b C-80 a^2 b^3 C+16 b^5 C\right ) \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{a}+\frac {2 \left (-a^3 A b^2-5 a A b^4-3 a^4 b B+9 a^2 b^3 B+15 a^5 C-29 a^3 b^2 C+8 a b^4 C\right ) \cos (2 (c+d x)) \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 {1-\cos ^2(c+d x)} \left (-1+2 \cos ^2(c+d x)\right )}}{16 (a-b)^2 b^3 (a+b)^2 d}+\frac {\sqrt {\cos (c+d x)} \left (\frac {-a A b^2 \sin (c+d x)+a^2 b B \sin (c+d x)-a^3 C \sin (c+d x)}{2 b^2 \left (-a^2+b^2\right ) (b+a \cos (c+d x))^2}+\frac {-a^3 A b^2 \sin (c+d x)-5 a A b^4 \sin (c+d x)-3 a^4 b B \sin (c+d x)+9 a^2 b^3 B \sin (c+d x)+7 a^5 C \sin (c+d x)-13 a^3 b^2 C \sin (c+d x)}{4 b^3 \left (-a^2+b^2\right )^2 (b+a \cos (c+d x))}+\frac {2 C \tan (c+d x)}{b^3}\right )}{d} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2021\) vs.
\(2(556)=1112\).
time = 0.81, size = 2022, normalized size = 4.08
method | result | size |
default | \(\text {Expression too large to display}\) | \(2022\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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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(-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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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 {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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