Optimal. Leaf size=229 \[ -\frac {(7 A-11 B+15 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(45 A-65 B+93 C) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(15 A-35 B+39 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d} \]
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Rubi [A]
time = 0.46, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4169, 4106,
4095, 4086, 3880, 209} \begin {gather*} -\frac {(7 A-11 B+15 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(15 A-35 B+39 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{30 a^2 d}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac {(5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt {a \sec (c+d x)+a}}+\frac {(45 A-65 B+93 C) \tan (c+d x)}{15 a d \sqrt {a \sec (c+d x)+a}} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 3880
Rule 4086
Rule 4095
Rule 4106
Rule 4169
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\sec ^3(c+d x) \left (-a (A-3 B+3 C)+\frac {1}{2} a (5 A-5 B+9 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\sec ^2(c+d x) \left (a^2 (5 A-5 B+9 C)-\frac {1}{4} a^2 (15 A-35 B+39 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(15 A-35 B+39 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}+\frac {2 \int \frac {\sec (c+d x) \left (-\frac {1}{8} a^3 (15 A-35 B+39 C)+\frac {1}{4} a^3 (45 A-65 B+93 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(45 A-65 B+93 C) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(15 A-35 B+39 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}-\frac {(7 A-11 B+15 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(45 A-65 B+93 C) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(15 A-35 B+39 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}+\frac {(7 A-11 B+15 C) \text {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{2 a d}\\ &=-\frac {(7 A-11 B+15 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(45 A-65 B+93 C) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(15 A-35 B+39 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in
optimal.
time = 25.05, size = 7166, normalized size = 31.29 \begin {gather*} \text {Result too large to show} \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. \(1151\) vs.
\(2(202)=404\).
time = 0.20, size = 1152, normalized size = 5.03
method | result | size |
default | \(\text {Expression too large to display}\) | \(1152\) |
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 [A]
time = 3.40, size = 534, normalized size = 2.33 \begin {gather*} \left [-\frac {15 \, \sqrt {2} {\left ({\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (75 \, A - 95 \, B + 147 \, C\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (5 \, A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (5 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 12 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{120 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}, \frac {15 \, \sqrt {2} {\left ({\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (75 \, A - 95 \, B + 147 \, C\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (5 \, A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (5 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 12 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{60 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}\right ] \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 )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 1.82, size = 303, normalized size = 1.32 \begin {gather*} \frac {\frac {15 \, \sqrt {2} {\left (7 \, A - 11 \, B + 15 \, C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {{\left ({\left ({\left (\frac {15 \, \sqrt {2} {\left (A a^{3} - B a^{3} + C a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left (165 \, A a^{3} - 245 \, B a^{3} + 381 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, \sqrt {2} {\left (57 \, A a^{3} - 73 \, B a^{3} + 105 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, \sqrt {2} {\left (9 \, A a^{3} - 9 \, B a^{3} + 17 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{60 \, d} \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 )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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