Optimal. Leaf size=159 \[ \frac {\sqrt {2} (B-C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 (5 B-7 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (5 B-C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 a d} \]
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Rubi [A]
time = 0.34, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4157, 4106,
4095, 4086, 3880, 209} \begin {gather*} \frac {\sqrt {2} (B-C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 (5 B-C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{15 a d}-\frac {4 (5 B-7 C) \tan (c+d x)}{15 d \sqrt {a \sec (c+d x)+a}}+\frac {2 C \tan (c+d x) \sec ^2(c+d x)}{5 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 4157
Rubi steps
\begin {align*} \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx &=\int \frac {\sec ^3(c+d x) (B+C \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \int \frac {\sec ^2(c+d x) \left (2 a C+\frac {1}{2} a (5 B-C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{5 a}\\ &=\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (5 B-C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+\frac {4 \int \frac {\sec (c+d x) \left (\frac {1}{4} a^2 (5 B-C)-\frac {1}{2} a^2 (5 B-7 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{15 a^2}\\ &=-\frac {4 (5 B-7 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (5 B-C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+(B-C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=-\frac {4 (5 B-7 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (5 B-C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 a d}-\frac {(2 (B-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 )}{d}\\ &=\frac {\sqrt {2} (B-C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 (5 B-7 C) \tan (c+d x)}{15 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (5 B-C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{15 a d}\\ \end {align*}
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Mathematica [A]
time = 0.40, size = 123, normalized size = 0.77 \begin {gather*} \frac {\left (15 \sqrt {2} (B-C) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+2 \sqrt {1-\sec (c+d x)} \left (-5 B+13 C+(5 B-C) \sec (c+d x)+3 C \sec ^2(c+d x)\right )\right ) \tan (c+d x)}{15 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \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. \(594\) vs.
\(2(138)=276\).
time = 13.48, size = 595, normalized size = 3.74
method | result | size |
default | \(\frac {\left (15 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-15 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right )+30 B \sin \left (d x +c \right ) \cos \left (d x +c \right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-30 C \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right )+15 B \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )-15 C \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )+40 B \left (\cos ^{3}\left (d x +c \right )\right )-104 C \left (\cos ^{3}\left (d x +c \right )\right )-80 B \left (\cos ^{2}\left (d x +c \right )\right )+112 C \left (\cos ^{2}\left (d x +c \right )\right )+40 B \cos \left (d x +c \right )-32 C \cos \left (d x +c \right )+24 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{60 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) a}\) | \(595\) |
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 [A]
time = 2.67, size = 397, normalized size = 2.50 \begin {gather*} \left [-\frac {15 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right )^{3} + {\left (B - C\right )} a \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (5 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (5 \, B - C\right )} \cos \left (d x + c\right ) - 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{30 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}, -\frac {2 \, {\left ({\left (5 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (5 \, B - C\right )} \cos \left (d x + c\right ) - 3 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac {15 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right )^{3} + {\left (B - C\right )} a \cos \left (d x + c\right )^{2}\right )} \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 )}{\sqrt {a}}}{15 \, {\left (a d \cos \left (d x + c\right )^{3} + a 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 (B + C \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 1.41, size = 229, normalized size = 1.44 \begin {gather*} -\frac {\frac {15 \, {\left (\sqrt {2} B - \sqrt {2} 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} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {2 \, {\left ({\left (10 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 20 \, \sqrt {2} C a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (10 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 17 \, \sqrt {2} C 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}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {15 \, \sqrt {2} C 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}}}{15 \, 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 = -0.01 \begin {gather*} \int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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