Optimal. Leaf size=145 \[ -\frac {75 \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {13 \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {9 \tan (c+d x)}{4 a^2 d \sqrt {a+a \sec (c+d x)}} \]
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Rubi [A]
time = 0.20, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3901, 4093,
4086, 3880, 209} \begin {gather*} -\frac {75 \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {9 \tan (c+d x)}{4 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {\tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac {13 \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 3880
Rule 3901
Rule 4086
Rule 4093
Rubi steps
\begin {align*} \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac {\sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {\int \frac {\sec ^2(c+d x) \left (2 a-\frac {9}{2} a \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {\sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {13 \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\sec (c+d x) \left (-\frac {39 a^2}{4}+9 a^2 \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {\sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {13 \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {9 \tan (c+d x)}{4 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {75 \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {\sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {13 \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {9 \tan (c+d x)}{4 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {75 \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 )}{16 a^2 d}\\ &=-\frac {75 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {13 \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {9 \tan (c+d x)}{4 a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 0.73, size = 125, normalized size = 0.86 \begin {gather*} \frac {\left (-150 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x)+\sqrt {1-\sec (c+d x)} \left (49+85 \sec (c+d x)+32 \sec ^2(c+d x)\right )\right ) \tan (c+d x)}{16 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \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. \(315\) vs.
\(2(122)=244\).
time = 0.12, size = 316, normalized size = 2.18
method | result | size |
default | \(-\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (75 \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+150 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right )+75 \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+98 \left (\cos ^{3}\left (d x +c \right )\right )+72 \left (\cos ^{2}\left (d x +c \right )\right )-106 \cos \left (d x +c \right )-64\right )}{32 d \sin \left (d x +c \right )^{5} a^{3}}\) | \(316\) |
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 = 2.38, size = 404, normalized size = 2.79 \begin {gather*} \left [-\frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\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 (49 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 32\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\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 (49 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 32\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\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 {\sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 1.22, size = 188, normalized size = 1.30 \begin {gather*} \frac {\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left ({\left (\frac {2 \, \sqrt {2} \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 {17 \, \sqrt {2}}{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 {83 \, \sqrt {2}}{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 )}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} + \frac {75 \, \sqrt {2} \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^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{32 \, 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 {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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