Optimal. Leaf size=100 \[ \frac {\left (a^2-2 b^2\right ) x}{2 a^3}-\frac {2 \sqrt {a-b} b \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d}+\frac {(2 b-a \cos (c+d x)) \sin (c+d x)}{2 a^2 d} \]
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Rubi [A]
time = 0.15, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3957, 2944,
2814, 2738, 214} \begin {gather*} -\frac {2 b \sqrt {a-b} \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d}+\frac {\sin (c+d x) (2 b-a \cos (c+d x))}{2 a^2 d}+\frac {x \left (a^2-2 b^2\right )}{2 a^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 214
Rule 2738
Rule 2814
Rule 2944
Rule 3957
Rubi steps
\begin {align*} \int \frac {\sin ^2(c+d x)}{a+b \sec (c+d x)} \, dx &=-\int \frac {\cos (c+d x) \sin ^2(c+d x)}{-b-a \cos (c+d x)} \, dx\\ &=\frac {(2 b-a \cos (c+d x)) \sin (c+d x)}{2 a^2 d}-\frac {\int \frac {-a b+\left (a^2-2 b^2\right ) \cos (c+d x)}{-b-a \cos (c+d x)} \, dx}{2 a^2}\\ &=\frac {\left (a^2-2 b^2\right ) x}{2 a^3}+\frac {(2 b-a \cos (c+d x)) \sin (c+d x)}{2 a^2 d}+\frac {\left (b \left (a^2-b^2\right )\right ) \int \frac {1}{-b-a \cos (c+d x)} \, dx}{a^3}\\ &=\frac {\left (a^2-2 b^2\right ) x}{2 a^3}+\frac {(2 b-a \cos (c+d x)) \sin (c+d x)}{2 a^2 d}+\frac {\left (2 b \left (a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 d}\\ &=\frac {\left (a^2-2 b^2\right ) x}{2 a^3}-\frac {2 \sqrt {a-b} b \sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d}+\frac {(2 b-a \cos (c+d x)) \sin (c+d x)}{2 a^2 d}\\ \end {align*}
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Mathematica [A]
time = 0.22, size = 96, normalized size = 0.96 \begin {gather*} \frac {2 \left (a^2-2 b^2\right ) (c+d x)+8 b \sqrt {a^2-b^2} \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+4 a b \sin (c+d x)-a^2 \sin (2 (c+d x))}{4 a^3 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.11, size = 142, normalized size = 1.42
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (a -b \right ) b \left (a +b \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\frac {2 \left (\left (\frac {1}{2} a^{2}+b a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (b a -\frac {1}{2} a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\left (a^{2}-2 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}}{d}\) | \(142\) |
default | \(\frac {-\frac {2 \left (a -b \right ) b \left (a +b \right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\frac {2 \left (\left (\frac {1}{2} a^{2}+b a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (b a -\frac {1}{2} a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\left (a^{2}-2 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}}{d}\) | \(142\) |
risch | \(\frac {x}{2 a}-\frac {x \,b^{2}}{a^{3}}-\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}+\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}-\frac {\sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {a^{2}-b^{2}}+b}{a}\right )}{d \,a^{3}}+\frac {\sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {a^{2}-b^{2}}-b}{a}\right )}{d \,a^{3}}-\frac {\sin \left (2 d x +2 c \right )}{4 a d}\) | \(176\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 3.90, size = 258, normalized size = 2.58 \begin {gather*} \left [\frac {{\left (a^{2} - 2 \, b^{2}\right )} d x + \sqrt {a^{2} - b^{2}} b \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - {\left (a^{2} \cos \left (d x + c\right ) - 2 \, a b\right )} \sin \left (d x + c\right )}{2 \, a^{3} d}, \frac {{\left (a^{2} - 2 \, b^{2}\right )} d x - 2 \, \sqrt {-a^{2} + b^{2}} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (a^{2} \cos \left (d x + c\right ) - 2 \, a b\right )} \sin \left (d x + c\right )}{2 \, a^{3} d}\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 {\sin ^{2}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 185 vs.
\(2 (86) = 172\).
time = 0.43, size = 185, normalized size = 1.85 \begin {gather*} \frac {\frac {{\left (a^{2} - 2 \, b^{2}\right )} {\left (d x + c\right )}}{a^{3}} - \frac {4 \, {\left (a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{3}} + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.48, size = 147, normalized size = 1.47 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {\sin \left (2\,c+2\,d\,x\right )}{4}}{a\,d}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^3\,d}+\frac {b\,\sin \left (c+d\,x\right )}{a^2\,d}-\frac {2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a+b\right )}\right )\,\sqrt {a^2-b^2}}{a^3\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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