Optimal. Leaf size=121 \[ \frac {2 a \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} (a c-b d) f}-\frac {2 d \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{\sqrt {c-d} \sqrt {c+d} (a c-b d) f} \]
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Rubi [A]
time = 0.19, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2907, 3080,
2738, 211, 214} \begin {gather*} \frac {2 a \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{f \sqrt {a-b} \sqrt {a+b} (a c-b d)}-\frac {2 d \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f \sqrt {c-d} \sqrt {c+d} (a c-b d)} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 214
Rule 2738
Rule 2907
Rule 3080
Rubi steps
\begin {align*} \int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))} \, dx &=\int \frac {\cos (e+f x)}{(a+b \cos (e+f x)) (d+c \cos (e+f x))} \, dx\\ &=\frac {a \int \frac {1}{a+b \cos (e+f x)} \, dx}{a c-b d}-\frac {d \int \frac {1}{d+c \cos (e+f x)} \, dx}{a c-b d}\\ &=\frac {(2 a) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{(a c-b d) f}-\frac {(2 d) \text {Subst}\left (\int \frac {1}{c+d+(-c+d) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{(a c-b d) f}\\ &=\frac {2 a \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} (a c-b d) f}-\frac {2 d \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{\sqrt {c-d} \sqrt {c+d} (a c-b d) f}\\ \end {align*}
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Mathematica [A]
time = 0.26, size = 106, normalized size = 0.88 \begin {gather*} \frac {-\frac {2 a \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {2 d \tanh ^{-1}\left (\frac {(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}}{a c f-b d f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.43, size = 108, normalized size = 0.89
method | result | size |
derivativedivides | \(\frac {\frac {2 a \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a c -b d \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 d \arctanh \left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (a c -b d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}}{f}\) | \(108\) |
default | \(\frac {\frac {2 a \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a c -b d \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 d \arctanh \left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (a c -b d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}}{f}\) | \(108\) |
risch | \(\frac {d \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i c^{2}-i d^{2}-d \sqrt {c^{2}-d^{2}}}{\sqrt {c^{2}-d^{2}}\, c}\right )}{\sqrt {c^{2}-d^{2}}\, \left (a c -b d \right ) f}-\frac {d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+d \sqrt {c^{2}-d^{2}}}{\sqrt {c^{2}-d^{2}}\, c}\right )}{\sqrt {c^{2}-d^{2}}\, \left (a c -b d \right ) f}-\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a c -b d \right ) f}+\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a c -b d \right ) f}\) | \(320\) |
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 = 1.87, size = 1054, normalized size = 8.71 \begin {gather*} \left [-\frac {{\left (a^{2} - b^{2}\right )} \sqrt {c^{2} - d^{2}} d \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) - {\left (a c^{2} - a d^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (f x + e\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (f x + e\right )^{2} + 2 \, a b \cos \left (f x + e\right ) + a^{2}}\right )}{2 \, {\left ({\left (a^{3} - a b^{2}\right )} c^{3} - {\left (a^{2} b - b^{3}\right )} c^{2} d - {\left (a^{3} - a b^{2}\right )} c d^{2} + {\left (a^{2} b - b^{3}\right )} d^{3}\right )} f}, -\frac {2 \, {\left (a^{2} - b^{2}\right )} \sqrt {-c^{2} + d^{2}} d \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) - {\left (a c^{2} - a d^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (f x + e\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (f x + e\right )^{2} + 2 \, a b \cos \left (f x + e\right ) + a^{2}}\right )}{2 \, {\left ({\left (a^{3} - a b^{2}\right )} c^{3} - {\left (a^{2} b - b^{3}\right )} c^{2} d - {\left (a^{3} - a b^{2}\right )} c d^{2} + {\left (a^{2} b - b^{3}\right )} d^{3}\right )} f}, -\frac {{\left (a^{2} - b^{2}\right )} \sqrt {c^{2} - d^{2}} d \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) - 2 \, {\left (a c^{2} - a d^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (f x + e\right )}\right )}{2 \, {\left ({\left (a^{3} - a b^{2}\right )} c^{3} - {\left (a^{2} b - b^{3}\right )} c^{2} d - {\left (a^{3} - a b^{2}\right )} c d^{2} + {\left (a^{2} b - b^{3}\right )} d^{3}\right )} f}, -\frac {{\left (a^{2} - b^{2}\right )} \sqrt {-c^{2} + d^{2}} d \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) - {\left (a c^{2} - a d^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (f x + e\right )}\right )}{{\left ({\left (a^{3} - a b^{2}\right )} c^{3} - {\left (a^{2} b - b^{3}\right )} c^{2} d - {\left (a^{3} - a b^{2}\right )} c d^{2} + {\left (a^{2} b - b^{3}\right )} d^{3}\right )} f}\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 {1}{\left (a + b \cos {\left (e + f x \right )}\right ) \left (c + d \sec {\left (e + f x \right )}\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. 511 vs.
\(2 (105) = 210\).
time = 0.53, size = 511, normalized size = 4.22 \begin {gather*} \frac {\frac {{\left (\sqrt {a^{2} - b^{2}} a c {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} {\left (2 \, a - b\right )} d {\left | a - b \right |} + \sqrt {a^{2} - b^{2}} {\left | a c - b d \right |} {\left | a - b \right |}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {\frac {b c - a d + \sqrt {{\left (a c + b c + a d + b d\right )} {\left (a c - b c - a d + b d\right )} + {\left (b c - a d\right )}^{2}}}{a c - b c - a d + b d}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (a c - b d\right )}^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} c {\left | a c - b d \right |} - {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} d {\left | a c - b d \right |}} + \frac {{\left (\sqrt {-c^{2} + d^{2}} a {\left (c - 2 \, d\right )} {\left | -c + d \right |} + \sqrt {-c^{2} + d^{2}} b d {\left | -c + d \right |} - \sqrt {-c^{2} + d^{2}} {\left | a c - b d \right |} {\left | -c + d \right |}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {\frac {b c - a d - \sqrt {{\left (a c + b c + a d + b d\right )} {\left (a c - b c - a d + b d\right )} + {\left (b c - a d\right )}^{2}}}{a c - b c - a d + b d}}}\right )\right )}}{{\left (a c - b d\right )}^{2} {\left (c^{2} - 2 \, c d + d^{2}\right )} + {\left (c^{2} d - 2 \, c d^{2} + d^{3}\right )} a {\left | a c - b d \right |} - {\left (c^{3} - 2 \, c^{2} d + c d^{2}\right )} b {\left | a c - b d \right |}}}{f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.79, size = 2665, normalized size = 22.02 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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