Optimal. Leaf size=141 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} \sqrt {d} f}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f \sqrt {c-d}} \]
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Rubi [A] time = 0.53, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {3985, 3983, 203, 3980, 206} \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} \sqrt {d} f}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} f \sqrt {c-d}} \]
Antiderivative was successfully verified.
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Rule 203
Rule 206
Rule 3980
Rule 3983
Rule 3985
Rubi steps
\begin {align*} \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}} \, dx &=\frac {\int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx}{a}-\int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}} \, dx\\ &=-\frac {2 \operatorname {Subst}\left (\int \frac {1}{1-a d x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{f}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{2+(a c-a d) x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{f}\\ &=-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {c-d} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} \sqrt {c-d} f}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c+d \sec (e+f x)}}\right )}{\sqrt {a} \sqrt {d} f}\\ \end {align*}
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Mathematica [A] time = 0.28, size = 171, normalized size = 1.21 \[ \frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {c \cos (e+f x)+d} \left (\sqrt {2} \sqrt {c-d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c \cos (e+f x)+d}}\right )-\sqrt {d} \tan ^{-1}\left (\frac {\sqrt {c-d} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c \cos (e+f x)+d}}\right )\right )}{\sqrt {d} f \sqrt {c-d} \sqrt {a (\sec (e+f x)+1)} \sqrt {c+d \sec (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.81, size = 1100, normalized size = 7.80 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{2}}{\sqrt {a \sec \left (f x + e\right ) + a} \sqrt {d \sec \left (f x + e\right ) + c}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 2.03, size = 403, normalized size = 2.86 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {d +c \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right ) \left (-\ln \left (-\frac {\sqrt {c -d}\, \cos \left (f x +e \right )-\sin \left (f x +e \right ) \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}-\sqrt {c -d}}{\sin \left (f x +e \right )}\right ) \sqrt {2}\, \sqrt {-d}+\ln \left (-\frac {2 \left (\sqrt {2}\, \sqrt {-d}\, \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )-c \sin \left (f x +e \right )-d \sin \left (f x +e \right )-c \cos \left (f x +e \right )+d \cos \left (f x +e \right )+c -d \right )}{-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}\right ) \sqrt {c -d}-\ln \left (\frac {2 \sqrt {2}\, \sqrt {-d}\, \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )-2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )+2 c \cos \left (f x +e \right )-2 d \cos \left (f x +e \right )-2 c +2 d}{1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}\right ) \sqrt {c -d}\right ) \sqrt {2}}{f \sin \left (f x +e \right )^{2} \sqrt {-\frac {2 \left (d +c \cos \left (f x +e \right )\right )}{1+\cos \left (f x +e \right )}}\, a \sqrt {c -d}\, \sqrt {-d}} \]
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 \[ \int \frac {\sec \left (f x + e\right )^{2}}{\sqrt {a \sec \left (f x + e\right ) + a} \sqrt {d \sec \left (f x + e\right ) + c}}\,{d x} \]
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 \[ \int \frac {1}{{\cos \left (e+f\,x\right )}^2\,\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,\sqrt {c+\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \]
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 \[ \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \sqrt {c + d \sec {\left (e + f x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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