Optimal. Leaf size=41 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} f} \]
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Rubi [A]
time = 0.04, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3751, 455, 65,
214} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f \sqrt {a-b}} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 214
Rule 455
Rule 3751
Rubi steps
\begin {align*} \int \frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx &=\frac {\text {Subst}\left (\int \frac {x}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\text {Subst}\left (\int \frac {1}{1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \tan ^2(e+f x)}\right )}{b f}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} f}\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 41, normalized size = 1.00 \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.09, size = 35, normalized size = 0.85
method | result | size |
derivativedivides | \(\frac {\arctan \left (\frac {\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}\) | \(35\) |
default | \(\frac {\arctan \left (\frac {\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {-a +b}}\right )}{f \sqrt {-a +b}}\) | \(35\) |
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 = 4.17, size = 193, normalized size = 4.71 \begin {gather*} \left [\frac {\log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right )}{4 \, \sqrt {a - b} f}, \frac {\sqrt {-a + b} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right )}{2 \, {\left (a - b\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 {\tan {\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.45, size = 35, normalized size = 0.85 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {-a + b}}\right )}{\sqrt {-a + b} f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 12.34, size = 35, normalized size = 0.85 \begin {gather*} -\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{\sqrt {a-b}}\right )}{f\,\sqrt {a-b}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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