Optimal. Leaf size=50 \[ \frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{\sqrt {b} f (a-b)}-\frac {x}{a-b} \]
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Rubi [A] time = 0.08, antiderivative size = 50, 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 = {3670, 481, 203, 205} \[ \frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{\sqrt {b} f (a-b)}-\frac {x}{a-b} \]
Antiderivative was successfully verified.
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Rule 203
Rule 205
Rule 481
Rule 3670
Rubi steps
\begin {align*} \int \frac {\tan ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b) f}+\frac {a \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{(a-b) f}\\ &=-\frac {x}{a-b}+\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{(a-b) \sqrt {b} f}\\ \end {align*}
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Mathematica [A] time = 0.03, size = 49, normalized size = 0.98 \[ \frac {\tan ^{-1}(\tan (e+f x))-\frac {\sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{\sqrt {b}}}{b f-a f} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 181, normalized size = 3.62 \[ \left [-\frac {4 \, f x + \sqrt {-\frac {a}{b}} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{4} - 6 \, a b \tan \left (f x + e\right )^{2} + a^{2} - 4 \, {\left (b^{2} \tan \left (f x + e\right )^{3} - a b \tan \left (f x + e\right )\right )} \sqrt {-\frac {a}{b}}}{b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}}\right )}{4 \, {\left (a - b\right )} f}, -\frac {2 \, f x - \sqrt {\frac {a}{b}} \arctan \left (\frac {{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt {\frac {a}{b}}}{2 \, a \tan \left (f x + e\right )}\right )}{2 \, {\left (a - b\right )} f}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.08, size = 67, normalized size = 1.34 \[ \frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} a}{\sqrt {a b} {\left (a - b\right )}} - \frac {f x + e}{a - b}}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 52, normalized size = 1.04 \[ \frac {a \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{f \left (a -b \right ) \sqrt {a b}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \left (a -b \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.74, size = 47, normalized size = 0.94 \[ \frac {\frac {a \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} {\left (a - b\right )}} - \frac {f x + e}{a - b}}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 11.57, size = 135, normalized size = 2.70 \[ -\frac {2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (2\,a^2\,b+2\,b^3\right )+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (-8\,a^3\,b^2+8\,a^2\,b^3+8\,a\,b^4-8\,b^5\right )}{{\left (2\,a-2\,b\right )}^2}}{a\,b\,\left (2\,a-2\,b\right )}\right )}{f\,\left (2\,a-2\,b\right )}-\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a\,b}}{a}\right )\,\sqrt {-a\,b}}{f\,\left (a\,b-b^2\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 2.29, size = 292, normalized size = 5.84 \[ \begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac {x}{b} & \text {for}\: a = 0 \\\frac {f x \tan ^{2}{\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} + \frac {f x}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} - \frac {\tan {\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} & \text {for}\: a = b \\\frac {x \tan ^{2}{\relax (e )}}{a + b \tan ^{2}{\relax (e )}} & \text {for}\: f = 0 \\\frac {- x + \frac {\tan {\left (e + f x \right )}}{f}}{a} & \text {for}\: b = 0 \\- \frac {2 i \sqrt {a} b f x \sqrt {\frac {1}{b}}}{2 i a^{\frac {3}{2}} b f \sqrt {\frac {1}{b}} - 2 i \sqrt {a} b^{2} f \sqrt {\frac {1}{b}}} + \frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \tan {\left (e + f x \right )} \right )}}{2 i a^{\frac {3}{2}} b f \sqrt {\frac {1}{b}} - 2 i \sqrt {a} b^{2} f \sqrt {\frac {1}{b}}} - \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \tan {\left (e + f x \right )} \right )}}{2 i a^{\frac {3}{2}} b f \sqrt {\frac {1}{b}} - 2 i \sqrt {a} b^{2} f \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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