Integrand size = 14, antiderivative size = 50 \[ \int \frac {1}{a+b \tan ^2(c+d x)} \, dx=\frac {x}{a-b}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b) d} \]
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Time = 0.09 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3741, 3756, 211} \[ \int \frac {1}{a+b \tan ^2(c+d x)} \, dx=\frac {x}{a-b}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} d (a-b)} \]
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Rule 211
Rule 3741
Rule 3756
Rubi steps \begin{align*} \text {integral}& = \frac {x}{a-b}-\frac {b \int \frac {\sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx}{a-b} \\ & = \frac {x}{a-b}-\frac {b \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{(a-b) d} \\ & = \frac {x}{a-b}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b) d} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98 \[ \int \frac {1}{a+b \tan ^2(c+d x)} \, dx=\frac {\arctan (\tan (c+d x))-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a}}}{a d-b d} \]
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Time = 0.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {-\frac {b \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{\left (a -b \right ) \sqrt {a b}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{a -b}}{d}\) | \(50\) |
default | \(\frac {-\frac {b \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{\left (a -b \right ) \sqrt {a b}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{a -b}}{d}\) | \(50\) |
risch | \(\frac {x}{a -b}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 a \left (a -b \right ) d}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 a \left (a -b \right ) d}\) | \(120\) |
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Time = 0.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 3.64 \[ \int \frac {1}{a+b \tan ^2(c+d x)} \, dx=\left [\frac {4 \, d x - \sqrt {-\frac {b}{a}} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{4} - 6 \, a b \tan \left (d x + c\right )^{2} + a^{2} + 4 \, {\left (a b \tan \left (d x + c\right )^{3} - a^{2} \tan \left (d x + c\right )\right )} \sqrt {-\frac {b}{a}}}{b^{2} \tan \left (d x + c\right )^{4} + 2 \, a b \tan \left (d x + c\right )^{2} + a^{2}}\right )}{4 \, {\left (a - b\right )} d}, \frac {2 \, d x - \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left (b \tan \left (d x + c\right )^{2} - a\right )} \sqrt {\frac {b}{a}}}{2 \, b \tan \left (d x + c\right )}\right )}{2 \, {\left (a - b\right )} d}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (37) = 74\).
Time = 1.22 (sec) , antiderivative size = 240, normalized size of antiderivative = 4.80 \[ \int \frac {1}{a+b \tan ^2(c+d x)} \, dx=\begin {cases} \frac {\tilde {\infty } x}{\tan ^{2}{\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {x}{a} & \text {for}\: b = 0 \\\frac {- x - \frac {1}{d \tan {\left (c + d x \right )}}}{b} & \text {for}\: a = 0 \\\frac {d x \tan ^{2}{\left (c + d x \right )}}{2 b d \tan ^{2}{\left (c + d x \right )} + 2 b d} + \frac {d x}{2 b d \tan ^{2}{\left (c + d x \right )} + 2 b d} + \frac {\tan {\left (c + d x \right )}}{2 b d \tan ^{2}{\left (c + d x \right )} + 2 b d} & \text {for}\: a = b \\\frac {x}{a + b \tan ^{2}{\left (c \right )}} & \text {for}\: d = 0 \\\frac {2 d x \sqrt {- \frac {a}{b}}}{2 a d \sqrt {- \frac {a}{b}} - 2 b d \sqrt {- \frac {a}{b}}} - \frac {\log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (c + d x \right )} \right )}}{2 a d \sqrt {- \frac {a}{b}} - 2 b d \sqrt {- \frac {a}{b}}} + \frac {\log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (c + d x \right )} \right )}}{2 a d \sqrt {- \frac {a}{b}} - 2 b d \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]
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Time = 0.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \frac {1}{a+b \tan ^2(c+d x)} \, dx=-\frac {\frac {b \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} {\left (a - b\right )}} - \frac {d x + c}{a - b}}{d} \]
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Time = 0.41 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.30 \[ \int \frac {1}{a+b \tan ^2(c+d x)} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} b}{\sqrt {a b} {\left (a - b\right )}} - \frac {d x + c}{a - b}}{d} \]
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Time = 11.04 (sec) , antiderivative size = 948, normalized size of antiderivative = 18.96 \[ \int \frac {1}{a+b \tan ^2(c+d x)} \, dx=-\frac {\mathrm {atan}\left (\frac {\frac {-4\,b^3\,\mathrm {tan}\left (c+d\,x\right )+\frac {\left (4\,b^4-8\,a\,b^3+4\,a^2\,b^2+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-8\,a^3\,b^2+8\,a^2\,b^3+8\,a\,b^4-8\,b^5\right )\,1{}\mathrm {i}}{2\,a-2\,b}\right )\,1{}\mathrm {i}}{2\,a-2\,b}}{2\,a-2\,b}+\frac {-4\,b^3\,\mathrm {tan}\left (c+d\,x\right )+\frac {\left (8\,a\,b^3-4\,b^4-4\,a^2\,b^2+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-8\,a^3\,b^2+8\,a^2\,b^3+8\,a\,b^4-8\,b^5\right )\,1{}\mathrm {i}}{2\,a-2\,b}\right )\,1{}\mathrm {i}}{2\,a-2\,b}}{2\,a-2\,b}}{\frac {\left (-4\,b^3\,\mathrm {tan}\left (c+d\,x\right )+\frac {\left (4\,b^4-8\,a\,b^3+4\,a^2\,b^2+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-8\,a^3\,b^2+8\,a^2\,b^3+8\,a\,b^4-8\,b^5\right )\,1{}\mathrm {i}}{2\,a-2\,b}\right )\,1{}\mathrm {i}}{2\,a-2\,b}\right )\,1{}\mathrm {i}}{2\,a-2\,b}-\frac {\left (-4\,b^3\,\mathrm {tan}\left (c+d\,x\right )+\frac {\left (8\,a\,b^3-4\,b^4-4\,a^2\,b^2+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-8\,a^3\,b^2+8\,a^2\,b^3+8\,a\,b^4-8\,b^5\right )\,1{}\mathrm {i}}{2\,a-2\,b}\right )\,1{}\mathrm {i}}{2\,a-2\,b}\right )\,1{}\mathrm {i}}{2\,a-2\,b}}\right )}{d\,\left (a-b\right )}+\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {-a\,b}\,\left (2\,b^3\,\mathrm {tan}\left (c+d\,x\right )-\frac {\sqrt {-a\,b}\,\left (2\,b^4-4\,a\,b^3+2\,a^2\,b^2+\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2+8\,a^2\,b^3+8\,a\,b^4-8\,b^5\right )}{4\,\left (a\,b-a^2\right )}\right )}{2\,\left (a\,b-a^2\right )}\right )\,1{}\mathrm {i}}{a\,b-a^2}+\frac {\sqrt {-a\,b}\,\left (2\,b^3\,\mathrm {tan}\left (c+d\,x\right )-\frac {\sqrt {-a\,b}\,\left (4\,a\,b^3-2\,b^4-2\,a^2\,b^2+\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2+8\,a^2\,b^3+8\,a\,b^4-8\,b^5\right )}{4\,\left (a\,b-a^2\right )}\right )}{2\,\left (a\,b-a^2\right )}\right )\,1{}\mathrm {i}}{a\,b-a^2}}{\frac {\sqrt {-a\,b}\,\left (2\,b^3\,\mathrm {tan}\left (c+d\,x\right )-\frac {\sqrt {-a\,b}\,\left (2\,b^4-4\,a\,b^3+2\,a^2\,b^2+\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2+8\,a^2\,b^3+8\,a\,b^4-8\,b^5\right )}{4\,\left (a\,b-a^2\right )}\right )}{2\,\left (a\,b-a^2\right )}\right )}{a\,b-a^2}-\frac {\sqrt {-a\,b}\,\left (2\,b^3\,\mathrm {tan}\left (c+d\,x\right )-\frac {\sqrt {-a\,b}\,\left (4\,a\,b^3-2\,b^4-2\,a^2\,b^2+\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2+8\,a^2\,b^3+8\,a\,b^4-8\,b^5\right )}{4\,\left (a\,b-a^2\right )}\right )}{2\,\left (a\,b-a^2\right )}\right )}{a\,b-a^2}}\right )\,\sqrt {-a\,b}\,1{}\mathrm {i}}{a\,d\,\left (a-b\right )} \]
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