Integrand size = 19, antiderivative size = 157 \[ \int \frac {\cos (c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {3 a b^2 \text {arctanh}\left (\frac {b-a \tan (c+d x)}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right ) \cos (c+d x) \sqrt {\sec ^2(c+d x)}}{\left (a^2+b^2\right )^{5/2} d}+\frac {b \left (a^2-2 b^2\right ) \sec (c+d x)}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))} \]
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Time = 0.16 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3593, 755, 821, 739, 212} \[ \int \frac {\cos (c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {3 a b^2 \cos (c+d x) \sqrt {\sec ^2(c+d x)} \text {arctanh}\left (\frac {b-a \tan (c+d x)}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right )}{d \left (a^2+b^2\right )^{5/2}}+\frac {\cos (c+d x) (a \tan (c+d x)+b)}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {b \left (a^2-2 b^2\right ) \sec (c+d x)}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))} \]
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Rule 212
Rule 739
Rule 755
Rule 821
Rule 3593
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(a+x)^2 \left (1+\frac {x^2}{b^2}\right )^{3/2}} \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = \frac {\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\left (b \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {-2-\frac {a x}{b^2}}{(a+x)^2 \sqrt {1+\frac {x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right ) d} \\ & = \frac {b \left (a^2-2 b^2\right ) \sec (c+d x)}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac {\left (3 a b \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(a+x) \sqrt {1+\frac {x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right )^2 d} \\ & = \frac {b \left (a^2-2 b^2\right ) \sec (c+d x)}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\left (3 a b \cos (c+d x) \sqrt {\sec ^2(c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a^2}{b^2}-x^2} \, dx,x,\frac {1-\frac {a \tan (c+d x)}{b}}{\sqrt {\sec ^2(c+d x)}}\right )}{\left (a^2+b^2\right )^2 d} \\ & = -\frac {3 a b^2 \text {arctanh}\left (\frac {b \left (1-\frac {a \tan (c+d x)}{b}\right )}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right ) \cos (c+d x) \sqrt {\sec ^2(c+d x)}}{\left (a^2+b^2\right )^{5/2} d}+\frac {b \left (a^2-2 b^2\right ) \sec (c+d x)}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))} \\ \end{align*}
Time = 0.94 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.97 \[ \int \frac {\cos (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\sec (c+d x) \left (12 a b^2 \sqrt {a^2+b^2} \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right ) (a \cos (c+d x)+b \sin (c+d x))+\left (a^2+b^2\right ) \left (3 b \left (a^2-b^2\right )+b \left (a^2+b^2\right ) \cos (2 (c+d x))+a \left (a^2+b^2\right ) \sin (2 (c+d x))\right )\right )}{2 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))} \]
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Time = 2.32 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (\left (-a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 a b \right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {2 b^{2} \left (\frac {-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-b}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a}-\frac {3 a \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\) | \(172\) |
default | \(\frac {-\frac {2 \left (\left (-a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 a b \right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {2 b^{2} \left (\frac {-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-b}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a}-\frac {3 a \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\) | \(172\) |
risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 \left (-2 i a b +a^{2}-b^{2}\right ) d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 \left (2 i a b +a^{2}-b^{2}\right ) d}-\frac {2 i b^{3} {\mathrm e}^{i \left (d x +c \right )}}{\left (-i a +b \right )^{2} d \left (i a +b \right )^{2} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )}+\frac {3 b^{2} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}-\frac {3 b^{2} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}\) | \(295\) |
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Time = 0.29 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.92 \[ \int \frac {\cos (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {2 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{2} b^{2} \cos \left (d x + c\right ) + a b^{3} \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right )}{2 \, {\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} d \sin \left (d x + c\right )\right )}} \]
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\[ \int \frac {\cos (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\cos {\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (151) = 302\).
Time = 0.42 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.22 \[ \int \frac {\cos (c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {3 \, a b^{2} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (2 \, a^{3} b - a b^{3} - \frac {3 \, a b^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {{\left (a^{4} + 3 \, a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {{\left (a^{4} - a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4} + \frac {2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}}{d} \]
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Time = 0.51 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.82 \[ \int \frac {\cos (c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {3 \, a b^{2} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{3} b + a b^{3}\right )}}{{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}}{d} \]
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Time = 6.54 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.82 \[ \int \frac {\cos (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {4\,a^2\,b-2\,b^3}{a^4+2\,a^2\,b^2+b^4}-\frac {6\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+3\,a^2\,b^2-b^4\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^4-2\,a^2\,b^2+2\,b^4\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {6\,a\,b^2\,\mathrm {atanh}\left (\frac {a^4\,b+b^5+2\,a^2\,b^3-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{{\left (a^2+b^2\right )}^{5/2}}\right )}{d\,{\left (a^2+b^2\right )}^{5/2}} \]
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