Integrand size = 29, antiderivative size = 90 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=-\frac {(A b-a B) x}{a^2}+\frac {2 b (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} d}+\frac {A \sin (c+d x)}{a d} \]
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Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4119, 12, 3868, 2738, 214} \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {2 b (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {x (A b-a B)}{a^2}+\frac {A \sin (c+d x)}{a d} \]
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Rule 12
Rule 214
Rule 2738
Rule 3868
Rule 4119
Rubi steps \begin{align*} \text {integral}& = \frac {A \sin (c+d x)}{a d}-\frac {\int \frac {A b-a B}{a+b \sec (c+d x)} \, dx}{a} \\ & = \frac {A \sin (c+d x)}{a d}-\frac {(A b-a B) \int \frac {1}{a+b \sec (c+d x)} \, dx}{a} \\ & = -\frac {(A b-a B) x}{a^2}+\frac {A \sin (c+d x)}{a d}+\frac {(A b-a B) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^2} \\ & = -\frac {(A b-a B) x}{a^2}+\frac {A \sin (c+d x)}{a d}+\frac {(2 (A b-a B)) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d} \\ & = -\frac {(A b-a B) x}{a^2}+\frac {2 b (A b-a B) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} d}+\frac {A \sin (c+d x)}{a d} \\ \end{align*}
Time = 0.68 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {(-A b+a B) (c+d x)-\frac {2 b (A b-a B) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+a A \sin (c+d x)}{a^2 d} \]
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Time = 1.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.23
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (-\frac {A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\left (A b -B a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a^{2}}+\frac {2 b \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(111\) |
default | \(\frac {-\frac {2 \left (-\frac {A a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\left (A b -B a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a^{2}}+\frac {2 b \left (A b -B a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(111\) |
risch | \(-\frac {x A b}{a^{2}}+\frac {x B}{a}-\frac {i A \,{\mathrm e}^{i \left (d x +c \right )}}{2 a d}+\frac {i A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a d}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right ) B}{\sqrt {a^{2}-b^{2}}\, d a}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d a}\) | \(348\) |
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Time = 0.30 (sec) , antiderivative size = 328, normalized size of antiderivative = 3.64 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\left [\frac {2 \, {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} d x - {\left (B a b - A b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \, {\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d}, \frac {{\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} d x - {\left (B a b - A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{{\left (a^{4} - a^{2} b^{2}\right )} d}\right ] \]
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\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
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Exception generated. \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.32 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.57 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\frac {{\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2}} + \frac {2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a} - \frac {2 \, {\left (B a b - A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{2}}}{d} \]
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Time = 15.99 (sec) , antiderivative size = 740, normalized size of antiderivative = 8.22 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {2\,A\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,\left (a^4-a^2\,b^2\right )}+\frac {2\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,\left (a^4-a^2\,b^2\right )}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{d\,\left (a^4-a^2\,b^2\right )}-\frac {A\,a\,b^2\,\sin \left (c+d\,x\right )}{d\,\left (a^4-a^2\,b^2\right )}+\frac {A\,b^2\,\mathrm {atan}\left (\frac {-a^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}\,2{}\mathrm {i}+b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}-a^2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,3{}\mathrm {i}+a^3\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+a^4\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^6-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^2+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^4}\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}}{d\,\left (a^4-a^2\,b^2\right )}-\frac {2\,A\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,\left (a^4-a^2\,b^2\right )}-\frac {2\,B\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,\left (a^4-a^2\,b^2\right )}-\frac {B\,a\,b\,\mathrm {atan}\left (\frac {-a^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}\,2{}\mathrm {i}+b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}-a^2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,3{}\mathrm {i}+a^3\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+a^4\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^6-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^2+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^4}\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}}{d\,\left (a^4-a^2\,b^2\right )} \]
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