Integrand size = 38, antiderivative size = 98 \[ \int \frac {\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{b^2 d}-\frac {2 a (b B-a C) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {C \tan (c+d x)}{b d} \]
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Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {4157, 4095, 12, 3874, 3855, 3916, 2738, 214} \[ \int \frac {\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{b^2 d}-\frac {2 a (b B-a C) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d \sqrt {a-b} \sqrt {a+b}}+\frac {C \tan (c+d x)}{b d} \]
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Rule 12
Rule 214
Rule 2738
Rule 3855
Rule 3874
Rule 3916
Rule 4095
Rule 4157
Rubi steps \begin{align*} \text {integral}& = \int \frac {\sec ^2(c+d x) (B+C \sec (c+d x))}{a+b \sec (c+d x)} \, dx \\ & = \frac {C \tan (c+d x)}{b d}+\frac {\int \frac {(b B-a C) \sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx}{b} \\ & = \frac {C \tan (c+d x)}{b d}+\frac {(b B-a C) \int \frac {\sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx}{b} \\ & = \frac {C \tan (c+d x)}{b d}+\frac {(b B-a C) \int \sec (c+d x) \, dx}{b^2}-\frac {(a (b B-a C)) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^2} \\ & = \frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{b^2 d}+\frac {C \tan (c+d x)}{b d}-\frac {(a (b B-a C)) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{b^3} \\ & = \frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{b^2 d}+\frac {C \tan (c+d x)}{b d}-\frac {(2 a (b B-a C)) \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 )}{b^3 d} \\ & = \frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{b^2 d}-\frac {2 a (b B-a C) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {C \tan (c+d x)}{b d} \\ \end{align*}
Time = 0.81 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.33 \[ \int \frac {\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {-\frac {2 a (-b B+a C) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-(b B-a C) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+b C \tan (c+d x)}{b^2 d} \]
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Time = 0.33 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.47
method | result | size |
derivativedivides | \(\frac {-\frac {C}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}-\frac {C}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B b -C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {2 a \left (B b -C a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(144\) |
default | \(\frac {-\frac {C}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}-\frac {C}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B b -C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {2 a \left (B b -C a \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(144\) |
risch | \(\frac {2 i C}{d b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{b d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{b^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{b d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{b^{2} d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, d b}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}\) | \(407\) |
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Leaf count of result is larger than twice the leaf count of optimal. 208 vs. \(2 (90) = 180\).
Time = 0.36 (sec) , antiderivative size = 472, normalized size of antiderivative = 4.82 \[ \int \frac {\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\left [-\frac {{\left (C a^{2} - B a b\right )} \sqrt {a^{2} - b^{2}} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )}, \frac {2 \, {\left (C a^{2} - B a b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )}\right ] \]
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\[ \int \frac {\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
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Exception generated. \[ \int \frac {\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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none
Time = 0.34 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.79 \[ \int \frac {\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=-\frac {\frac {{\left (C a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac {{\left (C a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} + \frac {2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} b} + \frac {2 \, {\left (C a^{2} - B a b\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} b^{2}}}{d} \]
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Time = 18.42 (sec) , antiderivative size = 719, normalized size of antiderivative = 7.34 \[ \int \frac {\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {2\,C\,a\,\mathrm {atanh}\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^2-b^2\right )}-\frac {2\,B\,b\,\mathrm {atanh}\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^2-b^2\right )}-\frac {C\,b\,\mathrm {tan}\left (c+d\,x\right )}{d\,\left (a^2-b^2\right )}-\frac {C\,a^2\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,{\left (a^2-b^2\right )}^{3/2}}+\frac {2\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d\,\left (a^2-b^2\right )}-\frac {2\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d\,\left (a^2-b^2\right )}-\frac {B\,a^3\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d\,{\left (a^2-b^2\right )}^{3/2}}+\frac {C\,a^4\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d\,{\left (a^2-b^2\right )}^{3/2}}+\frac {B\,a\,b\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,{\left (a^2-b^2\right )}^{3/2}}+\frac {C\,a^2\,\mathrm {tan}\left (c+d\,x\right )}{b\,d\,\left (a^2-b^2\right )}-\frac {C\,a^2\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{b^2\,d\,\left (a^2-b^2\right )}+\frac {B\,a\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {\left (a+b\right )\,\left (a-b\right )}}{b\,d\,\left (a^2-b^2\right )} \]
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