Integrand size = 39, antiderivative size = 299 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=-\frac {\left (4 a^2 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \]
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Time = 0.98 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4165, 4088, 12, 3916, 2738, 214} \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}-\frac {\left (-\left (a^3 (2 A+C)\right )+4 a^2 b B-a b^2 (3 A+4 C)+b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {\tan (c+d x) \left (a^3 C+2 a^2 b B-a b^2 (5 A+6 C)+3 b^3 B\right )}{6 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac {\tan (c+d x) \left (a^4 C+2 a^3 b B-a^2 b^2 (11 A+10 C)+13 a b^3 B-2 b^4 (2 A+3 C)\right )}{6 b d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))} \]
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Rule 12
Rule 214
Rule 2738
Rule 3916
Rule 4088
Rule 4165
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {\int \frac {\sec (c+d x) \left (3 b (b B-a (A+C))+\left (2 A b^2-2 a b B-a^2 C+3 b^2 C\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )} \\ & = -\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\int \frac {\sec (c+d x) \left (-2 b \left (5 a b B-a^2 (3 A+2 C)-b^2 (2 A+3 C)\right )+\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b \left (a^2-b^2\right )^2} \\ & = -\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\int \frac {3 b \left (4 a^2 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b \left (a^2-b^2\right )^3} \\ & = -\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (4 a^2 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3} \\ & = -\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (4 a^2 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^3} \\ & = -\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (4 a^2 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right ) \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 \left (a^2-b^2\right )^3 d} \\ & = \frac {\left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B+a^3 C+4 a b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \\ \end{align*}
Result contains complex when optimal does not.
Time = 8.67 (sec) , antiderivative size = 538, normalized size of antiderivative = 1.80 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\frac {(b+a \cos (c+d x)) \sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {6 i \left (-4 a^2 b B-b^3 B+a^3 (2 A+C)+a b^2 (3 A+4 C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x))^3 (\cos (c)-i \sin (c))}{\left (a^2-b^2\right )^{7/2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {2 b \left (A b^2+a (-b B+a C)\right ) \sec (c) (b \sin (c)-a \sin (d x))}{a^5-a^3 b^2}+\frac {(b+a \cos (c+d x)) \sec (c) \left (b \left (-11 a^2 A b^2+6 A b^4+8 a^3 b B-3 a b^3 B-5 a^4 C\right ) \sin (c)+a \left (-4 A b^4-6 a^3 b B+a b^3 B+3 a^4 C+a^2 b^2 (9 A+2 C)\right ) \sin (d x)\right )}{a^3 \left (a^2-b^2\right )^2}+\frac {(b+a \cos (c+d x))^2 \sec (c) \left (3 \left (-6 a^2 A b^4+2 A b^6-4 a^5 b B-a^3 b^3 B+a^6 C+a^4 b^2 (9 A+4 C)\right ) \sin (c)+a \left (-2 A b^5+6 a^5 B+10 a^3 b^2 B-a b^4 B+a^2 b^3 (5 A-2 C)-a^4 b (18 A+13 C)\right ) \sin (d x)\right )}{\left (a^3-a b^2\right )^3}\right )}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))^4} \]
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Time = 0.68 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.51
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (-\frac {\left (6 A \,a^{2} b +3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-2 B \,a^{2} b -6 B a \,b^{2}-B \,b^{3}+a^{3} C +6 a^{2} b C +2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (9 A \,a^{2} b +A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}+7 a^{2} b C +3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}+2 B \,a^{2} b -6 B a \,b^{2}+B \,b^{3}-a^{3} C +6 a^{2} b C -2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}+\frac {\left (2 a^{3} A +3 a A \,b^{2}-4 B \,a^{2} b -B \,b^{3}+a^{3} C +4 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(452\) |
default | \(\frac {-\frac {2 \left (-\frac {\left (6 A \,a^{2} b +3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-2 B \,a^{2} b -6 B a \,b^{2}-B \,b^{3}+a^{3} C +6 a^{2} b C +2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {2 \left (9 A \,a^{2} b +A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}+7 a^{2} b C +3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}+2 B \,a^{2} b -6 B a \,b^{2}+B \,b^{3}-a^{3} C +6 a^{2} b C -2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}+\frac {\left (2 a^{3} A +3 a A \,b^{2}-4 B \,a^{2} b -B \,b^{3}+a^{3} C +4 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(452\) |
risch | \(\text {Expression too large to display}\) | \(1968\) |
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Leaf count of result is larger than twice the leaf count of optimal. 678 vs. \(2 (281) = 562\).
Time = 0.39 (sec) , antiderivative size = 1414, normalized size of antiderivative = 4.73 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \]
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\[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{4}}\, dx \]
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Exception generated. \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Exception raised: ValueError} \]
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Leaf count of result is larger than twice the leaf count of optimal. 968 vs. \(2 (281) = 562\).
Time = 0.39 (sec) , antiderivative size = 968, normalized size of antiderivative = 3.24 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \]
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Time = 20.42 (sec) , antiderivative size = 516, normalized size of antiderivative = 1.73 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (2\,A\,a^3-B\,b^3+C\,a^3+3\,A\,a\,b^2-4\,B\,a^2\,b+4\,C\,a\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,b^3-2\,B\,a^3+B\,b^3-C\,a^3+2\,C\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2+2\,B\,a^2\,b-2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A\,b^3-3\,B\,a^3+3\,C\,b^3+9\,A\,a^2\,b-7\,B\,a\,b^2+7\,C\,a^2\,b\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,b^3-2\,B\,a^3-B\,b^3+C\,a^3+2\,C\,b^3+3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2-2\,B\,a^2\,b+2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \]
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