Integrand size = 41, antiderivative size = 314 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\frac {\left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \]
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Time = 1.22 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4175, 4165, 4088, 12, 3916, 2738, 214} \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}+\frac {\left (a^3 B-a^2 b (4 A+3 C)+4 a b^2 B-b^3 (A+2 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {\tan (c+d x) \left (-4 a^4 C+a^3 b B+a^2 b^2 (2 A+9 C)-6 a b^3 B+3 A b^4\right )}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac {\tan (c+d x) \left (2 a^5 C+a^4 b B+a^3 b^2 (2 A-5 C)-10 a^2 b^3 B+a b^4 (13 A+18 C)-6 b^5 B\right )}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))} \]
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Rule 12
Rule 214
Rule 2738
Rule 3916
Rule 4088
Rule 4165
Rule 4175
Rubi steps \begin{align*} \text {integral}& = \frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\int \frac {\sec (c+d x) \left (-3 b \left (A b^2-a (b B-a C)\right )+\left (a^2 b B-3 b^3 B-a^3 C+a b^2 (2 A+3 C)\right ) \sec (c+d x)+3 b \left (a^2-b^2\right ) C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b^2 \left (a^2-b^2\right )} \\ & = \frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \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^2 \left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right )-b \left (a^3 b B-6 a b^3 B+a^2 b^2 (2 A-3 C)+2 a^4 C+3 b^4 (A+2 C)\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b^3 \left (a^2-b^2\right )^2} \\ & = \frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\int \frac {3 b^3 \left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )^3} \\ & = \frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3} \\ & = \frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^3} \\ & = \frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 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 (4 a^2 A b+A b^3-a^3 B-4 a b^2 B+3 a^2 b C+2 b^3 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 {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \\ \end{align*}
Time = 1.68 (sec) , antiderivative size = 299, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\frac {\frac {24 \left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {2 \left (6 a^5 A+14 a^3 A b^2+25 a A b^4-11 a^4 b B-22 a^2 b^3 B-12 b^5 B+8 a^5 C+a^3 b^2 C+36 a b^4 C+6 \left (-A b^5+a^5 B-9 a^3 b^2 B-2 a b^4 B+9 a^2 b^3 (A+C)+a^4 b (2 A+C)\right ) \cos (c+d x)+a \left (-A b^4-13 a^3 b B-2 a b^3 B+a^4 (6 A+4 C)+a^2 b^2 (10 A+11 C)\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{(b+a \cos (c+d x))^3}}{24 \left (-a^2+b^2\right )^3 d} \]
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Time = 0.72 (sec) , antiderivative size = 453, normalized size of antiderivative = 1.44
method | result | size |
derivativedivides | \(\frac {\frac {-\frac {\left (2 a^{3} A +2 A \,a^{2} b +6 a A \,b^{2}+A \,b^{3}-B \,a^{3}-6 B \,a^{2} b -2 B a \,b^{2}-2 B \,b^{3}+2 a^{3} C +3 a^{2} b C +6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {4 \left (3 a^{3} A +7 a A \,b^{2}-7 B \,a^{2} b -3 B \,b^{3}+a^{3} C +9 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 a^{3} A -2 A \,a^{2} b +6 a A \,b^{2}-A \,b^{3}+B \,a^{3}-6 B \,a^{2} b +2 B a \,b^{2}-2 B \,b^{3}+2 a^{3} C -3 a^{2} b C +6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}-\frac {\left (4 A \,a^{2} b +A \,b^{3}-B \,a^{3}-4 B a \,b^{2}+3 a^{2} b C +2 C \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(453\) |
default | \(\frac {\frac {-\frac {\left (2 a^{3} A +2 A \,a^{2} b +6 a A \,b^{2}+A \,b^{3}-B \,a^{3}-6 B \,a^{2} b -2 B a \,b^{2}-2 B \,b^{3}+2 a^{3} C +3 a^{2} b C +6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {4 \left (3 a^{3} A +7 a A \,b^{2}-7 B \,a^{2} b -3 B \,b^{3}+a^{3} C +9 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 a^{3} A -2 A \,a^{2} b +6 a A \,b^{2}-A \,b^{3}+B \,a^{3}-6 B \,a^{2} b +2 B a \,b^{2}-2 B \,b^{3}+2 a^{3} C -3 a^{2} b C +6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{3}}-\frac {\left (4 A \,a^{2} b +A \,b^{3}-B \,a^{3}-4 B a \,b^{2}+3 a^{2} b C +2 C \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(453\) |
risch | \(\text {Expression too large to display}\) | \(1845\) |
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Leaf count of result is larger than twice the leaf count of optimal. 681 vs. \(2 (300) = 600\).
Time = 0.38 (sec) , antiderivative size = 1420, normalized size of antiderivative = 4.52 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \]
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\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{4}}\, dx \]
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Exception generated. \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Exception raised: ValueError} \]
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Leaf count of result is larger than twice the leaf count of optimal. 970 vs. \(2 (300) = 600\).
Time = 0.39 (sec) , antiderivative size = 970, normalized size of antiderivative = 3.09 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\text {Too large to display} \]
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Time = 20.51 (sec) , antiderivative size = 516, normalized size of antiderivative = 1.64 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx=\frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^3-A\,b^3+B\,a^3-2\,B\,b^3+2\,C\,a^3+6\,A\,a\,b^2-2\,A\,a^2\,b+2\,B\,a\,b^2-6\,B\,a^2\,b+6\,C\,a\,b^2-3\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A\,a^3-3\,B\,b^3+C\,a^3+7\,A\,a\,b^2-7\,B\,a^2\,b+9\,C\,a\,b^2\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,a^3+A\,b^3-B\,a^3-2\,B\,b^3+2\,C\,a^3+6\,A\,a\,b^2+2\,A\,a^2\,b-2\,B\,a\,b^2-6\,B\,a^2\,b+6\,C\,a\,b^2+3\,C\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (A\,b^3-B\,a^3+2\,C\,b^3+4\,A\,a^2\,b-4\,B\,a\,b^2+3\,C\,a^2\,b\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]
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