Integrand size = 28, antiderivative size = 119 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=\frac {\left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2} d}-\frac {b \left (\left (4 a^2+b^2\right ) \cos (c+d x)+3 a b \sin (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d (a \cos (c+d x)+b \sin (c+d x))^2} \]
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Time = 0.70 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.89, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1674, 12, 632, 212} \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=-\frac {\left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}}+\frac {2 b^2 \left (\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+a b\right )}{a^3 d \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b \left (4 a^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )+3 a^2 b^2+2 b^4\right )}{a^3 d \left (a^2+b^2\right )^2 \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+2 b \tan \left (\frac {1}{2} (c+d x)\right )\right )} \]
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Rule 12
Rule 212
Rule 632
Rule 1674
Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{\left (a+2 b x-a x^2\right )^3} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d} \\ & = \frac {2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {\text {Subst}\left (\int \frac {-\frac {8 \left (a^4+2 b^4\right )}{a^3}+16 b \left (1+\frac {b^2}{a^2}\right ) x+8 \left (a+\frac {b^2}{a}\right ) x^2}{\left (a+2 b x-a x^2\right )^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 \left (a^2+b^2\right ) d} \\ & = \frac {2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\text {Subst}\left (\int \frac {16 \left (2 a^2-b^2\right )}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{16 \left (a^2+b^2\right )^2 d} \\ & = \frac {2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\left (2 a^2-b^2\right ) \text {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^2 d} \\ & = \frac {2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}-\frac {\left (2 \left (2 a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^2 d} \\ & = -\frac {\left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {b-a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2} d}+\frac {2 b^2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right ) d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b \left (4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac {1}{2} (c+d x)\right )-a \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )} \\ \end{align*}
Time = 1.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=\frac {\frac {2 \left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {b \left (\left (4 a^2+b^2\right ) \cos (c+d x)+3 a b \sin (c+d x)\right )}{\left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2}}{2 d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(279\) vs. \(2(112)=224\).
Time = 0.88 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.35
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (-\frac {b^{2} \left (5 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (4 a^{4}-7 a^{2} b^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{2}+b^{2}\right )}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d}\) | \(280\) |
default | \(\frac {-\frac {2 \left (-\frac {b^{2} \left (5 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (4 a^{4}-7 a^{2} b^{2}-2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{2}+b^{2}\right )}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d}\) | \(280\) |
risch | \(\frac {b \,{\mathrm e}^{i \left (d x +c \right )} \left (-3 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 i b a +4 a^{2}+b^{2}\right )}{\left (-i a +b \right )^{2} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} d \left (i a +b \right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) a^{2}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) b^{2}}{2 \left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) a^{2}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) b^{2}}{2 \left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}\) | \(452\) |
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Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (112) = 224\).
Time = 0.27 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.96 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=-\frac {{\left (2 \, a^{2} b^{2} - b^{4} + {\left (2 \, a^{4} - 3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \, {\left (4 \, a^{4} b + 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} + 2 \, a^{6} b^{2} - 2 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d\right )}} \]
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Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (112) = 224\).
Time = 0.33 (sec) , antiderivative size = 412, normalized size of antiderivative = 3.46 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=-\frac {\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (4 \, a^{4} b + a^{2} b^{3} + \frac {{\left (11 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {{\left (4 \, a^{4} b - 7 \, a^{2} b^{3} - 2 \, b^{5}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {{\left (5 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4} + \frac {4 \, {\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, {\left (a^{8} - 3 \, a^{4} b^{4} - 2 \, a^{2} b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, {\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {{\left (a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}}{2 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (112) = 224\).
Time = 0.41 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.46 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=-\frac {\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{4} b - a^{2} b^{3}\right )}}{{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}^{2}}}{2 \, d} \]
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Time = 23.98 (sec) , antiderivative size = 443, normalized size of antiderivative = 3.72 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^3} \, dx=\frac {\ln \left ({\left (a^2+b^2\right )}^{5/2}-a^4\,b-b^5-2\,a^2\,b^3+a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-\frac {b^2}{2}\right )}{d\,{\left (a^2+b^2\right )}^{5/2}}-\frac {\ln \left ({\left (a^2+b^2\right )}^{5/2}+a^4\,b+b^5+2\,a^2\,b^3-a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-b^2\right )}{2\,d\,{\left (a^2+b^2\right )}^{5/2}}-\frac {\frac {4\,a^2\,b+b^3}{a^4+2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2-2\,b^2\right )\,\left (4\,a^2\,b+b^3\right )}{a^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (11\,a^2\,b+2\,b^3\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,a^2\,b+2\,b^3\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-4\,b^2\right )+a^2-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]
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