Integrand size = 21, antiderivative size = 138 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{\left (a^2+b^2\right ) d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{\left (a^2+b^2\right ) d \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}} \]
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Time = 0.07 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3155, 3157, 2719} \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 \sqrt {a \cos (c+d x)+b \sin (c+d x)} E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{d \left (a^2+b^2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{d \left (a^2+b^2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}} \]
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Rule 2719
Rule 3155
Rule 3157
Rubi steps \begin{align*} \text {integral}& = -\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{\left (a^2+b^2\right ) d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {\int \sqrt {a \cos (c+d x)+b \sin (c+d x)} \, dx}{a^2+b^2} \\ & = -\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{\left (a^2+b^2\right ) d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {\sqrt {a \cos (c+d x)+b \sin (c+d x)} \int \sqrt {\cos \left (c+d x-\tan ^{-1}(a,b)\right )} \, dx}{\left (a^2+b^2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}} \\ & = -\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{\left (a^2+b^2\right ) d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{\left (a^2+b^2\right ) d \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.38 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx=\frac {-\frac {2 b \cos (c+d x)}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}}+\frac {2 a \sin (c+d x)}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\sqrt {a \sqrt {1+\frac {b^2}{a^2}} \cos \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )} \tan \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )+\frac {\sqrt {a \sqrt {1+\frac {b^2}{a^2}} \cos \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )} \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (c+d x-\arctan \left (\frac {b}{a}\right )\right )\right ) \tan \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )}{\sqrt {\sin ^2\left (c+d x-\arctan \left (\frac {b}{a}\right )\right )}}}{\left (a^2+b^2\right ) d} \]
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Time = 1.08 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.68
method | result | size |
default | \(\frac {2 \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-2 \cos \left (d x +c -\arctan \left (-a , b\right )\right )^{2}}{\sqrt {a^{2}+b^{2}}\, \cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d}\) | \(232\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.01 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx=\frac {{\left (-i \, \sqrt {2} a \cos \left (d x + c\right ) - i \, \sqrt {2} b \sin \left (d x + c\right )\right )} \sqrt {a - i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (i \, \sqrt {2} a \cos \left (d x + c\right ) + i \, \sqrt {2} b \sin \left (d x + c\right )\right )} \sqrt {a + i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, \sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{{\left (a^{3} + a b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{2} b + b^{3}\right )} d \sin \left (d x + c\right )} \]
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\[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx=\int \frac {1}{\left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
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\[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx=\int \frac {1}{{\left (a\,\cos \left (c+d\,x\right )+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]
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