Integrand size = 15, antiderivative size = 115 \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}-\sqrt {b}} d}+\frac {\arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}+\sqrt {b}} d} \]
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Time = 0.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3288, 1180, 211} \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt {\sqrt {a}-\sqrt {b}}}+\frac {\arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt {\sqrt {a}+\sqrt {b}}} \]
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Rule 211
Rule 1180
Rule 3288
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1+x^2}{a+2 a x^2+(a-b) x^4} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (1-\frac {\sqrt {b}}{\sqrt {a}}\right ) \text {Subst}\left (\int \frac {1}{a-\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (1+\frac {\sqrt {b}}{\sqrt {a}}\right ) \text {Subst}\left (\int \frac {1}{a+\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = \frac {\arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}-\sqrt {b}} d}+\frac {\arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {\sqrt {a}+\sqrt {b}} d} \\ \end{align*}
Time = 0.97 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11 \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {\arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}-\frac {\text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}}{2 \sqrt {a} d} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.57 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11
method | result | size |
risch | \(\munderset {\textit {\_R} =\operatorname {RootOf}\left (1+\left (256 a^{4} d^{4}-256 a^{3} b \,d^{4}\right ) \textit {\_Z}^{4}+32 a^{2} d^{2} \textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {128 i d^{3} a^{4}}{b}-128 i a^{3} d^{3}\right ) \textit {\_R}^{3}+\left (-\frac {32 d^{2} a^{3}}{b}+32 a^{2} d^{2}\right ) \textit {\_R}^{2}+\left (\frac {8 i a^{2} d}{b}+8 i a d \right ) \textit {\_R} -\frac {2 a}{b}-1\right )\) | \(128\) |
derivativedivides | \(\frac {\left (a -b \right ) \left (\frac {\left (\sqrt {a b}+b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{d}\) | \(145\) |
default | \(\frac {\left (a -b \right ) \left (\frac {\left (\sqrt {a b}+b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (\sqrt {a b}-b \right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{d}\) | \(145\) |
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Leaf count of result is larger than twice the leaf count of optimal. 1079 vs. \(2 (79) = 158\).
Time = 0.41 (sec) , antiderivative size = 1079, normalized size of antiderivative = 9.38 \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {1}{b \sin \left (d x + c\right )^{4} - a} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (79) = 158\).
Time = 0.45 (sec) , antiderivative size = 361, normalized size of antiderivative = 3.14 \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {{\left (3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a^{2} - 6 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a b - \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a + \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} - 12 \, a^{4} b + 14 \, a^{3} b^{2} - 4 \, a^{2} b^{3} - a b^{4}} + \frac {{\left (3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a^{2} - 6 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a b - \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a - \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} - 12 \, a^{4} b + 14 \, a^{3} b^{2} - 4 \, a^{2} b^{3} - a b^{4}}}{2 \, d} \]
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Time = 15.42 (sec) , antiderivative size = 671, normalized size of antiderivative = 5.83 \[ \int \frac {1}{a-b \sin ^4(c+d x)} \, dx=\frac {\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}+a^5\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}+a^3\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}+a^2\,b\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}-a^4\,b\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}+a\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}+b\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}-a^2\,b\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}}{a\,b+\sqrt {a^3\,b}}\right )\,\sqrt {-\frac {1}{16\,a^2+16\,\sqrt {a^3\,b}}}\,2{}\mathrm {i}}{d}+\frac {\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}+a^5\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}-a^3\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}+a^2\,b\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,4{}\mathrm {i}-a^4\,b\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,64{}\mathrm {i}-a\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}-b\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,\sqrt {a^3\,b}\,4{}\mathrm {i}+a^2\,b\,\mathrm {tan}\left (c+d\,x\right )\,{\left (-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}\right )}^{3/2}\,\sqrt {a^3\,b}\,64{}\mathrm {i}}{a\,b-\sqrt {a^3\,b}}\right )\,\sqrt {-\frac {1}{16\,a^2-16\,\sqrt {a^3\,b}}}\,2{}\mathrm {i}}{d} \]
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