Integrand size = 23, antiderivative size = 273 \[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {x}{2 b}-\frac {2 a \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-b^{2/3}} b^{5/3} d}+\frac {2 a \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}} b^{5/3} d}+\frac {2 a \text {arctanh}\left (\frac {\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}} b^{5/3} d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d} \]
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Time = 0.41 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3299, 2715, 8, 2739, 632, 210, 212} \[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=-\frac {2 a \arctan \left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 a \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 b^{5/3} d \sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}+\frac {2 a \text {arctanh}\left (\frac {(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}+\frac {x}{2 b} \]
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Rule 8
Rule 210
Rule 212
Rule 632
Rule 2715
Rule 2739
Rule 3299
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\sin ^2(c+d x)}{b}-\frac {a \sin ^2(c+d x)}{b \left (a+b \sin ^3(c+d x)\right )}\right ) \, dx \\ & = \frac {\int \sin ^2(c+d x) \, dx}{b}-\frac {a \int \frac {\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx}{b} \\ & = -\frac {\cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\int 1 \, dx}{2 b}-\frac {a \int \left (\frac {1}{3 b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}+\frac {1}{3 b^{2/3} \left (-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}+\frac {1}{3 b^{2/3} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}\right ) \, dx}{b} \\ & = \frac {x}{2 b}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d}-\frac {a \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^{5/3}}-\frac {a \int \frac {1}{-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^{5/3}}-\frac {a \int \frac {1}{(-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)} \, dx}{3 b^{5/3}} \\ & = \frac {x}{2 b}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d}-\frac {(2 a) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}+2 \sqrt [3]{b} x+\sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^{5/3} d}-\frac {(2 a) \text {Subst}\left (\int \frac {1}{-\sqrt [3]{-1} \sqrt [3]{a}+2 \sqrt [3]{b} x-\sqrt [3]{-1} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^{5/3} d}-\frac {(2 a) \text {Subst}\left (\int \frac {1}{(-1)^{2/3} \sqrt [3]{a}+2 \sqrt [3]{b} x+(-1)^{2/3} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^{5/3} d} \\ & = \frac {x}{2 b}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d}+\frac {(4 a) \text {Subst}\left (\int \frac {1}{-4 \left (a^{2/3}-b^{2/3}\right )-x^2} \, dx,x,2 \sqrt [3]{b}+2 \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^{5/3} d}+\frac {(4 a) \text {Subst}\left (\int \frac {1}{-4 \left ((-1)^{2/3} a^{2/3}-b^{2/3}\right )-x^2} \, dx,x,2 \sqrt [3]{b}-2 \sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^{5/3} d}+\frac {(4 a) \text {Subst}\left (\int \frac {1}{4 \left (\sqrt [3]{-1} a^{2/3}+b^{2/3}\right )-x^2} \, dx,x,2 \sqrt [3]{b}+2 (-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^{5/3} d} \\ & = \frac {x}{2 b}-\frac {2 a \arctan \left (\frac {\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 \sqrt {a^{2/3}-b^{2/3}} b^{5/3} d}+\frac {2 a \text {arctanh}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {-(-1)^{2/3} a^{2/3}+b^{2/3}} b^{5/3} d}+\frac {2 a \text {arctanh}\left (\frac {\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}} b^{5/3} d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d} \\ \end{align*}
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.33 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {6 (c+d x)-2 i a \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-4 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+2 i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2+2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-i \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4}{b \text {$\#$1}-4 i a \text {$\#$1}^2-2 b \text {$\#$1}^3+b \text {$\#$1}^5}\&\right ]-3 \sin (2 (c+d x))}{12 b d} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.52
method | result | size |
derivativedivides | \(\frac {\frac {\frac {8 \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {4 a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b}}{d}\) | \(141\) |
default | \(\frac {\frac {\frac {8 \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}-\frac {4 a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 b}}{d}\) | \(141\) |
risch | \(\frac {x}{2 b}-\frac {i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (729 a^{2} b^{10} d^{6}-729 b^{12} d^{6}\right ) \textit {\_Z}^{6}-248832 a^{2} b^{8} d^{4} \textit {\_Z}^{4}-28311552 a^{4} b^{4} d^{2} \textit {\_Z}^{2}-1073741824 a^{6}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\left (\frac {243 i d^{5} b^{8}}{33554432 a^{3}}-\frac {243 i d^{5} b^{10}}{33554432 a^{5}}\right ) \textit {\_R}^{5}+\left (-\frac {81 i d^{4} b^{7}}{1048576 a^{3}}+\frac {81 i d^{4} b^{9}}{1048576 a^{5}}\right ) \textit {\_R}^{4}-\frac {81 i d^{3} b^{6} \textit {\_R}^{3}}{32768 a^{3}}+\left (\frac {9 i d^{2} b^{3}}{1024 a}+\frac {9 i d^{2} b^{5}}{512 a^{3}}\right ) \textit {\_R}^{2}-\frac {9 i d \,b^{2} \textit {\_R}}{32 a}+\frac {i b}{a}\right )\right )}{32}-\frac {\sin \left (2 d x +2 c \right )}{4 d b}\) | \(223\) |
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Result contains complex when optimal does not.
Time = 1.28 (sec) , antiderivative size = 29175, normalized size of antiderivative = 106.87 \[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{5}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
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\[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{5}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \]
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Time = 14.51 (sec) , antiderivative size = 1962, normalized size of antiderivative = 7.19 \[ \int \frac {\sin ^5(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \]
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