3.3.0 ∫ 1 ______ 1+sin8(x) dx [257]

Integrand size = 8, antiderivative size = 218 \[ \int \frac {1}{1+\sin ^8(x)} \, dx=\frac {1}{8} \left (\sqrt {1+\sqrt {4-2 \sqrt {2}}}+\sqrt {2+2 \sqrt [4]{2}+2 \sqrt {1+\sqrt {2}}+2 \sqrt {2+\sqrt {2}}}+\sqrt {1+\sqrt {4+2 \sqrt {2}}}\right ) (x-\arctan (\tan (x)))+\frac {\arctan \left (\sqrt {1-\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt {1-\sqrt [4]{-1}}}+\frac {\arctan \left (\sqrt {1+\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt {1+\sqrt [4]{-1}}}+\frac {\arctan \left (\sqrt {1-(-1)^{3/4}} \tan (x)\right )}{4 \sqrt {1-(-1)^{3/4}}}+\frac {\arctan \left (\sqrt {1+(-1)^{3/4}} \tan (x)\right )}{4 \sqrt {1+(-1)^{3/4}}} \]

Rubi [A] (veriﬁed)

Time = 0.23 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.59, number of steps used = 9, number of rules used = 3, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.375, Rules used = {3290, 3260, 209} \[ \int \frac {1}{1+\sin ^8(x)} \, dx=\frac {\arctan \left (\sqrt {1-\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt {1-\sqrt [4]{-1}}}+\frac {\arctan \left (\sqrt {1+\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt {1+\sqrt [4]{-1}}}+\frac {\arctan \left (\sqrt {1-(-1)^{3/4}} \tan (x)\right )}{4 \sqrt {1-(-1)^{3/4}}}+\frac {\arctan \left (\sqrt {1+(-1)^{3/4}} \tan (x)\right )}{4 \sqrt {1+(-1)^{3/4}}} \]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {1}{1-\sqrt [4]{-1} \sin ^2(x)} \, dx+\frac {1}{4} \int \frac {1}{1+\sqrt [4]{-1} \sin ^2(x)} \, dx+\frac {1}{4} \int \frac {1}{1-(-1)^{3/4} \sin ^2(x)} \, dx+\frac {1}{4} \int \frac {1}{1+(-1)^{3/4} \sin ^2(x)} \, dx \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {1}{1+\left (1-\sqrt [4]{-1}\right ) x^2} \, dx,x,\tan (x)\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+\left (1+\sqrt [4]{-1}\right ) x^2} \, dx,x,\tan (x)\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+\left (1-(-1)^{3/4}\right ) x^2} \, dx,x,\tan (x)\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+\left (1+(-1)^{3/4}\right ) x^2} \, dx,x,\tan (x)\right ) \\ & = \frac {\arctan \left (\sqrt {1-\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt {1-\sqrt [4]{-1}}}+\frac {\arctan \left (\sqrt {1+\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt {1+\sqrt [4]{-1}}}+\frac {\arctan \left (\sqrt {1-(-1)^{3/4}} \tan (x)\right )}{4 \sqrt {1-(-1)^{3/4}}}+\frac {\arctan \left (\sqrt {1+(-1)^{3/4}} \tan (x)\right )}{4 \sqrt {1+(-1)^{3/4}}} \\ \end{align*}

Mathematica [C] (veriﬁed)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 5.05 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.65 \[ \int \frac {1}{1+\sin ^8(x)} \, dx=8 \text {RootSum}\left [1-8 \text {$\#$1}+28 \text {$\#$1}^2-56 \text {$\#$1}^3+326 \text {$\#$1}^4-56 \text {$\#$1}^5+28 \text {$\#$1}^6-8 \text {$\#$1}^7+\text {$\#$1}^8\&,\frac {2 \arctan \left (\frac {\sin (2 x)}{\cos (2 x)-\text {$\#$1}}\right ) \text {$\#$1}^3-i \log \left (1-2 \cos (2 x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^3}{-1+7 \text {$\#$1}-21 \text {$\#$1}^2+163 \text {$\#$1}^3-35 \text {$\#$1}^4+21 \text {$\#$1}^5-7 \text {$\#$1}^6+\text {$\#$1}^7}\&\right ] \]

Maple [C] (veriﬁed)

Time = 4.45 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.33


method	result	size

default	$\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}+4 \textit {\_Z}^{6}+6 \textit {\_Z}^{4}+4 \textit {\_Z}^{2}+1\right )}{\sum }\frac {\left (\textit {\_R}^{6}+3 \textit {\_R}^{4}+3 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (x \right )-\textit {\_R} \right )}{2 \textit {\_R}^{7}+3 \textit {\_R}^{5}+3 \textit {\_R}^{3}+\textit {\_R}}\right )}{8}$	$71$
risch	$\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (8192 \textit {\_Z}^{4}+\left (128+128 i\right ) \textit {\_Z}^{2}+1+i\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}+\left (1024-1024 i\right ) \textit {\_R}^{3}+\left (128+128 i\right ) \textit {\_R}^{2}+\left (16+16 i\right ) \textit {\_R} -1+2 i\right )\right )+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (8192 \textit {\_Z}^{4}+\left (128-128 i\right ) \textit {\_Z}^{2}+1-i\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}+\left (-1024-1024 i\right ) \textit {\_R}^{3}+\left (128-128 i\right ) \textit {\_R}^{2}+\left (-16+16 i\right ) \textit {\_R} -1-2 i\right )\right )$	$104$

Fricas [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 893 vs. $2 (152) = 304$.

Time = 0.43 (sec) , antiderivative size = 893, normalized size of antiderivative = 4.10 \[ \int \frac {1}{1+\sin ^8(x)} \, dx=\text {Too large to display} \]

Sympy [F]

\[ \int \frac {1}{1+\sin ^8(x)} \, dx=\int \frac {1}{\sin ^{8}{\left (x \right )} + 1}\, dx \]

Maxima [F]

\[ \int \frac {1}{1+\sin ^8(x)} \, dx=\int { \frac {1}{\sin \left (x\right )^{8} + 1} \,d x } \]

Giac [F]

\[ \int \frac {1}{1+\sin ^8(x)} \, dx=\int { \frac {1}{\sin \left (x\right )^{8} + 1} \,d x } \]

Mupad [B] (veriﬁcation not implemented)

Time = 14.96 (sec) , antiderivative size = 945, normalized size of antiderivative = 4.33 \[ \int \frac {1}{1+\sin ^8(x)} \, dx=\text {Too large to display} \]

\(\int \frac {1}{1+\sin ^8(x)} \, dx\) [257]

Optimal result