3.251 \(\int \frac {1}{a+b \sin ^8(x)} \, dx\)

Optimal. Leaf size=245 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}-\sqrt [4]{b}} \tan (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}-\sqrt [4]{b}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}} \tan (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}} \tan (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+\sqrt [4]{b}} \tan (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+\sqrt [4]{b}}} \]

[Out]

-1/4*arctan(((-a)^(1/4)-b^(1/4))^(1/2)*tan(x)/(-a)^(1/8))/(-a)^(7/8)/((-a)^(1/4)-b^(1/4))^(1/2)-1/4*arctan(((-
a)^(1/4)-I*b^(1/4))^(1/2)*tan(x)/(-a)^(1/8))/(-a)^(7/8)/((-a)^(1/4)-I*b^(1/4))^(1/2)-1/4*arctan(((-a)^(1/4)+I*
b^(1/4))^(1/2)*tan(x)/(-a)^(1/8))/(-a)^(7/8)/((-a)^(1/4)+I*b^(1/4))^(1/2)-1/4*arctan(((-a)^(1/4)+b^(1/4))^(1/2
)*tan(x)/(-a)^(1/8))/(-a)^(7/8)/((-a)^(1/4)+b^(1/4))^(1/2)

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Rubi [A]  time = 0.53, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3211, 3181, 203} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}} \tan (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}} \tan (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+\sqrt [4]{b}} \tan (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+\sqrt [4]{b}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {a \sqrt [4]{b}+(-a)^{5/4}} \tan (x)}{(-a)^{5/8}}\right )}{4 (-a)^{3/8} \sqrt {a \sqrt [4]{b}+(-a)^{5/4}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[x]^8)^(-1),x]

[Out]

-ArcTan[(Sqrt[(-a)^(1/4) - I*b^(1/4)]*Tan[x])/(-a)^(1/8)]/(4*(-a)^(7/8)*Sqrt[(-a)^(1/4) - I*b^(1/4)]) - ArcTan
[(Sqrt[(-a)^(1/4) + I*b^(1/4)]*Tan[x])/(-a)^(1/8)]/(4*(-a)^(7/8)*Sqrt[(-a)^(1/4) + I*b^(1/4)]) - ArcTan[(Sqrt[
(-a)^(1/4) + b^(1/4)]*Tan[x])/(-a)^(1/8)]/(4*(-a)^(7/8)*Sqrt[(-a)^(1/4) + b^(1/4)]) - ArcTan[(Sqrt[(-a)^(5/4)
+ a*b^(1/4)]*Tan[x])/(-a)^(5/8)]/(4*(-a)^(3/8)*Sqrt[(-a)^(5/4) + a*b^(1/4)])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 3211

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(-1), x_Symbol] :> Module[{k}, Dist[2/(a*n), Sum[Int[1/(1 - Si
n[e + f*x]^2/((-1)^((4*k)/n)*Rt[-(a/b), n/2])), x], {k, 1, n/2}], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[n/
2]

Rubi steps

\begin {align*} \int \frac {1}{a+b \sin ^8(x)} \, dx &=\frac {\int \frac {1}{1-\frac {\sqrt [4]{b} \sin ^2(x)}{\sqrt [4]{-a}}} \, dx}{4 a}+\frac {\int \frac {1}{1-\frac {i \sqrt [4]{b} \sin ^2(x)}{\sqrt [4]{-a}}} \, dx}{4 a}+\frac {\int \frac {1}{1+\frac {i \sqrt [4]{b} \sin ^2(x)}{\sqrt [4]{-a}}} \, dx}{4 a}+\frac {\int \frac {1}{1+\frac {\sqrt [4]{b} \sin ^2(x)}{\sqrt [4]{-a}}} \, dx}{4 a}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{1+\left (1-\frac {\sqrt [4]{b}}{\sqrt [4]{-a}}\right ) x^2} \, dx,x,\tan (x)\right )}{4 a}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+\left (1-\frac {i \sqrt [4]{b}}{\sqrt [4]{-a}}\right ) x^2} \, dx,x,\tan (x)\right )}{4 a}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+\left (1+\frac {i \sqrt [4]{b}}{\sqrt [4]{-a}}\right ) x^2} \, dx,x,\tan (x)\right )}{4 a}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+\left (1+\frac {\sqrt [4]{b}}{\sqrt [4]{-a}}\right ) x^2} \, dx,x,\tan (x)\right )}{4 a}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}} \tan (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}} \tan (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+\sqrt [4]{b}} \tan (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+\sqrt [4]{b}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {(-a)^{5/4}+a \sqrt [4]{b}} \tan (x)}{(-a)^{5/8}}\right )}{4 (-a)^{3/8} \sqrt {(-a)^{5/4}+a \sqrt [4]{b}}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 174, normalized size = 0.71 \[ 8 \text {RootSum}\left [\text {$\#$1}^8 b-8 \text {$\#$1}^7 b+28 \text {$\#$1}^6 b-56 \text {$\#$1}^5 b+256 \text {$\#$1}^4 a+70 \text {$\#$1}^4 b-56 \text {$\#$1}^3 b+28 \text {$\#$1}^2 b-8 \text {$\#$1} b+b\& ,\frac {2 \text {$\#$1}^3 \tan ^{-1}\left (\frac {\sin (2 x)}{\cos (2 x)-\text {$\#$1}}\right )-i \text {$\#$1}^3 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (2 x)+1\right )}{\text {$\#$1}^7 b-7 \text {$\#$1}^6 b+21 \text {$\#$1}^5 b-35 \text {$\#$1}^4 b+128 \text {$\#$1}^3 a+35 \text {$\#$1}^3 b-21 \text {$\#$1}^2 b+7 \text {$\#$1} b-b}\& \right ] \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Sin[x]^8)^(-1),x]

[Out]

8*RootSum[b - 8*b*#1 + 28*b*#1^2 - 56*b*#1^3 + 256*a*#1^4 + 70*b*#1^4 - 56*b*#1^5 + 28*b*#1^6 - 8*b*#1^7 + b*#
1^8 & , (2*ArcTan[Sin[2*x]/(Cos[2*x] - #1)]*#1^3 - I*Log[1 - 2*Cos[2*x]*#1 + #1^2]*#1^3)/(-b + 7*b*#1 - 21*b*#
1^2 + 128*a*#1^3 + 35*b*#1^3 - 35*b*#1^4 + 21*b*#1^5 - 7*b*#1^6 + b*#1^7) & ]

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)^8),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b \sin \relax (x)^{8} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)^8),x, algorithm="giac")

[Out]

integrate(1/(b*sin(x)^8 + a), x)

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maple [C]  time = 0.26, size = 85, normalized size = 0.35 \[ \frac {\left (\munderset {\textit {\_R} =\RootOf \left (\left (a +b \right ) \textit {\_Z}^{8}+4 a \,\textit {\_Z}^{6}+6 a \,\textit {\_Z}^{4}+4 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{6}+3 \textit {\_R}^{4}+3 \textit {\_R}^{2}+1\right ) \ln \left (\tan \relax (x )-\textit {\_R} \right )}{\textit {\_R}^{7} a +\textit {\_R}^{7} b +3 \textit {\_R}^{5} a +3 \textit {\_R}^{3} a +\textit {\_R} a}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(x)^8),x)

[Out]

1/8*sum((_R^6+3*_R^4+3*_R^2+1)/(_R^7*a+_R^7*b+3*_R^5*a+3*_R^3*a+_R*a)*ln(tan(x)-_R),_R=RootOf((a+b)*_Z^8+4*a*_
Z^6+6*a*_Z^4+4*a*_Z^2+a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b \sin \relax (x)^{8} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)^8),x, algorithm="maxima")

[Out]

integrate(1/(b*sin(x)^8 + a), x)

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mupad [B]  time = 16.95, size = 816, normalized size = 3.33 \[ \sum _{k=1}^8\ln \left (-b^5\,\left (a+b\right )\,\left ({\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^2\,a^2\,800+{\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^4\,a^4\,43008+{\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^6\,a^6\,786432+\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )\,b\,\mathrm {tan}\relax (x)\,4-{\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^4\,a^3\,b\,6144+{\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^6\,a^5\,b\,786432-{\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^3\,a^3\,\mathrm {tan}\relax (x)\,9984-{\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^5\,a^5\,\mathrm {tan}\relax (x)\,557056-{\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^7\,a^7\,\mathrm {tan}\relax (x)\,10485760+{\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^2\,a\,b\,32-\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )\,a\,\mathrm {tan}\relax (x)\,60-{\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^3\,a^2\,b\,\mathrm {tan}\relax (x)\,768+{\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^5\,a^4\,b\,\mathrm {tan}\relax (x)\,98304-{\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right )}^7\,a^6\,b\,\mathrm {tan}\relax (x)\,10485760+5\right )\,2\right )\,\mathrm {root}\left (16777216\,a^7\,b\,d^8+16777216\,a^8\,d^8+1048576\,a^6\,d^6+24576\,a^4\,d^4+256\,a^2\,d^2+1,d,k\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(x)^8),x)

[Out]

symsum(log(-2*b^5*(a + b)*(800*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 24576*a^4*d^4 +
256*a^2*d^2 + 1, d, k)^2*a^2 + 43008*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 24576*a^4*
d^4 + 256*a^2*d^2 + 1, d, k)^4*a^4 + 786432*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 245
76*a^4*d^4 + 256*a^2*d^2 + 1, d, k)^6*a^6 + 4*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 2
4576*a^4*d^4 + 256*a^2*d^2 + 1, d, k)*b*tan(x) - 6144*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6
*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k)^4*a^3*b + 786432*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 10
48576*a^6*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k)^6*a^5*b - 9984*root(16777216*a^7*b*d^8 + 16777216*a^8*d
^8 + 1048576*a^6*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k)^3*a^3*tan(x) - 557056*root(16777216*a^7*b*d^8 +
16777216*a^8*d^8 + 1048576*a^6*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k)^5*a^5*tan(x) - 10485760*root(16777
216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k)^7*a^7*tan(x) + 32*
root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k)^2*a*b -
60*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k)*a*tan
(x) - 768*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k
)^3*a^2*b*tan(x) + 98304*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 24576*a^4*d^4 + 256*a^
2*d^2 + 1, d, k)^5*a^4*b*tan(x) - 10485760*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a^6*d^6 + 2457
6*a^4*d^4 + 256*a^2*d^2 + 1, d, k)^7*a^6*b*tan(x) + 5))*root(16777216*a^7*b*d^8 + 16777216*a^8*d^8 + 1048576*a
^6*d^6 + 24576*a^4*d^4 + 256*a^2*d^2 + 1, d, k), k, 1, 8)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \sin ^{8}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)**8),x)

[Out]

Integral(1/(a + b*sin(x)**8), x)

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