3.249 \(\int \frac{1}{a+b \sin ^5(x)} \, dx\)

Optimal. Leaf size=384 \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt [5]{a} \tan \left (\frac{x}{2}\right )+\sqrt [5]{b}}{\sqrt{a^{2/5}-b^{2/5}}}\right )}{5 a^{4/5} \sqrt{a^{2/5}-b^{2/5}}}+\frac{2 \tan ^{-1}\left (\frac{\sqrt [5]{a} \tan \left (\frac{x}{2}\right )+(-1)^{2/5} \sqrt [5]{b}}{\sqrt{a^{2/5}-(-1)^{4/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt{a^{2/5}-(-1)^{4/5} b^{2/5}}}+\frac{2 \tan ^{-1}\left (\frac{\sqrt [5]{a} \tan \left (\frac{x}{2}\right )+(-1)^{4/5} \sqrt [5]{b}}{\sqrt{a^{2/5}+(-1)^{3/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt{a^{2/5}+(-1)^{3/5} b^{2/5}}}-\frac{2 \tan ^{-1}\left (\frac{(-1)^{3/5} \left ((-1)^{2/5} \sqrt [5]{a} \tan \left (\frac{x}{2}\right )+\sqrt [5]{b}\right )}{\sqrt{a^{2/5}+\sqrt [5]{-1} b^{2/5}}}\right )}{5 a^{4/5} \sqrt{a^{2/5}+\sqrt [5]{-1} b^{2/5}}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt [5]{-1} \left ((-1)^{4/5} \sqrt [5]{a} \tan \left (\frac{x}{2}\right )+\sqrt [5]{b}\right )}{\sqrt{a^{2/5}-(-1)^{2/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt{a^{2/5}-(-1)^{2/5} b^{2/5}}} \]

[Out]

(2*ArcTan[(b^(1/5) + a^(1/5)*Tan[x/2])/Sqrt[a^(2/5) - b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) - b^(2/5)]) + (2*ArcT
an[((-1)^(2/5)*b^(1/5) + a^(1/5)*Tan[x/2])/Sqrt[a^(2/5) - (-1)^(4/5)*b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) - (-1)
^(4/5)*b^(2/5)]) + (2*ArcTan[((-1)^(4/5)*b^(1/5) + a^(1/5)*Tan[x/2])/Sqrt[a^(2/5) + (-1)^(3/5)*b^(2/5)]])/(5*a
^(4/5)*Sqrt[a^(2/5) + (-1)^(3/5)*b^(2/5)]) - (2*ArcTan[((-1)^(3/5)*(b^(1/5) + (-1)^(2/5)*a^(1/5)*Tan[x/2]))/Sq
rt[a^(2/5) + (-1)^(1/5)*b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) + (-1)^(1/5)*b^(2/5)]) - (2*ArcTan[((-1)^(1/5)*(b^(
1/5) + (-1)^(4/5)*a^(1/5)*Tan[x/2]))/Sqrt[a^(2/5) - (-1)^(2/5)*b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) - (-1)^(2/5)
*b^(2/5)])

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Rubi [A]  time = 0.714324, antiderivative size = 384, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3213, 2660, 618, 204} \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt [5]{a} \tan \left (\frac{x}{2}\right )+\sqrt [5]{b}}{\sqrt{a^{2/5}-b^{2/5}}}\right )}{5 a^{4/5} \sqrt{a^{2/5}-b^{2/5}}}+\frac{2 \tan ^{-1}\left (\frac{\sqrt [5]{a} \tan \left (\frac{x}{2}\right )+(-1)^{2/5} \sqrt [5]{b}}{\sqrt{a^{2/5}-(-1)^{4/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt{a^{2/5}-(-1)^{4/5} b^{2/5}}}+\frac{2 \tan ^{-1}\left (\frac{\sqrt [5]{a} \tan \left (\frac{x}{2}\right )+(-1)^{4/5} \sqrt [5]{b}}{\sqrt{a^{2/5}+(-1)^{3/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt{a^{2/5}+(-1)^{3/5} b^{2/5}}}-\frac{2 \tan ^{-1}\left (\frac{(-1)^{3/5} \left ((-1)^{2/5} \sqrt [5]{a} \tan \left (\frac{x}{2}\right )+\sqrt [5]{b}\right )}{\sqrt{a^{2/5}+\sqrt [5]{-1} b^{2/5}}}\right )}{5 a^{4/5} \sqrt{a^{2/5}+\sqrt [5]{-1} b^{2/5}}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt [5]{-1} \left ((-1)^{4/5} \sqrt [5]{a} \tan \left (\frac{x}{2}\right )+\sqrt [5]{b}\right )}{\sqrt{a^{2/5}-(-1)^{2/5} b^{2/5}}}\right )}{5 a^{4/5} \sqrt{a^{2/5}-(-1)^{2/5} b^{2/5}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[x]^5)^(-1),x]

[Out]

(2*ArcTan[(b^(1/5) + a^(1/5)*Tan[x/2])/Sqrt[a^(2/5) - b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) - b^(2/5)]) + (2*ArcT
an[((-1)^(2/5)*b^(1/5) + a^(1/5)*Tan[x/2])/Sqrt[a^(2/5) - (-1)^(4/5)*b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) - (-1)
^(4/5)*b^(2/5)]) + (2*ArcTan[((-1)^(4/5)*b^(1/5) + a^(1/5)*Tan[x/2])/Sqrt[a^(2/5) + (-1)^(3/5)*b^(2/5)]])/(5*a
^(4/5)*Sqrt[a^(2/5) + (-1)^(3/5)*b^(2/5)]) - (2*ArcTan[((-1)^(3/5)*(b^(1/5) + (-1)^(2/5)*a^(1/5)*Tan[x/2]))/Sq
rt[a^(2/5) + (-1)^(1/5)*b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) + (-1)^(1/5)*b^(2/5)]) - (2*ArcTan[((-1)^(1/5)*(b^(
1/5) + (-1)^(4/5)*a^(1/5)*Tan[x/2]))/Sqrt[a^(2/5) - (-1)^(2/5)*b^(2/5)]])/(5*a^(4/5)*Sqrt[a^(2/5) - (-1)^(2/5)
*b^(2/5)])

Rule 3213

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*
x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f, n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

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

Mathematica [C]  time = 0.211827, size = 149, normalized size = 0.39 \[ \frac{8}{5} i \text{RootSum}\left [32 \text{$\#$1}^5 a-i \text{$\#$1}^{10} b+5 i \text{$\#$1}^8 b-10 i \text{$\#$1}^6 b+10 i \text{$\#$1}^4 b-5 i \text{$\#$1}^2 b+i b\& ,\frac{2 \text{$\#$1}^3 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )-i \text{$\#$1}^3 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )}{16 i \text{$\#$1}^3 a+\text{$\#$1}^8 b-4 \text{$\#$1}^6 b+6 \text{$\#$1}^4 b-4 \text{$\#$1}^2 b+b}\& \right ] \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Sin[x]^5)^(-1),x]

[Out]

((8*I)/5)*RootSum[I*b - (5*I)*b*#1^2 + (10*I)*b*#1^4 + 32*a*#1^5 - (10*I)*b*#1^6 + (5*I)*b*#1^8 - I*b*#1^10 &
, (2*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^3 - I*Log[1 - 2*Cos[x]*#1 + #1^2]*#1^3)/(b - 4*b*#1^2 + (16*I)*a*#1^3 + 6
*b*#1^4 - 4*b*#1^6 + b*#1^8) & ]

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Maple [C]  time = 0.123, size = 109, normalized size = 0.3 \begin{align*}{\frac{1}{5}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{10}+5\,a{{\it \_Z}}^{8}+10\,a{{\it \_Z}}^{6}+32\,b{{\it \_Z}}^{5}+10\,a{{\it \_Z}}^{4}+5\,a{{\it \_Z}}^{2}+a \right ) }{\frac{{{\it \_R}}^{8}+4\,{{\it \_R}}^{6}+6\,{{\it \_R}}^{4}+4\,{{\it \_R}}^{2}+1}{{{\it \_R}}^{9}a+4\,{{\it \_R}}^{7}a+6\,{{\it \_R}}^{5}a+16\,{{\it \_R}}^{4}b+4\,{{\it \_R}}^{3}a+{\it \_R}\,a}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(x)^5),x)

[Out]

1/5*sum((_R^8+4*_R^6+6*_R^4+4*_R^2+1)/(_R^9*a+4*_R^7*a+6*_R^5*a+16*_R^4*b+4*_R^3*a+_R*a)*ln(tan(1/2*x)-_R),_R=
RootOf(_Z^10*a+5*_Z^8*a+10*_Z^6*a+32*_Z^5*b+10*_Z^4*a+5*_Z^2*a+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{b \sin \left (x\right )^{5} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)^5),x, algorithm="maxima")

[Out]

integrate(1/(b*sin(x)^5 + a), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)^5),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)**5),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{b \sin \left (x\right )^{5} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)^5),x, algorithm="giac")

[Out]

integrate(1/(b*sin(x)^5 + a), x)