3.237 \(\int \frac {1}{a+b \sin ^4(x)} \, dx\)

Optimal. Leaf size=487 \[ -\frac {\left (\sqrt {a+b}+\sqrt {a}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b}-\sqrt {2} (a+b)^{3/4} \tan (x)}{\sqrt [4]{a} \sqrt {\sqrt {a} \sqrt {a+b}+a+b}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{a+b} \sqrt {\sqrt {a} \sqrt {a+b}+a+b}}+\frac {\left (\sqrt {a+b}+\sqrt {a}\right ) \tan ^{-1}\left (\frac {\sqrt {2} (a+b)^{3/4} \tan (x)+\sqrt [4]{a} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b}}{\sqrt [4]{a} \sqrt {\sqrt {a} \sqrt {a+b}+a+b}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{a+b} \sqrt {\sqrt {a} \sqrt {a+b}+a+b}}+\frac {\left (\sqrt {a}-\sqrt {a+b}\right ) \log \left ((a+b)^{3/4} \tan ^2(x)-\sqrt {2} \sqrt [4]{a} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b} \tan (x)+\sqrt {a} \sqrt [4]{a+b}\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{a+b} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b}}-\frac {\left (\sqrt {a}-\sqrt {a+b}\right ) \log \left ((a+b)^{3/4} \tan ^2(x)+\sqrt {2} \sqrt [4]{a} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b} \tan (x)+\sqrt {a} \sqrt [4]{a+b}\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{a+b} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b}} \]

[Out]

1/8*ln((a+b)^(1/4)*a^(1/2)-a^(1/4)*2^(1/2)*(a+b-a^(1/2)*(a+b)^(1/2))^(1/2)*tan(x)+(a+b)^(3/4)*tan(x)^2)*(a^(1/
2)-(a+b)^(1/2))/a^(3/4)/(a+b)^(1/4)*2^(1/2)/(a+b-a^(1/2)*(a+b)^(1/2))^(1/2)-1/8*ln((a+b)^(1/4)*a^(1/2)+a^(1/4)
*2^(1/2)*(a+b-a^(1/2)*(a+b)^(1/2))^(1/2)*tan(x)+(a+b)^(3/4)*tan(x)^2)*(a^(1/2)-(a+b)^(1/2))/a^(3/4)/(a+b)^(1/4
)*2^(1/2)/(a+b-a^(1/2)*(a+b)^(1/2))^(1/2)-1/4*arctan((a^(1/4)*(a+b-a^(1/2)*(a+b)^(1/2))^(1/2)-(a+b)^(3/4)*2^(1
/2)*tan(x))/a^(1/4)/(a+b+a^(1/2)*(a+b)^(1/2))^(1/2))*(a^(1/2)+(a+b)^(1/2))/a^(3/4)/(a+b)^(1/4)*2^(1/2)/(a+b+a^
(1/2)*(a+b)^(1/2))^(1/2)+1/4*arctan((a^(1/4)*(a+b-a^(1/2)*(a+b)^(1/2))^(1/2)+(a+b)^(3/4)*2^(1/2)*tan(x))/a^(1/
4)/(a+b+a^(1/2)*(a+b)^(1/2))^(1/2))*(a^(1/2)+(a+b)^(1/2))/a^(3/4)/(a+b)^(1/4)*2^(1/2)/(a+b+a^(1/2)*(a+b)^(1/2)
)^(1/2)

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Rubi [A]  time = 1.14, antiderivative size = 487, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3209, 1169, 634, 618, 204, 628} \[ -\frac {\left (\sqrt {a+b}+\sqrt {a}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b}-\sqrt {2} (a+b)^{3/4} \tan (x)}{\sqrt [4]{a} \sqrt {\sqrt {a} \sqrt {a+b}+a+b}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{a+b} \sqrt {\sqrt {a} \sqrt {a+b}+a+b}}+\frac {\left (\sqrt {a+b}+\sqrt {a}\right ) \tan ^{-1}\left (\frac {\sqrt {2} (a+b)^{3/4} \tan (x)+\sqrt [4]{a} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b}}{\sqrt [4]{a} \sqrt {\sqrt {a} \sqrt {a+b}+a+b}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{a+b} \sqrt {\sqrt {a} \sqrt {a+b}+a+b}}+\frac {\left (\sqrt {a}-\sqrt {a+b}\right ) \log \left ((a+b)^{3/4} \tan ^2(x)-\sqrt {2} \sqrt [4]{a} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b} \tan (x)+\sqrt {a} \sqrt [4]{a+b}\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{a+b} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b}}-\frac {\left (\sqrt {a}-\sqrt {a+b}\right ) \log \left ((a+b)^{3/4} \tan ^2(x)+\sqrt {2} \sqrt [4]{a} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b} \tan (x)+\sqrt {a} \sqrt [4]{a+b}\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{a+b} \sqrt {-\sqrt {a} \sqrt {a+b}+a+b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[x]^4)^(-1),x]

[Out]

-((Sqrt[a] + Sqrt[a + b])*ArcTan[(a^(1/4)*Sqrt[a + b - Sqrt[a]*Sqrt[a + b]] - Sqrt[2]*(a + b)^(3/4)*Tan[x])/(a
^(1/4)*Sqrt[a + b + Sqrt[a]*Sqrt[a + b]])])/(2*Sqrt[2]*a^(3/4)*(a + b)^(1/4)*Sqrt[a + b + Sqrt[a]*Sqrt[a + b]]
) + ((Sqrt[a] + Sqrt[a + b])*ArcTan[(a^(1/4)*Sqrt[a + b - Sqrt[a]*Sqrt[a + b]] + Sqrt[2]*(a + b)^(3/4)*Tan[x])
/(a^(1/4)*Sqrt[a + b + Sqrt[a]*Sqrt[a + b]])])/(2*Sqrt[2]*a^(3/4)*(a + b)^(1/4)*Sqrt[a + b + Sqrt[a]*Sqrt[a +
b]]) + ((Sqrt[a] - Sqrt[a + b])*Log[Sqrt[a]*(a + b)^(1/4) - Sqrt[2]*a^(1/4)*Sqrt[a + b - Sqrt[a]*Sqrt[a + b]]*
Tan[x] + (a + b)^(3/4)*Tan[x]^2])/(4*Sqrt[2]*a^(3/4)*(a + b)^(1/4)*Sqrt[a + b - Sqrt[a]*Sqrt[a + b]]) - ((Sqrt
[a] - Sqrt[a + b])*Log[Sqrt[a]*(a + b)^(1/4) + Sqrt[2]*a^(1/4)*Sqrt[a + b - Sqrt[a]*Sqrt[a + b]]*Tan[x] + (a +
b)^(3/4)*Tan[x]^2])/(4*Sqrt[2]*a^(3/4)*(a + b)^(1/4)*Sqrt[a + b - Sqrt[a]*Sqrt[a + b]])

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])

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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rule 1169

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r =
Rt[2*q - b/c, 2]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(
d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rule 3209

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dis
t[ff/f, Subst[Int[(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(2*p + 1), x], x, Tan[e + f*x]/ff], x
]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[p]

Rubi steps

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

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Mathematica [C]  time = 0.31, size = 148, normalized size = 0.30 \[ \frac {\left (\sqrt {a}-i \sqrt {b}\right ) \sqrt {a+i \sqrt {a} \sqrt {b}} \tan ^{-1}\left (\frac {\sqrt {a+i \sqrt {a} \sqrt {b}} \tan (x)}{\sqrt {a}}\right )-\left (\sqrt {a}+i \sqrt {b}\right ) \sqrt {-a+i \sqrt {a} \sqrt {b}} \tanh ^{-1}\left (\frac {\sqrt {-a+i \sqrt {a} \sqrt {b}} \tan (x)}{\sqrt {a}}\right )}{2 a (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[x]^4)^(-1),x]

[Out]

((Sqrt[a] - I*Sqrt[b])*Sqrt[a + I*Sqrt[a]*Sqrt[b]]*ArcTan[(Sqrt[a + I*Sqrt[a]*Sqrt[b]]*Tan[x])/Sqrt[a]] - (Sqr
t[a] + I*Sqrt[b])*Sqrt[-a + I*Sqrt[a]*Sqrt[b]]*ArcTanh[(Sqrt[-a + I*Sqrt[a]*Sqrt[b]]*Tan[x])/Sqrt[a]])/(2*a*(a
+ b))

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fricas [B]  time = 0.57, size = 823, normalized size = 1.69 \[ -\frac {1}{8} \, \sqrt {-\frac {{\left (a^{2} + a b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} + 1}{a^{2} + a b}} \log \left (\frac {1}{4} \, b \cos \relax (x)^{2} + \frac {1}{2} \, {\left (a b \cos \relax (x) \sin \relax (x) + {\left (a^{4} + a^{3} b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} \cos \relax (x) \sin \relax (x)\right )} \sqrt {-\frac {{\left (a^{2} + a b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} + 1}{a^{2} + a b}} - \frac {1}{4} \, {\left (a^{3} + a^{2} b - 2 \, {\left (a^{3} + a^{2} b\right )} \cos \relax (x)^{2}\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} - \frac {1}{4} \, b\right ) + \frac {1}{8} \, \sqrt {-\frac {{\left (a^{2} + a b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} + 1}{a^{2} + a b}} \log \left (\frac {1}{4} \, b \cos \relax (x)^{2} - \frac {1}{2} \, {\left (a b \cos \relax (x) \sin \relax (x) + {\left (a^{4} + a^{3} b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} \cos \relax (x) \sin \relax (x)\right )} \sqrt {-\frac {{\left (a^{2} + a b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} + 1}{a^{2} + a b}} - \frac {1}{4} \, {\left (a^{3} + a^{2} b - 2 \, {\left (a^{3} + a^{2} b\right )} \cos \relax (x)^{2}\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} - \frac {1}{4} \, b\right ) + \frac {1}{8} \, \sqrt {\frac {{\left (a^{2} + a b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} - 1}{a^{2} + a b}} \log \left (-\frac {1}{4} \, b \cos \relax (x)^{2} + \frac {1}{2} \, {\left (a b \cos \relax (x) \sin \relax (x) - {\left (a^{4} + a^{3} b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} \cos \relax (x) \sin \relax (x)\right )} \sqrt {\frac {{\left (a^{2} + a b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} - 1}{a^{2} + a b}} - \frac {1}{4} \, {\left (a^{3} + a^{2} b - 2 \, {\left (a^{3} + a^{2} b\right )} \cos \relax (x)^{2}\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} + \frac {1}{4} \, b\right ) - \frac {1}{8} \, \sqrt {\frac {{\left (a^{2} + a b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} - 1}{a^{2} + a b}} \log \left (-\frac {1}{4} \, b \cos \relax (x)^{2} - \frac {1}{2} \, {\left (a b \cos \relax (x) \sin \relax (x) - {\left (a^{4} + a^{3} b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} \cos \relax (x) \sin \relax (x)\right )} \sqrt {\frac {{\left (a^{2} + a b\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} - 1}{a^{2} + a b}} - \frac {1}{4} \, {\left (a^{3} + a^{2} b - 2 \, {\left (a^{3} + a^{2} b\right )} \cos \relax (x)^{2}\right )} \sqrt {-\frac {b}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}}} + \frac {1}{4} \, b\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)^4),x, algorithm="fricas")

[Out]

-1/8*sqrt(-((a^2 + a*b)*sqrt(-b/(a^5 + 2*a^4*b + a^3*b^2)) + 1)/(a^2 + a*b))*log(1/4*b*cos(x)^2 + 1/2*(a*b*cos
(x)*sin(x) + (a^4 + a^3*b)*sqrt(-b/(a^5 + 2*a^4*b + a^3*b^2))*cos(x)*sin(x))*sqrt(-((a^2 + a*b)*sqrt(-b/(a^5 +
2*a^4*b + a^3*b^2)) + 1)/(a^2 + a*b)) - 1/4*(a^3 + a^2*b - 2*(a^3 + a^2*b)*cos(x)^2)*sqrt(-b/(a^5 + 2*a^4*b +
a^3*b^2)) - 1/4*b) + 1/8*sqrt(-((a^2 + a*b)*sqrt(-b/(a^5 + 2*a^4*b + a^3*b^2)) + 1)/(a^2 + a*b))*log(1/4*b*co
s(x)^2 - 1/2*(a*b*cos(x)*sin(x) + (a^4 + a^3*b)*sqrt(-b/(a^5 + 2*a^4*b + a^3*b^2))*cos(x)*sin(x))*sqrt(-((a^2
+ a*b)*sqrt(-b/(a^5 + 2*a^4*b + a^3*b^2)) + 1)/(a^2 + a*b)) - 1/4*(a^3 + a^2*b - 2*(a^3 + a^2*b)*cos(x)^2)*sqr
t(-b/(a^5 + 2*a^4*b + a^3*b^2)) - 1/4*b) + 1/8*sqrt(((a^2 + a*b)*sqrt(-b/(a^5 + 2*a^4*b + a^3*b^2)) - 1)/(a^2
+ a*b))*log(-1/4*b*cos(x)^2 + 1/2*(a*b*cos(x)*sin(x) - (a^4 + a^3*b)*sqrt(-b/(a^5 + 2*a^4*b + a^3*b^2))*cos(x)
*sin(x))*sqrt(((a^2 + a*b)*sqrt(-b/(a^5 + 2*a^4*b + a^3*b^2)) - 1)/(a^2 + a*b)) - 1/4*(a^3 + a^2*b - 2*(a^3 +
a^2*b)*cos(x)^2)*sqrt(-b/(a^5 + 2*a^4*b + a^3*b^2)) + 1/4*b) - 1/8*sqrt(((a^2 + a*b)*sqrt(-b/(a^5 + 2*a^4*b +
a^3*b^2)) - 1)/(a^2 + a*b))*log(-1/4*b*cos(x)^2 - 1/2*(a*b*cos(x)*sin(x) - (a^4 + a^3*b)*sqrt(-b/(a^5 + 2*a^4*
b + a^3*b^2))*cos(x)*sin(x))*sqrt(((a^2 + a*b)*sqrt(-b/(a^5 + 2*a^4*b + a^3*b^2)) - 1)/(a^2 + a*b)) - 1/4*(a^3
+ a^2*b - 2*(a^3 + a^2*b)*cos(x)^2)*sqrt(-b/(a^5 + 2*a^4*b + a^3*b^2)) + 1/4*b)

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giac [A]  time = 0.39, size = 318, normalized size = 0.65 \[ \frac {{\left (3 \, \sqrt {a^{2} + a b + \sqrt {-a b} {\left (a + b\right )}} a^{2} + 6 \, \sqrt {a^{2} + a b + \sqrt {-a b} {\left (a + b\right )}} a b - \sqrt {a^{2} + a b + \sqrt {-a b} {\left (a + b\right )}} b^{2}\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \relax (x)}{\sqrt {\frac {4 \, a + \sqrt {-16 \, {\left (a + b\right )} a + 16 \, a^{2}}}{a + b}}}\right )\right )} {\left | a + b \right |}}{2 \, {\left (3 \, a^{5} + 12 \, a^{4} b + 14 \, a^{3} b^{2} + 4 \, a^{2} b^{3} - a b^{4}\right )}} + \frac {{\left (3 \, \sqrt {a^{2} + a b - \sqrt {-a b} {\left (a + b\right )}} a^{2} + 6 \, \sqrt {a^{2} + a b - \sqrt {-a b} {\left (a + b\right )}} a b - \sqrt {a^{2} + a b - \sqrt {-a b} {\left (a + b\right )}} b^{2}\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \relax (x)}{\sqrt {\frac {4 \, a - \sqrt {-16 \, {\left (a + b\right )} a + 16 \, a^{2}}}{a + b}}}\right )\right )} {\left | a + b \right |}}{2 \, {\left (3 \, a^{5} + 12 \, a^{4} b + 14 \, a^{3} b^{2} + 4 \, a^{2} b^{3} - a b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)^4),x, algorithm="giac")

[Out]

1/2*(3*sqrt(a^2 + a*b + sqrt(-a*b)*(a + b))*a^2 + 6*sqrt(a^2 + a*b + sqrt(-a*b)*(a + b))*a*b - sqrt(a^2 + a*b
+ sqrt(-a*b)*(a + b))*b^2)*(pi*floor(x/pi + 1/2) + arctan(2*tan(x)/sqrt((4*a + sqrt(-16*(a + b)*a + 16*a^2))/(
a + b))))*abs(a + b)/(3*a^5 + 12*a^4*b + 14*a^3*b^2 + 4*a^2*b^3 - a*b^4) + 1/2*(3*sqrt(a^2 + a*b - sqrt(-a*b)*
(a + b))*a^2 + 6*sqrt(a^2 + a*b - sqrt(-a*b)*(a + b))*a*b - sqrt(a^2 + a*b - sqrt(-a*b)*(a + b))*b^2)*(pi*floo
r(x/pi + 1/2) + arctan(2*tan(x)/sqrt((4*a - sqrt(-16*(a + b)*a + 16*a^2))/(a + b))))*abs(a + b)/(3*a^5 + 12*a^
4*b + 14*a^3*b^2 + 4*a^2*b^3 - a*b^4)

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maple [B]  time = 0.41, size = 1677, normalized size = 3.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(x)^4),x)

[Out]

1/8/b/(a+b)^(1/2)*ln(-(a+b)^(1/2)*tan(x)^2+tan(x)*(2*(a*(a+b))^(1/2)-2*a)^(1/2)-a^(1/2))*(2*(a^2+a*b)^(1/2)-2*
a)^(1/2)+1/8/a/b/(a+b)^(1/2)*ln(-(a+b)^(1/2)*tan(x)^2+tan(x)*(2*(a*(a+b))^(1/2)-2*a)^(1/2)-a^(1/2))*(2*(a^2+a*
b)^(1/2)-2*a)^(1/2)*(a^2+a*b)^(1/2)-1/8/a^(3/2)/b*ln(-(a+b)^(1/2)*tan(x)^2+tan(x)*(2*(a*(a+b))^(1/2)-2*a)^(1/2
)-a^(1/2))*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)*(a^2+a*b)^(1/2)-1/8/a^(1/2)/b*ln(-(a+b)^(1/2)*tan(x)^2+tan(x)*(2*(a*(
a+b))^(1/2)-2*a)^(1/2)-a^(1/2))*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)-1/a^(1/2)/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/
2)+2*a)^(1/2)*arctan((-2*(a+b)^(1/2)*tan(x)+(2*(a*(a+b))^(1/2)-2*a)^(1/2))/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^
(1/2)+2*a)^(1/2))-1/4/b/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2)*arctan((-2*(a+b)^(1/2)*tan(x)+(2*(
a*(a+b))^(1/2)-2*a)^(1/2))/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2))*(2*(a*(a+b))^(1/2)-2*a)^(1/2)/
(a+b)^(1/2)*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)-1/4/a/b/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2)*arctan((
-2*(a+b)^(1/2)*tan(x)+(2*(a*(a+b))^(1/2)-2*a)^(1/2))/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2))*(2*(
a*(a+b))^(1/2)-2*a)^(1/2)/(a+b)^(1/2)*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)*(a^2+a*b)^(1/2)+1/4/a^(3/2)/b/(4*a^(1/2)*(
a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2)*arctan((-2*(a+b)^(1/2)*tan(x)+(2*(a*(a+b))^(1/2)-2*a)^(1/2))/(4*a^(1/2
)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2))*(2*(a*(a+b))^(1/2)-2*a)^(1/2)*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)*(a^2+a
*b)^(1/2)+1/4/a^(1/2)/b/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2)*arctan((-2*(a+b)^(1/2)*tan(x)+(2*(
a*(a+b))^(1/2)-2*a)^(1/2))/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2))*(2*(a*(a+b))^(1/2)-2*a)^(1/2)*
(2*(a^2+a*b)^(1/2)-2*a)^(1/2)-1/8/b/(a+b)^(1/2)*ln((a+b)^(1/2)*tan(x)^2+tan(x)*(2*(a*(a+b))^(1/2)-2*a)^(1/2)+a
^(1/2))*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)-1/8/a/b/(a+b)^(1/2)*ln((a+b)^(1/2)*tan(x)^2+tan(x)*(2*(a*(a+b))^(1/2)-2*
a)^(1/2)+a^(1/2))*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)*(a^2+a*b)^(1/2)+1/8/a^(3/2)/b*ln((a+b)^(1/2)*tan(x)^2+tan(x)*(
2*(a*(a+b))^(1/2)-2*a)^(1/2)+a^(1/2))*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)*(a^2+a*b)^(1/2)+1/8/a^(1/2)/b*ln((a+b)^(1/
2)*tan(x)^2+tan(x)*(2*(a*(a+b))^(1/2)-2*a)^(1/2)+a^(1/2))*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)+1/a^(1/2)/(4*a^(1/2)*(
a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2)*arctan((2*(a+b)^(1/2)*tan(x)+(2*(a*(a+b))^(1/2)-2*a)^(1/2))/(4*a^(1/2)
*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2))+1/4/b/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2)*arctan((2
*(a+b)^(1/2)*tan(x)+(2*(a*(a+b))^(1/2)-2*a)^(1/2))/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2))*(2*(a*
(a+b))^(1/2)-2*a)^(1/2)/(a+b)^(1/2)*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)+1/4/a/b/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(
1/2)+2*a)^(1/2)*arctan((2*(a+b)^(1/2)*tan(x)+(2*(a*(a+b))^(1/2)-2*a)^(1/2))/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))
^(1/2)+2*a)^(1/2))*(2*(a*(a+b))^(1/2)-2*a)^(1/2)/(a+b)^(1/2)*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)*(a^2+a*b)^(1/2)-1/4
/a^(3/2)/b/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2)*arctan((2*(a+b)^(1/2)*tan(x)+(2*(a*(a+b))^(1/2)
-2*a)^(1/2))/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2))*(2*(a*(a+b))^(1/2)-2*a)^(1/2)*(2*(a^2+a*b)^(
1/2)-2*a)^(1/2)*(a^2+a*b)^(1/2)-1/4/a^(1/2)/b/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2)*arctan((2*(a
+b)^(1/2)*tan(x)+(2*(a*(a+b))^(1/2)-2*a)^(1/2))/(4*a^(1/2)*(a+b)^(1/2)-2*(a*(a+b))^(1/2)+2*a)^(1/2))*(2*(a*(a+
b))^(1/2)-2*a)^(1/2)*(2*(a^2+a*b)^(1/2)-2*a)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b \sin \relax (x)^{4} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x)^4),x, algorithm="maxima")

[Out]

integrate(1/(b*sin(x)^4 + a), x)

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mupad [B]  time = 15.18, size = 407, normalized size = 0.84 \[ \mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\relax (x)\,\sqrt {-\frac {a^2-\sqrt {-a^3\,b}}{16\,a^4+16\,b\,a^3}}\,4{}\mathrm {i}+a^5\,\mathrm {tan}\relax (x)\,{\left (-\frac {a^2-\sqrt {-a^3\,b}}{16\,a^4+16\,b\,a^3}\right )}^{3/2}\,64{}\mathrm {i}-a^2\,b\,\mathrm {tan}\relax (x)\,\sqrt {-\frac {a^2-\sqrt {-a^3\,b}}{16\,a^4+16\,b\,a^3}}\,4{}\mathrm {i}+a^4\,b\,\mathrm {tan}\relax (x)\,{\left (-\frac {a^2-\sqrt {-a^3\,b}}{16\,a^4+16\,b\,a^3}\right )}^{3/2}\,64{}\mathrm {i}}{\sqrt {-a^3\,b}}\right )\,\sqrt {-\frac {a^2-\sqrt {-a^3\,b}}{16\,a^4+16\,b\,a^3}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\relax (x)\,\sqrt {-\frac {a^2+\sqrt {-a^3\,b}}{16\,a^4+16\,b\,a^3}}\,4{}\mathrm {i}+a^5\,\mathrm {tan}\relax (x)\,{\left (-\frac {a^2+\sqrt {-a^3\,b}}{16\,a^4+16\,b\,a^3}\right )}^{3/2}\,64{}\mathrm {i}-a^2\,b\,\mathrm {tan}\relax (x)\,\sqrt {-\frac {a^2+\sqrt {-a^3\,b}}{16\,a^4+16\,b\,a^3}}\,4{}\mathrm {i}+a^4\,b\,\mathrm {tan}\relax (x)\,{\left (-\frac {a^2+\sqrt {-a^3\,b}}{16\,a^4+16\,b\,a^3}\right )}^{3/2}\,64{}\mathrm {i}}{\sqrt {-a^3\,b}}\right )\,\sqrt {-\frac {a^2+\sqrt {-a^3\,b}}{16\,a^4+16\,b\,a^3}}\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(x)^4),x)

[Out]

atan((a^3*tan(x)*(-(a^2 - (-a^3*b)^(1/2))/(16*a^3*b + 16*a^4))^(1/2)*4i + a^5*tan(x)*(-(a^2 - (-a^3*b)^(1/2))/
(16*a^3*b + 16*a^4))^(3/2)*64i - a^2*b*tan(x)*(-(a^2 - (-a^3*b)^(1/2))/(16*a^3*b + 16*a^4))^(1/2)*4i + a^4*b*t
an(x)*(-(a^2 - (-a^3*b)^(1/2))/(16*a^3*b + 16*a^4))^(3/2)*64i)/(-a^3*b)^(1/2))*(-(a^2 - (-a^3*b)^(1/2))/(16*a^
3*b + 16*a^4))^(1/2)*2i - atan((a^3*tan(x)*(-(a^2 + (-a^3*b)^(1/2))/(16*a^3*b + 16*a^4))^(1/2)*4i + a^5*tan(x)
*(-(a^2 + (-a^3*b)^(1/2))/(16*a^3*b + 16*a^4))^(3/2)*64i - a^2*b*tan(x)*(-(a^2 + (-a^3*b)^(1/2))/(16*a^3*b + 1
6*a^4))^(1/2)*4i + a^4*b*tan(x)*(-(a^2 + (-a^3*b)^(1/2))/(16*a^3*b + 16*a^4))^(3/2)*64i)/(-a^3*b)^(1/2))*(-(a^
2 + (-a^3*b)^(1/2))/(16*a^3*b + 16*a^4))^(1/2)*2i

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sympy [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)**4),x)

[Out]

Timed out

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