3.207 \(\int \frac{1}{a-b \sin ^4(c+d x)} \, dx\)
Optimal. Leaf size=115 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt{\sqrt{a}+\sqrt{b}}} \]
[Out]
ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)]/(2*a^(3/4)*Sqrt[Sqrt[a] - Sqrt[b]]*d) + ArcTan[(Sqrt[Sq
rt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)]/(2*a^(3/4)*Sqrt[Sqrt[a] + Sqrt[b]]*d)
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Rubi [A] time = 0.0904601, antiderivative size = 115, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3209, 1166, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} d \sqrt{\sqrt{a}+\sqrt{b}}} \]
Antiderivative was successfully verified.
[In]
Int[(a - b*Sin[c + d*x]^4)^(-1),x]
[Out]
ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)]/(2*a^(3/4)*Sqrt[Sqrt[a] - Sqrt[b]]*d) + ArcTan[(Sqrt[Sq
rt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)]/(2*a^(3/4)*Sqrt[Sqrt[a] + Sqrt[b]]*d)
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]
Rule 1166
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{1}{a-b \sin ^4(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{a+2 a x^2+(a-b) x^4} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (1-\frac{\sqrt{b}}{\sqrt{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{a-\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left (1+\frac{\sqrt{b}}{\sqrt{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{a+\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt{\sqrt{a}-\sqrt{b}} d}+\frac{\tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt{\sqrt{a}+\sqrt{b}} d}\\ \end{align*}
Mathematica [A] time = 0.256142, size = 128, normalized size = 1.11 \[ \frac{\frac{\tan ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}+a}}-\frac{\tanh ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}-a}}}{2 \sqrt{a} d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a - b*Sin[c + d*x]^4)^(-1),x]
[Out]
(ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]]/Sqrt[a + Sqrt[a]*Sqrt[b]] - ArcTanh[((Sq
rt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]]/Sqrt[-a + Sqrt[a]*Sqrt[b]])/(2*Sqrt[a]*d)
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Maple [B] time = 0.094, size = 492, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a-b*sin(d*x+c)^4),x)
[Out]
1/2/d*a/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^(1/2))
*b+1/2/d*a/(a-b)/(((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^(1/2))+1/2/d*a/
(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))-1/2/d*a/(a*b)^(1/
2)/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))*b-1/2/d/(a*b)^
(1/2)/(a-b)/(((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^(1/2))*b^2-1/2/d*b/(
a-b)/(((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^(1/2))-1/2/d*b/(a-b)/(((a*b
)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))+1/2/d/(a*b)^(1/2)/(a-b)/(((a*
b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))*b^2
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{1}{b \sin \left (d x + c\right )^{4} - a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a-b*sin(d*x+c)^4),x, algorithm="maxima")
[Out]
-integrate(1/(b*sin(d*x + c)^4 - a), x)
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Fricas [B] time = 3.06246, size = 2388, normalized size = 20.77 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a-b*sin(d*x+c)^4),x, algorithm="fricas")
[Out]
1/8*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/((a^2 - a*b)*d^2))*log(1/4*b*cos(d*x +
c)^2 + 1/2*((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4))*cos(d*x + c)*sin(d*x + c) - a*b*d*cos(d
*x + c)*sin(d*x + c))*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/((a^2 - a*b)*d^2)) -
1/4*(2*(a^3 - a^2*b)*d^2*cos(d*x + c)^2 - (a^3 - a^2*b)*d^2)*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1/4*b)
- 1/8*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/((a^2 - a*b)*d^2))*log(1/4*b*cos(d*
x + c)^2 - 1/2*((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4))*cos(d*x + c)*sin(d*x + c) - a*b*d*co
s(d*x + c)*sin(d*x + c))*sqrt(-((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1)/((a^2 - a*b)*d^2)
) - 1/4*(2*(a^3 - a^2*b)*d^2*cos(d*x + c)^2 - (a^3 - a^2*b)*d^2)*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1/4
*b) + 1/8*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*d^2))*log(-1/4*b*cos
(d*x + c)^2 + 1/2*((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4))*cos(d*x + c)*sin(d*x + c) + a*b*d
*cos(d*x + c)*sin(d*x + c))*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*d^
2)) - 1/4*(2*(a^3 - a^2*b)*d^2*cos(d*x + c)^2 - (a^3 - a^2*b)*d^2)*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) + 1
/4*b) - 1/8*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*d^2))*log(-1/4*b*c
os(d*x + c)^2 - 1/2*((a^4 - a^3*b)*d^3*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4))*cos(d*x + c)*sin(d*x + c) + a*b
*d*cos(d*x + c)*sin(d*x + c))*sqrt(((a^2 - a*b)*d^2*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) - 1)/((a^2 - a*b)*
d^2)) - 1/4*(2*(a^3 - a^2*b)*d^2*cos(d*x + c)^2 - (a^3 - a^2*b)*d^2)*sqrt(b/((a^5 - 2*a^4*b + a^3*b^2)*d^4)) +
1/4*b)
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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(d*x+c)**4),x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a-b*sin(d*x+c)^4),x, algorithm="giac")
[Out]
Exception raised: NotImplementedError