3.199 \(\int \frac{\sin (c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

Optimal. Leaf size=125 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 \sqrt{a} \sqrt [4]{b} d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 \sqrt{a} \sqrt [4]{b} d \sqrt{\sqrt{a}+\sqrt{b}}} \]

[Out]

-ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]]/(2*Sqrt[a]*Sqrt[Sqrt[a] - Sqrt[b]]*b^(1/4)*d) - ArcTan
h[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]]/(2*Sqrt[a]*Sqrt[Sqrt[a] + Sqrt[b]]*b^(1/4)*d)

________________________________________________________________________________________

Rubi [A]  time = 0.102077, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3215, 1093, 205, 208} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}-\sqrt{b}}}\right )}{2 \sqrt{a} \sqrt [4]{b} d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt{\sqrt{a}+\sqrt{b}}}\right )}{2 \sqrt{a} \sqrt [4]{b} d \sqrt{\sqrt{a}+\sqrt{b}}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]/(a - b*Sin[c + d*x]^4),x]

[Out]

-ArcTan[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] - Sqrt[b]]]/(2*Sqrt[a]*Sqrt[Sqrt[a] - Sqrt[b]]*b^(1/4)*d) - ArcTan
h[(b^(1/4)*Cos[c + d*x])/Sqrt[Sqrt[a] + Sqrt[b]]]/(2*Sqrt[a]*Sqrt[Sqrt[a] + Sqrt[b]]*b^(1/4)*d)

Rule 3215

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4
)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 1093

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

Rule 208

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

Rubi steps

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

Mathematica [C]  time = 0.162016, size = 183, normalized size = 1.46 \[ \frac{i \text{RootSum}\left [-16 \text{$\#$1}^4 a+\text{$\#$1}^8 b-4 \text{$\#$1}^6 b+6 \text{$\#$1}^4 b-4 \text{$\#$1}^2 b+b\& ,\frac{-i \text{$\#$1}^3 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+i \text{$\#$1} \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^3 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-2 \text{$\#$1} \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{-8 \text{$\#$1}^2 a+\text{$\#$1}^6 b-3 \text{$\#$1}^4 b+3 \text{$\#$1}^2 b-b}\& \right ]}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]/(a - b*Sin[c + d*x]^4),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.095, size = 90, normalized size = 0.7 \begin{align*} -{\frac{b}{2\,d}\arctan \left ({b\cos \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}-b \right ) b}}}} \right ){\frac{1}{\sqrt{ab}}}{\frac{1}{\sqrt{ \left ( \sqrt{ab}-b \right ) b}}}}-{\frac{b}{2\,d}{\it Artanh} \left ({b\cos \left ( dx+c \right ){\frac{1}{\sqrt{ \left ( \sqrt{ab}+b \right ) b}}}} \right ){\frac{1}{\sqrt{ab}}}{\frac{1}{\sqrt{ \left ( \sqrt{ab}+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)/(a-b*sin(d*x+c)^4),x)

[Out]

-1/2*b/d/(a*b)^(1/2)/(((a*b)^(1/2)-b)*b)^(1/2)*arctan(cos(d*x+c)*b/(((a*b)^(1/2)-b)*b)^(1/2))-1/2*b/d/(a*b)^(1
/2)/(((a*b)^(1/2)+b)*b)^(1/2)*arctanh(cos(d*x+c)*b/(((a*b)^(1/2)+b)*b)^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{\sin \left (d x + c\right )}{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(sin(d*x+c)/(a-b*sin(d*x+c)^4),x, algorithm="maxima")

[Out]

-integrate(sin(d*x + c)/(b*sin(d*x + c)^4 - a), x)

________________________________________________________________________________________

Fricas [B]  time = 2.56133, size = 1480, normalized size = 11.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(-((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + 1)/((a^2 - a*b)*d^2))*log(-((a^2*b - a
*b^2)*d^3*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - a*d)*sqrt(-((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 +
a*b^3)*d^4)) + 1)/((a^2 - a*b)*d^2)) + cos(d*x + c)) + 1/4*sqrt(-((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2
+ a*b^3)*d^4)) + 1)/((a^2 - a*b)*d^2))*log(-((a^2*b - a*b^2)*d^3*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - a
*d)*sqrt(-((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + 1)/((a^2 - a*b)*d^2)) - cos(d*x + c)) +
1/4*sqrt(((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - 1)/((a^2 - a*b)*d^2))*log(-((a^2*b - a*
b^2)*d^3*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + a*d)*sqrt(((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a
*b^3)*d^4)) - 1)/((a^2 - a*b)*d^2)) + cos(d*x + c)) - 1/4*sqrt(((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a
*b^3)*d^4)) - 1)/((a^2 - a*b)*d^2))*log(-((a^2*b - a*b^2)*d^3*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) + a*d)
*sqrt(((a^2 - a*b)*d^2*sqrt(1/((a^3*b - 2*a^2*b^2 + a*b^3)*d^4)) - 1)/((a^2 - a*b)*d^2)) - cos(d*x + c))

________________________________________________________________________________________

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(sin(d*x+c)/(a-b*sin(d*x+c)**4),x)

[Out]

Timed out

________________________________________________________________________________________

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(sin(d*x+c)/(a-b*sin(d*x+c)^4),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError