3.203 \(\int \frac{\sin ^8(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

Optimal. Leaf size=184 \[ \frac{a^{5/4} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^2 d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{a^{5/4} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^2 d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{x (a+b)}{b^2}-\frac{\sin (c+d x) \cos ^3(c+d x)}{4 b d}+\frac{5 \sin (c+d x) \cos (c+d x)}{8 b d}+\frac{5 x}{8 b} \]

[Out]

(5*x)/(8*b) - ((a + b)*x)/b^2 + (a^(5/4)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[
a] - Sqrt[b]]*b^2*d) + (a^(5/4)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] + Sqrt
[b]]*b^2*d) + (5*Cos[c + d*x]*Sin[c + d*x])/(8*b*d) - (Cos[c + d*x]^3*Sin[c + d*x])/(4*b*d)

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Rubi [A]  time = 0.292675, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3217, 1287, 199, 203, 1166, 205} \[ \frac{a^{5/4} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^2 d \sqrt{\sqrt{a}-\sqrt{b}}}+\frac{a^{5/4} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^2 d \sqrt{\sqrt{a}+\sqrt{b}}}-\frac{x (a+b)}{b^2}-\frac{\sin (c+d x) \cos ^3(c+d x)}{4 b d}+\frac{5 \sin (c+d x) \cos (c+d x)}{8 b d}+\frac{5 x}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^8/(a - b*Sin[c + d*x]^4),x]

[Out]

(5*x)/(8*b) - ((a + b)*x)/b^2 + (a^(5/4)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[
a] - Sqrt[b]]*b^2*d) + (a^(5/4)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] + Sqrt
[b]]*b^2*d) + (5*Cos[c + d*x]*Sin[c + d*x])/(8*b*d) - (Cos[c + d*x]^3*Sin[c + d*x])/(4*b*d)

Rule 3217

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

Rule 1287

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[Ex
pandIntegrand[((f*x)^m*(d + e*x^2)^q)/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^
2 - 4*a*c, 0] && IntegerQ[q] && IntegerQ[m]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

Mathematica [A]  time = 0.891659, size = 172, normalized size = 0.93 \[ -\frac{-\frac{16 a^{3/2} \tan ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}+a}}+\frac{16 a^{3/2} \tanh ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}-a}}+4 (8 a+3 b) (c+d x)-8 b \sin (2 (c+d x))+b \sin (4 (c+d x))}{32 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^8/(a - b*Sin[c + d*x]^4),x]

[Out]

-(4*(8*a + 3*b)*(c + d*x) - (16*a^(3/2)*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/
Sqrt[a + Sqrt[a]*Sqrt[b]] + (16*a^(3/2)*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]]
)/Sqrt[-a + Sqrt[a]*Sqrt[b]] - 8*b*Sin[2*(c + d*x)] + b*Sin[4*(c + d*x)])/(32*b^2*d)

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Maple [B]  time = 0.119, size = 605, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^8/(a-b*sin(d*x+c)^4),x)

[Out]

1/2/d*a^3/b/(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))+1/2/d*a^3/b^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))+
1/2/d*a^3/b^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))-1/2
/d*a^3/b/(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))-1/2/d*a^2/(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))-1/2/d*a^2/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*a^2/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^2/(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))+
5/8/d/b/(tan(d*x+c)^2+1)^2*tan(d*x+c)^3+3/8/d/b/(tan(d*x+c)^2+1)^2*tan(d*x+c)-3/8/d/b*arctan(tan(d*x+c))-1/d/b
^2*arctan(tan(d*x+c))*a

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Maxima [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)^8/(a-b*sin(d*x+c)^4),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 3.42395, size = 2712, normalized size = 14.74 \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)^8/(a-b*sin(d*x+c)^4),x, algorithm="fricas")

[Out]

-1/8*(b^2*d*sqrt(-((a*b^4 - b^5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) + a^3)/((a*b^4 - b^5)*d^2))*log
(1/4*a^3*cos(d*x + c)^2 - 1/4*a^3 - 1/4*(2*(a^2*b^3 - a*b^4)*d^2*cos(d*x + c)^2 - (a^2*b^3 - a*b^4)*d^2)*sqrt(
a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) + 1/2*(a^2*b^2*d*cos(d*x + c)*sin(d*x + c) - (a*b^5 - b^6)*d^3*sqrt(a^5/(
(a^2*b^7 - 2*a*b^8 + b^9)*d^4))*cos(d*x + c)*sin(d*x + c))*sqrt(-((a*b^4 - b^5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b
^8 + b^9)*d^4)) + a^3)/((a*b^4 - b^5)*d^2))) - b^2*d*sqrt(-((a*b^4 - b^5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b
^9)*d^4)) + a^3)/((a*b^4 - b^5)*d^2))*log(1/4*a^3*cos(d*x + c)^2 - 1/4*a^3 - 1/4*(2*(a^2*b^3 - a*b^4)*d^2*cos(
d*x + c)^2 - (a^2*b^3 - a*b^4)*d^2)*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) - 1/2*(a^2*b^2*d*cos(d*x + c)*si
n(d*x + c) - (a*b^5 - b^6)*d^3*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4))*cos(d*x + c)*sin(d*x + c))*sqrt(-((a*
b^4 - b^5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) + a^3)/((a*b^4 - b^5)*d^2))) - b^2*d*sqrt(((a*b^4 - b
^5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) - a^3)/((a*b^4 - b^5)*d^2))*log(-1/4*a^3*cos(d*x + c)^2 + 1/
4*a^3 - 1/4*(2*(a^2*b^3 - a*b^4)*d^2*cos(d*x + c)^2 - (a^2*b^3 - a*b^4)*d^2)*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^
9)*d^4)) + 1/2*(a^2*b^2*d*cos(d*x + c)*sin(d*x + c) + (a*b^5 - b^6)*d^3*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^
4))*cos(d*x + c)*sin(d*x + c))*sqrt(((a*b^4 - b^5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) - a^3)/((a*b^
4 - b^5)*d^2))) + b^2*d*sqrt(((a*b^4 - b^5)*d^2*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) - a^3)/((a*b^4 - b^5
)*d^2))*log(-1/4*a^3*cos(d*x + c)^2 + 1/4*a^3 - 1/4*(2*(a^2*b^3 - a*b^4)*d^2*cos(d*x + c)^2 - (a^2*b^3 - a*b^4
)*d^2)*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4)) - 1/2*(a^2*b^2*d*cos(d*x + c)*sin(d*x + c) + (a*b^5 - b^6)*d^
3*sqrt(a^5/((a^2*b^7 - 2*a*b^8 + b^9)*d^4))*cos(d*x + c)*sin(d*x + c))*sqrt(((a*b^4 - b^5)*d^2*sqrt(a^5/((a^2*
b^7 - 2*a*b^8 + b^9)*d^4)) - a^3)/((a*b^4 - b^5)*d^2))) + (8*a + 3*b)*d*x + (2*b*cos(d*x + c)^3 - 5*b*cos(d*x
+ c))*sin(d*x + c))/(b^2*d)

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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(sin(d*x+c)**8/(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(sin(d*x+c)^8/(a-b*sin(d*x+c)^4),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError