3.204 \(\int \frac{\sin ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

Optimal. Leaf size=155 \[ \frac{a^{3/4} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^{3/2} d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{a^{3/4} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^{3/2} d \sqrt{\sqrt{a}+\sqrt{b}}}+\frac{\sin (c+d x) \cos (c+d x)}{2 b d}-\frac{x}{2 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.204312, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3217, 1287, 199, 203, 1130, 205} \[ \frac{a^{3/4} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^{3/2} d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{a^{3/4} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^{3/2} d \sqrt{\sqrt{a}+\sqrt{b}}}+\frac{\sin (c+d x) \cos (c+d x)}{2 b d}-\frac{x}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^6/(a - b*Sin[c + d*x]^4),x]

[Out]

-x/(2*b) + (a^(3/4)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^(3/2)
*d) - (a^(3/4)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^(3/2)*d) +
(Cos[c + d*x]*Sin[c + d*x])/(2*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 1130

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

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 ^6(c+d x)}{a-b \sin ^4(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^2 \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 )^2}-\frac{1}{b \left (1+x^2\right )}+\frac{a x^2}{b \left (a+2 a x^2+(a-b) x^4\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{b d}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{b d}+\frac{a \operatorname{Subst}\left (\int \frac{x^2}{a+2 a x^2+(a-b) x^4} \, dx,x,\tan (c+d x)\right )}{b d}\\ &=-\frac{x}{b}+\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\left (a \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^{3/2} d}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 b d}-\frac{\left (a \left (-1+\frac{\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 d}\\ &=-\frac{x}{2 b}+\frac{a^{3/4} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{\sqrt{a}-\sqrt{b}} b^{3/2} d}-\frac{a^{3/4} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{\sqrt{a}+\sqrt{b}} b^{3/2} d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b d}\\ \end{align*}

Mathematica [A]  time = 0.805675, size = 157, normalized size = 1.01 \[ \frac{-\frac{2 a \tan ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}+a}}-\frac{2 a \tanh ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}-a}}-2 \sqrt{b} (c+d x)+\sqrt{b} \sin (2 (c+d x))}{4 b^{3/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^6/(a - b*Sin[c + d*x]^4),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.097, size = 551, normalized size = 3.6 \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)^6/(a-b*sin(d*x+c)^4),x)

[Out]

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
^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^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
^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/(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/(
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/
(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))+1/2/d/b*tan(
d*x+c)/(tan(d*x+c)^2+1)-1/2/d/b*arctan(tan(d*x+c))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \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)^6/(a-b*sin(d*x+c)^4),x, algorithm="maxima")

[Out]

1/4*(4*b*d*integrate(-4*(4*a*b*cos(6*d*x + 6*c)^2 + 4*a*b*cos(2*d*x + 2*c)^2 + 4*a*b*sin(6*d*x + 6*c)^2 + 4*a*
b*sin(2*d*x + 2*c)^2 - 4*(8*a^2 - 3*a*b)*cos(4*d*x + 4*c)^2 - a*b*cos(2*d*x + 2*c) - 4*(8*a^2 - 3*a*b)*sin(4*d
*x + 4*c)^2 + 2*(8*a^2 - 7*a*b)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) - (a*b*cos(6*d*x + 6*c) - 2*a*b*cos(4*d*x +
4*c) + a*b*cos(2*d*x + 2*c))*cos(8*d*x + 8*c) + (8*a*b*cos(2*d*x + 2*c) - a*b + 2*(8*a^2 - 7*a*b)*cos(4*d*x +
4*c))*cos(6*d*x + 6*c) + 2*(a*b + (8*a^2 - 7*a*b)*cos(2*d*x + 2*c))*cos(4*d*x + 4*c) - (a*b*sin(6*d*x + 6*c) -
2*a*b*sin(4*d*x + 4*c) + a*b*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 2*(4*a*b*sin(2*d*x + 2*c) + (8*a^2 - 7*a*b)
*sin(4*d*x + 4*c))*sin(6*d*x + 6*c))/(b^3*cos(8*d*x + 8*c)^2 + 16*b^3*cos(6*d*x + 6*c)^2 + 16*b^3*cos(2*d*x +
2*c)^2 + b^3*sin(8*d*x + 8*c)^2 + 16*b^3*sin(6*d*x + 6*c)^2 + 16*b^3*sin(2*d*x + 2*c)^2 - 8*b^3*cos(2*d*x + 2*
c) + b^3 + 4*(64*a^2*b - 48*a*b^2 + 9*b^3)*cos(4*d*x + 4*c)^2 + 4*(64*a^2*b - 48*a*b^2 + 9*b^3)*sin(4*d*x + 4*
c)^2 + 16*(8*a*b^2 - 3*b^3)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) - 2*(4*b^3*cos(6*d*x + 6*c) + 4*b^3*cos(2*d*x +
2*c) - b^3 + 2*(8*a*b^2 - 3*b^3)*cos(4*d*x + 4*c))*cos(8*d*x + 8*c) + 8*(4*b^3*cos(2*d*x + 2*c) - b^3 + 2*(8*a
*b^2 - 3*b^3)*cos(4*d*x + 4*c))*cos(6*d*x + 6*c) - 4*(8*a*b^2 - 3*b^3 - 4*(8*a*b^2 - 3*b^3)*cos(2*d*x + 2*c))*
cos(4*d*x + 4*c) - 4*(2*b^3*sin(6*d*x + 6*c) + 2*b^3*sin(2*d*x + 2*c) + (8*a*b^2 - 3*b^3)*sin(4*d*x + 4*c))*si
n(8*d*x + 8*c) + 16*(2*b^3*sin(2*d*x + 2*c) + (8*a*b^2 - 3*b^3)*sin(4*d*x + 4*c))*sin(6*d*x + 6*c)), x) - 2*d*
x + sin(2*d*x + 2*c))/(b*d)

________________________________________________________________________________________

Fricas [B]  time = 3.63702, size = 2641, normalized size = 17.04 \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)^6/(a-b*sin(d*x+c)^4),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)**6/(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)^6/(a-b*sin(d*x+c)^4),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError