3.85 \(\int \frac{\sin ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=163 \[ \frac{a^{7/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{b^4 d \sqrt{a+b}}-\frac{\left (8 a^2-6 a b+5 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 b^3 d}-\frac{x \left (-8 a^2 b+16 a^3+6 a b^2-5 b^3\right )}{16 b^4}+\frac{(6 a-5 b) \sin ^3(c+d x) \cos (c+d x)}{24 b^2 d}-\frac{\sin ^5(c+d x) \cos (c+d x)}{6 b d} \]

[Out]

-((16*a^3 - 8*a^2*b + 6*a*b^2 - 5*b^3)*x)/(16*b^4) + (a^(7/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(b^4
*Sqrt[a + b]*d) - ((8*a^2 - 6*a*b + 5*b^2)*Cos[c + d*x]*Sin[c + d*x])/(16*b^3*d) + ((6*a - 5*b)*Cos[c + d*x]*S
in[c + d*x]^3)/(24*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^5)/(6*b*d)

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Rubi [A]  time = 0.364735, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3187, 470, 578, 522, 203, 205} \[ \frac{a^{7/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{b^4 d \sqrt{a+b}}-\frac{\left (8 a^2-6 a b+5 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 b^3 d}-\frac{x \left (-8 a^2 b+16 a^3+6 a b^2-5 b^3\right )}{16 b^4}+\frac{(6 a-5 b) \sin ^3(c+d x) \cos (c+d x)}{24 b^2 d}-\frac{\sin ^5(c+d x) \cos (c+d x)}{6 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^8/(a + b*Sin[c + d*x]^2),x]

[Out]

-((16*a^3 - 8*a^2*b + 6*a*b^2 - 5*b^3)*x)/(16*b^4) + (a^(7/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(b^4
*Sqrt[a + b]*d) - ((8*a^2 - 6*a*b + 5*b^2)*Cos[c + d*x]*Sin[c + d*x])/(16*b^3*d) + ((6*a - 5*b)*Cos[c + d*x]*S
in[c + d*x]^3)/(24*b^2*d) - (Cos[c + d*x]*Sin[c + d*x]^5)/(6*b*d)

Rule 3187

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

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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 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 ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^8}{\left (1+x^2\right )^4 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\cos (c+d x) \sin ^5(c+d x)}{6 b d}+\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (5 a+(-a+5 b) x^2\right )}{\left (1+x^2\right )^3 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 b d}\\ &=\frac{(6 a-5 b) \cos (c+d x) \sin ^3(c+d x)}{24 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x)}{6 b d}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a (6 a-5 b)-3 \left (2 a^2-a b+5 b^2\right ) x^2\right )}{\left (1+x^2\right )^2 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{24 b^2 d}\\ &=-\frac{\left (8 a^2-6 a b+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 b^3 d}+\frac{(6 a-5 b) \cos (c+d x) \sin ^3(c+d x)}{24 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x)}{6 b d}+\frac{\operatorname{Subst}\left (\int \frac{3 a \left (8 a^2-6 a b+5 b^2\right )-3 \left (8 a^3-2 a^2 b+a b^2-5 b^3\right ) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{48 b^3 d}\\ &=-\frac{\left (8 a^2-6 a b+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 b^3 d}+\frac{(6 a-5 b) \cos (c+d x) \sin ^3(c+d x)}{24 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x)}{6 b d}+\frac{a^4 \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{b^4 d}-\frac{\left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{16 b^4 d}\\ &=-\frac{\left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) x}{16 b^4}+\frac{a^{7/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{b^4 \sqrt{a+b} d}-\frac{\left (8 a^2-6 a b+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 b^3 d}+\frac{(6 a-5 b) \cos (c+d x) \sin ^3(c+d x)}{24 b^2 d}-\frac{\cos (c+d x) \sin ^5(c+d x)}{6 b d}\\ \end{align*}

Mathematica [A]  time = 1.19546, size = 133, normalized size = 0.82 \[ -\frac{12 \left (-8 a^2 b+16 a^3+6 a b^2-5 b^3\right ) (c+d x)+3 b \left (16 a^2-16 a b+15 b^2\right ) \sin (2 (c+d x))-\frac{192 a^{7/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{a+b}}+3 b^2 (2 a-3 b) \sin (4 (c+d x))+b^3 \sin (6 (c+d x))}{192 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^8/(a + b*Sin[c + d*x]^2),x]

[Out]

-(12*(16*a^3 - 8*a^2*b + 6*a*b^2 - 5*b^3)*(c + d*x) - (192*a^(7/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])
/Sqrt[a + b] + 3*b*(16*a^2 - 16*a*b + 15*b^2)*Sin[2*(c + d*x)] + 3*(2*a - 3*b)*b^2*Sin[4*(c + d*x)] + b^3*Sin[
6*(c + d*x)])/(192*b^4*d)

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Maple [B]  time = 0.102, size = 361, normalized size = 2.2 \begin{align*} -{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}{a}^{2}}{2\,d{b}^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{5\, \left ( \tan \left ( dx+c \right ) \right ) ^{5}a}{8\,{b}^{2}d \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{11\, \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{16\,bd \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}{a}^{2}}{d{b}^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}a}{{b}^{2}d \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{5\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{6\,bd \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{{a}^{2}\tan \left ( dx+c \right ) }{2\,d{b}^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{3\,a\tan \left ( dx+c \right ) }{8\,{b}^{2}d \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{5\,\tan \left ( dx+c \right ) }{16\,bd \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{3}}{d{b}^{4}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{2\,d{b}^{3}}}-{\frac{3\,\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{8\,{b}^{2}d}}+{\frac{5\,\arctan \left ( \tan \left ( dx+c \right ) \right ) }{16\,bd}}+{\frac{{a}^{4}}{d{b}^{4}}\arctan \left ({ \left ( a+b \right ) \tan \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^8/(a+sin(d*x+c)^2*b),x)

[Out]

-1/2/d/b^3/(tan(d*x+c)^2+1)^3*tan(d*x+c)^5*a^2+5/8/d/b^2/(tan(d*x+c)^2+1)^3*tan(d*x+c)^5*a-11/16/d/b/(tan(d*x+
c)^2+1)^3*tan(d*x+c)^5-1/d/b^3/(tan(d*x+c)^2+1)^3*tan(d*x+c)^3*a^2+1/d/b^2/(tan(d*x+c)^2+1)^3*tan(d*x+c)^3*a-5
/6/d/b/(tan(d*x+c)^2+1)^3*tan(d*x+c)^3-1/2/d/b^3/(tan(d*x+c)^2+1)^3*tan(d*x+c)*a^2+3/8/d/b^2/(tan(d*x+c)^2+1)^
3*tan(d*x+c)*a-5/16/d/b/(tan(d*x+c)^2+1)^3*tan(d*x+c)-1/d/b^4*arctan(tan(d*x+c))*a^3+1/2/d/b^3*arctan(tan(d*x+
c))*a^2-3/8/d/b^2*arctan(tan(d*x+c))*a+5/16/d/b*arctan(tan(d*x+c))+1/d*a^4/b^4/(a*(a+b))^(1/2)*arctan((a+b)*ta
n(d*x+c)/(a*(a+b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.93242, size = 1077, normalized size = 6.61 \begin{align*} \left [\frac{12 \, a^{3} \sqrt{-\frac{a}{a + b}} \log \left (\frac{{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt{-\frac{a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 3 \,{\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} d x -{\left (8 \, b^{3} \cos \left (d x + c\right )^{5} + 2 \,{\left (6 \, a b^{2} - 13 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (8 \, a^{2} b - 10 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, b^{4} d}, -\frac{24 \, a^{3} \sqrt{\frac{a}{a + b}} \arctan \left (\frac{{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{\frac{a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 3 \,{\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} d x +{\left (8 \, b^{3} \cos \left (d x + c\right )^{5} + 2 \,{\left (6 \, a b^{2} - 13 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (8 \, a^{2} b - 10 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, b^{4} d}\right ] \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)^2),x, algorithm="fricas")

[Out]

[1/48*(12*a^3*sqrt(-a/(a + b))*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c
)^2 - 4*((2*a^2 + 3*a*b + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-a/(a + b))*sin(d*x + c
) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)) - 3*(16*a^3 -
8*a^2*b + 6*a*b^2 - 5*b^3)*d*x - (8*b^3*cos(d*x + c)^5 + 2*(6*a*b^2 - 13*b^3)*cos(d*x + c)^3 + 3*(8*a^2*b - 10
*a*b^2 + 11*b^3)*cos(d*x + c))*sin(d*x + c))/(b^4*d), -1/48*(24*a^3*sqrt(a/(a + b))*arctan(1/2*((2*a + b)*cos(
d*x + c)^2 - a - b)*sqrt(a/(a + b))/(a*cos(d*x + c)*sin(d*x + c))) + 3*(16*a^3 - 8*a^2*b + 6*a*b^2 - 5*b^3)*d*
x + (8*b^3*cos(d*x + c)^5 + 2*(6*a*b^2 - 13*b^3)*cos(d*x + c)^3 + 3*(8*a^2*b - 10*a*b^2 + 11*b^3)*cos(d*x + c)
)*sin(d*x + c))/(b^4*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)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.17834, size = 315, normalized size = 1.93 \begin{align*} \frac{\frac{48 \,{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt{a^{2} + a b}}\right )\right )} a^{4}}{\sqrt{a^{2} + a b} b^{4}} - \frac{3 \,{\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )}{\left (d x + c\right )}}{b^{4}} - \frac{24 \, a^{2} \tan \left (d x + c\right )^{5} - 30 \, a b \tan \left (d x + c\right )^{5} + 33 \, b^{2} \tan \left (d x + c\right )^{5} + 48 \, a^{2} \tan \left (d x + c\right )^{3} - 48 \, a b \tan \left (d x + c\right )^{3} + 40 \, b^{2} \tan \left (d x + c\right )^{3} + 24 \, a^{2} \tan \left (d x + c\right ) - 18 \, a b \tan \left (d x + c\right ) + 15 \, b^{2} \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{3} b^{3}}}{48 \, d} \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)^2),x, algorithm="giac")

[Out]

1/48*(48*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*
b)))*a^4/(sqrt(a^2 + a*b)*b^4) - 3*(16*a^3 - 8*a^2*b + 6*a*b^2 - 5*b^3)*(d*x + c)/b^4 - (24*a^2*tan(d*x + c)^5
- 30*a*b*tan(d*x + c)^5 + 33*b^2*tan(d*x + c)^5 + 48*a^2*tan(d*x + c)^3 - 48*a*b*tan(d*x + c)^3 + 40*b^2*tan(
d*x + c)^3 + 24*a^2*tan(d*x + c) - 18*a*b*tan(d*x + c) + 15*b^2*tan(d*x + c))/((tan(d*x + c)^2 + 1)^3*b^3))/d