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

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

[Out]

-1/16*(16*a^3-8*a^2*b+6*a*b^2-5*b^3)*x/b^4-1/16*(8*a^2-6*a*b+5*b^2)*cos(d*x+c)*sin(d*x+c)/b^3/d+1/24*(6*a-5*b)
*cos(d*x+c)*sin(d*x+c)^3/b^2/d-1/6*cos(d*x+c)*sin(d*x+c)^5/b/d+a^(7/2)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/
b^4/d/(a+b)^(1/2)

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Rubi [A]  time = 0.36, antiderivative size = 163, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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]

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*}

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Mathematica [A]  time = 1.18, size = 133, normalized size = 0.82 \[ -\frac {-\frac {192 a^{7/2} \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a+b}}+3 b \left (16 a^2-16 a b+15 b^2\right ) \sin (2 (c+d x))+12 \left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) (c+d x)+3 b^2 (2 a-3 b) \sin (4 (c+d x))+b^3 \sin (6 (c+d x))}{192 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/192*(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])/Sqr
t[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)])/(b^4*d)

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fricas [A]  time = 0.49, size = 453, normalized size = 2.78 \[ \left [\frac {12 \, a^{3} \sqrt {-\frac {a}{a + b}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} d x - {\left (8 \, b^{3} \cos \left (d x + c\right )^{5} + 2 \, {\left (6 \, a b^{2} - 13 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} b - 10 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, b^{4} d}, -\frac {24 \, a^{3} \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} d x + {\left (8 \, b^{3} \cos \left (d x + c\right )^{5} + 2 \, {\left (6 \, a b^{2} - 13 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} b - 10 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, b^{4} d}\right ] \]

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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giac [A]  time = 0.18, size = 233, normalized size = 1.43 \[ \frac {\frac {48 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} a^{4}}{\sqrt {a^{2} + a b} b^{4}} - \frac {3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {24 \, a^{2} \tan \left (d x + c\right )^{5} - 30 \, a b \tan \left (d x + c\right )^{5} + 33 \, b^{2} \tan \left (d x + c\right )^{5} + 48 \, a^{2} \tan \left (d x + c\right )^{3} - 48 \, a b \tan \left (d x + c\right )^{3} + 40 \, b^{2} \tan \left (d x + c\right )^{3} + 24 \, a^{2} \tan \left (d x + c\right ) - 18 \, a b \tan \left (d x + c\right ) + 15 \, b^{2} \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{3} b^{3}}}{48 \, d} \]

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

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maple [B]  time = 0.31, size = 361, normalized size = 2.21 \[ \frac {a^{4} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{d \,b^{4} \sqrt {a \left (a +b \right )}}-\frac {\left (\tan ^{5}\left (d x +c \right )\right ) a^{2}}{2 d \,b^{3} \left (\tan ^{2}\left (d x +c \right )+1\right )^{3}}+\frac {5 \left (\tan ^{5}\left (d x +c \right )\right ) a}{8 d \,b^{2} \left (\tan ^{2}\left (d x +c \right )+1\right )^{3}}-\frac {11 \left (\tan ^{5}\left (d x +c \right )\right )}{16 d b \left (\tan ^{2}\left (d x +c \right )+1\right )^{3}}-\frac {\left (\tan ^{3}\left (d x +c \right )\right ) a^{2}}{d \,b^{3} \left (\tan ^{2}\left (d x +c \right )+1\right )^{3}}+\frac {\left (\tan ^{3}\left (d x +c \right )\right ) a}{d \,b^{2} \left (\tan ^{2}\left (d x +c \right )+1\right )^{3}}-\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{6 d b \left (\tan ^{2}\left (d x +c \right )+1\right )^{3}}-\frac {\tan \left (d x +c \right ) a^{2}}{2 d \,b^{3} \left (\tan ^{2}\left (d x +c \right )+1\right )^{3}}+\frac {3 \tan \left (d x +c \right ) a}{8 d \,b^{2} \left (\tan ^{2}\left (d x +c \right )+1\right )^{3}}-\frac {5 \tan \left (d x +c \right )}{16 d b \left (\tan ^{2}\left (d x +c \right )+1\right )^{3}}-\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{d \,b^{4}}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{2}}{2 d \,b^{3}}-\frac {3 \arctan \left (\tan \left (d x +c \right )\right ) a}{8 d \,b^{2}}+\frac {5 \arctan \left (\tan \left (d x +c \right )\right )}{16 d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^4/b^4/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))-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+5/1
6/d/b*arctan(tan(d*x+c))

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maxima [A]  time = 0.44, size = 192, normalized size = 1.18 \[ \frac {\frac {48 \, a^{4} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b^{4}} - \frac {3 \, {\left (8 \, a^{2} - 10 \, a b + 11 \, b^{2}\right )} \tan \left (d x + c\right )^{5} + 8 \, {\left (6 \, a^{2} - 6 \, a b + 5 \, b^{2}\right )} \tan \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} - 6 \, a b + 5 \, b^{2}\right )} \tan \left (d x + c\right )}{b^{3} \tan \left (d x + c\right )^{6} + 3 \, b^{3} \tan \left (d x + c\right )^{4} + 3 \, b^{3} \tan \left (d x + c\right )^{2} + b^{3}} - \frac {3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} {\left (d x + c\right )}}{b^{4}}}{48 \, d} \]

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]

1/48*(48*a^4*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a)*b^4) - (3*(8*a^2 - 10*a*b + 11*b^2)
*tan(d*x + c)^5 + 8*(6*a^2 - 6*a*b + 5*b^2)*tan(d*x + c)^3 + 3*(8*a^2 - 6*a*b + 5*b^2)*tan(d*x + c))/(b^3*tan(
d*x + c)^6 + 3*b^3*tan(d*x + c)^4 + 3*b^3*tan(d*x + c)^2 + b^3) - 3*(16*a^3 - 8*a^2*b + 6*a*b^2 - 5*b^3)*(d*x
+ c)/b^4)/d

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mupad [B]  time = 15.32, size = 2244, normalized size = 13.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan(((((tan(c + d*x)*(15*a*b^8 + 768*a^8*b + 512*a^9 + 25*b^9 + 11*a^2*b^7 - 63*a^3*b^6 - 224*a^4*b^5 - 140*
a^5*b^4 + 256*a^7*b^2))/(128*b^6) - ((((5*a*b^12)/4 + a^2*b^11 + (a^3*b^10)/4 + (5*a^4*b^9)/2 + 2*a^5*b^8)/b^9
- (tan(c + d*x)*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i)*(4096*a*b^10 + 1024*b^11 + 5120*a^2*b^9 + 2048*a^3*b
^8))/(4096*b^10))*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i))/(32*b^4))*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i)
*1i)/(32*b^4) + (((tan(c + d*x)*(15*a*b^8 + 768*a^8*b + 512*a^9 + 25*b^9 + 11*a^2*b^7 - 63*a^3*b^6 - 224*a^4*b
^5 - 140*a^5*b^4 + 256*a^7*b^2))/(128*b^6) + ((((5*a*b^12)/4 + a^2*b^11 + (a^3*b^10)/4 + (5*a^4*b^9)/2 + 2*a^5
*b^8)/b^9 + (tan(c + d*x)*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i)*(4096*a*b^10 + 1024*b^11 + 5120*a^2*b^9 + 2
048*a^3*b^8))/(4096*b^10))*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i))/(32*b^4))*(a*b^2*6i - a^2*b*8i + a^3*16i
- b^3*5i)*1i)/(32*b^4))/(((a^10*b)/4 + a^11 + (25*a^4*b^7)/128 - (5*a^5*b^6)/64 + (21*a^6*b^5)/128 - (21*a^7*b
^4)/32 - (15*a^8*b^3)/32 - (a^9*b^2)/8)/b^9 - (((tan(c + d*x)*(15*a*b^8 + 768*a^8*b + 512*a^9 + 25*b^9 + 11*a^
2*b^7 - 63*a^3*b^6 - 224*a^4*b^5 - 140*a^5*b^4 + 256*a^7*b^2))/(128*b^6) - ((((5*a*b^12)/4 + a^2*b^11 + (a^3*b
^10)/4 + (5*a^4*b^9)/2 + 2*a^5*b^8)/b^9 - (tan(c + d*x)*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i)*(4096*a*b^10
+ 1024*b^11 + 5120*a^2*b^9 + 2048*a^3*b^8))/(4096*b^10))*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i))/(32*b^4))*(
a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i))/(32*b^4) + (((tan(c + d*x)*(15*a*b^8 + 768*a^8*b + 512*a^9 + 25*b^9 +
11*a^2*b^7 - 63*a^3*b^6 - 224*a^4*b^5 - 140*a^5*b^4 + 256*a^7*b^2))/(128*b^6) + ((((5*a*b^12)/4 + a^2*b^11 +
(a^3*b^10)/4 + (5*a^4*b^9)/2 + 2*a^5*b^8)/b^9 + (tan(c + d*x)*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i)*(4096*a
*b^10 + 1024*b^11 + 5120*a^2*b^9 + 2048*a^3*b^8))/(4096*b^10))*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i))/(32*b
^4))*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i))/(32*b^4)))*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i)*1i)/(16*b^4
*d) - ((tan(c + d*x)*(8*a^2 - 6*a*b + 5*b^2))/(16*b^3) + (tan(c + d*x)^3*(6*a^2 - 6*a*b + 5*b^2))/(6*b^3) + (t
an(c + d*x)^5*(8*a^2 - 10*a*b + 11*b^2))/(16*b^3))/(d*(3*tan(c + d*x)^2 + 3*tan(c + d*x)^4 + tan(c + d*x)^6 +
1)) + (atan((((-a^7*(a + b))^(1/2)*((tan(c + d*x)*(15*a*b^8 + 768*a^8*b + 512*a^9 + 25*b^9 + 11*a^2*b^7 - 63*a
^3*b^6 - 224*a^4*b^5 - 140*a^5*b^4 + 256*a^7*b^2))/(128*b^6) - ((-a^7*(a + b))^(1/2)*((320*a*b^12 + 256*a^2*b^
11 + 64*a^3*b^10 + 640*a^4*b^9 + 512*a^5*b^8)/(256*b^9) - (tan(c + d*x)*(-a^7*(a + b))^(1/2)*(4096*a*b^10 + 10
24*b^11 + 5120*a^2*b^9 + 2048*a^3*b^8))/(256*b^6*(a*b^4 + b^5))))/(2*(a*b^4 + b^5)))*1i)/(2*(a*b^4 + b^5)) + (
(-a^7*(a + b))^(1/2)*((tan(c + d*x)*(15*a*b^8 + 768*a^8*b + 512*a^9 + 25*b^9 + 11*a^2*b^7 - 63*a^3*b^6 - 224*a
^4*b^5 - 140*a^5*b^4 + 256*a^7*b^2))/(128*b^6) + ((-a^7*(a + b))^(1/2)*((320*a*b^12 + 256*a^2*b^11 + 64*a^3*b^
10 + 640*a^4*b^9 + 512*a^5*b^8)/(256*b^9) + (tan(c + d*x)*(-a^7*(a + b))^(1/2)*(4096*a*b^10 + 1024*b^11 + 5120
*a^2*b^9 + 2048*a^3*b^8))/(256*b^6*(a*b^4 + b^5))))/(2*(a*b^4 + b^5)))*1i)/(2*(a*b^4 + b^5)))/((32*a^10*b + 12
8*a^11 + 25*a^4*b^7 - 10*a^5*b^6 + 21*a^6*b^5 - 84*a^7*b^4 - 60*a^8*b^3 - 16*a^9*b^2)/(128*b^9) - ((-a^7*(a +
b))^(1/2)*((tan(c + d*x)*(15*a*b^8 + 768*a^8*b + 512*a^9 + 25*b^9 + 11*a^2*b^7 - 63*a^3*b^6 - 224*a^4*b^5 - 14
0*a^5*b^4 + 256*a^7*b^2))/(128*b^6) - ((-a^7*(a + b))^(1/2)*((320*a*b^12 + 256*a^2*b^11 + 64*a^3*b^10 + 640*a^
4*b^9 + 512*a^5*b^8)/(256*b^9) - (tan(c + d*x)*(-a^7*(a + b))^(1/2)*(4096*a*b^10 + 1024*b^11 + 5120*a^2*b^9 +
2048*a^3*b^8))/(256*b^6*(a*b^4 + b^5))))/(2*(a*b^4 + b^5))))/(2*(a*b^4 + b^5)) + ((-a^7*(a + b))^(1/2)*((tan(c
+ d*x)*(15*a*b^8 + 768*a^8*b + 512*a^9 + 25*b^9 + 11*a^2*b^7 - 63*a^3*b^6 - 224*a^4*b^5 - 140*a^5*b^4 + 256*a
^7*b^2))/(128*b^6) + ((-a^7*(a + b))^(1/2)*((320*a*b^12 + 256*a^2*b^11 + 64*a^3*b^10 + 640*a^4*b^9 + 512*a^5*b
^8)/(256*b^9) + (tan(c + d*x)*(-a^7*(a + b))^(1/2)*(4096*a*b^10 + 1024*b^11 + 5120*a^2*b^9 + 2048*a^3*b^8))/(2
56*b^6*(a*b^4 + b^5))))/(2*(a*b^4 + b^5))))/(2*(a*b^4 + b^5))))*(-a^7*(a + b))^(1/2)*1i)/(d*(a*b^4 + b^5))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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