3.1226 \(\int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=235 \[ -\frac {4 a \left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin ^4(c+d x)}{4 b^4 d}+\frac {a^2 \left (7 a^4-10 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{b^8 d}-\frac {2 a \left (3 a^4-4 a^2 b^2+b^4\right ) \sin (c+d x)}{b^7 d}+\frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \sin ^2(c+d x)}{2 b^6 d}+\frac {a^3 \left (a^2-b^2\right )^2}{b^8 d (a+b \sin (c+d x))}-\frac {2 a \sin ^5(c+d x)}{5 b^3 d}+\frac {\sin ^6(c+d x)}{6 b^2 d} \]

[Out]

a^2*(7*a^4-10*a^2*b^2+3*b^4)*ln(a+b*sin(d*x+c))/b^8/d-2*a*(3*a^4-4*a^2*b^2+b^4)*sin(d*x+c)/b^7/d+1/2*(5*a^4-6*
a^2*b^2+b^4)*sin(d*x+c)^2/b^6/d-4/3*a*(a^2-b^2)*sin(d*x+c)^3/b^5/d+1/4*(3*a^2-2*b^2)*sin(d*x+c)^4/b^4/d-2/5*a*
sin(d*x+c)^5/b^3/d+1/6*sin(d*x+c)^6/b^2/d+a^3*(a^2-b^2)^2/b^8/d/(a+b*sin(d*x+c))

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Rubi [A]  time = 0.28, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2837, 12, 948} \[ \frac {\left (3 a^2-2 b^2\right ) \sin ^4(c+d x)}{4 b^4 d}-\frac {4 a \left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^5 d}+\frac {\left (-6 a^2 b^2+5 a^4+b^4\right ) \sin ^2(c+d x)}{2 b^6 d}-\frac {2 a \left (-4 a^2 b^2+3 a^4+b^4\right ) \sin (c+d x)}{b^7 d}+\frac {a^3 \left (a^2-b^2\right )^2}{b^8 d (a+b \sin (c+d x))}+\frac {a^2 \left (-10 a^2 b^2+7 a^4+3 b^4\right ) \log (a+b \sin (c+d x))}{b^8 d}-\frac {2 a \sin ^5(c+d x)}{5 b^3 d}+\frac {\sin ^6(c+d x)}{6 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + b*Sin[c + d*x])^2,x]

[Out]

(a^2*(7*a^4 - 10*a^2*b^2 + 3*b^4)*Log[a + b*Sin[c + d*x]])/(b^8*d) - (2*a*(3*a^4 - 4*a^2*b^2 + b^4)*Sin[c + d*
x])/(b^7*d) + ((5*a^4 - 6*a^2*b^2 + b^4)*Sin[c + d*x]^2)/(2*b^6*d) - (4*a*(a^2 - b^2)*Sin[c + d*x]^3)/(3*b^5*d
) + ((3*a^2 - 2*b^2)*Sin[c + d*x]^4)/(4*b^4*d) - (2*a*Sin[c + d*x]^5)/(5*b^3*d) + Sin[c + d*x]^6/(6*b^2*d) + (
a^3*(a^2 - b^2)^2)/(b^8*d*(a + b*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 948

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^3 \left (b^2-x^2\right )^2}{b^3 (a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^3 \left (b^2-x^2\right )^2}{(a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^8 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-2 a \left (3 a^4-4 a^2 b^2+b^4\right )+\left (5 a^4-6 a^2 b^2+b^4\right ) x-4 a \left (a^2-b^2\right ) x^2+\left (3 a^2-2 b^2\right ) x^3-2 a x^4+x^5-\frac {a^3 \left (a^2-b^2\right )^2}{(a+x)^2}+\frac {7 a^6-10 a^4 b^2+3 a^2 b^4}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^8 d}\\ &=\frac {a^2 \left (7 a^4-10 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{b^8 d}-\frac {2 a \left (3 a^4-4 a^2 b^2+b^4\right ) \sin (c+d x)}{b^7 d}+\frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \sin ^2(c+d x)}{2 b^6 d}-\frac {4 a \left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin ^4(c+d x)}{4 b^4 d}-\frac {2 a \sin ^5(c+d x)}{5 b^3 d}+\frac {\sin ^6(c+d x)}{6 b^2 d}+\frac {a^3 \left (a^2-b^2\right )^2}{b^8 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 2.19, size = 264, normalized size = 1.12 \[ \frac {\left (50 a b^6-35 a^3 b^4\right ) \sin ^4(c+d x)+60 a^2 b \left (a^2-b^2\right ) \sin (c+d x) \left (\left (7 a^2-3 b^2\right ) \log (a+b \sin (c+d x))-6 a^2+2 b^2\right )+3 b^5 \left (7 a^2-10 b^2\right ) \sin ^5(c+d x)-30 a b^2 \left (7 a^4-10 a^2 b^2+3 b^4\right ) \sin ^2(c+d x)+10 b^3 \left (7 a^4-10 a^2 b^2+3 b^4\right ) \sin ^3(c+d x)+60 a^3 \left (a^2-b^2\right ) \left (\left (7 a^2-3 b^2\right ) \log (a+b \sin (c+d x))+a^2-b^2\right )-14 a b^6 \sin ^6(c+d x)+10 b^7 \sin ^7(c+d x)}{60 b^8 d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + b*Sin[c + d*x])^2,x]

[Out]

(60*a^3*(a^2 - b^2)*(a^2 - b^2 + (7*a^2 - 3*b^2)*Log[a + b*Sin[c + d*x]]) + 60*a^2*b*(a^2 - b^2)*(-6*a^2 + 2*b
^2 + (7*a^2 - 3*b^2)*Log[a + b*Sin[c + d*x]])*Sin[c + d*x] - 30*a*b^2*(7*a^4 - 10*a^2*b^2 + 3*b^4)*Sin[c + d*x
]^2 + 10*b^3*(7*a^4 - 10*a^2*b^2 + 3*b^4)*Sin[c + d*x]^3 + (-35*a^3*b^4 + 50*a*b^6)*Sin[c + d*x]^4 + 3*b^5*(7*
a^2 - 10*b^2)*Sin[c + d*x]^5 - 14*a*b^6*Sin[c + d*x]^6 + 10*b^7*Sin[c + d*x]^7)/(60*b^8*d*(a + b*Sin[c + d*x])
)

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fricas [A]  time = 0.86, size = 279, normalized size = 1.19 \[ \frac {112 \, a b^{6} \cos \left (d x + c\right )^{6} + 480 \, a^{7} - 3240 \, a^{5} b^{2} + 3185 \, a^{3} b^{4} - 487 \, a b^{6} - 8 \, {\left (35 \, a^{3} b^{4} - 8 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} + 16 \, {\left (105 \, a^{5} b^{2} - 115 \, a^{3} b^{4} + 16 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + 480 \, {\left (7 \, a^{7} - 10 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + {\left (7 \, a^{6} b - 10 \, a^{4} b^{3} + 3 \, a^{2} b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (80 \, b^{7} \cos \left (d x + c\right )^{6} - 168 \, a^{2} b^{5} \cos \left (d x + c\right )^{4} + 2880 \, a^{6} b - 3800 \, a^{4} b^{3} + 1007 \, a^{2} b^{5} - 25 \, b^{7} + 16 \, {\left (35 \, a^{4} b^{3} - 29 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, {\left (b^{9} d \sin \left (d x + c\right ) + a b^{8} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/480*(112*a*b^6*cos(d*x + c)^6 + 480*a^7 - 3240*a^5*b^2 + 3185*a^3*b^4 - 487*a*b^6 - 8*(35*a^3*b^4 - 8*a*b^6)
*cos(d*x + c)^4 + 16*(105*a^5*b^2 - 115*a^3*b^4 + 16*a*b^6)*cos(d*x + c)^2 + 480*(7*a^7 - 10*a^5*b^2 + 3*a^3*b
^4 + (7*a^6*b - 10*a^4*b^3 + 3*a^2*b^5)*sin(d*x + c))*log(b*sin(d*x + c) + a) - (80*b^7*cos(d*x + c)^6 - 168*a
^2*b^5*cos(d*x + c)^4 + 2880*a^6*b - 3800*a^4*b^3 + 1007*a^2*b^5 - 25*b^7 + 16*(35*a^4*b^3 - 29*a^2*b^5)*cos(d
*x + c)^2)*sin(d*x + c))/(b^9*d*sin(d*x + c) + a*b^8*d)

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giac [A]  time = 0.23, size = 300, normalized size = 1.28 \[ \frac {\frac {60 \, {\left (7 \, a^{6} - 10 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{8}} - \frac {60 \, {\left (7 \, a^{6} b \sin \left (d x + c\right ) - 10 \, a^{4} b^{3} \sin \left (d x + c\right ) + 3 \, a^{2} b^{5} \sin \left (d x + c\right ) + 6 \, a^{7} - 8 \, a^{5} b^{2} + 2 \, a^{3} b^{4}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{8}} + \frac {10 \, b^{10} \sin \left (d x + c\right )^{6} - 24 \, a b^{9} \sin \left (d x + c\right )^{5} + 45 \, a^{2} b^{8} \sin \left (d x + c\right )^{4} - 30 \, b^{10} \sin \left (d x + c\right )^{4} - 80 \, a^{3} b^{7} \sin \left (d x + c\right )^{3} + 80 \, a b^{9} \sin \left (d x + c\right )^{3} + 150 \, a^{4} b^{6} \sin \left (d x + c\right )^{2} - 180 \, a^{2} b^{8} \sin \left (d x + c\right )^{2} + 30 \, b^{10} \sin \left (d x + c\right )^{2} - 360 \, a^{5} b^{5} \sin \left (d x + c\right ) + 480 \, a^{3} b^{7} \sin \left (d x + c\right ) - 120 \, a b^{9} \sin \left (d x + c\right )}{b^{12}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/60*(60*(7*a^6 - 10*a^4*b^2 + 3*a^2*b^4)*log(abs(b*sin(d*x + c) + a))/b^8 - 60*(7*a^6*b*sin(d*x + c) - 10*a^4
*b^3*sin(d*x + c) + 3*a^2*b^5*sin(d*x + c) + 6*a^7 - 8*a^5*b^2 + 2*a^3*b^4)/((b*sin(d*x + c) + a)*b^8) + (10*b
^10*sin(d*x + c)^6 - 24*a*b^9*sin(d*x + c)^5 + 45*a^2*b^8*sin(d*x + c)^4 - 30*b^10*sin(d*x + c)^4 - 80*a^3*b^7
*sin(d*x + c)^3 + 80*a*b^9*sin(d*x + c)^3 + 150*a^4*b^6*sin(d*x + c)^2 - 180*a^2*b^8*sin(d*x + c)^2 + 30*b^10*
sin(d*x + c)^2 - 360*a^5*b^5*sin(d*x + c) + 480*a^3*b^7*sin(d*x + c) - 120*a*b^9*sin(d*x + c))/b^12)/d

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maple [A]  time = 0.55, size = 342, normalized size = 1.46 \[ \frac {\sin ^{6}\left (d x +c \right )}{6 b^{2} d}-\frac {2 a \left (\sin ^{5}\left (d x +c \right )\right )}{5 b^{3} d}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right ) a^{2}}{4 d \,b^{4}}-\frac {\sin ^{4}\left (d x +c \right )}{2 b^{2} d}-\frac {4 \left (\sin ^{3}\left (d x +c \right )\right ) a^{3}}{3 d \,b^{5}}+\frac {4 a \left (\sin ^{3}\left (d x +c \right )\right )}{3 b^{3} d}+\frac {5 \left (\sin ^{2}\left (d x +c \right )\right ) a^{4}}{2 d \,b^{6}}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}}{d \,b^{4}}+\frac {\sin ^{2}\left (d x +c \right )}{2 b^{2} d}-\frac {6 a^{5} \sin \left (d x +c \right )}{d \,b^{7}}+\frac {8 a^{3} \sin \left (d x +c \right )}{d \,b^{5}}-\frac {2 a \sin \left (d x +c \right )}{b^{3} d}+\frac {7 a^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{8}}-\frac {10 a^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{6}}+\frac {3 a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{4}}+\frac {a^{7}}{d \,b^{8} \left (a +b \sin \left (d x +c \right )\right )}-\frac {2 a^{5}}{d \,b^{6} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a^{3}}{d \,b^{4} \left (a +b \sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c))^2,x)

[Out]

1/6*sin(d*x+c)^6/b^2/d-2/5*a*sin(d*x+c)^5/b^3/d+3/4/d/b^4*sin(d*x+c)^4*a^2-1/2*sin(d*x+c)^4/b^2/d-4/3/d/b^5*si
n(d*x+c)^3*a^3+4/3*a*sin(d*x+c)^3/b^3/d+5/2/d/b^6*sin(d*x+c)^2*a^4-3/d/b^4*sin(d*x+c)^2*a^2+1/2*sin(d*x+c)^2/b
^2/d-6/d/b^7*a^5*sin(d*x+c)+8/d/b^5*a^3*sin(d*x+c)-2*a*sin(d*x+c)/b^3/d+7/d*a^6/b^8*ln(a+b*sin(d*x+c))-10/d*a^
4/b^6*ln(a+b*sin(d*x+c))+3/d*a^2/b^4*ln(a+b*sin(d*x+c))+1/d*a^7/b^8/(a+b*sin(d*x+c))-2/d*a^5/b^6/(a+b*sin(d*x+
c))+1/d*a^3/b^4/(a+b*sin(d*x+c))

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maxima [A]  time = 0.32, size = 218, normalized size = 0.93 \[ \frac {\frac {60 \, {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )}}{b^{9} \sin \left (d x + c\right ) + a b^{8}} + \frac {10 \, b^{5} \sin \left (d x + c\right )^{6} - 24 \, a b^{4} \sin \left (d x + c\right )^{5} + 15 \, {\left (3 \, a^{2} b^{3} - 2 \, b^{5}\right )} \sin \left (d x + c\right )^{4} - 80 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{3} + 30 \, {\left (5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{2} - 120 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{b^{7}} + \frac {60 \, {\left (7 \, a^{6} - 10 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{8}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/60*(60*(a^7 - 2*a^5*b^2 + a^3*b^4)/(b^9*sin(d*x + c) + a*b^8) + (10*b^5*sin(d*x + c)^6 - 24*a*b^4*sin(d*x +
c)^5 + 15*(3*a^2*b^3 - 2*b^5)*sin(d*x + c)^4 - 80*(a^3*b^2 - a*b^4)*sin(d*x + c)^3 + 30*(5*a^4*b - 6*a^2*b^3 +
b^5)*sin(d*x + c)^2 - 120*(3*a^5 - 4*a^3*b^2 + a*b^4)*sin(d*x + c))/b^7 + 60*(7*a^6 - 10*a^4*b^2 + 3*a^2*b^4)
*log(b*sin(d*x + c) + a)/b^8)/d

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mupad [B]  time = 0.14, size = 375, normalized size = 1.60 \[ \frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {2\,a^3}{3\,b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{3\,b}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^4\,\left (\frac {1}{2\,b^2}-\frac {3\,a^2}{4\,b^4}\right )}{d}+\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{2\,b^2}+\frac {a^2\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{2\,b^2}-\frac {a\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{b}\right )}{d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {a^2\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{b^2}+\frac {2\,a\,\left (\frac {1}{b^2}+\frac {a^2\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b^2}-\frac {2\,a\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{b}\right )}{b}\right )}{d}+\frac {{\sin \left (c+d\,x\right )}^6}{6\,b^2\,d}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^5}{5\,b^3\,d}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (7\,a^6-10\,a^4\,b^2+3\,a^2\,b^4\right )}{b^8\,d}+\frac {a^7-2\,a^5\,b^2+a^3\,b^4}{b\,d\,\left (\sin \left (c+d\,x\right )\,b^8+a\,b^7\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*sin(c + d*x)^3)/(a + b*sin(c + d*x))^2,x)

[Out]

(sin(c + d*x)^3*((2*a^3)/(3*b^5) + (2*a*(2/b^2 - (3*a^2)/b^4))/(3*b)))/d - (sin(c + d*x)^4*(1/(2*b^2) - (3*a^2
)/(4*b^4)))/d + (sin(c + d*x)^2*(1/(2*b^2) + (a^2*(2/b^2 - (3*a^2)/b^4))/(2*b^2) - (a*((2*a^3)/b^5 + (2*a*(2/b
^2 - (3*a^2)/b^4))/b))/b))/d - (sin(c + d*x)*((a^2*((2*a^3)/b^5 + (2*a*(2/b^2 - (3*a^2)/b^4))/b))/b^2 + (2*a*(
1/b^2 + (a^2*(2/b^2 - (3*a^2)/b^4))/b^2 - (2*a*((2*a^3)/b^5 + (2*a*(2/b^2 - (3*a^2)/b^4))/b))/b))/b))/d + sin(
c + d*x)^6/(6*b^2*d) - (2*a*sin(c + d*x)^5)/(5*b^3*d) + (log(a + b*sin(c + d*x))*(7*a^6 + 3*a^2*b^4 - 10*a^4*b
^2))/(b^8*d) + (a^7 + a^3*b^4 - 2*a^5*b^2)/(b*d*(a*b^7 + b^8*sin(c + d*x)))

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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(cos(d*x+c)**5*sin(d*x+c)**3/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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