Integrand size = 27, antiderivative size = 142 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{c+d \sin (e+f x)} \, dx=-\frac {2 a^{5/2} (c-d)^2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{d^{5/2} \sqrt {c+d} f}+\frac {2 a^3 (3 c-7 d) \cos (e+f x)}{3 d^2 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{3 d f} \] Output:
-2*a^(5/2)*(c-d)^2*arctanh(a^(1/2)*d^(1/2)*cos(f*x+e)/(c+d)^(1/2)/(a+a*sin (f*x+e))^(1/2))/d^(5/2)/(c+d)^(1/2)/f+2/3*a^3*(3*c-7*d)*cos(f*x+e)/d^2/f/( a+a*sin(f*x+e))^(1/2)-2/3*a^2*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/d/f
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 7.88 (sec) , antiderivative size = 801, normalized size of antiderivative = 5.64 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{c+d \sin (e+f x)} \, dx =\text {Too large to display} \] Input:
Integrate[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x]),x]
Output:
((a*(1 + Sin[e + f*x]))^(5/2)*(6*(2*c - 5*d)*Sqrt[d]*Cos[(e + f*x)/2] - 2* d^(3/2)*Cos[(3*(e + f*x))/2] + (3*(c - d)^2*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^(3/2 )*Log[-#1 + Tan[(e + f*x)/4]] - d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - 2*c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1 - c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d]*L og[-#1 + Tan[(e + f*x)/4]]*#1^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + 3*d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Sqrt[c + d]*Log[-# 1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ])/(c + d) + (3*(c - d)^2*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(c*S qrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]] + d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - 2*c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[c + d]* Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^ 2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - 3*d*Sqrt[c + d]*Log[-#1 + T an[(e + f*x)/4]]*#1^2 + c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(- d - c*#1 + 3*d*#1^2 - c*#1^3) & ])/(c + d) + 6*Sqrt[d]*(-2*c + 5*d)*Sin[(e + f*x)/2] - 2*d^(3/2)*Sin[(3*(e + f*x))/2]))/(6*d^(5/2)*f*(Cos[(e + f*x)/ 2] + Sin[(e + f*x)/2])^5)
Time = 0.76 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3242, 27, 3042, 3460, 3042, 3252, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{c+d \sin (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{c+d \sin (e+f x)}dx\) |
\(\Big \downarrow \) 3242 |
\(\displaystyle \frac {2 \int \frac {\sqrt {\sin (e+f x) a+a} \left (a^2 (c+3 d)-a^2 (3 c-7 d) \sin (e+f x)\right )}{2 (c+d \sin (e+f x))}dx}{3 d}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin (e+f x) a+a} \left (a^2 (c+3 d)-a^2 (3 c-7 d) \sin (e+f x)\right )}{c+d \sin (e+f x)}dx}{3 d}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin (e+f x) a+a} \left (a^2 (c+3 d)-a^2 (3 c-7 d) \sin (e+f x)\right )}{c+d \sin (e+f x)}dx}{3 d}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}\) |
\(\Big \downarrow \) 3460 |
\(\displaystyle \frac {\frac {3 a^2 (c-d)^2 \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{d}+\frac {2 a^3 (3 c-7 d) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}}{3 d}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 a^2 (c-d)^2 \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{d}+\frac {2 a^3 (3 c-7 d) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}}{3 d}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}\) |
\(\Big \downarrow \) 3252 |
\(\displaystyle \frac {\frac {2 a^3 (3 c-7 d) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}-\frac {6 a^3 (c-d)^2 \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{d f}}{3 d}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {2 a^3 (3 c-7 d) \cos (e+f x)}{d f \sqrt {a \sin (e+f x)+a}}-\frac {6 a^{5/2} (c-d)^2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{d^{3/2} f \sqrt {c+d}}}{3 d}-\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 d f}\) |
Input:
Int[(a + a*Sin[e + f*x])^(5/2)/(c + d*Sin[e + f*x]),x]
Output:
(-2*a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*d*f) + ((-6*a^(5/2)*(c - d)^2*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(d^(3/2)*Sqrt[c + d]*f) + (2*a^3*(3*c - 7*d)*Cos[e + f*x])/(d* f*Sqrt[a + a*Sin[e + f*x]]))/(3*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* (m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !LtQ[ n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ c, 0]))
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
Time = 1.50 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.61
method | result | size |
default | \(\frac {2 a \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \left (\left (-a \left (-1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {a \left (c +d \right ) d}\, d -3 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} c^{2}+6 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} c d -3 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) a^{2} d^{2}+3 \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a \left (c +d \right ) d}\, a c -9 \sqrt {-a \left (-1+\sin \left (f x +e \right )\right )}\, \sqrt {a \left (c +d \right ) d}\, a d \right )}{3 d^{2} \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\) | \(228\) |
Input:
int((a+sin(f*x+e)*a)^(5/2)/(c+d*sin(f*x+e)),x,method=_RETURNVERBOSE)
Output:
2/3*a*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*((-a*(-1+sin(f*x+e)))^(3/2 )*(a*(c+d)*d)^(1/2)*d-3*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^( 1/2))*a^2*c^2+6*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^ 2*c*d-3*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^2*d^2+3* (-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2)*a*c-9*(-a*(-1+sin(f*x+e)))^(1 /2)*(a*(c+d)*d)^(1/2)*a*d)/d^2/(a*(c+d)*d)^(1/2)/cos(f*x+e)/(a+sin(f*x+e)* a)^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (120) = 240\).
Time = 0.25 (sec) , antiderivative size = 868, normalized size of antiderivative = 6.11 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{c+d \sin (e+f x)} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="fricas")
Output:
[1/6*(3*(a^2*c^2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*c os(f*x + e) + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*sin(f*x + e))*sqrt(a/(c*d + d^2))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a *d^2)*cos(f*x + e)^2 + 4*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2 + d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d + 4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt (a/(c*d + d^2)) - (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f* x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*cos(f*x + e))*s in(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2* c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) - 4*(a^2*d*cos(f*x + e)^2 - 3*a^ 2*c + 7*a^2*d - (3*a^2*c - 8*a^2*d)*cos(f*x + e) + (a^2*d*cos(f*x + e) + 3 *a^2*c - 7*a^2*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(d^2*f*cos(f*x + e) + d^2*f*sin(f*x + e) + d^2*f), -1/3*(3*(a^2*c^2 - 2*a^2*c*d + a^2*d^2 + (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*cos(f*x + e) + (a^2*c^2 - 2*a^2*c*d + a^ 2*d^2)*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(-a/(c*d + d^2))/(a*cos(f*x + e))) + 2* (a^2*d*cos(f*x + e)^2 - 3*a^2*c + 7*a^2*d - (3*a^2*c - 8*a^2*d)*cos(f*x + e) + (a^2*d*cos(f*x + e) + 3*a^2*c - 7*a^2*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(d^2*f*cos(f*x + e) + d^2*f*sin(f*x + e) + d^2*f)]
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{c+d \sin (e+f x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**(5/2)/(c+d*sin(f*x+e)),x)
Output:
Timed out
\[ \int \frac {(a+a \sin (e+f x))^{5/2}}{c+d \sin (e+f x)} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{d \sin \left (f x + e\right ) + c} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^(5/2)/(d*sin(f*x + e) + c), x)
Time = 0.15 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.61 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{c+d \sin (e+f x)} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {3 \, \sqrt {2} {\left (a^{2} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{\sqrt {-c d - d^{2}} d^{2}} + \frac {2 \, {\left (2 \, a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, a^{2} c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, a^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{d^{3}}\right )}}{3 \, f} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="giac")
Output:
-1/3*sqrt(2)*sqrt(a)*(3*sqrt(2)*(a^2*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e )) - 2*a^2*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + a^2*d^2*sgn(cos(-1/4* pi + 1/2*f*x + 1/2*e)))*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)/sq rt(-c*d - d^2))/(sqrt(-c*d - d^2)*d^2) + 2*(2*a^2*d^2*sgn(cos(-1/4*pi + 1/ 2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 3*a^2*c*d*sgn(cos(-1/4* pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 9*a^2*d^2*sgn(cos( -1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e))/d^3)/f
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{c+d \sin (e+f x)} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{c+d\,\sin \left (e+f\,x\right )} \,d x \] Input:
int((a + a*sin(e + f*x))^(5/2)/(c + d*sin(e + f*x)),x)
Output:
int((a + a*sin(e + f*x))^(5/2)/(c + d*sin(e + f*x)), x)
\[ \int \frac {(a+a \sin (e+f x))^{5/2}}{c+d \sin (e+f x)} \, dx=\sqrt {a}\, a^{2} \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right ) d +c}d x +\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right ) d +c}d x +2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right ) d +c}d x \right )\right ) \] Input:
int((a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x)
Output:
sqrt(a)*a**2*(int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)*d + c),x) + int((sq rt(sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)*d + c),x) + 2*int((sqr t(sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)*d + c),x))