\(\int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx\) [569]

Optimal result

Integrand size = 29, antiderivative size = 89 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {2 \cos (e+f x)}{(c+d) f \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac {4 \cos (e+f x)}{(c+d)^2 f \sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]

[Out]

-2/3*a*cos(f*x+e)/(c+d)/f/(c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2)-4/3*a*cos(f*x+e)/(c+d)^2/f/(a+a*sin(f*
x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2851, 2850} \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {4 a \cos (e+f x)}{3 f (c+d)^2 \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}-\frac {2 a \cos (e+f x)}{3 f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}} \]

[In]

Int[Sqrt[a + a*Sin[e + f*x]]/(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(-2*a*Cos[e + f*x])/(3*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(3/2)) - (4*a*Cos[e + f*x])/(3*
(c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])

Rule 2850

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*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]

Rule 2851

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x]
+ Dist[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a \cos (e+f x)}{3 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac {2 \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 (c+d)} \\ & = -\frac {2 a \cos (e+f x)}{3 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac {4 a \cos (e+f x)}{3 (c+d)^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {1+\sin (e+f x)} (3 c+d+2 d \sin (e+f x))}{\sqrt {3} (c+d)^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))^{3/2}} \]

[In]

Integrate[Sqrt[3 + 3*Sin[e + f*x]]/(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(-2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[1 + Sin[e + f*x]]*(3*c + d + 2*d*Sin[e + f*x]))/(Sqrt[3]*(c + d
)^2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c + d*Sin[e + f*x])^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(196\) vs. \(2(83)=166\).

Time = 3.87 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.21

method	result	size
default	\(-\frac {2 \sec \left (f x +e \right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {c +d \sin \left (f x +e \right )}\, \left (2 \left (\sin ^{2}\left (f x +e \right )\right ) \left (\cos ^{2}\left (f x +e \right )\right ) d^{3}+\left (\sin ^{3}\left (f x +e \right )\right ) c \,d^{2}+\left (\sin ^{3}\left (f x +e \right )\right ) d^{3}+\left (\sin ^{2}\left (f x +e \right )\right ) c \,d^{2}-d^{3} \left (\sin ^{2}\left (f x +e \right )\right )-4 \left (\cos ^{2}\left (f x +e \right )\right ) c^{2} d -3 c^{3} \sin \left (f x +e \right )-5 c^{2} d \sin \left (f x +e \right )-2 \sin \left (f x +e \right ) c \,d^{2}+3 c^{3}+5 c^{2} d \right )}{3 f {\left (\left (\cos ^{2}\left (f x +e \right )\right ) d^{2}+c^{2}-d^{2}\right )}^{2} \left (c +d \right )^{2}}\)	\(197\)

[In]

int((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/f*sec(f*x+e)*(a*(sin(f*x+e)+1))^(1/2)*(c+d*sin(f*x+e))^(1/2)*(2*sin(f*x+e)^2*cos(f*x+e)^2*d^3+sin(f*x+e)^
3*c*d^2+sin(f*x+e)^3*d^3+sin(f*x+e)^2*c*d^2-d^3*sin(f*x+e)^2-4*cos(f*x+e)^2*c^2*d-3*c^3*sin(f*x+e)-5*c^2*d*sin
(f*x+e)-2*sin(f*x+e)*c*d^2+3*c^3+5*c^2*d)/(cos(f*x+e)^2*d^2+c^2-d^2)^2/(c+d)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (83) = 166\).

Time = 0.29 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.37 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (2 \, d \cos \left (f x + e\right )^{2} + {\left (3 \, c + d\right )} \cos \left (f x + e\right ) + {\left (2 \, d \cos \left (f x + e\right ) - 3 \, c + d\right )} \sin \left (f x + e\right ) + 3 \, c - d\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c}}{3 \, {\left ({\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{3} + {\left (2 \, c^{3} d + 5 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (c^{4} + 2 \, c^{3} d + 2 \, c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right ) - {\left (c^{4} + 4 \, c^{3} d + 6 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f + {\left ({\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{3} d + 2 \, c^{2} d^{2} + c d^{3}\right )} f \cos \left (f x + e\right ) - {\left (c^{4} + 4 \, c^{3} d + 6 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

2/3*(2*d*cos(f*x + e)^2 + (3*c + d)*cos(f*x + e) + (2*d*cos(f*x + e) - 3*c + d)*sin(f*x + e) + 3*c - d)*sqrt(a
*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/((c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e)^3 + (2*c^3*d + 5*c^2*d^2
+ 4*c*d^3 + d^4)*f*cos(f*x + e)^2 - (c^4 + 2*c^3*d + 2*c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e) - (c^4 + 4*c^3
*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f + ((c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e)^2 - 2*(c^3*d + 2*c^2*d^2 + c*d^3
)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f)*sin(f*x + e))

Sympy [F]

\[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )}}{\left (c + d \sin {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((a+a*sin(f*x+e))**(1/2)/(c+d*sin(f*x+e))**(5/2),x)

[Out]

Integral(sqrt(a*(sin(e + f*x) + 1))/(c + d*sin(e + f*x))**(5/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (83) = 166\).

Time = 0.34 (sec) , antiderivative size = 340, normalized size of antiderivative = 3.82 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left ({\left (3 \, c^{2} + c d\right )} \sqrt {a} - \frac {{\left (3 \, c^{2} - 9 \, c d - 2 \, d^{2}\right )} \sqrt {a} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, {\left (3 \, c^{2} - 4 \, c d + 3 \, d^{2}\right )} \sqrt {a} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, {\left (3 \, c^{2} - 4 \, c d + 3 \, d^{2}\right )} \sqrt {a} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {{\left (3 \, c^{2} - 9 \, c d - 2 \, d^{2}\right )} \sqrt {a} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {{\left (3 \, c^{2} + c d\right )} \sqrt {a} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{2}}{3 \, {\left (c^{2} + 2 \, c d + d^{2} + \frac {2 \, {\left (c^{2} + 2 \, c d + d^{2}\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {{\left (c^{2} + 2 \, c d + d^{2}\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} {\left (c + \frac {2 \, d \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}^{\frac {5}{2}} f} \]

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-2/3*((3*c^2 + c*d)*sqrt(a) - (3*c^2 - 9*c*d - 2*d^2)*sqrt(a)*sin(f*x + e)/(cos(f*x + e) + 1) + 2*(3*c^2 - 4*c
*d + 3*d^2)*sqrt(a)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*(3*c^2 - 4*c*d + 3*d^2)*sqrt(a)*sin(f*x + e)^3/(co
s(f*x + e) + 1)^3 + (3*c^2 - 9*c*d - 2*d^2)*sqrt(a)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - (3*c^2 + c*d)*sqrt(a
)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^2/((c^2 + 2*c*d + d^2 + 2*(c^
2 + 2*c*d + d^2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (c^2 + 2*c*d + d^2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4
)*(c + 2*d*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)^(5/2)*f)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 471 vs. \(2 (83) = 166\).

Time = 0.83 (sec) , antiderivative size = 471, normalized size of antiderivative = 5.29 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx=\frac {4 \, \sqrt {2} {\left ({\left (\frac {3 \, {\left (c^{6} d^{4} - 2 \, c^{5} d^{5} - c^{4} d^{6} + 4 \, c^{3} d^{7} - c^{2} d^{8} - 2 \, c d^{9} + d^{10}\right )} \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2}}{c^{7} d^{4} - c^{6} d^{5} - 3 \, c^{5} d^{6} + 3 \, c^{4} d^{7} + 3 \, c^{3} d^{8} - 3 \, c^{2} d^{9} - c d^{10} + d^{11}} + \frac {2 \, {\left (3 \, c^{6} d^{4} - 14 \, c^{5} d^{5} + 21 \, c^{4} d^{6} - 4 \, c^{3} d^{7} - 19 \, c^{2} d^{8} + 18 \, c d^{9} - 5 \, d^{10}\right )}}{c^{7} d^{4} - c^{6} d^{5} - 3 \, c^{5} d^{6} + 3 \, c^{4} d^{7} + 3 \, c^{3} d^{8} - 3 \, c^{2} d^{9} - c d^{10} + d^{11}}\right )} \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2} + \frac {3 \, {\left (c^{6} d^{4} - 2 \, c^{5} d^{5} - c^{4} d^{6} + 4 \, c^{3} d^{7} - c^{2} d^{8} - 2 \, c d^{9} + d^{10}\right )}}{c^{7} d^{4} - c^{6} d^{5} - 3 \, c^{5} d^{6} + 3 \, c^{4} d^{7} + 3 \, c^{3} d^{8} - 3 \, c^{2} d^{9} - c d^{10} + d^{11}}\right )} \sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )}{3 \, {\left (c \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{4} + d \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{4} + 2 \, c \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2} - 6 \, d \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2} + c + d\right )}^{\frac {3}{2}} f} \]

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

4/3*sqrt(2)*((3*(c^6*d^4 - 2*c^5*d^5 - c^4*d^6 + 4*c^3*d^7 - c^2*d^8 - 2*c*d^9 + d^10)*tan(-1/8*pi + 1/4*f*x +
1/4*e)^2/(c^7*d^4 - c^6*d^5 - 3*c^5*d^6 + 3*c^4*d^7 + 3*c^3*d^8 - 3*c^2*d^9 - c*d^10 + d^11) + 2*(3*c^6*d^4 -
14*c^5*d^5 + 21*c^4*d^6 - 4*c^3*d^7 - 19*c^2*d^8 + 18*c*d^9 - 5*d^10)/(c^7*d^4 - c^6*d^5 - 3*c^5*d^6 + 3*c^4*
d^7 + 3*c^3*d^8 - 3*c^2*d^9 - c*d^10 + d^11))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 3*(c^6*d^4 - 2*c^5*d^5 - c^4*
d^6 + 4*c^3*d^7 - c^2*d^8 - 2*c*d^9 + d^10)/(c^7*d^4 - c^6*d^5 - 3*c^5*d^6 + 3*c^4*d^7 + 3*c^3*d^8 - 3*c^2*d^9
- c*d^10 + d^11))*sqrt(a)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)/((c*tan(-1/8*pi
+ 1/4*f*x + 1/4*e)^4 + d*tan(-1/8*pi + 1/4*f*x + 1/4*e)^4 + 2*c*tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 - 6*d*tan(-1/
8*pi + 1/4*f*x + 1/4*e)^2 + c + d)^(3/2)*f)

Mupad [B] (verification not implemented)

Time = 15.15 (sec) , antiderivative size = 353, normalized size of antiderivative = 3.97 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {c+d\,\sin \left (e+f\,x\right )}\,\left (\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,8{}\mathrm {i}}{3\,d\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}+\frac {8\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,d\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}+\frac {8\,c\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{d^2\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}+\frac {c\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,8{}\mathrm {i}}{d^2\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}\right )}{{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}-\frac {{\left (c+d\right )}^2\,1{}\mathrm {i}}{{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}-\frac {2\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\left (2\,c^2+2\,c\,d+d^2\right )}{d^2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left (4\,c+d\right )}{d}+\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,{\left (c+d\right )}^2\,\left (2\,c^2+2\,c\,d+d^2\right )\,2{}\mathrm {i}}{d^2\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}-\frac {{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,{\left (c+d\right )}^2\,\left (4\,c+d\right )\,1{}\mathrm {i}}{d\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}} \]

[In]

int((a + a*sin(e + f*x))^(1/2)/(c + d*sin(e + f*x))^(5/2),x)

[Out]

-((c + d*sin(e + f*x))^(1/2)*((exp(e*1i + f*x*1i)*(a + a*sin(e + f*x))^(1/2)*8i)/(3*d*f*(c*1i + d*1i)^2) + (8*
exp(e*4i + f*x*4i)*(a + a*sin(e + f*x))^(1/2))/(3*d*f*(c*1i + d*1i)^2) + (8*c*exp(e*2i + f*x*2i)*(a + a*sin(e
+ f*x))^(1/2))/(d^2*f*(c*1i + d*1i)^2) + (c*exp(e*3i + f*x*3i)*(a + a*sin(e + f*x))^(1/2)*8i)/(d^2*f*(c*1i + d
*1i)^2)))/(exp(e*5i + f*x*5i) - ((c + d)^2*1i)/(c*1i + d*1i)^2 - (2*exp(e*3i + f*x*3i)*(2*c*d + 2*c^2 + d^2))/
d^2 + (exp(e*1i + f*x*1i)*(4*c + d))/d + (exp(e*2i + f*x*2i)*(c + d)^2*(2*c*d + 2*c^2 + d^2)*2i)/(d^2*(c*1i +
d*1i)^2) - (exp(e*4i + f*x*4i)*(c + d)^2*(4*c + d)*1i)/(d*(c*1i + d*1i)^2))