\(\int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx\) [158]

Optimal result

Integrand size = 40, antiderivative size = 146 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {(A-4 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac {(A-4 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{240 c^2 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

1/10*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(11/2)+1/40*(A-4*B)*cos(f*x+e)*(a+a*sin(f*x+e)
)^(5/2)/c/f/(c-c*sin(f*x+e))^(9/2)+1/240*(A-4*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/c^2/f/(c-c*sin(f*x+e))^(7/2
)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {3051, 2822, 2821} \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {(A-4 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{240 c^2 f (c-c \sin (e+f x))^{7/2}}+\frac {(A-4 B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}} \]

[In]

Int[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(10*f*(c - c*Sin[e + f*x])^(11/2)) + ((A - 4*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(5/2))/(40*c*f*(c - c*Sin[e + f*x])^(9/2)) + ((A - 4*B)*Cos[e + f*x]*(a + a*Sin[e + f*x]
)^(5/2))/(240*c^2*f*(c - c*Sin[e + f*x])^(7/2))

Rule 2821

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

Rule 2822

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] + Dist[(m + n + 1)/(a*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m
, 1] ||  !SumSimplerQ[n, 1])

Rule 3051

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] + Dist[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2*m + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {(A-4 B) \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx}{5 c} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {(A-4 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac {(A-4 B) \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{7/2}} \, dx}{40 c^2} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {(A-4 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac {(A-4 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{240 c^2 f (c-c \sin (e+f x))^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 13.16 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (-36 A-6 B+10 (2 A+B) \cos (2 (e+f x))-5 (8 A+13 B) \sin (e+f x)+15 B \sin (3 (e+f x)))}{120 c^5 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^5 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(-36*A - 6*B + 10*(2*A + B)*Cos[2*(e + f
*x)] - 5*(8*A + 13*B)*Sin[e + f*x] + 15*B*Sin[3*(e + f*x)]))/(120*c^5*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*
(-1 + Sin[e + f*x])^5*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 3.47 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {a^{2} \tan \left (f x +e \right ) \left (4 A \left (\cos ^{4}\left (f x +e \right )\right )+B \left (\sin ^{2}\left (f x +e \right )\right ) \left (\cos ^{2}\left (f x +e \right )\right )+20 A \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+5 B \left (\sin ^{3}\left (f x +e \right )\right )-48 A \left (\cos ^{2}\left (f x +e \right )\right )+4 B \left (\sin ^{2}\left (f x +e \right )\right )-50 A \sin \left (f x +e \right )+15 B \sin \left (f x +e \right )+74 A \right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{30 c^{5} f \left (\cos ^{4}\left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(190\)
parts	\(\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2} \left (2 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-10 \left (\cos ^{3}\left (f x +e \right )\right )-24 \cos \left (f x +e \right ) \sin \left (f x +e \right )+35 \cos \left (f x +e \right )+37 \tan \left (f x +e \right )-25 \sec \left (f x +e \right )\right )}{15 f \left (\cos ^{4}\left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{5}}-\frac {B \sec \left (f x +e \right ) \left (\cos \left (f x +e \right )-1\right ) \left (1+\cos \left (f x +e \right )\right ) \left (\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-5 \left (\cos ^{2}\left (f x +e \right )\right )+4 \sin \left (f x +e \right )+20\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}{30 f \left (\cos ^{4}\left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{5}}\)	\(290\)

[In]

int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/30*a^2/c^5/f*tan(f*x+e)*(4*A*cos(f*x+e)^4+B*sin(f*x+e)^2*cos(f*x+e)^2+20*A*sin(f*x+e)*cos(f*x+e)^2+5*B*sin(f
*x+e)^3-48*A*cos(f*x+e)^2+4*B*sin(f*x+e)^2-50*A*sin(f*x+e)+15*B*sin(f*x+e)+74*A)*(a*(1+sin(f*x+e)))^(1/2)/(cos
(f*x+e)^4+4*cos(f*x+e)^2*sin(f*x+e)-8*cos(f*x+e)^2-8*sin(f*x+e)+8)/(-c*(sin(f*x+e)-1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.25 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {{\left (5 \, {\left (2 \, A + B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (7 \, A + 2 \, B\right )} a^{2} + 5 \, {\left (3 \, B a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + 2 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{30 \, {\left (5 \, c^{6} f \cos \left (f x + e\right )^{5} - 20 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right ) - {\left (c^{6} f \cos \left (f x + e\right )^{5} - 12 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

-1/30*(5*(2*A + B)*a^2*cos(f*x + e)^2 - 2*(7*A + 2*B)*a^2 + 5*(3*B*a^2*cos(f*x + e)^2 - 2*(A + 2*B)*a^2)*sin(f
*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(5*c^6*f*cos(f*x + e)^5 - 20*c^6*f*cos(f*x + e)^3
+ 16*c^6*f*cos(f*x + e) - (c^6*f*cos(f*x + e)^5 - 12*c^6*f*cos(f*x + e)^3 + 16*c^6*f*cos(f*x + e))*sin(f*x + e
))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(11/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {11}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(5/2)/(-c*sin(f*x + e) + c)^(11/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (128) = 256\).

Time = 0.44 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.84 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {{\left (30 \, B a^{2} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 10 \, A a^{2} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 40 \, B a^{2} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, A a^{2} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 20 \, B a^{2} \sqrt {c} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + A a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 4 \, B a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{240 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{5} c^{6} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(11/2),x, algorithm="giac")

[Out]

1/240*(30*B*a^2*sqrt(c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 10*A*a^2*sqrt(c
)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 40*B*a^2*sqrt(c)*cos(-1/4*pi + 1/2*f*
x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 5*A*a^2*sqrt(c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-1
/4*pi + 1/2*f*x + 1/2*e)) + 20*B*a^2*sqrt(c)*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*
e)) + A*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 4*B*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))
*sqrt(a)/((cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^5*c^6*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))

Mupad [B] (verification not implemented)

Time = 21.07 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.34 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {16\,a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (A\,6{}\mathrm {i}+B\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{5\,c^6\,f}-\frac {16\,a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\left (A\,2{}\mathrm {i}+B\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^6\,f}+\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\left (8\,A+13\,B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,8{}\mathrm {i}}{3\,c^6\,f}-\frac {B\,a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,8{}\mathrm {i}}{c^6\,f}\right )}{\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,264{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )\,220{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )\,20{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )\,330{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )\,88{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (6\,e+6\,f\,x\right )\,2{}\mathrm {i}} \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2))/(c - c*sin(e + f*x))^(11/2),x)

[Out]

((c - c*sin(e + f*x))^(1/2)*((16*a^2*exp(e*6i + f*x*6i)*(A*6i + B*1i)*(a + a*sin(e + f*x))^(1/2))/(5*c^6*f) -
(16*a^2*exp(e*6i + f*x*6i)*cos(2*e + 2*f*x)*(A*2i + B*1i)*(a + a*sin(e + f*x))^(1/2))/(3*c^6*f) + (a^2*exp(e*6
i + f*x*6i)*sin(e + f*x)*(8*A + 13*B)*(a + a*sin(e + f*x))^(1/2)*8i)/(3*c^6*f) - (B*a^2*exp(e*6i + f*x*6i)*sin
(3*e + 3*f*x)*(a + a*sin(e + f*x))^(1/2)*8i)/(c^6*f)))/(cos(e + f*x)*exp(e*6i + f*x*6i)*264i - exp(e*6i + f*x*
6i)*cos(3*e + 3*f*x)*220i + exp(e*6i + f*x*6i)*cos(5*e + 5*f*x)*20i - exp(e*6i + f*x*6i)*sin(2*e + 2*f*x)*330i
+ exp(e*6i + f*x*6i)*sin(4*e + 4*f*x)*88i - exp(e*6i + f*x*6i)*sin(6*e + 6*f*x)*2i)