3.474 \(\int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=125 \[ -\frac {\left (2 c^2+6 c d+7 d^2\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a \sin (e+f x)+a)^3}-\frac {(c-d) (2 c+5 d) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2} \]

[Out]

-1/15*(c-d)*(2*c+5*d)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^2-1/15*(2*c^2+6*c*d+7*d^2)*cos(f*x+e)/f/(a^3+a^3*sin(f*x
+e))-1/5*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))/f/(a+a*sin(f*x+e))^3

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Rubi [A]  time = 0.18, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2760, 2750, 2648} \[ -\frac {\left (2 c^2+6 c d+7 d^2\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a \sin (e+f x)+a)^3}-\frac {(c-d) (2 c+5 d) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Sin[e + f*x])^2/(a + a*Sin[e + f*x])^3,x]

[Out]

-((c - d)*(2*c + 5*d)*Cos[e + f*x])/(15*a*f*(a + a*Sin[e + f*x])^2) - ((2*c^2 + 6*c*d + 7*d^2)*Cos[e + f*x])/(
15*f*(a^3 + a^3*Sin[e + f*x])) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x]))/(5*f*(a + a*Sin[e + f*x])^3)

Rule 2648

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

Rule 2750

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

Rule 2760

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

Rubi steps

\begin {align*} \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx &=-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a+a \sin (e+f x))^3}-\frac {\int \frac {-a \left (2 c^2+4 c d-d^2\right )-a d (c+4 d) \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac {(c-d) (2 c+5 d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a+a \sin (e+f x))^3}+\frac {\left (2 c^2+6 c d+7 d^2\right ) \int \frac {1}{a+a \sin (e+f x)} \, dx}{15 a^2}\\ &=-\frac {(c-d) (2 c+5 d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2+6 c d+7 d^2\right ) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a+a \sin (e+f x))^3}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 84, normalized size = 0.67 \[ -\frac {\cos (e+f x) \left (\left (2 c^2+6 c d+7 d^2\right ) \sin ^2(e+f x)+6 \left (c^2+3 c d+d^2\right ) \sin (e+f x)+7 c^2+6 c d+2 d^2\right )}{15 a^3 f (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(Cos[e + f*x]*(7*c^2 + 6*c*d + 2*d^2 + 6*(c^2 + 3*c*d + d^2)*Sin[e + f*x] + (2*c^2 + 6*c*d + 7*d^2)*Sin[
e + f*x]^2))/(a^3*f*(1 + Sin[e + f*x])^3)

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fricas [B]  time = 0.44, size = 242, normalized size = 1.94 \[ -\frac {{\left (2 \, c^{2} + 6 \, c d + 7 \, d^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (4 \, c^{2} + 12 \, c d - d^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, c^{2} + 6 \, c d - 3 \, d^{2} - 3 \, {\left (3 \, c^{2} + 4 \, c d + 3 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left ({\left (2 \, c^{2} + 6 \, c d + 7 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, c^{2} + 6 \, c d - 3 \, d^{2} + 6 \, {\left (c^{2} + 3 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/15*((2*c^2 + 6*c*d + 7*d^2)*cos(f*x + e)^3 - (4*c^2 + 12*c*d - d^2)*cos(f*x + e)^2 - 3*c^2 + 6*c*d - 3*d^2
- 3*(3*c^2 + 4*c*d + 3*d^2)*cos(f*x + e) - ((2*c^2 + 6*c*d + 7*d^2)*cos(f*x + e)^2 - 3*c^2 + 6*c*d - 3*d^2 + 6
*(c^2 + 3*c*d + d^2)*cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(
f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))

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giac [A]  time = 2.21, size = 181, normalized size = 1.45 \[ -\frac {2 \, {\left (15 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 30 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 30 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 10 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, c^{2} + 6 \, c d + 2 \, d^{2}\right )}}{15 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-2/15*(15*c^2*tan(1/2*f*x + 1/2*e)^4 + 30*c^2*tan(1/2*f*x + 1/2*e)^3 + 30*c*d*tan(1/2*f*x + 1/2*e)^3 + 40*c^2*
tan(1/2*f*x + 1/2*e)^2 + 30*c*d*tan(1/2*f*x + 1/2*e)^2 + 20*d^2*tan(1/2*f*x + 1/2*e)^2 + 20*c^2*tan(1/2*f*x +
1/2*e) + 30*c*d*tan(1/2*f*x + 1/2*e) + 10*d^2*tan(1/2*f*x + 1/2*e) + 7*c^2 + 6*c*d + 2*d^2)/(a^3*f*(tan(1/2*f*
x + 1/2*e) + 1)^5)

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maple [A]  time = 0.24, size = 139, normalized size = 1.11 \[ \frac {-\frac {-8 c^{2}+16 c d -8 d^{2}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 c^{2}-8 c d +4 d^{2}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 \left (8 c^{2}-12 c d +4 d^{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 c^{2}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {4 c \left (c -d \right )}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x)

[Out]

2/f/a^3*(-1/4*(-8*c^2+16*c*d-8*d^2)/(tan(1/2*f*x+1/2*e)+1)^4-1/5*(4*c^2-8*c*d+4*d^2)/(tan(1/2*f*x+1/2*e)+1)^5-
1/3*(8*c^2-12*c*d+4*d^2)/(tan(1/2*f*x+1/2*e)+1)^3-c^2/(tan(1/2*f*x+1/2*e)+1)+2*c*(c-d)/(tan(1/2*f*x+1/2*e)+1)^
2)

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maxima [B]  time = 0.34, size = 553, normalized size = 4.42 \[ -\frac {2 \, {\left (\frac {c^{2} {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 7\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {2 \, d^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {6 \, c d {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-2/15*(c^2*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 2*d^2*(5*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) +
1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 6*c*d*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/
(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e
)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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mupad [B]  time = 7.49, size = 218, normalized size = 1.74 \[ \frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (6\,c\,d-4\,c^2\,\cos \left (e+f\,x\right )+d^2\,\cos \left (e+f\,x\right )+\frac {25\,c^2\,\sin \left (e+f\,x\right )}{2}+\frac {5\,d^2\,\sin \left (e+f\,x\right )}{2}+\frac {53\,c^2}{4}+\frac {13\,d^2}{4}-\frac {9\,c^2\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {9\,d^2\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {5\,c^2\,\sin \left (2\,e+2\,f\,x\right )}{4}+\frac {5\,d^2\,\sin \left (2\,e+2\,f\,x\right )}{4}+3\,c\,d\,\cos \left (e+f\,x\right )+15\,c\,d\,\sin \left (e+f\,x\right )-3\,c\,d\,\cos \left (2\,e+2\,f\,x\right )\right )}{15\,a^3\,f\,\left (\frac {5\,\sqrt {2}\,\cos \left (\frac {3\,e}{2}+\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{4}-\frac {5\,\sqrt {2}\,\cos \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f\,x}{2}\right )}{2}+\frac {\sqrt {2}\,\cos \left (\frac {5\,e}{2}-\frac {\pi }{4}+\frac {5\,f\,x}{2}\right )}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*sin(e + f*x))^2/(a + a*sin(e + f*x))^3,x)

[Out]

(2*cos(e/2 + (f*x)/2)*(6*c*d - 4*c^2*cos(e + f*x) + d^2*cos(e + f*x) + (25*c^2*sin(e + f*x))/2 + (5*d^2*sin(e
+ f*x))/2 + (53*c^2)/4 + (13*d^2)/4 - (9*c^2*cos(2*e + 2*f*x))/4 - (9*d^2*cos(2*e + 2*f*x))/4 - (5*c^2*sin(2*e
+ 2*f*x))/4 + (5*d^2*sin(2*e + 2*f*x))/4 + 3*c*d*cos(e + f*x) + 15*c*d*sin(e + f*x) - 3*c*d*cos(2*e + 2*f*x))
)/(15*a^3*f*((5*2^(1/2)*cos((3*e)/2 + pi/4 + (3*f*x)/2))/4 - (5*2^(1/2)*cos(e/2 - pi/4 + (f*x)/2))/2 + (2^(1/2
)*cos((5*e)/2 - pi/4 + (5*f*x)/2))/4))

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sympy [A]  time = 15.74, size = 1365, normalized size = 10.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**3,x)

[Out]

Piecewise((-30*c**2*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a
**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*c**2
*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x
/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 80*c**2*tan(e/2 + f*x/2)**
2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f
*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 40*c**2*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 +
f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 +
75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 14*c**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)
**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f
) - 60*c*d*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan
(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*c*d*tan(e/2 +
f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 1
50*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*c*d*tan(e/2 + f*x/2)/(15*a**3*f*t
an(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x
/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 12*c*d/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 +
f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15
*a**3*f) - 40*d**2*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a*
*3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 20*d**2*
tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)*
*3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 4*d**2/(15*a**3*f*tan(e/2 + f*
x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75
*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f), Ne(f, 0)), (x*(c + d*sin(e))**2/(a*sin(e) + a)**3, True))

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