[next] [prev] [prev-tail] [tail] [up]

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

Optimal. Leaf size=111 \[ \frac {f \sin (c+d x)}{a d^2}+\frac {2 f \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )\right )}{a d^2}-\frac {(e+f x) \cos (c+d x)}{a d}-\frac {(e+f x) \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{a d}-\frac {e x}{a}-\frac {f x^2}{2 a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4515, 3296, 2637, 3318, 4184, 3475} \[ \frac {f \sin (c+d x)}{a d^2}+\frac {2 f \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )\right )}{a d^2}-\frac {(e+f x) \cos (c+d x)}{a d}-\frac {(e+f x) \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{a d}-\frac {e x}{a}-\frac {f x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e*x)/a) - (f*x^2)/(2*a) - ((e + f*x)*Cos[c + d*x])/(a*d) - ((e + f*x)*Cot[c/2 + Pi/4 + (d*x)/2])/(a*d) + (2
*f*Log[Sin[c/2 + Pi/4 + (d*x)/2]])/(a*d^2) + (f*Sin[c + d*x])/(a*d^2)

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3318

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4515

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> Dist[1/b, Int[(e + f*x)^m*Sin[c + d*x]^(n - 1), x], x] - Dist[a/b, Int[((e + f*x)^m*Sin[c + d*x]^(n - 1)
)/(a + b*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(e+f x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int (e+f x) \sin (c+d x) \, dx}{a}-\int \frac {(e+f x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx\\ &=-\frac {(e+f x) \cos (c+d x)}{a d}-\frac {\int (e+f x) \, dx}{a}+\frac {f \int \cos (c+d x) \, dx}{a d}+\int \frac {e+f x}{a+a \sin (c+d x)} \, dx\\ &=-\frac {e x}{a}-\frac {f x^2}{2 a}-\frac {(e+f x) \cos (c+d x)}{a d}+\frac {f \sin (c+d x)}{a d^2}+\frac {\int (e+f x) \csc ^2\left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {d x}{2}\right ) \, dx}{2 a}\\ &=-\frac {e x}{a}-\frac {f x^2}{2 a}-\frac {(e+f x) \cos (c+d x)}{a d}-\frac {(e+f x) \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {f \sin (c+d x)}{a d^2}+\frac {f \int \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \, dx}{a d}\\ &=-\frac {e x}{a}-\frac {f x^2}{2 a}-\frac {(e+f x) \cos (c+d x)}{a d}-\frac {(e+f x) \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {2 f \log \left (\sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )\right )}{a d^2}+\frac {f \sin (c+d x)}{a d^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.83, size = 236, normalized size = 2.13 \[ -\frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right ) \left (c^2 (-f)+2 d (e+f x) \cos (c+d x)+2 c d e-2 f \sin (c+d x)-4 f \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 c f+2 d^2 e x+d^2 f x^2-4 d e-2 d f x\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \left (c^2 (-f)+2 d (e+f x) \cos (c+d x)+2 c d e-2 f \sin (c+d x)-4 f \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 c f+2 d^2 e x+d^2 f x^2+2 d f x\right )\right )}{2 a d^2 (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(Sin[(c + d*x)/2]*(-4*d*e + 2*c*d*e + 2*c*f - c^2*f + 2*d^2*e*x -
2*d*f*x + d^2*f*x^2 + 2*d*(e + f*x)*Cos[c + d*x] - 4*f*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 2*f*Sin[c +
d*x]) + Cos[(c + d*x)/2]*(2*c*d*e + 2*c*f - c^2*f + 2*d^2*e*x + 2*d*f*x + d^2*f*x^2 + 2*d*(e + f*x)*Cos[c + d*
x] - 4*f*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 2*f*Sin[c + d*x])))/(a*d^2*(1 + Sin[c + d*x]))

________________________________________________________________________________________

fricas [B]  time = 0.48, size = 196, normalized size = 1.77 \[ -\frac {d^{2} f x^{2} + 2 \, {\left (d f x + d e + f\right )} \cos \left (d x + c\right )^{2} + 2 \, d e + 2 \, {\left (d^{2} e + d f\right )} x + {\left (d^{2} f x^{2} + 4 \, d e + 2 \, {\left (d^{2} e + 2 \, d f\right )} x\right )} \cos \left (d x + c\right ) - 2 \, {\left (f \cos \left (d x + c\right ) + f \sin \left (d x + c\right ) + f\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (d^{2} f x^{2} - 2 \, d e + 2 \, {\left (d^{2} e - d f\right )} x + 2 \, {\left (d f x + d e - f\right )} \cos \left (d x + c\right ) - 2 \, f\right )} \sin \left (d x + c\right ) - 2 \, f}{2 \, {\left (a d^{2} \cos \left (d x + c\right ) + a d^{2} \sin \left (d x + c\right ) + a d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.24, size = 216, normalized size = 1.95 \[ -\frac {2 e}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {x f}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) d}+\frac {x f \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) d}+\frac {2 f \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a \,d^{2}}-\frac {f \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,d^{2}}-\frac {2 e}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {2 e \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {f \sin \left (d x +c \right )}{a \,d^{2}}-\frac {f \cos \left (d x +c \right ) x}{a d}-\frac {f \,x^{2}}{2 a}+\frac {f \,c^{2}}{2 a \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/a*e/d/(tan(1/2*d*x+1/2*c)+1)-1/a/(tan(1/2*d*x+1/2*c)+1)/d*x*f+1/a/(tan(1/2*d*x+1/2*c)+1)/d*x*f*tan(1/2*d*x+
1/2*c)+2/a*f/d^2*ln(tan(1/2*d*x+1/2*c)+1)-1/a*f/d^2*ln(1+tan(1/2*d*x+1/2*c)^2)-2/a*e/d/(1+tan(1/2*d*x+1/2*c)^2
)-2/a*e/d*arctan(tan(1/2*d*x+1/2*c))+f*sin(d*x+c)/a/d^2-1/a*f/d*cos(d*x+c)*x-1/2*f*x^2/a+1/2/a*f/d^2*c^2

________________________________________________________________________________________

maxima [B]  time = 1.12, size = 1762, normalized size = 15.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*c*f*((sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2)/(a*d + a*d*sin(d*x + c
)/(cos(d*x + c) + 1) + a*d*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*d*sin(d*x + c)^3/(cos(d*x + c) + 1)^3) + ar
ctan(sin(d*x + c)/(cos(d*x + c) + 1))/(a*d)) - 4*e*((sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x
 + c) + 1)^2 + 2)/(a + a*sin(d*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x +
 c)^3/(cos(d*x + c) + 1)^3) + arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a) - (((d*x + c)^2 - 1)*cos(d*x + c)^4 +
 ((d*x + c)^2 - 1)*sin(d*x + c)^4 + ((d*x + c)*cos(d*x + c) + sin(d*x + c) + 1)*cos(2*d*x + 2*c)^3 + 7*(d*x +
c)*cos(d*x + c)^3 + (d*x + (d*x + c)*sin(d*x + c) + c - cos(d*x + c))*sin(2*d*x + 2*c)^3 + (2*(d*x + c)^2 - 3)
*sin(d*x + c)^3 + (((d*x + c)^2 - 1)*cos(d*x + c)^2 + ((d*x + c)^2 - 3)*sin(d*x + c)^2 + (d*x + c)^2 + 6*(d*x
+ c)*cos(d*x + c) + 2*((d*x + c)^2 - (d*x + c)*cos(d*x + c) - 2)*sin(d*x + c) - 1)*cos(2*d*x + 2*c)^2 + ((d*x
+ c)^2 - 1)*cos(d*x + c)^2 + (((d*x + c)^2 - 3)*cos(d*x + c)^2 + ((d*x + c)^2 - 1)*sin(d*x + c)^2 + (d*x + c)^
2 + ((d*x + c)*cos(d*x + c) + sin(d*x + c) + 1)*cos(2*d*x + 2*c) + 8*(d*x + c)*cos(d*x + c) + 2*((d*x + c)^2 +
 (d*x + c)*cos(d*x + c) - 1)*sin(d*x + c) - 1)*sin(2*d*x + 2*c)^2 + (2*((d*x + c)^2 - 1)*cos(d*x + c)^2 + (d*x
 + c)^2 + 7*(d*x + c)*cos(d*x + c) - 3)*sin(d*x + c)^2 + ((d*x + c)*cos(d*x + c)^3 - (2*(d*x + c)^2 - 3)*sin(d
*x + c)^3 - (4*(d*x + c)^2 - (d*x + c)*cos(d*x + c) - 6)*sin(d*x + c)^2 + 2*cos(d*x + c)^2 - ((2*(d*x + c)^2 -
 3)*cos(d*x + c)^2 + 2*(d*x + c)^2 + 12*(d*x + c)*cos(d*x + c) - 4)*sin(d*x + c) + 1)*cos(2*d*x + 2*c) + (d*x
+ c)*cos(d*x + c) - 2*(cos(d*x + c)^4 + sin(d*x + c)^4 + (cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1
)*cos(2*d*x + 2*c)^2 + (cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1)*sin(2*d*x + 2*c)^2 + 2*cos(d*x +
 c)^2*sin(d*x + c) + (2*cos(d*x + c)^2 + 1)*sin(d*x + c)^2 + 2*sin(d*x + c)^3 - 2*(sin(d*x + c)^3 + (cos(d*x +
 c)^2 + 1)*sin(d*x + c) + 2*sin(d*x + c)^2)*cos(2*d*x + 2*c) + cos(d*x + c)^2 + 2*(cos(d*x + c)^3 + cos(d*x +
c)*sin(d*x + c)^2 + 2*cos(d*x + c)*sin(d*x + c) + cos(d*x + c))*sin(2*d*x + 2*c))*log(cos(d*x + c)^2 + sin(d*x
 + c)^2 + 2*sin(d*x + c) + 1) + ((2*(d*x + c)^2 - 3)*cos(d*x + c)^3 + (d*x + c)*sin(d*x + c)^3 + (d*x + (d*x +
 c)*sin(d*x + c) + c - cos(d*x + c))*cos(2*d*x + 2*c)^2 + 14*(d*x + c)*cos(d*x + c)^2 + (2*d*x + (2*(d*x + c)^
2 - 3)*cos(d*x + c) + 2*c)*sin(d*x + c)^2 + d*x + 2*((d*x + c)*cos(d*x + c)^2 - (d*x + c)*sin(d*x + c)^2 - (d*
x + c - 2*cos(d*x + c))*sin(d*x + c) + cos(d*x + c))*cos(2*d*x + 2*c) + 2*((d*x + c)^2 - 1)*cos(d*x + c) + ((d
*x + c)*cos(d*x + c)^2 + 2*d*x + 4*((d*x + c)^2 - 1)*cos(d*x + c) + 2*c)*sin(d*x + c) + c)*sin(2*d*x + 2*c) +
((2*(d*x + c)^2 - 3)*cos(d*x + c)^2 + 2*(d*x + c)*cos(d*x + c) - 1)*sin(d*x + c))*f/(a*d*cos(d*x + c)^4 + a*d*
sin(d*x + c)^4 + 2*a*d*cos(d*x + c)^2*sin(d*x + c) + 2*a*d*sin(d*x + c)^3 + a*d*cos(d*x + c)^2 + (a*d*cos(d*x
+ c)^2 + a*d*sin(d*x + c)^2 + 2*a*d*sin(d*x + c) + a*d)*cos(2*d*x + 2*c)^2 + (a*d*cos(d*x + c)^2 + a*d*sin(d*x
 + c)^2 + 2*a*d*sin(d*x + c) + a*d)*sin(2*d*x + 2*c)^2 + (2*a*d*cos(d*x + c)^2 + a*d)*sin(d*x + c)^2 - 2*(a*d*
sin(d*x + c)^3 + 2*a*d*sin(d*x + c)^2 + (a*d*cos(d*x + c)^2 + a*d)*sin(d*x + c))*cos(2*d*x + 2*c) + 2*(a*d*cos
(d*x + c)^3 + a*d*cos(d*x + c)*sin(d*x + c)^2 + 2*a*d*cos(d*x + c)*sin(d*x + c) + a*d*cos(d*x + c))*sin(2*d*x
+ 2*c)))/d

________________________________________________________________________________________

mupad [B]  time = 1.77, size = 164, normalized size = 1.48 \[ -{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {d\,e+f\,1{}\mathrm {i}}{2\,a\,d^2}+\frac {f\,x}{2\,a\,d}\right )+{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\left (\frac {-d\,e+f\,1{}\mathrm {i}}{2\,a\,d^2}-\frac {f\,x}{2\,a\,d}\right )-\frac {f\,x^2}{2\,a}+\frac {2\,f\,\ln \left ({\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}{a\,d^2}-\frac {x\,\left (d\,e+f\,2{}\mathrm {i}\right )}{a\,d}-\frac {\left (e+f\,x\right )\,2{}\mathrm {i}}{a\,d\,\left (-1+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(- c*1i - d*x*1i)*((f*1i - d*e)/(2*a*d^2) - (f*x)/(2*a*d)) - exp(c*1i + d*x*1i)*((f*1i + d*e)/(2*a*d^2) + (
f*x)/(2*a*d)) - (f*x^2)/(2*a) + (2*f*log(exp(c*1i)*exp(d*x*1i) + 1i))/(a*d^2) - (x*(f*2i + d*e))/(a*d) - ((e +
 f*x)*2i)/(a*d*(exp(c*1i + d*x*1i)*1i - 1))

________________________________________________________________________________________

sympy [A]  time = 3.94, size = 1867, normalized size = 16.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*d**2*e*x*tan(c/2 + d*x/2)**3/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*
d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 2*d**2*e*x*tan(c/2 + d*x/2)**2/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*ta
n(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 2*d**2*e*x*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*
x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 2*d**2*e*x/(2*a*d**2*tan(c/2
+ d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - d**2*f*x**2*tan(c/2 + d*x
/2)**3/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) -
d**2*f*x**2*tan(c/2 + d*x/2)**2/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/
2 + d*x/2) + 2*a*d**2) - d**2*f*x**2*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2
)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - d**2*f*x**2/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 +
d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 4*d*e*tan(c/2 + d*x/2)**2/(2*a*d**2*tan(c/2 + d*x/2)**3 +
2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 4*d*e*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/
2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 8*d*e/(2*a*d**2*tan(c/2
 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + 4*d*f*x*tan(c/2 + d*x/2)
**3/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 4*d
*f*x/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + 4*
f*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**3/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 +
 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + 4*f*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(2*a*d**2*tan(c/2 +
 d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + 4*f*log(tan(c/2 + d*x/2) +
 1)*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2)
+ 2*a*d**2) + 4*f*log(tan(c/2 + d*x/2) + 1)/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a
*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 2*f*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**3/(2*a*d**2*tan(c/2 +
d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 2*f*log(tan(c/2 + d*x/2)**2
 + 1)*tan(c/2 + d*x/2)**2/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*
x/2) + 2*a*d**2) - 2*f*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*
tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) - 2*f*log(tan(c/2 + d*x/2)**2 + 1)/(2*a*d**2*tan(c
/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + 4*f*tan(c/2 + d*x/2)**
2/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a*d**2) + 4*f*t
an(c/2 + d*x/2)/(2*a*d**2*tan(c/2 + d*x/2)**3 + 2*a*d**2*tan(c/2 + d*x/2)**2 + 2*a*d**2*tan(c/2 + d*x/2) + 2*a
*d**2), Ne(d, 0)), ((e*x + f*x**2/2)*sin(c)**2/(a*sin(c) + a), True))

________________________________________________________________________________________

[next] [prev] [prev-tail] [front] [up]