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

Optimal. Leaf size=188 \[ -\frac{4 i f^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}+\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{2 f (e+f x) \sin (c+d x)}{a d^2}+\frac{2 f^2 \cos (c+d x)}{a d^3}-\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{i (e+f x)^2}{a d}-\frac{(e+f x)^3}{3 a f} \]

[Out]

((-I)*(e + f*x)^2)/(a*d) - (e + f*x)^3/(3*a*f) + (2*f^2*Cos[c + d*x])/(a*d^3) - ((e + f*x)^2*Cos[c + d*x])/(a*
d) - ((e + f*x)^2*Cot[c/2 + Pi/4 + (d*x)/2])/(a*d) + (4*f*(e + f*x)*Log[1 - I*E^(I*(c + d*x))])/(a*d^2) - ((4*
I)*f^2*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^3) + (2*f*(e + f*x)*Sin[c + d*x])/(a*d^2)

________________________________________________________________________________________

Rubi [A]  time = 0.347819, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4515, 3296, 2638, 32, 3318, 4184, 3717, 2190, 2279, 2391} \[ -\frac{4 i f^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}+\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{2 f (e+f x) \sin (c+d x)}{a d^2}+\frac{2 f^2 \cos (c+d x)}{a d^3}-\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{i (e+f x)^2}{a d}-\frac{(e+f x)^3}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(e + f*x)^2)/(a*d) - (e + f*x)^3/(3*a*f) + (2*f^2*Cos[c + d*x])/(a*d^3) - ((e + f*x)^2*Cos[c + d*x])/(a*
d) - ((e + f*x)^2*Cot[c/2 + Pi/4 + (d*x)/2])/(a*d) + (4*f*(e + f*x)*Log[1 - I*E^(I*(c + d*x))])/(a*d^2) - ((4*
I)*f^2*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^3) + (2*f*(e + f*x)*Sin[c + d*x])/(a*d^2)

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]

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 2638

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{(e+f x)^2 \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int (e+f x)^2 \sin (c+d x) \, dx}{a}-\int \frac{(e+f x)^2 \sin (c+d x)}{a+a \sin (c+d x)} \, dx\\ &=-\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{\int (e+f x)^2 \, dx}{a}+\frac{(2 f) \int (e+f x) \cos (c+d x) \, dx}{a d}+\int \frac{(e+f x)^2}{a+a \sin (c+d x)} \, dx\\ &=-\frac{(e+f x)^3}{3 a f}-\frac{(e+f x)^2 \cos (c+d x)}{a d}+\frac{2 f (e+f x) \sin (c+d x)}{a d^2}+\frac{\int (e+f x)^2 \csc ^2\left (\frac{1}{2} \left (c+\frac{\pi }{2}\right )+\frac{d x}{2}\right ) \, dx}{2 a}-\frac{\left (2 f^2\right ) \int \sin (c+d x) \, dx}{a d^2}\\ &=-\frac{(e+f x)^3}{3 a f}+\frac{2 f^2 \cos (c+d x)}{a d^3}-\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{2 f (e+f x) \sin (c+d x)}{a d^2}+\frac{(2 f) \int (e+f x) \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right ) \, dx}{a d}\\ &=-\frac{i (e+f x)^2}{a d}-\frac{(e+f x)^3}{3 a f}+\frac{2 f^2 \cos (c+d x)}{a d^3}-\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{2 f (e+f x) \sin (c+d x)}{a d^2}+\frac{(4 f) \int \frac{e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )} (e+f x)}{1-i e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}} \, dx}{a d}\\ &=-\frac{i (e+f x)^2}{a d}-\frac{(e+f x)^3}{3 a f}+\frac{2 f^2 \cos (c+d x)}{a d^3}-\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{2 f (e+f x) \sin (c+d x)}{a d^2}-\frac{\left (4 f^2\right ) \int \log \left (1-i e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}\right ) \, dx}{a d^2}\\ &=-\frac{i (e+f x)^2}{a d}-\frac{(e+f x)^3}{3 a f}+\frac{2 f^2 \cos (c+d x)}{a d^3}-\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{2 f (e+f x) \sin (c+d x)}{a d^2}+\frac{\left (4 i f^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}\right )}{a d^3}\\ &=-\frac{i (e+f x)^2}{a d}-\frac{(e+f x)^3}{3 a f}+\frac{2 f^2 \cos (c+d x)}{a d^3}-\frac{(e+f x)^2 \cos (c+d x)}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}-\frac{4 i f^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}+\frac{2 f (e+f x) \sin (c+d x)}{a d^2}\\ \end{align*}

Mathematica [A]  time = 2.56786, size = 295, normalized size = 1.57 \[ -\frac{\frac{12 f (\cos (c)+i \sin (c)) \left (\frac{f (\cos (c)-i (\sin (c)+1)) \text{PolyLog}(2,-\sin (c+d x)-i \cos (c+d x))}{d^2}-\frac{(\sin (c)+i \cos (c)+1) (e+f x) \log (\sin (c+d x)+i \cos (c+d x)+1)}{d}+\frac{(\cos (c)-i \sin (c)) (e+f x)^2}{2 f}\right )}{d (\cos (c)+i (\sin (c)+1))}+\frac{3 \cos (d x) \left (\cos (c) \left (d^2 (e+f x)^2-2 f^2\right )-2 d f \sin (c) (e+f x)\right )}{d^3}-\frac{3 \sin (d x) \left (\sin (c) \left (d^2 (e+f x)^2-2 f^2\right )+2 d f \cos (c) (e+f x)\right )}{d^3}-\frac{6 \sin \left (\frac{d x}{2}\right ) (e+f x)^2}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*(3*e^2 + 3*e*f*x + f^2*x^2) + (3*Cos[d*x]*((-2*f^2 + d^2*(e + f*x)^2)*Cos[c] - 2*d*f*(e + f*x)*Sin[c]))/d^
3 + (12*f*(Cos[c] + I*Sin[c])*(((e + f*x)^2*(Cos[c] - I*Sin[c]))/(2*f) - ((e + f*x)*Log[1 + I*Cos[c + d*x] + S
in[c + d*x]]*(1 + I*Cos[c] + Sin[c]))/d + (f*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*(1 + Sin
[c])))/d^2))/(d*(Cos[c] + I*(1 + Sin[c]))) - (3*(2*d*f*(e + f*x)*Cos[c] + (-2*f^2 + d^2*(e + f*x)^2)*Sin[c])*S
in[d*x])/d^3 - (6*(e + f*x)^2*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(
3*a)

________________________________________________________________________________________

Maple [B]  time = 0.336, size = 408, normalized size = 2.2 \begin{align*} -{\frac{{f}^{2}{x}^{3}}{3\,a}}-{\frac{fe{x}^{2}}{a}}-{\frac{{e}^{2}x}{a}}-{\frac{ \left ({f}^{2}{x}^{2}{d}^{2}+2\,id{f}^{2}x+2\,{d}^{2}efx+2\,idef+{d}^{2}{e}^{2}-2\,{f}^{2} \right ){{\rm e}^{i \left ( dx+c \right ) }}}{2\,{d}^{3}a}}-{\frac{ \left ({f}^{2}{x}^{2}{d}^{2}-2\,id{f}^{2}x+2\,{d}^{2}efx-2\,idef+{d}^{2}{e}^{2}-2\,{f}^{2} \right ){{\rm e}^{-i \left ( dx+c \right ) }}}{2\,{d}^{3}a}}-2\,{\frac{{f}^{2}{x}^{2}+2\,fex+{e}^{2}}{da \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \right ) }}+4\,{\frac{\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \right ) ef}{a{d}^{2}}}-4\,{\frac{\ln \left ({{\rm e}^{i \left ( dx+c \right ) }} \right ) ef}{a{d}^{2}}}-{\frac{4\,i{f}^{2}cx}{a{d}^{2}}}-{\frac{2\,i{f}^{2}{x}^{2}}{da}}-{\frac{4\,i{f}^{2}{\it polylog} \left ( 2,i{{\rm e}^{i \left ( dx+c \right ) }} \right ) }{{d}^{3}a}}+4\,{\frac{{f}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( dx+c \right ) }} \right ) x}{a{d}^{2}}}+4\,{\frac{{f}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( dx+c \right ) }} \right ) c}{{d}^{3}a}}-{\frac{2\,i{f}^{2}{c}^{2}}{{d}^{3}a}}-4\,{\frac{c{f}^{2}\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \right ) }{{d}^{3}a}}+4\,{\frac{c{f}^{2}\ln \left ({{\rm e}^{i \left ( dx+c \right ) }} \right ) }{{d}^{3}a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/a*f^2*x^3-1/a*f*e*x^2-1/a*e^2*x-1/2*(f^2*x^2*d^2+2*I*d*f^2*x+2*d^2*e*f*x+2*I*d*e*f+d^2*e^2-2*f^2)/d^3/a*e
xp(I*(d*x+c))-1/2*(f^2*x^2*d^2-2*I*d*f^2*x+2*d^2*e*f*x-2*I*d*e*f+d^2*e^2-2*f^2)/d^3/a*exp(-I*(d*x+c))-2*(f^2*x
^2+2*e*f*x+e^2)/d/a/(exp(I*(d*x+c))+I)+4*f/d^2/a*ln(exp(I*(d*x+c))+I)*e-4*f/d^2/a*ln(exp(I*(d*x+c)))*e-4*I/d^2
/a*f^2*c*x-2*I/d/a*f^2*x^2-4*I*f^2*polylog(2,I*exp(I*(d*x+c)))/a/d^3+4*f^2/d^2/a*ln(1-I*exp(I*(d*x+c)))*x+4*f^
2/d^3/a*ln(1-I*exp(I*(d*x+c)))*c-2*I/d^3/a*f^2*c^2-4*f^2/d^3/a*c*ln(exp(I*(d*x+c))+I)+4*f^2/d^3/a*c*ln(exp(I*(
d*x+c)))

________________________________________________________________________________________

Maxima [B]  time = 2.45847, size = 815, normalized size = 4.34 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*d^3*f^2*x^3 - 15*I*d^2*e^2 - 6*d*e*f + 3*(2*d^3*e*f - I*d^2*f^2)*x^2 + 6*I*f^2 + (6*d^3*e^2 - 6*I*d^2*e*f
- 6*d*f^2)*x - (24*d*e*f*cos(d*x + c) + 24*I*d*e*f*sin(d*x + c) + 24*I*d*e*f)*arctan2(sin(d*x + c) + 1, cos(d*
x + c)) + (24*d*f^2*x*cos(d*x + c) + 24*I*d*f^2*x*sin(d*x + c) + 24*I*d*f^2*x)*arctan2(cos(d*x + c), sin(d*x +
c) + 1) - (3*I*d^2*f^2*x^2 + 3*I*d^2*e^2 - 6*d*e*f - 6*I*f^2 + (6*I*d^2*e*f - 6*d*f^2)*x)*cos(2*d*x + 2*c) -
(2*I*d^3*f^2*x^3 - 3*d^2*e^2 - 6*I*d*e*f + (6*I*d^3*e*f - 15*d^2*f^2)*x^2 + 6*f^2 + (6*I*d^3*e^2 - 30*d^2*e*f
- 6*I*d*f^2)*x)*cos(d*x + c) + (24*f^2*cos(d*x + c) + 24*I*f^2*sin(d*x + c) + 24*I*f^2)*dilog(I*e^(I*d*x + I*c
)) - (12*d*f^2*x + 12*d*e*f + (-12*I*d*f^2*x - 12*I*d*e*f)*cos(d*x + c) + 12*(d*f^2*x + d*e*f)*sin(d*x + c))*l
og(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + (3*d^2*f^2*x^2 + 3*d^2*e^2 + 6*I*d*e*f - 6*f^2 + 6*
(d^2*e*f + I*d*f^2)*x)*sin(2*d*x + 2*c) + (2*d^3*f^2*x^3 + 3*I*d^2*e^2 - 6*d*e*f + 3*(2*d^3*e*f + 5*I*d^2*f^2)
*x^2 - 6*I*f^2 + (6*d^3*e^2 + 30*I*d^2*e*f - 6*d*f^2)*x)*sin(d*x + c))/(-6*I*a*d^3*cos(d*x + c) + 6*a*d^3*sin(
d*x + c) + 6*a*d^3)

________________________________________________________________________________________

Fricas [B]  time = 2.18301, size = 1674, normalized size = 8.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(d^3*f^2*x^3 + 3*d^2*e^2 - 6*d*e*f + 3*(d^3*e*f + d^2*f^2)*x^2 + 3*(d^2*f^2*x^2 + d^2*e^2 + 2*d*e*f - 2*f
^2 + 2*(d^2*e*f + d*f^2)*x)*cos(d*x + c)^2 + 3*(d^3*e^2 + 2*d^2*e*f - 2*d*f^2)*x + (d^3*f^2*x^3 + 6*d^2*e^2 +
3*(d^3*e*f + 2*d^2*f^2)*x^2 - 6*f^2 + 3*(d^3*e^2 + 4*d^2*e*f)*x)*cos(d*x + c) - (-6*I*f^2*cos(d*x + c) - 6*I*f
^2*sin(d*x + c) - 6*I*f^2)*dilog(I*cos(d*x + c) - sin(d*x + c)) - (6*I*f^2*cos(d*x + c) + 6*I*f^2*sin(d*x + c)
+ 6*I*f^2)*dilog(-I*cos(d*x + c) - sin(d*x + c)) - 6*(d*e*f - c*f^2 + (d*e*f - c*f^2)*cos(d*x + c) + (d*e*f -
c*f^2)*sin(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) - 6*(d*f^2*x + c*f^2 + (d*f^2*x + c*f^2)*cos(d*x
+ c) + (d*f^2*x + c*f^2)*sin(d*x + c))*log(I*cos(d*x + c) + sin(d*x + c) + 1) - 6*(d*f^2*x + c*f^2 + (d*f^2*x
+ c*f^2)*cos(d*x + c) + (d*f^2*x + c*f^2)*sin(d*x + c))*log(-I*cos(d*x + c) + sin(d*x + c) + 1) - 6*(d*e*f - c
*f^2 + (d*e*f - c*f^2)*cos(d*x + c) + (d*e*f - c*f^2)*sin(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + I) +
(d^3*f^2*x^3 - 3*d^2*e^2 - 6*d*e*f + 3*(d^3*e*f - d^2*f^2)*x^2 + 3*(d^3*e^2 - 2*d^2*e*f - 2*d*f^2)*x + 3*(d^2*
f^2*x^2 + d^2*e^2 - 2*d*e*f - 2*f^2 + 2*(d^2*e*f - d*f^2)*x)*cos(d*x + c))*sin(d*x + c))/(a*d^3*cos(d*x + c) +
a*d^3*sin(d*x + c) + a*d^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{e^{2} \sin ^{2}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx + \int \frac{f^{2} x^{2} \sin ^{2}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx + \int \frac{2 e f x \sin ^{2}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{2} \sin \left (d x + c\right )^{2}}{a \sin \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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