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

Optimal. Leaf size=278 \[ \frac{4 i f^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{f (e+f x) \sin ^2(c+d x)}{2 a d^2}-\frac{2 f (e+f x) \sin (c+d x)}{a d^2}-\frac{2 f^2 \cos (c+d x)}{a d^3}+\frac{f^2 \sin (c+d x) \cos (c+d x)}{4 a d^3}+\frac{(e+f x)^2 \cos (c+d x)}{a d}+\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 a d}-\frac{f^2 x}{4 a d^2}+\frac{i (e+f x)^2}{a d}+\frac{(e+f x)^3}{2 a f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.492938, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 13, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.464, Rules used = {4515, 3311, 32, 2635, 8, 3296, 2638, 3318, 4184, 3717, 2190, 2279, 2391} \[ \frac{4 i f^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{f (e+f x) \sin ^2(c+d x)}{2 a d^2}-\frac{2 f (e+f x) \sin (c+d x)}{a d^2}-\frac{2 f^2 \cos (c+d x)}{a d^3}+\frac{f^2 \sin (c+d x) \cos (c+d x)}{4 a d^3}+\frac{(e+f x)^2 \cos (c+d x)}{a d}+\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 a d}-\frac{f^2 x}{4 a d^2}+\frac{i (e+f x)^2}{a d}+\frac{(e+f x)^3}{2 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(f^2*x)/(4*a*d^2) + (I*(e + f*x)^2)/(a*d) + (e + f*x)^3/(2*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) + (f^2*
Cos[c + d*x]*Sin[c + d*x])/(4*a*d^3) - ((e + f*x)^2*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) + (f*(e + f*x)*Sin[c +
d*x]^2)/(2*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 3311

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

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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2638

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

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

Mathematica [B]  time = 2.86717, size = 830, normalized size = 2.99 \[ -\frac{-8 f^2 x^3 \sin \left (\frac{1}{2} (c+d x)\right ) d^3-24 e f x^2 \sin \left (\frac{1}{2} (c+d x)\right ) d^3-24 e^2 x \sin \left (\frac{1}{2} (c+d x)\right ) d^3-6 e^2 \cos \left (\frac{3}{2} (c+d x)\right ) d^2-6 f^2 x^2 \cos \left (\frac{3}{2} (c+d x)\right ) d^2-12 e f x \cos \left (\frac{3}{2} (c+d x)\right ) d^2-2 e^2 \cos \left (\frac{5}{2} (c+d x)\right ) d^2-2 f^2 x^2 \cos \left (\frac{5}{2} (c+d x)\right ) d^2-4 e f x \cos \left (\frac{5}{2} (c+d x)\right ) d^2+(24+16 i) e^2 \sin \left (\frac{1}{2} (c+d x)\right ) d^2+(24+16 i) f^2 x^2 \sin \left (\frac{1}{2} (c+d x)\right ) d^2+(48+32 i) e f x \sin \left (\frac{1}{2} (c+d x)\right ) d^2-6 e^2 \sin \left (\frac{3}{2} (c+d x)\right ) d^2-6 f^2 x^2 \sin \left (\frac{3}{2} (c+d x)\right ) d^2-12 e f x \sin \left (\frac{3}{2} (c+d x)\right ) d^2+2 e^2 \sin \left (\frac{5}{2} (c+d x)\right ) d^2+2 f^2 x^2 \sin \left (\frac{5}{2} (c+d x)\right ) d^2+4 e f x \sin \left (\frac{5}{2} (c+d x)\right ) d^2-14 e f \cos \left (\frac{3}{2} (c+d x)\right ) d-14 f^2 x \cos \left (\frac{3}{2} (c+d x)\right ) d+2 e f \cos \left (\frac{5}{2} (c+d x)\right ) d+2 f^2 x \cos \left (\frac{5}{2} (c+d x)\right ) d+16 e f \sin \left (\frac{1}{2} (c+d x)\right ) d+16 f^2 x \sin \left (\frac{1}{2} (c+d x)\right ) d+64 e f \log (i \cos (c+d x)+\sin (c+d x)+1) \sin \left (\frac{1}{2} (c+d x)\right ) d+64 f^2 x \log (i \cos (c+d x)+\sin (c+d x)+1) \sin \left (\frac{1}{2} (c+d x)\right ) d+14 e f \sin \left (\frac{3}{2} (c+d x)\right ) d+14 f^2 x \sin \left (\frac{3}{2} (c+d x)\right ) d+2 e f \sin \left (\frac{5}{2} (c+d x)\right ) d+2 f^2 x \sin \left (\frac{5}{2} (c+d x)\right ) d+15 f^2 \cos \left (\frac{3}{2} (c+d x)\right )+f^2 \cos \left (\frac{5}{2} (c+d x)\right )-8 \cos \left (\frac{1}{2} (c+d x)\right ) \left (x \left (3 e^2+3 f x e+f^2 x^2\right ) d^3+(3-2 i) (e+f x)^2 d^2-2 f (e+f x) d-8 f (e+f x) \log (i \cos (c+d x)+\sin (c+d x)+1) d-2 f^2\right )-16 f^2 \sin \left (\frac{1}{2} (c+d x)\right )+64 i f^2 \text{PolyLog}(2,-i \cos (c+d x)-\sin (c+d x)) \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )+15 f^2 \sin \left (\frac{3}{2} (c+d x)\right )-f^2 \sin \left (\frac{5}{2} (c+d x)\right )}{16 a d^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-6*d^2*e^2*Cos[(3*(c + d*x))/2] - 14*d*e*f*Cos[(3*(c + d*x))/2] + 15*f^2*Cos[(3*(c + d*x))/2] - 12*d^2*e*f*x
*Cos[(3*(c + d*x))/2] - 14*d*f^2*x*Cos[(3*(c + d*x))/2] - 6*d^2*f^2*x^2*Cos[(3*(c + d*x))/2] - 2*d^2*e^2*Cos[(
5*(c + d*x))/2] + 2*d*e*f*Cos[(5*(c + d*x))/2] + f^2*Cos[(5*(c + d*x))/2] - 4*d^2*e*f*x*Cos[(5*(c + d*x))/2] +
2*d*f^2*x*Cos[(5*(c + d*x))/2] - 2*d^2*f^2*x^2*Cos[(5*(c + d*x))/2] - 8*Cos[(c + d*x)/2]*(-2*f^2 - 2*d*f*(e +
f*x) + (3 - 2*I)*d^2*(e + f*x)^2 + d^3*x*(3*e^2 + 3*e*f*x + f^2*x^2) - 8*d*f*(e + f*x)*Log[1 + I*Cos[c + d*x]
+ Sin[c + d*x]]) + (24 + 16*I)*d^2*e^2*Sin[(c + d*x)/2] + 16*d*e*f*Sin[(c + d*x)/2] - 16*f^2*Sin[(c + d*x)/2]
- 24*d^3*e^2*x*Sin[(c + d*x)/2] + (48 + 32*I)*d^2*e*f*x*Sin[(c + d*x)/2] + 16*d*f^2*x*Sin[(c + d*x)/2] - 24*d
^3*e*f*x^2*Sin[(c + d*x)/2] + (24 + 16*I)*d^2*f^2*x^2*Sin[(c + d*x)/2] - 8*d^3*f^2*x^3*Sin[(c + d*x)/2] + 64*d
*e*f*Log[1 + I*Cos[c + d*x] + Sin[c + d*x]]*Sin[(c + d*x)/2] + 64*d*f^2*x*Log[1 + I*Cos[c + d*x] + Sin[c + d*x
]]*Sin[(c + d*x)/2] + (64*I)*f^2*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2]) - 6*d^2*e^2*Sin[(3*(c + d*x))/2] + 14*d*e*f*Sin[(3*(c + d*x))/2] + 15*f^2*Sin[(3*(c + d*x))/2] - 12*d^2*
e*f*x*Sin[(3*(c + d*x))/2] + 14*d*f^2*x*Sin[(3*(c + d*x))/2] - 6*d^2*f^2*x^2*Sin[(3*(c + d*x))/2] + 2*d^2*e^2*
Sin[(5*(c + d*x))/2] + 2*d*e*f*Sin[(5*(c + d*x))/2] - f^2*Sin[(5*(c + d*x))/2] + 4*d^2*e*f*x*Sin[(5*(c + d*x))
/2] + 2*d*f^2*x*Sin[(5*(c + d*x))/2] + 2*d^2*f^2*x^2*Sin[(5*(c + d*x))/2])/(16*a*d^3*(Cos[(c + d*x)/2] + Sin[(
c + d*x)/2]))

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Maple [B]  time = 0.447, size = 538, normalized size = 1.9 \begin{align*}{\frac{{f}^{2}{x}^{3}}{2\,a}}+{\frac{3\,fe{x}^{2}}{2\,a}}+{\frac{3\,{e}^{2}x}{2\,a}}+{\frac{2\,i{f}^{2}{x}^{2}}{da}}+{\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}}+{\frac{4\,i{f}^{2}cx}{a{d}^{2}}}+2\,{\frac{{f}^{2}{x}^{2}+2\,fex+{e}^{2}}{da \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \right ) }}+4\,{\frac{f\ln \left ({{\rm e}^{i \left ( dx+c \right ) }} \right ) e}{a{d}^{2}}}-4\,{\frac{f\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \right ) e}{a{d}^{2}}}-{\frac{{\frac{i}{16}} \left ( 2\,{f}^{2}{x}^{2}{d}^{2}-2\,id{f}^{2}x+4\,{d}^{2}efx-2\,idef+2\,{d}^{2}{e}^{2}-{f}^{2} \right ){{\rm e}^{-2\,i \left ( dx+c \right ) }}}{{d}^{3}a}}+{\frac{{\frac{i}{16}} \left ( 2\,{f}^{2}{x}^{2}{d}^{2}+2\,id{f}^{2}x+4\,{d}^{2}efx+2\,idef+2\,{d}^{2}{e}^{2}-{f}^{2} \right ){{\rm e}^{2\,i \left ( dx+c \right ) }}}{{d}^{3}a}}+{\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 ) }} \right ) }{{d}^{3}a}}+4\,{\frac{c{f}^{2}\ln \left ({{\rm e}^{i \left ( dx+c \right ) }}+i \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)^3/(a+a*sin(d*x+c)),x)

[Out]

1/2/a*f^2*x^3+3/2/a*f*e*x^2+3/2/a*e^2*x+2*I*f^2/d/a*x^2+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))+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))+4*I*f^2/d^2/a*c*x+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)))*e-4*f
/d^2/a*ln(exp(I*(d*x+c))+I)*e-1/16*I*(2*f^2*x^2*d^2-2*I*d*f^2*x+4*d^2*e*f*x-2*I*d*e*f+2*d^2*e^2-f^2)/d^3/a*exp
(-2*I*(d*x+c))+1/16*I*(2*f^2*x^2*d^2+2*I*d*f^2*x+4*d^2*e*f*x+2*I*d*e*f+2*d^2*e^2-f^2)/d^3/a*exp(2*I*(d*x+c))+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*f^2/d^3/a*c^2-4*f^2/d^3/a*c*ln(exp(I*(d*x+c)))+4*f^2/d^3/a*c*ln(exp(I*(d*x+c))+I)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.3222, size = 1960, normalized size = 7.05 \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)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/4*(2*d^3*f^2*x^3 + 4*d^2*e^2 + (2*d^2*f^2*x^2 + 2*d^2*e^2 - 2*d*e*f - f^2 + 2*(2*d^2*e*f - d*f^2)*x)*cos(d*x
+ c)^3 - 7*d*e*f + 2*(3*d^3*e*f + 2*d^2*f^2)*x^2 + 2*(2*d^2*f^2*x^2 + 2*d^2*e^2 + 3*d*e*f - 4*f^2 + (4*d^2*e*
f + 3*d*f^2)*x)*cos(d*x + c)^2 + (6*d^3*e^2 + 8*d^2*e*f - 7*d*f^2)*x + (2*d^3*f^2*x^3 + 6*d^2*e^2 + d*e*f + 6*
(d^3*e*f + d^2*f^2)*x^2 - 7*f^2 + (6*d^3*e^2 + 12*d^2*e*f + d*f^2)*x)*cos(d*x + c) + (8*I*f^2*cos(d*x + c) + 8
*I*f^2*sin(d*x + c) + 8*I*f^2)*dilog(I*cos(d*x + c) - sin(d*x + c)) + (-8*I*f^2*cos(d*x + c) - 8*I*f^2*sin(d*x
+ c) - 8*I*f^2)*dilog(-I*cos(d*x + c) - sin(d*x + c)) - 8*(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) - 8*(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) - 8*(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) - 8*(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) + (2*d^3*f^2*x^3 - 4*d^2*e^2 - 7*d*e*f + 2*(3*d^3*e*f - 2*d^2*f^2)*x^2 - (2*d^2*f^2*x^2 + 2*d^2*e^2 + 2*d*e
*f - f^2 + 2*(2*d^2*e*f + d*f^2)*x)*cos(d*x + c)^2 + (6*d^3*e^2 - 8*d^2*e*f - 7*d*f^2)*x + (2*d^2*f^2*x^2 + 2*
d^2*e^2 - 8*d*e*f - 7*f^2 + 4*(d^2*e*f - 2*d*f^2)*x)*cos(d*x + c))*sin(d*x + c))/(a*d^3*cos(d*x + c) + a*d^3*s
in(d*x + c) + a*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{e^{2} \sin ^{3}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx + \int \frac{f^{2} x^{2} \sin ^{3}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx + \int \frac{2 e f x \sin ^{3}{\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)**3/(a+a*sin(d*x+c)),x)

[Out]

(Integral(e**2*sin(c + d*x)**3/(sin(c + d*x) + 1), x) + Integral(f**2*x**2*sin(c + d*x)**3/(sin(c + d*x) + 1),
x) + Integral(2*e*f*x*sin(c + d*x)**3/(sin(c + d*x) + 1), x))/a

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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 )^{3}}{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)^3/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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