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

Optimal. Leaf size=278 \[ \frac {4 i f^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}-\frac {2 f^2 \cos (c+d x)}{a d^3}+\frac {f^2 \sin (c+d x) \cos (c+d x)}{4 a d^3}-\frac {4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 a d^2}-\frac {2 f (e+f x) \sin (c+d x)}{a d^2}+\frac {(e+f x)^2 \cos (c+d x)}{a d}+\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{a d}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {f^2 x}{4 a d^2}+\frac {i (e+f x)^2}{a d}+\frac {(e+f x)^3}{2 a f} \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 278, normalized size of antiderivative = 1.00, 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 8

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

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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 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 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 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 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)^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*}

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Mathematica [B]  time = 3.11, 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 {Li}_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]

-1/16*(-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])/(a*d^3*(Cos[(c + d*x)/2] + Sin
[(c + d*x)/2]))

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fricas [B]  time = 0.57, size = 844, normalized size = 3.04 \[ \frac {2 \, d^{3} f^{2} x^{3} + 4 \, d^{2} e^{2} + {\left (2 \, d^{2} f^{2} x^{2} + 2 \, d^{2} e^{2} - 2 \, d e f - f^{2} + 2 \, {\left (2 \, d^{2} e f - d f^{2}\right )} x\right )} \cos \left (d x + c\right )^{3} - 7 \, d e f + 2 \, {\left (3 \, d^{3} e f + 2 \, d^{2} f^{2}\right )} x^{2} + 2 \, {\left (2 \, d^{2} f^{2} x^{2} + 2 \, d^{2} e^{2} + 3 \, d e f - 4 \, f^{2} + {\left (4 \, d^{2} e f + 3 \, d f^{2}\right )} x\right )} \cos \left (d x + c\right )^{2} + {\left (6 \, d^{3} e^{2} + 8 \, d^{2} e f - 7 \, d f^{2}\right )} x + {\left (2 \, d^{3} f^{2} x^{3} + 6 \, d^{2} e^{2} + d e f + 6 \, {\left (d^{3} e f + d^{2} f^{2}\right )} x^{2} - 7 \, f^{2} + {\left (6 \, d^{3} e^{2} + 12 \, d^{2} e f + d f^{2}\right )} x\right )} \cos \left (d x + c\right ) + {\left (8 i \, f^{2} \cos \left (d x + c\right ) + 8 i \, f^{2} \sin \left (d x + c\right ) + 8 i \, f^{2}\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + {\left (-8 i \, f^{2} \cos \left (d x + c\right ) - 8 i \, f^{2} \sin \left (d x + c\right ) - 8 i \, f^{2}\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - 8 \, {\left (d e f - c f^{2} + {\left (d e f - c f^{2}\right )} \cos \left (d x + c\right ) + {\left (d e f - c f^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) - 8 \, {\left (d f^{2} x + c f^{2} + {\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right ) + {\left (d f^{2} x + c f^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) - 8 \, {\left (d f^{2} x + c f^{2} + {\left (d f^{2} x + c f^{2}\right )} \cos \left (d x + c\right ) + {\left (d f^{2} x + c f^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) - 8 \, {\left (d e f - c f^{2} + {\left (d e f - c f^{2}\right )} \cos \left (d x + c\right ) + {\left (d e f - c f^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) + {\left (2 \, d^{3} f^{2} x^{3} - 4 \, d^{2} e^{2} - 7 \, d e f + 2 \, {\left (3 \, d^{3} e f - 2 \, d^{2} f^{2}\right )} x^{2} - {\left (2 \, d^{2} f^{2} x^{2} + 2 \, d^{2} e^{2} + 2 \, d e f - f^{2} + 2 \, {\left (2 \, d^{2} e f + d f^{2}\right )} x\right )} \cos \left (d x + c\right )^{2} + {\left (6 \, d^{3} e^{2} - 8 \, d^{2} e f - 7 \, d f^{2}\right )} x + {\left (2 \, d^{2} f^{2} x^{2} + 2 \, d^{2} e^{2} - 8 \, d e f - 7 \, f^{2} + 4 \, {\left (d^{2} e f - 2 \, d f^{2}\right )} x\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a d^{3} \cos \left (d x + c\right ) + a d^{3} \sin \left (d x + c\right ) + a d^{3}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{2} \sin \left (d x + c\right )^{3}}{a \sin \left (d x + c\right ) + a}\,{d x} \]

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)

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maple [A]  time = 0.71, size = 481, normalized size = 1.73 \[ \frac {f^{2} x^{3}}{2 a}+\frac {3 f e \,x^{2}}{2 a}+\frac {3 e^{2} x}{2 a}+\frac {\left (f^{2} x^{2} d^{2}+2 d^{2} e f x +2 i d \,f^{2} x +d^{2} e^{2}+2 i d e f -2 f^{2}\right ) {\mathrm e}^{i \left (d x +c \right )}}{2 d^{3} a}+\frac {\left (f^{2} x^{2} d^{2}+2 d^{2} e f x -2 i d \,f^{2} x +d^{2} e^{2}-2 i d e f -2 f^{2}\right ) {\mathrm e}^{-i \left (d x +c \right )}}{2 d^{3} a}+\frac {2 f^{2} x^{2}+4 f e x +2 e^{2}}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}-\frac {4 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e}{a \,d^{2}}+\frac {4 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right ) e}{a \,d^{2}}+\frac {4 i f^{2} c x}{a \,d^{2}}+\frac {4 i f^{2} \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {2 i f^{2} c^{2}}{a \,d^{3}}-\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}-\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{a \,d^{3}}+\frac {2 i f^{2} x^{2}}{a d}+\frac {4 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a \,d^{3}}-\frac {4 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}-\frac {f \left (f x +e \right ) \cos \left (2 d x +2 c \right )}{4 d^{2} a}-\frac {\left (2 f^{2} x^{2} d^{2}+4 d^{2} e f x +2 d^{2} e^{2}-f^{2}\right ) \sin \left (2 d x +2 c \right )}{8 d^{3} a} \]

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+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))+2*(f^
2*x^2+2*e*f*x+e^2)/d/a/(exp(I*(d*x+c))+I)-4/a/d^2*f*ln(exp(I*(d*x+c))+I)*e+4/a/d^2*f*ln(exp(I*(d*x+c)))*e+4*I/
a/d^2*f^2*c*x+2*I/a/d^3*f^2*c^2+4*I*f^2*polylog(2,I*exp(I*(d*x+c)))/a/d^3-4/a/d^2*f^2*ln(1-I*exp(I*(d*x+c)))*x
-4/a/d^3*f^2*ln(1-I*exp(I*(d*x+c)))*c+2*I/a/d*f^2*x^2+4/a/d^3*f^2*c*ln(exp(I*(d*x+c))+I)-4/a/d^3*f^2*c*ln(exp(
I*(d*x+c)))-1/4/d^2*f*(f*x+e)/a*cos(2*d*x+2*c)-1/8*(2*d^2*f^2*x^2+4*d^2*e*f*x+2*d^2*e^2-f^2)/d^3/a*sin(2*d*x+2
*c)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (c+d\,x\right )}^3\,{\left (e+f\,x\right )}^2}{a+a\,\sin \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {e^{2} \sin ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{2} x^{2} \sin ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {2 e f x \sin ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

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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