3.2.92 \(\int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [192]

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

[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 = 0.32, 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 = {4611, 3392, 32, 2715, 8, 3377, 2718, 3399, 4269, 3798, 2221, 2317, 2438} \begin {gather*} \frac {4 i f^2 \text {PolyLog}\left (2,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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(f^2*x)/(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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 2715

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] && Integ
erQ[2*n]

Rule 2718

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

Rule 3377

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 3392

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 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + 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 3798

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 4269

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 4611

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 ) \text {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] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(931\) vs. \(2(278)=556\).
time = 1.52, size = 931, normalized size = 3.35 \begin {gather*} \frac {8 d^2 e^2 \cos \left (c+\frac {d x}{2}\right )-16 f^2 \cos \left (c+\frac {d x}{2}\right )+48 d^2 e f x \cos \left (c+\frac {d x}{2}\right )+24 d^2 f^2 x^2 \cos \left (c+\frac {d x}{2}\right )+6 d^2 e^2 \cos \left (c+\frac {3 d x}{2}\right )-15 f^2 \cos \left (c+\frac {3 d x}{2}\right )+12 d^2 e f x \cos \left (c+\frac {3 d x}{2}\right )+6 d^2 f^2 x^2 \cos \left (c+\frac {3 d x}{2}\right )+14 d e f \cos \left (2 c+\frac {3 d x}{2}\right )+14 d f^2 x \cos \left (2 c+\frac {3 d x}{2}\right )-2 d e f \cos \left (2 c+\frac {5 d x}{2}\right )-2 d f^2 x \cos \left (2 c+\frac {5 d x}{2}\right )+2 d^2 e^2 \cos \left (3 c+\frac {5 d x}{2}\right )-f^2 \cos \left (3 c+\frac {5 d x}{2}\right )+4 d^2 e f x \cos \left (3 c+\frac {5 d x}{2}\right )+2 d^2 f^2 x^2 \cos \left (3 c+\frac {5 d x}{2}\right )+8 d \cos \left (\frac {d x}{2}\right ) \left (3 d^2 e^2 x+f^2 x \left (-2+2 i d x+d^2 x^2\right )+e f \left (-2+4 i d x+3 d^2 x^2\right )-8 f (e+f x) \log (1-i \cos (c+d x)+\sin (c+d x))\right )-40 d^2 e^2 \sin \left (\frac {d x}{2}\right )+16 f^2 \sin \left (\frac {d x}{2}\right )-48 d^2 e f x \sin \left (\frac {d x}{2}\right )-24 d^2 f^2 x^2 \sin \left (\frac {d x}{2}\right )-16 d e f \sin \left (c+\frac {d x}{2}\right )+24 d^3 e^2 x \sin \left (c+\frac {d x}{2}\right )+32 i d^2 e f x \sin \left (c+\frac {d x}{2}\right )-16 d f^2 x \sin \left (c+\frac {d x}{2}\right )+24 d^3 e f x^2 \sin \left (c+\frac {d x}{2}\right )+16 i d^2 f^2 x^2 \sin \left (c+\frac {d x}{2}\right )+8 d^3 f^2 x^3 \sin \left (c+\frac {d x}{2}\right )-64 d e f \log (1-i \cos (c+d x)+\sin (c+d x)) \sin \left (c+\frac {d x}{2}\right )-64 d f^2 x \log (1-i \cos (c+d x)+\sin (c+d x)) \sin \left (c+\frac {d x}{2}\right )+64 i f^2 \text {Li}_2(i \cos (c+d x)-\sin (c+d x)) \left (\cos \left (\frac {d x}{2}\right )+\sin \left (c+\frac {d x}{2}\right )\right )-14 d e f \sin \left (c+\frac {3 d x}{2}\right )-14 d f^2 x \sin \left (c+\frac {3 d x}{2}\right )+6 d^2 e^2 \sin \left (2 c+\frac {3 d x}{2}\right )-15 f^2 \sin \left (2 c+\frac {3 d x}{2}\right )+12 d^2 e f x \sin \left (2 c+\frac {3 d x}{2}\right )+6 d^2 f^2 x^2 \sin \left (2 c+\frac {3 d x}{2}\right )-2 d^2 e^2 \sin \left (2 c+\frac {5 d x}{2}\right )+f^2 \sin \left (2 c+\frac {5 d x}{2}\right )-4 d^2 e f x \sin \left (2 c+\frac {5 d x}{2}\right )-2 d^2 f^2 x^2 \sin \left (2 c+\frac {5 d x}{2}\right )-2 d e f \sin \left (3 c+\frac {5 d x}{2}\right )-2 d f^2 x \sin \left (3 c+\frac {5 d x}{2}\right )}{16 a d^3 \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*d^2*e^2*Cos[c + (d*x)/2] - 16*f^2*Cos[c + (d*x)/2] + 48*d^2*e*f*x*Cos[c + (d*x)/2] + 24*d^2*f^2*x^2*Cos[c +
(d*x)/2] + 6*d^2*e^2*Cos[c + (3*d*x)/2] - 15*f^2*Cos[c + (3*d*x)/2] + 12*d^2*e*f*x*Cos[c + (3*d*x)/2] + 6*d^2
*f^2*x^2*Cos[c + (3*d*x)/2] + 14*d*e*f*Cos[2*c + (3*d*x)/2] + 14*d*f^2*x*Cos[2*c + (3*d*x)/2] - 2*d*e*f*Cos[2*
c + (5*d*x)/2] - 2*d*f^2*x*Cos[2*c + (5*d*x)/2] + 2*d^2*e^2*Cos[3*c + (5*d*x)/2] - f^2*Cos[3*c + (5*d*x)/2] +
4*d^2*e*f*x*Cos[3*c + (5*d*x)/2] + 2*d^2*f^2*x^2*Cos[3*c + (5*d*x)/2] + 8*d*Cos[(d*x)/2]*(3*d^2*e^2*x + f^2*x*
(-2 + (2*I)*d*x + d^2*x^2) + e*f*(-2 + (4*I)*d*x + 3*d^2*x^2) - 8*f*(e + f*x)*Log[1 - I*Cos[c + d*x] + Sin[c +
d*x]]) - 40*d^2*e^2*Sin[(d*x)/2] + 16*f^2*Sin[(d*x)/2] - 48*d^2*e*f*x*Sin[(d*x)/2] - 24*d^2*f^2*x^2*Sin[(d*x)
/2] - 16*d*e*f*Sin[c + (d*x)/2] + 24*d^3*e^2*x*Sin[c + (d*x)/2] + (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] + (16*I)*d^2*f^2*x^2*Sin[c + (d*x)/2] + 8*d^3*f^2*x^3*Si
n[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[(d*x)/2]
+ Sin[c + (d*x)/2]) - 14*d*e*f*Sin[c + (3*d*x)/2] - 14*d*f^2*x*Sin[c + (3*d*x)/2] + 6*d^2*e^2*Sin[2*c + (3*d*x
)/2] - 15*f^2*Sin[2*c + (3*d*x)/2] + 12*d^2*e*f*x*Sin[2*c + (3*d*x)/2] + 6*d^2*f^2*x^2*Sin[2*c + (3*d*x)/2] -
2*d^2*e^2*Sin[2*c + (5*d*x)/2] + f^2*Sin[2*c + (5*d*x)/2] - 4*d^2*e*f*x*Sin[2*c + (5*d*x)/2] - 2*d^2*f^2*x^2*S
in[2*c + (5*d*x)/2] - 2*d*e*f*Sin[3*c + (5*d*x)/2] - 2*d*f^2*x*Sin[3*c + (5*d*x)/2])/(16*a*d^3*(Cos[c/2] + Sin
[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]
time = 0.52, size = 492, normalized size = 1.77

method	result	size
risch	\(\frac {f^{2} x^{3}}{2 a}+\frac {3 f e \,x^{2}}{2 a}+\frac {3 e^{2} x}{2 a}+\frac {e^{3}}{2 a f}+\frac {\left (d^{2} x^{2} f^{2}+2 d^{2} e f x +2 i d \,f^{2} x +d^{2} e^{2}+2 i d e f -2 f^{2}\right ) {\mathrm e}^{i \left (d x +c \right )}}{2 a \,d^{3}}+\frac {\left (d^{2} x^{2} f^{2}+2 d^{2} e f x -2 i d \,f^{2} x +d^{2} e^{2}-2 i d e f -2 f^{2}\right ) {\mathrm e}^{-i \left (d x +c \right )}}{2 a \,d^{3}}+\frac {2 x^{2} f^{2}+4 e f x +2 e^{2}}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}-\frac {4 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e}{a \,d^{2}}+\frac {4 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right ) e}{a \,d^{2}}+\frac {4 i f^{2} c x}{a \,d^{2}}+\frac {2 i f^{2} c^{2}}{a \,d^{3}}+\frac {2 i f^{2} x^{2}}{a d}-\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}-\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{a \,d^{3}}+\frac {4 i f^{2} \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {4 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a \,d^{3}}-\frac {4 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}-\frac {f \left (f x +e \right ) \cos \left (2 d x +2 c \right )}{4 d^{2} a}-\frac {\left (2 d^{2} x^{2} f^{2}+4 d^{2} e f x +2 d^{2} e^{2}-f^{2}\right ) \sin \left (2 d x +2 c \right )}{8 a \,d^{3}}\)	\(492\)

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,method=_RETURNVERBOSE)

[Out]

1/2/a*f^2*x^3+3/2/a*f*e*x^2+3/2/a*e^2*x+1/2/a/f*e^3+1/2*(d^2*x^2*f^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)/a/d^3*exp(I*(d*x+c))+1/2*(d^2*x^2*f^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)/a/d^3*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)/a/d^3
*sin(2*d*x+2*c)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 851 vs. \(2 (257) = 514\).
time = 0.41, size = 851, normalized size = 3.06 \begin {gather*} \frac {2 \, d^{3} f^{2} x^{3} + 4 \, d^{2} f^{2} x^{2} - 7 \, d f^{2} x + {\left (2 \, d^{2} f^{2} x^{2} - 2 \, d f^{2} x + 2 \, d^{2} e^{2} - f^{2} + 2 \, {\left (2 \, d^{2} f x - d f\right )} e\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, d^{2} f^{2} x^{2} + 3 \, d f^{2} x + 2 \, d^{2} e^{2} - 4 \, f^{2} + {\left (4 \, d^{2} f x + 3 \, d f\right )} e\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, d^{3} f^{2} x^{3} + 6 \, d^{2} f^{2} x^{2} + d f^{2} x - 7 \, f^{2} + 6 \, {\left (d^{3} x + d^{2}\right )} e^{2} + {\left (6 \, d^{3} f x^{2} + 12 \, d^{2} f x + d f\right )} e\right )} \cos \left (d x + c\right ) - 8 \, {\left (-i \, f^{2} \cos \left (d x + c\right ) - i \, f^{2} \sin \left (d x + c\right ) - i \, f^{2}\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - 8 \, {\left (i \, f^{2} \cos \left (d x + c\right ) + i \, f^{2} \sin \left (d x + c\right ) + i \, f^{2}\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, d^{3} x + 2 \, d^{2}\right )} e^{2} + {\left (6 \, d^{3} f x^{2} + 8 \, d^{2} f x - 7 \, d f\right )} e + 8 \, {\left (c f^{2} - d f e + {\left (c f^{2} - d f e\right )} \cos \left (d x + c\right ) + {\left (c f^{2} - d f e\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 (c f^{2} - d f e + {\left (c f^{2} - d f e\right )} \cos \left (d x + c\right ) + {\left (c f^{2} - d f e\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} f^{2} x^{2} - 7 \, d f^{2} x - {\left (2 \, d^{2} f^{2} x^{2} + 2 \, d f^{2} x + 2 \, d^{2} e^{2} - f^{2} + 2 \, {\left (2 \, d^{2} f x + d f\right )} e\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, d^{2} f^{2} x^{2} - 8 \, d f^{2} x + 2 \, d^{2} e^{2} - 7 \, f^{2} + 4 \, {\left (d^{2} f x - 2 \, d f\right )} e\right )} \cos \left (d x + c\right ) + 2 \, {\left (3 \, d^{3} x - 2 \, d^{2}\right )} e^{2} + {\left (6 \, d^{3} f x^{2} - 8 \, d^{2} f x - 7 \, d f\right )} e\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 )}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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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