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

Optimal. Leaf size=151 \[ \frac{12 f^2 (e+f x) \text{PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^3}-\frac{6 i f (e+f x)^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^2}+\frac{12 i f^3 \text{PolyLog}\left (4,i e^{i (c+d x)}\right )}{a d^4}+\frac{2 (e+f x)^3 \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac{i (e+f x)^4}{4 a f} \]

[Out]

((-I/4)*(e + f*x)^4)/(a*f) + (2*(e + f*x)^3*Log[1 - I*E^(I*(c + d*x))])/(a*d) - ((6*I)*f*(e + f*x)^2*PolyLog[2
, I*E^(I*(c + d*x))])/(a*d^2) + (12*f^2*(e + f*x)*PolyLog[3, I*E^(I*(c + d*x))])/(a*d^3) + ((12*I)*f^3*PolyLog
[4, I*E^(I*(c + d*x))])/(a*d^4)

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Rubi [A]  time = 0.233964, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {4517, 2190, 2531, 6609, 2282, 6589} \[ \frac{12 f^2 (e+f x) \text{PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^3}-\frac{6 i f (e+f x)^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^2}+\frac{12 i f^3 \text{PolyLog}\left (4,i e^{i (c+d x)}\right )}{a d^4}+\frac{2 (e+f x)^3 \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac{i (e+f x)^4}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/4)*(e + f*x)^4)/(a*f) + (2*(e + f*x)^3*Log[1 - I*E^(I*(c + d*x))])/(a*d) - ((6*I)*f*(e + f*x)^2*PolyLog[2
, I*E^(I*(c + d*x))])/(a*d^2) + (12*f^2*(e + f*x)*PolyLog[3, I*E^(I*(c + d*x))])/(a*d^3) + ((12*I)*f^3*PolyLog
[4, I*E^(I*(c + d*x))])/(a*d^4)

Rule 4517

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + Dist[2, Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - I*b*E^(I*(c + d
*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[a^2 - b^2, 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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6609

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{(e+f x)^3 \cos (c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac{i (e+f x)^4}{4 a f}+2 \int \frac{e^{i (c+d x)} (e+f x)^3}{a-i a e^{i (c+d x)}} \, dx\\ &=-\frac{i (e+f x)^4}{4 a f}+\frac{2 (e+f x)^3 \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac{(6 f) \int (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right ) \, dx}{a d}\\ &=-\frac{i (e+f x)^4}{4 a f}+\frac{2 (e+f x)^3 \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac{6 i f (e+f x)^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{a d^2}+\frac{\left (12 i f^2\right ) \int (e+f x) \text{Li}_2\left (i e^{i (c+d x)}\right ) \, dx}{a d^2}\\ &=-\frac{i (e+f x)^4}{4 a f}+\frac{2 (e+f x)^3 \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac{6 i f (e+f x)^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{a d^2}+\frac{12 f^2 (e+f x) \text{Li}_3\left (i e^{i (c+d x)}\right )}{a d^3}-\frac{\left (12 f^3\right ) \int \text{Li}_3\left (i e^{i (c+d x)}\right ) \, dx}{a d^3}\\ &=-\frac{i (e+f x)^4}{4 a f}+\frac{2 (e+f x)^3 \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac{6 i f (e+f x)^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{a d^2}+\frac{12 f^2 (e+f x) \text{Li}_3\left (i e^{i (c+d x)}\right )}{a d^3}+\frac{\left (12 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^4}\\ &=-\frac{i (e+f x)^4}{4 a f}+\frac{2 (e+f x)^3 \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac{6 i f (e+f x)^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{a d^2}+\frac{12 f^2 (e+f x) \text{Li}_3\left (i e^{i (c+d x)}\right )}{a d^3}+\frac{12 i f^3 \text{Li}_4\left (i e^{i (c+d x)}\right )}{a d^4}\\ \end{align*}

Mathematica [A]  time = 1.40661, size = 276, normalized size = 1.83 \[ \frac{x \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (6 e^2 f x+4 e^3+4 e f^2 x^2+f^3 x^3\right )}{4 a \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right )}-\frac{2 (\cos (c)+i \sin (c)) \left (\frac{3 f (\cos (c)-i \sin (c)) (\sin (c)-i \cos (c)+1) \left (d^2 (e+f x)^2 \text{PolyLog}(2,-\sin (c+d x)-i \cos (c+d x))-2 i d f (e+f x) \text{PolyLog}(3,-\sin (c+d x)-i \cos (c+d x))-2 f^2 \text{PolyLog}(4,-\sin (c+d x)-i \cos (c+d x))\right )}{d^4}-\frac{(\sin (c)+i \cos (c)+1) (e+f x)^3 \log (\sin (c+d x)+i \cos (c+d x)+1)}{d}+\frac{(\cos (c)-i \sin (c)) (e+f x)^4}{4 f}\right )}{a (\cos (c)+i (\sin (c)+1))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3)*(Cos[c/2] - Sin[c/2]))/(4*a*(Cos[c/2] + Sin[c/2])) - (2*(Cos[c]
+ I*Sin[c])*(((e + f*x)^4*(Cos[c] - I*Sin[c]))/(4*f) + (3*f*(d^2*(e + f*x)^2*PolyLog[2, (-I)*Cos[c + d*x] - S
in[c + d*x]] - (2*I)*d*f*(e + f*x)*PolyLog[3, (-I)*Cos[c + d*x] - Sin[c + d*x]] - 2*f^2*PolyLog[4, (-I)*Cos[c
+ d*x] - Sin[c + d*x]])*(Cos[c] - I*Sin[c])*(1 - I*Cos[c] + Sin[c]))/d^4 - ((e + f*x)^3*Log[1 + I*Cos[c + d*x]
+ Sin[c + d*x]]*(1 + I*Cos[c] + Sin[c]))/d))/(a*(Cos[c] + I*(1 + Sin[c])))

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Maple [B]  time = 0.184, size = 679, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I/a*e^3*x-1/4*I/a*f^3*x^4-2/a/d*ln(exp(I*(d*x+c)))*e^3+2/a/d*ln(exp(I*(d*x+c))+I)*e^3+2/a/d^4*f^3*c^3*ln(exp(I
*(d*x+c)))-I/a*e*f^2*x^3-2/a/d^4*f^3*c^3*ln(exp(I*(d*x+c))+I)+12/a/d^3*e*f^2*polylog(3,I*exp(I*(d*x+c)))+12/a/
d^3*f^3*polylog(3,I*exp(I*(d*x+c)))*x-3/2*I/a/d^4*f^3*c^4-3/2*I/a*e^2*f*x^2+12*I*f^3*polylog(4,I*exp(I*(d*x+c)
))/a/d^4+2/a/d^4*f^3*c^3*ln(1-I*exp(I*(d*x+c)))+2/a/d*f^3*ln(1-I*exp(I*(d*x+c)))*x^3+4*I/a/d^3*e*f^2*c^3-3*I/a
/d^2*e^2*f*c^2-6*I/a/d^2*e^2*f*polylog(2,I*exp(I*(d*x+c)))-6*I/a/d^2*f^3*polylog(2,I*exp(I*(d*x+c)))*x^2-2*I/a
/d^3*f^3*c^3*x+6*I/a/d^2*e*f^2*c^2*x+6/a/d*e*f^2*ln(1-I*exp(I*(d*x+c)))*x^2-12*I/a/d^2*e*f^2*polylog(2,I*exp(I
*(d*x+c)))*x-6*I/a/d*e^2*f*c*x-6/a/d^3*e*f^2*c^2*ln(1-I*exp(I*(d*x+c)))+6/a/d*e^2*f*ln(1-I*exp(I*(d*x+c)))*x+6
/a/d^2*e^2*f*ln(1-I*exp(I*(d*x+c)))*c-6/a/d^2*e^2*f*c*ln(exp(I*(d*x+c))+I)+6/a/d^3*e*f^2*c^2*ln(exp(I*(d*x+c))
+I)+6/a/d^2*e^2*f*c*ln(exp(I*(d*x+c)))-6/a/d^3*e*f^2*c^2*ln(exp(I*(d*x+c)))

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Maxima [B]  time = 1.39536, size = 689, normalized size = 4.56 \begin{align*} -\frac{\frac{12 \, c e^{2} f \log \left (a d \sin \left (d x + c\right ) + a d\right )}{a d} - \frac{4 \, e^{3} \log \left (a \sin \left (d x + c\right ) + a\right )}{a} - \frac{-i \,{\left (d x + c\right )}^{4} f^{3} +{\left (-4 i \, d e f^{2} + 4 i \, c f^{3}\right )}{\left (d x + c\right )}^{3} + 48 i \, f^{3}{\rm Li}_{4}(i \, e^{\left (i \, d x + i \, c\right )}) +{\left (-6 i \, d^{2} e^{2} f + 12 i \, c d e f^{2} - 6 i \, c^{2} f^{3}\right )}{\left (d x + c\right )}^{2} +{\left (-12 i \, c^{2} d e f^{2} + 4 i \, c^{3} f^{3}\right )}{\left (d x + c\right )} +{\left (24 i \, c^{2} d e f^{2} - 8 i \, c^{3} f^{3}\right )} \arctan \left (\sin \left (d x + c\right ) + 1, \cos \left (d x + c\right )\right ) +{\left (-8 i \,{\left (d x + c\right )}^{3} f^{3} +{\left (-24 i \, d e f^{2} + 24 i \, c f^{3}\right )}{\left (d x + c\right )}^{2} +{\left (-24 i \, d^{2} e^{2} f + 48 i \, c d e f^{2} - 24 i \, c^{2} f^{3}\right )}{\left (d x + c\right )}\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) +{\left (-24 i \, d^{2} e^{2} f + 48 i \, c d e f^{2} - 24 i \,{\left (d x + c\right )}^{2} f^{3} - 24 i \, c^{2} f^{3} +{\left (-48 i \, d e f^{2} + 48 i \, c f^{3}\right )}{\left (d x + c\right )}\right )}{\rm Li}_2\left (i \, e^{\left (i \, d x + i \, c\right )}\right ) + 4 \,{\left (3 \, c^{2} d e f^{2} +{\left (d x + c\right )}^{3} f^{3} - c^{3} f^{3} + 3 \,{\left (d e f^{2} - c f^{3}\right )}{\left (d x + c\right )}^{2} + 3 \,{\left (d^{2} e^{2} f - 2 \, c d e f^{2} + c^{2} f^{3}\right )}{\left (d x + c\right )}\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) + 48 \,{\left (d e f^{2} +{\left (d x + c\right )} f^{3} - c f^{3}\right )}{\rm Li}_{3}(i \, e^{\left (i \, d x + i \, c\right )})}{a d^{3}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(12*c*e^2*f*log(a*d*sin(d*x + c) + a*d)/(a*d) - 4*e^3*log(a*sin(d*x + c) + a)/a - (-I*(d*x + c)^4*f^3 + (
-4*I*d*e*f^2 + 4*I*c*f^3)*(d*x + c)^3 + 48*I*f^3*polylog(4, I*e^(I*d*x + I*c)) + (-6*I*d^2*e^2*f + 12*I*c*d*e*
f^2 - 6*I*c^2*f^3)*(d*x + c)^2 + (-12*I*c^2*d*e*f^2 + 4*I*c^3*f^3)*(d*x + c) + (24*I*c^2*d*e*f^2 - 8*I*c^3*f^3
)*arctan2(sin(d*x + c) + 1, cos(d*x + c)) + (-8*I*(d*x + c)^3*f^3 + (-24*I*d*e*f^2 + 24*I*c*f^3)*(d*x + c)^2 +
(-24*I*d^2*e^2*f + 48*I*c*d*e*f^2 - 24*I*c^2*f^3)*(d*x + c))*arctan2(cos(d*x + c), sin(d*x + c) + 1) + (-24*I
*d^2*e^2*f + 48*I*c*d*e*f^2 - 24*I*(d*x + c)^2*f^3 - 24*I*c^2*f^3 + (-48*I*d*e*f^2 + 48*I*c*f^3)*(d*x + c))*di
log(I*e^(I*d*x + I*c)) + 4*(3*c^2*d*e*f^2 + (d*x + c)^3*f^3 - c^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d
^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + 48*(d
*e*f^2 + (d*x + c)*f^3 - c*f^3)*polylog(3, I*e^(I*d*x + I*c)))/(a*d^3))/d

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Fricas [C]  time = 2.03033, size = 1189, normalized size = 7.87 \begin{align*} \frac{6 i \, f^{3}{\rm polylog}\left (4, i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - 6 i \, f^{3}{\rm polylog}\left (4, -i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) +{\left (-3 i \, d^{2} f^{3} x^{2} - 6 i \, d^{2} e f^{2} x - 3 i \, d^{2} e^{2} f\right )}{\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) +{\left (3 i \, d^{2} f^{3} x^{2} + 6 i \, d^{2} e f^{2} x + 3 i \, d^{2} e^{2} f\right )}{\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) +{\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \log \left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) +{\left (d^{3} f^{3} x^{3} + 3 \, d^{3} e f^{2} x^{2} + 3 \, d^{3} e^{2} f x + 3 \, c d^{2} e^{2} f - 3 \, c^{2} d e f^{2} + c^{3} f^{3}\right )} \log \left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) +{\left (d^{3} f^{3} x^{3} + 3 \, d^{3} e f^{2} x^{2} + 3 \, d^{3} e^{2} f x + 3 \, c d^{2} e^{2} f - 3 \, c^{2} d e f^{2} + c^{3} f^{3}\right )} \log \left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) +{\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \log \left (-\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) + 6 \,{\left (d f^{3} x + d e f^{2}\right )}{\rm polylog}\left (3, i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + 6 \,{\left (d f^{3} x + d e f^{2}\right )}{\rm polylog}\left (3, -i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )}{a d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

(Integral(e**3*cos(c + d*x)/(sin(c + d*x) + 1), x) + Integral(f**3*x**3*cos(c + d*x)/(sin(c + d*x) + 1), x) +
Integral(3*e*f**2*x**2*cos(c + d*x)/(sin(c + d*x) + 1), x) + Integral(3*e**2*f*x*cos(c + d*x)/(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 )}^{3} \cos \left (d x + c\right )}{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)^3*cos(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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