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

Optimal. Leaf size=151 \[ \frac {12 i f^3 \text {Li}_4\left (i e^{i (c+d x)}\right )}{a d^4}+\frac {12 f^2 (e+f x) \text {Li}_3\left (i e^{i (c+d x)}\right )}{a d^3}-\frac {6 i f (e+f x)^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^2}+\frac {2 (e+f x)^3 \log \left (1-i e^{i (c+d x)}\right )}{a d}-\frac {i (e+f x)^4}{4 a f} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 151, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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]

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*}

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Mathematica [A]  time = 1.50, size = 276, normalized size = 1.83 \[ \frac {x \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right )}{4 a \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right )}-\frac {2 (\cos (c)+i \sin (c)) \left (\frac {3 f (\cos (c)-i \sin (c)) (\sin (c)-i \cos (c)+1) \left (d^2 (e+f x)^2 \text {Li}_2(-i \cos (c+d x)-\sin (c+d x))-2 i d f (e+f x) \text {Li}_3(-i \cos (c+d x)-\sin (c+d x))-2 f^2 \text {Li}_4(-i \cos (c+d x)-\sin (c+d x))\right )}{d^4}-\frac {(\sin (c)+i \cos (c)+1) (e+f x)^3 \log (\sin (c+d x)+i \cos (c+d x)+1)}{d}+\frac {(\cos (c)-i \sin (c)) (e+f x)^4}{4 f}\right )}{a (\cos (c)+i (\sin (c)+1))} \]

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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fricas [C]  time = 0.50, size = 490, normalized size = 3.25 \[ \frac {6 i \, f^{3} {\rm polylog}\left (4, i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) - 6 i \, f^{3} {\rm polylog}\left (4, -i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + {\left (-3 i \, d^{2} f^{3} x^{2} - 6 i \, d^{2} e f^{2} x - 3 i \, d^{2} e^{2} f\right )} {\rm Li}_2\left (i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + {\left (3 i \, d^{2} f^{3} x^{2} + 6 i \, d^{2} e f^{2} x + 3 i \, d^{2} e^{2} f\right )} {\rm Li}_2\left (-i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + {\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \log \left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) + {\left (d^{3} f^{3} x^{3} + 3 \, d^{3} e f^{2} x^{2} + 3 \, d^{3} e^{2} f x + 3 \, c d^{2} e^{2} f - 3 \, c^{2} d e f^{2} + c^{3} f^{3}\right )} \log \left (i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + {\left (d^{3} f^{3} x^{3} + 3 \, d^{3} e f^{2} x^{2} + 3 \, d^{3} e^{2} f x + 3 \, c d^{2} e^{2} f - 3 \, c^{2} d e f^{2} + c^{3} f^{3}\right )} \log \left (-i \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right ) + {\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \log \left (-\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right ) + i\right ) + 6 \, {\left (d f^{3} x + d e f^{2}\right )} {\rm polylog}\left (3, i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right ) + 6 \, {\left (d f^{3} x + d e f^{2}\right )} {\rm polylog}\left (3, -i \, \cos \left (d x + c\right ) - \sin \left (d x + c\right )\right )}{a d^{4}} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^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)

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maple [B]  time = 0.27, size = 679, normalized size = 4.50 \[ \frac {i e^{3} x}{a}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right ) e^{3}}{d a}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e^{3}}{d a}-\frac {i f^{3} x^{4}}{4 a}-\frac {6 i e^{2} f c x}{d a}+\frac {6 i c^{2} e \,f^{2} x}{d^{2} a}-\frac {12 i e \,f^{2} \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{d^{2} a}-\frac {6 e \,f^{2} c^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right )}{d^{3} a}-\frac {6 e^{2} f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d^{2} a}+\frac {6 e \,f^{2} c^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d^{3} a}+\frac {6 e^{2} f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} a}-\frac {6 e \,f^{2} c^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{d^{3} a}+\frac {6 e^{2} f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{d a}+\frac {6 e^{2} f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{d^{2} a}+\frac {6 e \,f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x^{2}}{d a}+\frac {2 f^{3} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x^{3}}{d a}+\frac {2 f^{3} c^{3} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right )}{d^{4} a}+\frac {2 f^{3} c^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{d^{4} a}-\frac {2 f^{3} c^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d^{4} a}-\frac {3 i f^{3} c^{4}}{2 d^{4} a}-\frac {i e \,f^{2} x^{3}}{a}-\frac {3 i e^{2} f \,x^{2}}{2 a}+\frac {12 f^{3} \polylog \left (3, i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{d^{3} a}+\frac {12 e \,f^{2} \polylog \left (3, i {\mathrm e}^{i \left (d x +c \right )}\right )}{d^{3} a}+\frac {12 i f^{3} \polylog \left (4, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{4}}+\frac {4 i c^{3} e \,f^{2}}{d^{3} a}-\frac {3 i e^{2} f \,c^{2}}{d^{2} a}-\frac {2 i f^{3} c^{3} x}{d^{3} a}-\frac {6 i f^{3} \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right ) x^{2}}{d^{2} a}-\frac {6 i e^{2} f \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} a} \]

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

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maxima [B]  time = 1.23, size = 510, normalized size = 3.38 \[ -\frac {\frac {12 \, c e^{2} f \log \left (a d \sin \left (d x + c\right ) + a d\right )}{a d} - \frac {4 \, e^{3} \log \left (a \sin \left (d x + c\right ) + a\right )}{a} - \frac {-i \, {\left (d x + c\right )}^{4} f^{3} + {\left (-4 i \, d e f^{2} + 4 i \, c f^{3}\right )} {\left (d x + c\right )}^{3} + 48 i \, f^{3} {\rm Li}_{4}(i \, e^{\left (i \, d x + i \, c\right )}) + {\left (-6 i \, d^{2} e^{2} f + 12 i \, c d e f^{2} - 6 i \, c^{2} f^{3}\right )} {\left (d x + c\right )}^{2} + {\left (-12 i \, c^{2} d e f^{2} + 4 i \, c^{3} f^{3}\right )} {\left (d x + c\right )} + {\left (24 i \, c^{2} d e f^{2} - 8 i \, c^{3} f^{3}\right )} \arctan \left (\sin \left (d x + c\right ) + 1, \cos \left (d x + c\right )\right ) + {\left (-8 i \, {\left (d x + c\right )}^{3} f^{3} + {\left (-24 i \, d e f^{2} + 24 i \, c f^{3}\right )} {\left (d x + c\right )}^{2} + {\left (-24 i \, d^{2} e^{2} f + 48 i \, c d e f^{2} - 24 i \, c^{2} f^{3}\right )} {\left (d x + c\right )}\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) + {\left (-24 i \, d^{2} e^{2} f + 48 i \, c d e f^{2} - 24 i \, {\left (d x + c\right )}^{2} f^{3} - 24 i \, c^{2} f^{3} + {\left (-48 i \, d e f^{2} + 48 i \, c f^{3}\right )} {\left (d x + c\right )}\right )} {\rm Li}_2\left (i \, e^{\left (i \, d x + i \, c\right )}\right ) + 4 \, {\left (3 \, c^{2} d e f^{2} + {\left (d x + c\right )}^{3} f^{3} - c^{3} f^{3} + 3 \, {\left (d e f^{2} - c f^{3}\right )} {\left (d x + c\right )}^{2} + 3 \, {\left (d^{2} e^{2} f - 2 \, c d e f^{2} + c^{2} f^{3}\right )} {\left (d x + c\right )}\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) + 48 \, {\left (d e f^{2} + {\left (d x + c\right )} f^{3} - c f^{3}\right )} {\rm Li}_{3}(i \, e^{\left (i \, d x + i \, c\right )})}{a d^{3}}}{4 \, d} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (c+d\,x\right )\,{\left (e+f\,x\right )}^3}{a+a\,\sin \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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