\(\int \frac {(c+d x)^3}{a+b \sin (e+f x)} \, dx\) [163]

Optimal result

Integrand size = 20, antiderivative size = 495 \[ \int \frac {(c+d x)^3}{a+b \sin (e+f x)} \, dx=-\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {6 d^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^4}-\frac {6 d^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 495, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3404, 2296, 2221, 2611, 6744, 2320, 6724} \[ \int \frac {(c+d x)^3}{a+b \sin (e+f x)} \, dx=-\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \sqrt {a^2-b^2}}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^3 \sqrt {a^2-b^2}}-\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}-\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f \sqrt {a^2-b^2}}+\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{\sqrt {a^2-b^2}+a}\right )}{f \sqrt {a^2-b^2}}+\frac {6 d^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^4 \sqrt {a^2-b^2}}-\frac {6 d^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^4 \sqrt {a^2-b^2}} \]

[In]

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

[Out]

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

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 2296

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[2*(c/q), Int[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Dist[2*(c/q), Int[(f + g
*x)^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2320

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 2611

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 3404

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

Rule 6724

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 6744

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*} \text {integral}& = 2 \int \frac {e^{i (e+f x)} (c+d x)^3}{i b+2 a e^{i (e+f x)}-i b e^{2 i (e+f x)}} \, dx \\ & = -\frac {(2 i b) \int \frac {e^{i (e+f x)} (c+d x)^3}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\sqrt {a^2-b^2}}+\frac {(2 i b) \int \frac {e^{i (e+f x)} (c+d x)^3}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (e+f x)}} \, dx}{\sqrt {a^2-b^2}} \\ & = -\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {(3 i d) \int (c+d x)^2 \log \left (1-\frac {2 i b e^{i (e+f x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f}-\frac {(3 i d) \int (c+d x)^2 \log \left (1-\frac {2 i b e^{i (e+f x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f} \\ & = -\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}+\frac {\left (6 d^2\right ) \int (c+d x) \operatorname {PolyLog}\left (2,\frac {2 i b e^{i (e+f x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f^2}-\frac {\left (6 d^2\right ) \int (c+d x) \operatorname {PolyLog}\left (2,\frac {2 i b e^{i (e+f x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f^2} \\ & = -\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {\left (6 i d^3\right ) \int \operatorname {PolyLog}\left (3,\frac {2 i b e^{i (e+f x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f^3}-\frac {\left (6 i d^3\right ) \int \operatorname {PolyLog}\left (3,\frac {2 i b e^{i (e+f x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f^3} \\ & = -\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {\left (6 d^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\sqrt {a^2-b^2} f^4}-\frac {\left (6 d^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (e+f x)}\right )}{\sqrt {a^2-b^2} f^4} \\ & = -\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {i (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {6 d^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^4}-\frac {6 d^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 401, normalized size of antiderivative = 0.81 \[ \int \frac {(c+d x)^3}{a+b \sin (e+f x)} \, dx=-\frac {i \left ((c+d x)^3 \log \left (1+\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )-(c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )+\frac {3 d \left (-i f^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )+2 d \left (f (c+d x) \operatorname {PolyLog}\left (3,-\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )+i d \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )\right )\right )}{f^3}+\frac {3 i d \left (f^2 (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )+2 i d f (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )-2 d^2 \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )}{f^3}\right )}{\sqrt {a^2-b^2} f} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (d x +c \right )^{3}}{a +b \sin \left (f x +e \right )}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2173 vs. \(2 (421) = 842\).

Time = 0.46 (sec) , antiderivative size = 2173, normalized size of antiderivative = 4.39 \[ \int \frac {(c+d x)^3}{a+b \sin (e+f x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(c+d x)^3}{a+b \sin (e+f x)} \, dx=\int \frac {\left (c + d x\right )^{3}}{a + b \sin {\left (e + f x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^3}{a+b \sin (e+f x)} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [F]

\[ \int \frac {(c+d x)^3}{a+b \sin (e+f x)} \, dx=\int { \frac {{\left (d x + c\right )}^{3}}{b \sin \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^3}{a+b \sin (e+f x)} \, dx=\int \frac {{\left (c+d\,x\right )}^3}{a+b\,\sin \left (e+f\,x\right )} \,d x \]

[In]

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

[Out]

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