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

Optimal. Leaf size=495 \[ -\frac{6 i d^2 (c+d x) \text{PolyLog}\left (3,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^3 \sqrt{a^2-b^2}}+\frac{6 i d^2 (c+d x) \text{PolyLog}\left (3,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^3 \sqrt{a^2-b^2}}-\frac{3 d (c+d x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^2 \sqrt{a^2-b^2}}+\frac{3 d (c+d x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^2 \sqrt{a^2-b^2}}+\frac{6 d^3 \text{PolyLog}\left (4,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^4 \sqrt{a^2-b^2}}-\frac{6 d^3 \text{PolyLog}\left (4,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^4 \sqrt{a^2-b^2}}-\frac{i (c+d x)^3 \log \left (1-\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2}}+\frac{i (c+d x)^3 \log \left (1-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f \sqrt{a^2-b^2}} \]

[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)

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Rubi [A]  time = 0.969198, antiderivative size = 495, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {3323, 2264, 2190, 2531, 6609, 2282, 6589} \[ -\frac{6 i d^2 (c+d x) \text{PolyLog}\left (3,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^3 \sqrt{a^2-b^2}}+\frac{6 i d^2 (c+d x) \text{PolyLog}\left (3,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^3 \sqrt{a^2-b^2}}-\frac{3 d (c+d x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^2 \sqrt{a^2-b^2}}+\frac{3 d (c+d x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^2 \sqrt{a^2-b^2}}+\frac{6 d^3 \text{PolyLog}\left (4,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f^4 \sqrt{a^2-b^2}}-\frac{6 d^3 \text{PolyLog}\left (4,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f^4 \sqrt{a^2-b^2}}-\frac{i (c+d x)^3 \log \left (1-\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )}{f \sqrt{a^2-b^2}}+\frac{i (c+d x)^3 \log \left (1-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )}{f \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[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 3323

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 2264

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 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{(c+d x)^3}{a+b \sin (e+f x)} \, dx &=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 \text{Li}_2\left (\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 \text{Li}_2\left (\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) \text{Li}_2\left (\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) \text{Li}_2\left (\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 \text{Li}_2\left (\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 \text{Li}_2\left (\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) \text{Li}_3\left (\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) \text{Li}_3\left (\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 \text{Li}_3\left (\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 \text{Li}_3\left (\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 \text{Li}_2\left (\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 \text{Li}_2\left (\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) \text{Li}_3\left (\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) \text{Li}_3\left (\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 ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\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 ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\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 \text{Li}_2\left (\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 \text{Li}_2\left (\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) \text{Li}_3\left (\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) \text{Li}_3\left (\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 \text{Li}_4\left (\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 \text{Li}_4\left (\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]  time = 0.234814, size = 401, normalized size = 0.81 \[ -\frac{i \left (\frac{3 i d \left (f^2 (c+d x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )+2 i d f (c+d x) \text{PolyLog}\left (3,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )-2 d^2 \text{PolyLog}\left (4,\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )\right )}{f^3}+\frac{3 d \left (2 d \left (f (c+d x) \text{PolyLog}\left (3,-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}-a}\right )+i d \text{PolyLog}\left (4,\frac{i b e^{i (e+f x)}}{a-\sqrt{a^2-b^2}}\right )\right )-i f^2 (c+d x)^2 \text{PolyLog}\left (2,-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}-a}\right )\right )}{f^3}+(c+d x)^3 \log \left (1+\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}-a}\right )-(c+d x)^3 \log \left (1-\frac{i b e^{i (e+f x)}}{\sqrt{a^2-b^2}+a}\right )\right )}{f \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 0.401, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{3}}{a+b\sin \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 3.67475, size = 5222, normalized size = 10.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(12*I*b*d^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, 1/2*(2*I*a*cos(f*x + e) - 2*a*sin(f*x + e) + 2*(b*cos(f*x +
e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))/b) - 12*I*b*d^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, 1/2*(2*I*a*c
os(f*x + e) - 2*a*sin(f*x + e) - 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))/b) - 12*I*b*d^3
*sqrt(-(a^2 - b^2)/b^2)*polylog(4, 1/2*(-2*I*a*cos(f*x + e) - 2*a*sin(f*x + e) + 2*(b*cos(f*x + e) - I*b*sin(f
*x + e))*sqrt(-(a^2 - b^2)/b^2))/b) + 12*I*b*d^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, 1/2*(-2*I*a*cos(f*x + e) -
2*a*sin(f*x + e) - 2*(b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))/b) + 2*(-3*I*b*d^3*f^2*x^2 -
6*I*b*c*d^2*f^2*x - 3*I*b*c^2*d*f^2)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(f*x + e) + 2*a*sin(f*x + e)
+ 2*(b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*(3*I*b*d^3*f^2*x^2 + 6*I*b*c*
d^2*f^2*x + 3*I*b*c^2*d*f^2)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(f*x + e) + 2*a*sin(f*x + e) - 2*(b*c
os(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*(3*I*b*d^3*f^2*x^2 + 6*I*b*c*d^2*f^2*
x + 3*I*b*c^2*d*f^2)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(f*x + e) + 2*a*sin(f*x + e) + 2*(b*cos(f*x
+ e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*(-3*I*b*d^3*f^2*x^2 - 6*I*b*c*d^2*f^2*x - 3*
I*b*c^2*d*f^2)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(f*x + e) + 2*a*sin(f*x + e) - 2*(b*cos(f*x + e) +
I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + 2*(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) + 2*(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) - 2*(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) - 2*(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) - 2*(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(1/2*(2*I*a*cos(f*x +
e) + 2*a*sin(f*x + e) + 2*(b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*(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(1/2*(2*I*a*cos(f*x + e) + 2*a*sin(f*x + e) - 2*(b*cos(f*x + e) - I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b
^2) + 2*b)/b) - 2*(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(1/2*(-2*I*a*cos(f*x + e) + 2*a*sin(f*x + e) + 2*(b*cos(f*x + e) + I*b*sin
(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*(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(1/2*(-2*I*a*cos(f*x + e) + 2*a*sin(f*x + e)
- 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 12*(b*d^3*f*x + b*c*d^2*f)*sqrt(-(
a^2 - b^2)/b^2)*polylog(3, 1/2*(2*I*a*cos(f*x + e) - 2*a*sin(f*x + e) + 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*
sqrt(-(a^2 - b^2)/b^2))/b) - 12*(b*d^3*f*x + b*c*d^2*f)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, 1/2*(2*I*a*cos(f*x +
e) - 2*a*sin(f*x + e) - 2*(b*cos(f*x + e) + I*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))/b) + 12*(b*d^3*f*x + b*
c*d^2*f)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, 1/2*(-2*I*a*cos(f*x + e) - 2*a*sin(f*x + e) + 2*(b*cos(f*x + e) - I
*b*sin(f*x + e))*sqrt(-(a^2 - b^2)/b^2))/b) - 12*(b*d^3*f*x + b*c*d^2*f)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, 1/2
*(-2*I*a*cos(f*x + e) - 2*a*sin(f*x + e) - 2*(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{3}}{b \sin \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)