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

Optimal. Leaf size=164 \[ -\frac {12 f^3 \text {Li}_3\left (i e^{i (c+d x)}\right )}{a d^4}+\frac {12 i f^2 (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}-\frac {6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {(e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{a d}+\frac {i (e+f x)^3}{a d}+\frac {(e+f x)^4}{4 a f} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {4515, 32, 3318, 4184, 3717, 2190, 2531, 2282, 6589} \[ \frac {12 i f^2 (e+f x) \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac {12 f^3 \text {PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^4}-\frac {6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {(e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{a d}+\frac {i (e+f x)^3}{a d}+\frac {(e+f x)^4}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 3318

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*(e + (Pi*a)/(2*b)))/2 + (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 3717

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 4184

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 4515

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]

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

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Mathematica [A]  time = 2.04, size = 261, normalized size = 1.59 \[ \frac {\frac {24 f (\cos (c)+i \sin (c)) \left (\frac {2 f (\cos (c)-i (\sin (c)+1)) (d (e+f x) \text {Li}_2(-i \cos (c+d x)-\sin (c+d x))-i f \text {Li}_3(-i \cos (c+d x)-\sin (c+d x)))}{d^3}-\frac {(\sin (c)+i \cos (c)+1) (e+f x)^2 \log (\sin (c+d x)+i \cos (c+d x)+1)}{d}+\frac {(\cos (c)-i \sin (c)) (e+f x)^3}{3 f}\right )}{d (\cos (c)+i (\sin (c)+1))}-\frac {8 \sin \left (\frac {d x}{2}\right ) (e+f x)^3}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right )}{4 a} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*Sin[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) + (24*f*(Cos[c] + I*Sin[c])*(((e + f*x)^3*(Cos[c] - I*Sin[c]))/
(3*f) - ((e + f*x)^2*Log[1 + I*Cos[c + d*x] + Sin[c + d*x]]*(1 + I*Cos[c] + Sin[c]))/d + (2*f*(d*(e + f*x)*Pol
yLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]] - I*f*PolyLog[3, (-I)*Cos[c + d*x] - Sin[c + d*x]])*(Cos[c] - I*(1 +
Sin[c])))/d^3))/(d*(Cos[c] + I*(1 + Sin[c]))) - (8*(e + f*x)^3*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])))/(4*a)

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fricas [C]  time = 0.51, size = 1042, normalized size = 6.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(d^4*f^3*x^4 + 4*d^3*e^3 + 4*(d^4*e*f^2 + d^3*f^3)*x^3 + 6*(d^4*e^2*f + 2*d^3*e*f^2)*x^2 + 4*(d^4*e^3 + 3*
d^3*e^2*f)*x + (d^4*f^3*x^4 + 4*d^3*e^3 + 4*(d^4*e*f^2 + d^3*f^3)*x^3 + 6*(d^4*e^2*f + 2*d^3*e*f^2)*x^2 + 4*(d
^4*e^3 + 3*d^3*e^2*f)*x)*cos(d*x + c) + (24*I*d*f^3*x + 24*I*d*e*f^2 + (24*I*d*f^3*x + 24*I*d*e*f^2)*cos(d*x +
c) + (24*I*d*f^3*x + 24*I*d*e*f^2)*sin(d*x + c))*dilog(I*cos(d*x + c) - sin(d*x + c)) + (-24*I*d*f^3*x - 24*I
*d*e*f^2 + (-24*I*d*f^3*x - 24*I*d*e*f^2)*cos(d*x + c) + (-24*I*d*f^3*x - 24*I*d*e*f^2)*sin(d*x + c))*dilog(-I
*cos(d*x + c) - sin(d*x + c)) - 12*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3 + (d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*co
s(d*x + c) + (d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*sin(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) - 12*(d^
2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3 + (d^2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*cos(
d*x + c) + (d^2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*sin(d*x + c))*log(I*cos(d*x + c) + sin(d*x +
c) + 1) - 12*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3 + (d^2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2
- c^2*f^3)*cos(d*x + c) + (d^2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*sin(d*x + c))*log(-I*cos(d*x
+ c) + sin(d*x + c) + 1) - 12*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3 + (d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*cos(d*x
+ c) + (d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*sin(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + I) - 24*(f^3*co
s(d*x + c) + f^3*sin(d*x + c) + f^3)*polylog(3, I*cos(d*x + c) - sin(d*x + c)) - 24*(f^3*cos(d*x + c) + f^3*si
n(d*x + c) + f^3)*polylog(3, -I*cos(d*x + c) - sin(d*x + c)) + (d^4*f^3*x^4 - 4*d^3*e^3 + 4*(d^4*e*f^2 - d^3*f
^3)*x^3 + 6*(d^4*e^2*f - 2*d^3*e*f^2)*x^2 + 4*(d^4*e^3 - 3*d^3*e^2*f)*x)*sin(d*x + c))/(a*d^4*cos(d*x + c) + a
*d^4*sin(d*x + c) + a*d^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.27, size = 526, normalized size = 3.21 \[ \frac {f^{3} x^{4}}{4 a}+\frac {e \,f^{2} x^{3}}{a}+\frac {3 e^{2} f \,x^{2}}{2 a}+\frac {e^{3} x}{a}+\frac {2 f^{3} x^{3}+6 e \,f^{2} x^{2}+6 e^{2} f x +2 e^{3}}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}+\frac {12 i f^{3} \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{3}}+\frac {6 i f^{2} e \,c^{2}}{a \,d^{3}}-\frac {6 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e^{2}}{a \,d^{2}}-\frac {12 f^{2} e \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}-\frac {12 f^{2} e \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{a \,d^{3}}-\frac {12 f^{2} e c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {6 i f^{2} e \,x^{2}}{a d}-\frac {6 f^{3} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x^{2}}{a \,d^{2}}+\frac {6 f^{3} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c^{2}}{a \,d^{4}}+\frac {6 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right ) e^{2}}{a \,d^{2}}-\frac {6 i f^{3} c^{2} x}{a \,d^{3}}+\frac {12 i f^{2} e \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {2 i f^{3} x^{3}}{a d}-\frac {12 f^{3} \polylog \left (3, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{4}}+\frac {6 f^{3} c^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{4}}-\frac {4 i f^{3} c^{3}}{a \,d^{4}}-\frac {6 f^{3} c^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a \,d^{4}}+\frac {12 i f^{2} e c x}{a \,d^{2}}+\frac {12 f^{2} e c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a \,d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.63, size = 1307, normalized size = 7.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((sin(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 \[ \frac {\int \frac {e^{3} \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{3} x^{3} \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e f^{2} x^{2} \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e^{2} f x \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(e**3*sin(c + d*x)/(sin(c + d*x) + 1), x) + Integral(f**3*x**3*sin(c + d*x)/(sin(c + d*x) + 1), x) +
Integral(3*e*f**2*x**2*sin(c + d*x)/(sin(c + d*x) + 1), x) + Integral(3*e**2*f*x*sin(c + d*x)/(sin(c + d*x) +
1), x))/a

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