3.107 \(\int \frac{(c+d x)^3}{a+a \sin (e+f x)} \, dx\)
Optimal. Leaf size=148 \[ -\frac{12 i d^2 (c+d x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{a f^3}+\frac{12 d^3 \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{a f^4}+\frac{6 d (c+d x)^2 \log \left (1-i e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^3 \cot \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )}{a f}-\frac{i (c+d x)^3}{a f} \]
[Out]
((-I)*(c + d*x)^3)/(a*f) - ((c + d*x)^3*Cot[e/2 + Pi/4 + (f*x)/2])/(a*f) + (6*d*(c + d*x)^2*Log[1 - I*E^(I*(e
+ f*x))])/(a*f^2) - ((12*I)*d^2*(c + d*x)*PolyLog[2, I*E^(I*(e + f*x))])/(a*f^3) + (12*d^3*PolyLog[3, I*E^(I*(
e + f*x))])/(a*f^4)
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Rubi [A] time = 0.306161, antiderivative size = 148, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used =
{3318, 4184, 3717, 2190, 2531, 2282, 6589} \[ -\frac{12 i d^2 (c+d x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{a f^3}+\frac{12 d^3 \text{PolyLog}\left (3,i e^{i (e+f x)}\right )}{a f^4}+\frac{6 d (c+d x)^2 \log \left (1-i e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^3 \cot \left (\frac{e}{2}+\frac{f x}{2}+\frac{\pi }{4}\right )}{a f}-\frac{i (c+d x)^3}{a f} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)^3/(a + a*Sin[e + f*x]),x]
[Out]
((-I)*(c + d*x)^3)/(a*f) - ((c + d*x)^3*Cot[e/2 + Pi/4 + (f*x)/2])/(a*f) + (6*d*(c + d*x)^2*Log[1 - I*E^(I*(e
+ f*x))])/(a*f^2) - ((12*I)*d^2*(c + d*x)*PolyLog[2, I*E^(I*(e + f*x))])/(a*f^3) + (12*d^3*PolyLog[3, I*E^(I*(
e + f*x))])/(a*f^4)
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 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 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 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 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+a \sin (e+f x)} \, dx &=\frac{\int (c+d x)^3 \csc ^2\left (\frac{1}{2} \left (e+\frac{\pi }{2}\right )+\frac{f x}{2}\right ) \, dx}{2 a}\\ &=-\frac{(c+d x)^3 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f}+\frac{(3 d) \int (c+d x)^2 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right ) \, dx}{a f}\\ &=-\frac{i (c+d x)^3}{a f}-\frac{(c+d x)^3 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f}+\frac{(6 d) \int \frac{e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )} (c+d x)^2}{1-i e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}} \, dx}{a f}\\ &=-\frac{i (c+d x)^3}{a f}-\frac{(c+d x)^3 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f}+\frac{6 d (c+d x)^2 \log \left (1-i e^{i (e+f x)}\right )}{a f^2}-\frac{\left (12 d^2\right ) \int (c+d x) \log \left (1-i e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}\right ) \, dx}{a f^2}\\ &=-\frac{i (c+d x)^3}{a f}-\frac{(c+d x)^3 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f}+\frac{6 d (c+d x)^2 \log \left (1-i e^{i (e+f x)}\right )}{a f^2}-\frac{12 i d^2 (c+d x) \text{Li}_2\left (i e^{i (e+f x)}\right )}{a f^3}+\frac{\left (12 i d^3\right ) \int \text{Li}_2\left (i e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}\right ) \, dx}{a f^3}\\ &=-\frac{i (c+d x)^3}{a f}-\frac{(c+d x)^3 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f}+\frac{6 d (c+d x)^2 \log \left (1-i e^{i (e+f x)}\right )}{a f^2}-\frac{12 i d^2 (c+d x) \text{Li}_2\left (i e^{i (e+f x)}\right )}{a f^3}+\frac{\left (12 d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )}\right )}{a f^4}\\ &=-\frac{i (c+d x)^3}{a f}-\frac{(c+d x)^3 \cot \left (\frac{e}{2}+\frac{\pi }{4}+\frac{f x}{2}\right )}{a f}+\frac{6 d (c+d x)^2 \log \left (1-i e^{i (e+f x)}\right )}{a f^2}-\frac{12 i d^2 (c+d x) \text{Li}_2\left (i e^{i (e+f x)}\right )}{a f^3}+\frac{12 d^3 \text{Li}_3\left (i e^{i (e+f x)}\right )}{a f^4}\\ \end{align*}
Mathematica [A] time = 1.03841, size = 126, normalized size = 0.85 \[ \frac{-12 i d^2 f (c+d x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )+12 d^3 \text{PolyLog}\left (3,i e^{i (e+f x)}\right )+f^2 (c+d x)^2 \left (f (c+d x) \tan \left (\frac{1}{4} (2 e+2 f x-\pi )\right )-i f (c+d x)+6 d \log \left (1-i e^{i (e+f x)}\right )\right )}{a f^4} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)^3/(a + a*Sin[e + f*x]),x]
[Out]
((-12*I)*d^2*f*(c + d*x)*PolyLog[2, I*E^(I*(e + f*x))] + 12*d^3*PolyLog[3, I*E^(I*(e + f*x))] + f^2*(c + d*x)^
2*((-I)*f*(c + d*x) + 6*d*Log[1 - I*E^(I*(e + f*x))] + f*(c + d*x)*Tan[(2*e - Pi + 2*f*x)/4]))/(a*f^4)
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Maple [B] time = 0.141, size = 484, normalized size = 3.3 \begin{align*} -2\,{\frac{{d}^{3}{x}^{3}+3\,c{d}^{2}{x}^{2}+3\,{c}^{2}dx+{c}^{3}}{af \left ({{\rm e}^{i \left ( fx+e \right ) }}+i \right ) }}+6\,{\frac{{d}^{3}\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ){x}^{2}}{a{f}^{2}}}-6\,{\frac{{d}^{3}\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ){e}^{2}}{{f}^{4}a}}+6\,{\frac{\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+i \right ){c}^{2}d}{a{f}^{2}}}+6\,{\frac{{d}^{3}{e}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+i \right ) }{{f}^{4}a}}+{\frac{6\,i{d}^{3}{e}^{2}x}{a{f}^{3}}}-{\frac{6\,ic{d}^{2}{e}^{2}}{a{f}^{3}}}-{\frac{12\,i{d}^{3}{\it polylog} \left ( 2,i{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{a{f}^{3}}}-6\,{\frac{{d}^{3}{e}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{4}a}}+12\,{\frac{{d}^{3}{\it polylog} \left ( 3,i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{4}a}}-{\frac{12\,ic{d}^{2}ex}{a{f}^{2}}}-{\frac{6\,ic{d}^{2}{x}^{2}}{af}}-6\,{\frac{\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ){c}^{2}d}{a{f}^{2}}}+12\,{\frac{c{d}^{2}e\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{a{f}^{3}}}-12\,{\frac{c{d}^{2}e\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+i \right ) }{a{f}^{3}}}-{\frac{2\,i{d}^{3}{x}^{3}}{af}}+{\frac{4\,i{d}^{3}{e}^{3}}{{f}^{4}a}}-{\frac{12\,ic{d}^{2}{\it polylog} \left ( 2,i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{a{f}^{3}}}+12\,{\frac{c{d}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{a{f}^{2}}}+12\,{\frac{c{d}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) e}{a{f}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x+c)^3/(a+a*sin(f*x+e)),x)
[Out]
-2*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)/f/a/(exp(I*(f*x+e))+I)+6/f^2/a*d^3*ln(1-I*exp(I*(f*x+e)))*x^2-6/f^4/a*d
^3*ln(1-I*exp(I*(f*x+e)))*e^2+6/f^2/a*ln(exp(I*(f*x+e))+I)*c^2*d+6/f^4/a*d^3*e^2*ln(exp(I*(f*x+e))+I)+6*I/f^3/
a*d^3*e^2*x-6*I/f^3/a*c*d^2*e^2-12*I/f^3/a*d^3*polylog(2,I*exp(I*(f*x+e)))*x-6/f^4/a*d^3*e^2*ln(exp(I*(f*x+e))
)+12*d^3*polylog(3,I*exp(I*(f*x+e)))/a/f^4-12*I/f^2/a*c*d^2*e*x-6*I/f/a*c*d^2*x^2-6/f^2/a*ln(exp(I*(f*x+e)))*c
^2*d+12/f^3/a*c*d^2*e*ln(exp(I*(f*x+e)))-12/f^3/a*c*d^2*e*ln(exp(I*(f*x+e))+I)-2*I/f/a*d^3*x^3+4*I/f^4/a*d^3*e
^3-12*I/f^3/a*c*d^2*polylog(2,I*exp(I*(f*x+e)))+12/f^2/a*c*d^2*ln(1-I*exp(I*(f*x+e)))*x+12/f^3/a*c*d^2*ln(1-I*
exp(I*(f*x+e)))*e
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Maxima [B] time = 1.47734, size = 1315, normalized size = 8.89 \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+a*sin(f*x+e)),x, algorithm="maxima")
[Out]
(6*(2*(f*x + e)*cos(f*x + e) - (cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1)*log(cos(f*x + e)^2 + sin
(f*x + e)^2 + 2*sin(f*x + e) + 1))*c*d^2*e/(a*f^2*cos(f*x + e)^2 + a*f^2*sin(f*x + e)^2 + 2*a*f^2*sin(f*x + e)
+ a*f^2) - 6*c*d^2*e^2/(a*f^2 + a*f^2*sin(f*x + e)/(cos(f*x + e) + 1)) - 3*(2*(f*x + e)*cos(f*x + e) - (cos(f
*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1)*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1))*c^
2*d/(a*f*cos(f*x + e)^2 + a*f*sin(f*x + e)^2 + 2*a*f*sin(f*x + e) + a*f) + 6*c^2*d*e/(a*f + a*f*sin(f*x + e)/(
cos(f*x + e) + 1)) - 2*c^3/(a + a*sin(f*x + e)/(cos(f*x + e) + 1)) + (-2*I*d^3*e^3 + (6*d^3*e^2*cos(f*x + e) +
6*I*d^3*e^2*sin(f*x + e) + 6*I*d^3*e^2)*arctan2(sin(f*x + e) + 1, cos(f*x + e)) + (-6*I*(f*x + e)^2*d^3 + (12
*I*d^3*e - 12*I*c*d^2*f)*(f*x + e) - 6*((f*x + e)^2*d^3 - 2*(d^3*e - c*d^2*f)*(f*x + e))*cos(f*x + e) + (-6*I*
(f*x + e)^2*d^3 + (12*I*d^3*e - 12*I*c*d^2*f)*(f*x + e))*sin(f*x + e))*arctan2(cos(f*x + e), sin(f*x + e) + 1)
- 2*((f*x + e)^3*d^3 + 3*(f*x + e)*d^3*e^2 - 3*(d^3*e - c*d^2*f)*(f*x + e)^2)*cos(f*x + e) + (-12*I*(f*x + e)
*d^3 + 12*I*d^3*e - 12*I*c*d^2*f - 12*((f*x + e)*d^3 - d^3*e + c*d^2*f)*cos(f*x + e) + (-12*I*(f*x + e)*d^3 +
12*I*d^3*e - 12*I*c*d^2*f)*sin(f*x + e))*dilog(I*e^(I*f*x + I*e)) + (3*(f*x + e)^2*d^3 + 3*d^3*e^2 - 6*(d^3*e
- c*d^2*f)*(f*x + e) + (-3*I*(f*x + e)^2*d^3 - 3*I*d^3*e^2 + (6*I*d^3*e - 6*I*c*d^2*f)*(f*x + e))*cos(f*x + e)
+ 3*((f*x + e)^2*d^3 + d^3*e^2 - 2*(d^3*e - c*d^2*f)*(f*x + e))*sin(f*x + e))*log(cos(f*x + e)^2 + sin(f*x +
e)^2 + 2*sin(f*x + e) + 1) + (-12*I*d^3*cos(f*x + e) + 12*d^3*sin(f*x + e) + 12*d^3)*polylog(3, I*e^(I*f*x + I
*e)) + (-2*I*(f*x + e)^3*d^3 - 6*I*(f*x + e)*d^3*e^2 + (6*I*d^3*e - 6*I*c*d^2*f)*(f*x + e)^2)*sin(f*x + e))/(-
I*a*f^3*cos(f*x + e) + a*f^3*sin(f*x + e) + a*f^3))/f
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Fricas [C] time = 2.02191, size = 2134, normalized size = 14.42 \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+a*sin(f*x+e)),x, algorithm="fricas")
[Out]
-(d^3*f^3*x^3 + 3*c*d^2*f^3*x^2 + 3*c^2*d*f^3*x + c^3*f^3 + (d^3*f^3*x^3 + 3*c*d^2*f^3*x^2 + 3*c^2*d*f^3*x + c
^3*f^3)*cos(f*x + e) - (-6*I*d^3*f*x - 6*I*c*d^2*f + (-6*I*d^3*f*x - 6*I*c*d^2*f)*cos(f*x + e) + (-6*I*d^3*f*x
- 6*I*c*d^2*f)*sin(f*x + e))*dilog(I*cos(f*x + e) - sin(f*x + e)) - (6*I*d^3*f*x + 6*I*c*d^2*f + (6*I*d^3*f*x
+ 6*I*c*d^2*f)*cos(f*x + e) + (6*I*d^3*f*x + 6*I*c*d^2*f)*sin(f*x + e))*dilog(-I*cos(f*x + e) - sin(f*x + e))
- 3*(d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2 + (d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*cos(f*x + e) + (d^3*e^2 - 2*c*d^
2*e*f + c^2*d*f^2)*sin(f*x + e))*log(cos(f*x + e) + I*sin(f*x + e) + I) - 3*(d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3
*e^2 + 2*c*d^2*e*f + (d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*cos(f*x + e) + (d^3*f^2*x^2 + 2*c*d
^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*sin(f*x + e))*log(I*cos(f*x + e) + sin(f*x + e) + 1) - 3*(d^3*f^2*x^2 + 2*c*
d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f + (d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*cos(f*x + e) + (d^3*
f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*sin(f*x + e))*log(-I*cos(f*x + e) + sin(f*x + e) + 1) - 3*(d^
3*e^2 - 2*c*d^2*e*f + c^2*d*f^2 + (d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*cos(f*x + e) + (d^3*e^2 - 2*c*d^2*e*f +
c^2*d*f^2)*sin(f*x + e))*log(-cos(f*x + e) + I*sin(f*x + e) + I) - 6*(d^3*cos(f*x + e) + d^3*sin(f*x + e) + d^
3)*polylog(3, I*cos(f*x + e) - sin(f*x + e)) - 6*(d^3*cos(f*x + e) + d^3*sin(f*x + e) + d^3)*polylog(3, -I*cos
(f*x + e) - sin(f*x + e)) - (d^3*f^3*x^3 + 3*c*d^2*f^3*x^2 + 3*c^2*d*f^3*x + c^3*f^3)*sin(f*x + e))/(a*f^4*cos
(f*x + e) + a*f^4*sin(f*x + e) + a*f^4)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c^{3}}{\sin{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{3} x^{3}}{\sin{\left (e + f x \right )} + 1}\, dx + \int \frac{3 c d^{2} x^{2}}{\sin{\left (e + f x \right )} + 1}\, dx + \int \frac{3 c^{2} d x}{\sin{\left (e + f x \right )} + 1}\, dx}{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)**3/(a+a*sin(f*x+e)),x)
[Out]
(Integral(c**3/(sin(e + f*x) + 1), x) + Integral(d**3*x**3/(sin(e + f*x) + 1), x) + Integral(3*c*d**2*x**2/(si
n(e + f*x) + 1), x) + Integral(3*c**2*d*x/(sin(e + f*x) + 1), x))/a
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{3}}{a \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+a*sin(f*x+e)),x, algorithm="giac")
[Out]
integrate((d*x + c)^3/(a*sin(f*x + e) + a), x)