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

Optimal. Leaf size=133 \[ -\frac{12 i d^2 (c+d x) \text{Li}_2\left (e^{i (e+f x)}\right )}{a f^3}+\frac{6 d (c+d x)^2 \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^3 \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}-\frac{i (c+d x)^3}{a f}+\frac{12 d^3 \text{Li}_3\left (e^{i (e+f x)}\right )}{a f^4} \]

[Out]

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

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Rubi [A]  time = 0.283706, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3318, 4184, 3717, 2190, 2531, 2282, 6589} \[ -\frac{12 i d^2 (c+d x) \text{Li}_2\left (e^{i (e+f x)}\right )}{a f^3}+\frac{6 d (c+d x)^2 \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac{(c+d x)^3 \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}-\frac{i (c+d x)^3}{a f}+\frac{12 d^3 \text{Li}_3\left (e^{i (e+f x)}\right )}{a f^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(c + d*x)^3)/(a*f) - ((c + d*x)^3*Cot[e/2 + (f*x)/2])/(a*f) + (6*d*(c + d*x)^2*Log[1 - E^(I*(e + f*x))])
/(a*f^2) - ((12*I)*d^2*(c + d*x)*PolyLog[2, E^(I*(e + f*x))])/(a*f^3) + (12*d^3*PolyLog[3, 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 \cos (e+f x)} \, dx &=\frac{\int (c+d x)^3 \csc ^2\left (\frac{e}{2}+\frac{f x}{2}\right ) \, dx}{2 a}\\ &=-\frac{(c+d x)^3 \cot \left (\frac{e}{2}+\frac{f x}{2}\right )}{a f}+\frac{(3 d) \int (c+d x)^2 \cot \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{f x}{2}\right )}{a f}-\frac{(6 i d) \int \frac{e^{2 i \left (\frac{e}{2}+\frac{f x}{2}\right )} (c+d x)^2}{1-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{f x}{2}\right )}{a f}+\frac{6 d (c+d x)^2 \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac{\left (12 d^2\right ) \int (c+d x) \log \left (1-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{f x}{2}\right )}{a f}+\frac{6 d (c+d x)^2 \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac{12 i d^2 (c+d x) \text{Li}_2\left (e^{i (e+f x)}\right )}{a f^3}+\frac{\left (12 i d^3\right ) \int \text{Li}_2\left (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{f x}{2}\right )}{a f}+\frac{6 d (c+d x)^2 \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac{12 i d^2 (c+d x) \text{Li}_2\left (e^{i (e+f x)}\right )}{a f^3}+\frac{\left (12 d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(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{f x}{2}\right )}{a f}+\frac{6 d (c+d x)^2 \log \left (1-e^{i (e+f x)}\right )}{a f^2}-\frac{12 i d^2 (c+d x) \text{Li}_2\left (e^{i (e+f x)}\right )}{a f^3}+\frac{12 d^3 \text{Li}_3\left (e^{i (e+f x)}\right )}{a f^4}\\ \end{align*}

Mathematica [A]  time = 1.23837, size = 164, normalized size = 1.23 \[ \frac{2 \sin \left (\frac{1}{2} (e+f x)\right ) \left (f^3 \csc \left (\frac{e}{2}\right ) (c+d x)^3 \sin \left (\frac{f x}{2}\right )+2 \sin \left (\frac{1}{2} (e+f x)\right ) \left (6 i d^2 f (c+d x) \text{Li}_2\left (e^{-i (e+f x)}\right )-\frac{i f^3 (c+d x)^3}{-1+e^{i e}}+3 d f^2 (c+d x)^2 \log \left (1-e^{-i (e+f x)}\right )+6 d^3 \text{Li}_3\left (e^{-i (e+f x)}\right )\right )\right )}{f^4 (a-a \cos (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sin[(e + f*x)/2]*(f^3*(c + d*x)^3*Csc[e/2]*Sin[(f*x)/2] + 2*(((-I)*f^3*(c + d*x)^3)/(-1 + E^(I*e)) + 3*d*f^
2*(c + d*x)^2*Log[1 - E^((-I)*(e + f*x))] + (6*I)*d^2*f*(c + d*x)*PolyLog[2, E^((-I)*(e + f*x))] + 6*d^3*PolyL
og[3, E^((-I)*(e + f*x))])*Sin[(e + f*x)/2]))/(f^4*(a - a*Cos[e + f*x]))

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Maple [B]  time = 0.46, size = 468, normalized size = 3.5 \begin{align*}{\frac{-12\,i{d}^{2}c{\it polylog} \left ( 2,{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{a{f}^{3}}}-6\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{a{f}^{2}}}+6\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}-1 \right ) }{a{f}^{2}}}+6\,{\frac{{d}^{3}{e}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}-1 \right ) }{{f}^{4}a}}-6\,{\frac{{d}^{3}{e}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{4}a}}-{\frac{6\,ic{d}^{2}{x}^{2}}{af}}-{\frac{2\,i{d}^{3}{x}^{3}}{af}}+{\frac{6\,i{d}^{3}{e}^{2}x}{a{f}^{3}}}-{\frac{12\,i{d}^{3}{\it polylog} \left ( 2,{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{a{f}^{3}}}+6\,{\frac{{d}^{3}\ln \left ( 1-{{\rm e}^{i \left ( fx+e \right ) }} \right ){x}^{2}}{a{f}^{2}}}-6\,{\frac{{d}^{3}\ln \left ( 1-{{\rm e}^{i \left ( fx+e \right ) }} \right ){e}^{2}}{{f}^{4}a}}+{\frac{4\,i{d}^{3}{e}^{3}}{{f}^{4}a}}+12\,{\frac{{d}^{3}{\it polylog} \left ( 3,{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{4}a}}-12\,{\frac{c{d}^{2}e\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}-1 \right ) }{a{f}^{3}}}+12\,{\frac{c{d}^{2}e\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{a{f}^{3}}}+12\,{\frac{c{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{a{f}^{2}}}+12\,{\frac{c{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( fx+e \right ) }} \right ) e}{a{f}^{3}}}-{\frac{12\,i{d}^{2}cex}{a{f}^{2}}}-{\frac{2\,i \left ({d}^{3}{x}^{3}+3\,c{d}^{2}{x}^{2}+3\,{c}^{2}dx+{c}^{3} \right ) }{af \left ({{\rm e}^{i \left ( fx+e \right ) }}-1 \right ) }}-{\frac{6\,ic{d}^{2}{e}^{2}}{a{f}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.06757, size = 1295, normalized size = 9.74 \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*cos(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 1.77678, size = 1173, normalized size = 8.82 \begin{align*} -\frac{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} - 6 \, d^{3}{\rm polylog}\left (3, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) - 6 \, d^{3}{\rm polylog}\left (3, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) -{\left (-6 i \, d^{3} f x - 6 i \, c d^{2} f\right )}{\rm Li}_2\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) -{\left (6 i \, d^{3} f x + 6 i \, c d^{2} f\right )}{\rm Li}_2\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) - 3 \,{\left (d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2} i \, \sin \left (f x + e\right ) + \frac{1}{2}\right ) \sin \left (f x + e\right ) - 3 \,{\left (d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) - \frac{1}{2} i \, \sin \left (f x + e\right ) + \frac{1}{2}\right ) \sin \left (f x + e\right ) - 3 \,{\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x - d^{3} e^{2} + 2 \, c d^{2} e f\right )} \log \left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 3 \,{\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x - d^{3} e^{2} + 2 \, c d^{2} e f\right )} \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) +{\left (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}\right )} \cos \left (f x + e\right )}{a f^{4} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{c^{3}}{\cos{\left (e + f x \right )} - 1}\, dx + \int \frac{d^{3} x^{3}}{\cos{\left (e + f x \right )} - 1}\, dx + \int \frac{3 c d^{2} x^{2}}{\cos{\left (e + f x \right )} - 1}\, dx + \int \frac{3 c^{2} d x}{\cos{\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*cos(f*x+e)),x)

[Out]

-(Integral(c**3/(cos(e + f*x) - 1), x) + Integral(d**3*x**3/(cos(e + f*x) - 1), x) + Integral(3*c*d**2*x**2/(c
os(e + f*x) - 1), x) + Integral(3*c**2*d*x/(cos(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 \cos \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*cos(f*x+e)),x, algorithm="giac")

[Out]

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