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

Optimal. Leaf size=134 \[ -\frac {12 i d^2 (c+d x) \text {Li}_2\left (-e^{i (e+f x)}\right )}{a f^3}+\frac {6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\frac {i (c+d x)^3}{a f}+\frac {12 d^3 \text {Li}_3\left (-e^{i (e+f x)}\right )}{a f^4} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3318, 4184, 3719, 2190, 2531, 2282, 6589} \[ -\frac {12 i d^2 (c+d x) \text {Li}_2\left (-e^{i (e+f x)}\right )}{a f^3}+\frac {6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\frac {i (c+d x)^3}{a f}+\frac {12 d^3 \text {Li}_3\left (-e^{i (e+f x)}\right )}{a f^4} \]

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) + (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) + ((c + d*x)^3*Tan[e/2 + (f*x)/2
])/(a*f)

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 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
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 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+\pi }{2}+\frac {f x}{2}\right ) \, dx}{2 a}\\ &=\frac {(c+d x)^3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\frac {(3 d) \int (c+d x)^2 \tan \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 \tan \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 {6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac {(c+d x)^3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}-\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 {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 {(c+d x)^3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\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 {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 {(c+d x)^3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}+\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 {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}+\frac {(c+d x)^3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 151, normalized size = 1.13 \[ \frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) \left ((c+d x)^3 \sin \left (\frac {1}{2} (e+f x)\right )-\frac {i \cos \left (\frac {1}{2} (e+f x)\right ) \left (12 d^2 f (c+d x) \text {Li}_2\left (-e^{i (e+f x)}\right )+f^2 (c+d x)^2 \left (f (c+d x)+6 i d \log \left (1+e^{i (e+f x)}\right )\right )+12 i d^3 \text {Li}_3\left (-e^{i (e+f x)}\right )\right )}{f^3}\right )}{a f (\cos (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.65, size = 418, normalized size = 3.12 \[ \frac {{\left (6 i \, d^{3} f x + 6 i \, c d^{2} f + {\left (6 i \, d^{3} f x + 6 i \, c d^{2} f\right )} \cos \left (f x + e\right )\right )} {\rm Li}_2\left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + {\left (-6 i \, d^{3} f x - 6 i \, c d^{2} f + {\left (-6 i \, d^{3} f x - 6 i \, c d^{2} f\right )} \cos \left (f x + e\right )\right )} {\rm Li}_2\left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 3 \, {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2} + {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) + 3 \, {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2} + {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) + 6 \, {\left (d^{3} \cos \left (f x + e\right ) + d^{3}\right )} {\rm polylog}\left (3, -\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 6 \, {\left (d^{3} \cos \left (f x + e\right ) + d^{3}\right )} {\rm polylog}\left (3, -\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + {\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + c^{3} f^{3}\right )} \sin \left (f x + e\right )}{a f^{4} \cos \left (f x + e\right ) + a f^{4}} \]

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]

((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))*dilog(-cos(f*x + e) + I*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))*dilog(-cos(f*x + e) - I*sin(f*x + e
)) + 3*(d^3*f^2*x^2 + 2*c*d^2*f^2*x + c^2*d*f^2 + (d^3*f^2*x^2 + 2*c*d^2*f^2*x + c^2*d*f^2)*cos(f*x + e))*log(
cos(f*x + e) + I*sin(f*x + e) + 1) + 3*(d^3*f^2*x^2 + 2*c*d^2*f^2*x + c^2*d*f^2 + (d^3*f^2*x^2 + 2*c*d^2*f^2*x
+ c^2*d*f^2)*cos(f*x + e))*log(cos(f*x + e) - I*sin(f*x + e) + 1) + 6*(d^3*cos(f*x + e) + d^3)*polylog(3, -co
s(f*x + e) + I*sin(f*x + e)) + 6*(d^3*cos(f*x + e) + d^3)*polylog(3, -cos(f*x + e) - I*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)

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

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)

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maple [B]  time = 0.16, size = 364, normalized size = 2.72 \[ \frac {2 i \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{f a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}+\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{a \,f^{2}}-\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{2}}-\frac {6 d^{3} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{4}}-\frac {12 i d^{3} \polylog \left (2, -{\mathrm e}^{i \left (f x +e \right )}\right ) x}{a \,f^{3}}+\frac {12 d^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{a \,f^{2}}+\frac {6 i d^{3} e^{2} x}{a \,f^{3}}+\frac {4 i d^{3} e^{3}}{a \,f^{4}}+\frac {6 d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) x^{2}}{a \,f^{2}}-\frac {12 i d^{2} c e x}{a \,f^{2}}+\frac {12 d^{3} \polylog \left (3, -{\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{4}}+\frac {12 d^{2} c e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {2 i d^{3} x^{3}}{a f}-\frac {6 i d^{2} c \,e^{2}}{a \,f^{3}}-\frac {12 i d^{2} c \polylog \left (2, -{\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {6 i d^{2} c \,x^{2}}{a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.16, size = 936, normalized size = 6.99 \[ \text {result too large to display} \]

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*sin(f*x + e)/(a*(cos(f*x + e) + 1)) - 3*c*d^2*e^2*sin(f*x + e)/(a*f^2*(cos(f*x + e) + 1)) +
3*c^2*d*e*sin(f*x + e)/(a*f*(cos(f*x + e) + 1)) + (2*d^3*e^3 - (6*(f*x + e)^2*d^3 + 6*d^3*e^2 - 12*(d^3*e - c*
d^2*f)*(f*x + e) + 6*((f*x + e)^2*d^3 + d^3*e^2 - 2*(d^3*e - c*d^2*f)*(f*x + e))*cos(f*x + e) + (6*I*(f*x + e)
^2*d^3 + 6*I*d^3*e^2 + (-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 + 1
2*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 + s
in(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*sin(f*x + e) - I*a*f^3))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {c^{3}}{\cos {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{3} x^{3}}{\cos {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c d^{2} x^{2}}{\cos {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c^{2} d x}{\cos {\left (e + f x \right )} + 1}\, dx}{a} \]

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

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