3.1.28 \(\int \cot ^3(c+d x) (a+a \sin (c+d x))^3 \, dx\) [28]

Optimal. Leaf size=98 \[ -\frac {3 a^3 \csc (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}+\frac {2 a^3 \log (\sin (c+d x))}{d}-\frac {2 a^3 \sin (c+d x)}{d}-\frac {3 a^3 \sin ^2(c+d x)}{2 d}-\frac {a^3 \sin ^3(c+d x)}{3 d} \]

[Out]

-3*a^3*csc(d*x+c)/d-1/2*a^3*csc(d*x+c)^2/d+2*a^3*ln(sin(d*x+c))/d-2*a^3*sin(d*x+c)/d-3/2*a^3*sin(d*x+c)^2/d-1/
3*a^3*sin(d*x+c)^3/d

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Rubi [A]
time = 0.05, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2786, 76} \begin {gather*} -\frac {a^3 \sin ^3(c+d x)}{3 d}-\frac {3 a^3 \sin ^2(c+d x)}{2 d}-\frac {2 a^3 \sin (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {3 a^3 \csc (c+d x)}{d}+\frac {2 a^3 \log (\sin (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + a*Sin[c + d*x])^3,x]

[Out]

(-3*a^3*Csc[c + d*x])/d - (a^3*Csc[c + d*x]^2)/(2*d) + (2*a^3*Log[Sin[c + d*x]])/d - (2*a^3*Sin[c + d*x])/d -
(3*a^3*Sin[c + d*x]^2)/(2*d) - (a^3*Sin[c + d*x]^3)/(3*d)

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rule 2786

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[x^p*((a + x)^(m - (p + 1)/2)/(a - x)^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int \cot ^3(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac {\text {Subst}\left (\int \frac {(a-x) (a+x)^4}{x^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (-2 a^2+\frac {a^5}{x^3}+\frac {3 a^4}{x^2}+\frac {2 a^3}{x}-3 a x-x^2\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {3 a^3 \csc (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}+\frac {2 a^3 \log (\sin (c+d x))}{d}-\frac {2 a^3 \sin (c+d x)}{d}-\frac {3 a^3 \sin ^2(c+d x)}{2 d}-\frac {a^3 \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 67, normalized size = 0.68 \begin {gather*} -\frac {a^3 \left (30+18 \csc (c+d x)+3 \csc ^2(c+d x)-12 \log (\sin (c+d x))+12 \sin (c+d x)+9 \sin ^2(c+d x)+2 \sin ^3(c+d x)\right )}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*(a + a*Sin[c + d*x])^3,x]

[Out]

-1/6*(a^3*(30 + 18*Csc[c + d*x] + 3*Csc[c + d*x]^2 - 12*Log[Sin[c + d*x]] + 12*Sin[c + d*x] + 9*Sin[c + d*x]^2
+ 2*Sin[c + d*x]^3))/d

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Maple [A]
time = 0.20, size = 116, normalized size = 1.18

method	result	size
derivativedivides	\(\frac {\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 a^{3} \left (\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+3 a^{3} \left (-\frac {\cos ^{4}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right )+a^{3} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(116\)
default	\(\frac {\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 a^{3} \left (\frac {\left (\cos ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+3 a^{3} \left (-\frac {\cos ^{4}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right )+a^{3} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(116\)
risch	\(-2 i a^{3} x -\frac {i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {3 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {9 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {9 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {i a^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {4 i a^{3} c}{d}-\frac {2 i a^{3} \left (i {\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {2 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(202\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/3*a^3*(2+cos(d*x+c)^2)*sin(d*x+c)+3*a^3*(1/2*cos(d*x+c)^2+ln(sin(d*x+c)))+3*a^3*(-1/sin(d*x+c)*cos(d*x+
c)^4-(2+cos(d*x+c)^2)*sin(d*x+c))+a^3*(-1/2*cot(d*x+c)^2-ln(sin(d*x+c))))

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Maxima [A]
time = 0.29, size = 80, normalized size = 0.82 \begin {gather*} -\frac {2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 12 \, a^{3} \sin \left (d x + c\right ) + \frac {3 \, {\left (6 \, a^{3} \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*a^3*sin(d*x + c)^3 + 9*a^3*sin(d*x + c)^2 - 12*a^3*log(sin(d*x + c)) + 12*a^3*sin(d*x + c) + 3*(6*a^3*
sin(d*x + c) + a^3)/sin(d*x + c)^2)/d

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Fricas [A]
time = 0.38, size = 118, normalized size = 1.20 \begin {gather*} \frac {18 \, a^{3} \cos \left (d x + c\right )^{4} - 27 \, a^{3} \cos \left (d x + c\right )^{2} + 15 \, a^{3} + 24 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 4 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 8 \, a^{3} \cos \left (d x + c\right )^{2} + 16 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(18*a^3*cos(d*x + c)^4 - 27*a^3*cos(d*x + c)^2 + 15*a^3 + 24*(a^3*cos(d*x + c)^2 - a^3)*log(1/2*sin(d*x +
c)) + 4*(a^3*cos(d*x + c)^4 - 8*a^3*cos(d*x + c)^2 + 16*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2 - d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int 3 \sin {\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx + \int \cot ^{3}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(a+a*sin(d*x+c))**3,x)

[Out]

a**3*(Integral(3*sin(c + d*x)*cot(c + d*x)**3, x) + Integral(3*sin(c + d*x)**2*cot(c + d*x)**3, x) + Integral(
sin(c + d*x)**3*cot(c + d*x)**3, x) + Integral(cot(c + d*x)**3, x))

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Giac [A]
time = 8.26, size = 94, normalized size = 0.96 \begin {gather*} -\frac {2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 12 \, a^{3} \sin \left (d x + c\right ) + \frac {3 \, {\left (6 \, a^{3} \sin \left (d x + c\right )^{2} + 6 \, a^{3} \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*a^3*sin(d*x + c)^3 + 9*a^3*sin(d*x + c)^2 - 12*a^3*log(abs(sin(d*x + c))) + 12*a^3*sin(d*x + c) + 3*(6
*a^3*sin(d*x + c)^2 + 6*a^3*sin(d*x + c) + a^3)/sin(d*x + c)^2)/d

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Mupad [B]
time = 6.76, size = 253, normalized size = 2.58 \begin {gather*} \frac {2\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {22\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {49\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {182\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+\frac {51\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+34\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+6\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^3}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {2\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^3*(a + a*sin(c + d*x))^3,x)

[Out]

(2*a^3*log(tan(c/2 + (d*x)/2)))/d - (a^3*tan(c/2 + (d*x)/2)^2)/(8*d) - ((3*a^3*tan(c/2 + (d*x)/2)^2)/2 + 34*a^
3*tan(c/2 + (d*x)/2)^3 + (51*a^3*tan(c/2 + (d*x)/2)^4)/2 + (182*a^3*tan(c/2 + (d*x)/2)^5)/3 + (49*a^3*tan(c/2
+ (d*x)/2)^6)/2 + 22*a^3*tan(c/2 + (d*x)/2)^7 + a^3/2 + 6*a^3*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 +
12*tan(c/2 + (d*x)/2)^4 + 12*tan(c/2 + (d*x)/2)^6 + 4*tan(c/2 + (d*x)/2)^8)) - (3*a^3*tan(c/2 + (d*x)/2))/(2*
d) - (2*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d

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