3.2.43 \(\int \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx\) [143]

Optimal. Leaf size=54 \[ -\frac {b \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \log (\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2800, 780} \begin {gather*} -\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \log (\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {b \csc (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 780

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2800

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)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int \cot ^3(c+d x) (a+b \sin (c+d x)) \, dx &=\frac {\text {Subst}\left (\int \frac {(a+x) \left (b^2-x^2\right )}{x^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (-1+\frac {a b^2}{x^3}+\frac {b^2}{x^2}-\frac {a}{x}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {b \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \log (\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 60, normalized size = 1.11 \begin {gather*} -\frac {b \csc (c+d x)}{d}-\frac {a \left (\cot ^2(c+d x)+2 \log (\cos (c+d x))+2 \log (\tan (c+d x))\right )}{2 d}-\frac {b \sin (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*Csc[c + d*x])/d) - (a*(Cot[c + d*x]^2 + 2*Log[Cos[c + d*x]] + 2*Log[Tan[c + d*x]]))/(2*d) - (b*Sin[c + d*
x])/d

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Maple [A]
time = 0.22, size = 67, normalized size = 1.24

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos ^{4}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right )}{d}\)	\(67\)
default	\(\frac {a \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos ^{4}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right )}{d}\)	\(67\)
risch	\(i a x +\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a c}{d}-\frac {2 i \left (i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 45, normalized size = 0.83 \begin {gather*} -\frac {2 \, a \log \left (\sin \left (d x + c\right )\right ) + 2 \, b \sin \left (d x + c\right ) + \frac {2 \, b \sin \left (d x + c\right ) + a}{\sin \left (d x + c\right )^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a*log(sin(d*x + c)) + 2*b*sin(d*x + c) + (2*b*sin(d*x + c) + a)/sin(d*x + c)^2)/d

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Fricas [A]
time = 0.37, size = 69, normalized size = 1.28 \begin {gather*} -\frac {2 \, {\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 2 \, {\left (b \cos \left (d x + c\right )^{2} - 2 \, b\right )} \sin \left (d x + c\right ) - a}{2 \, {\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+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(2*(a*cos(d*x + c)^2 - a)*log(1/2*sin(d*x + c)) + 2*(b*cos(d*x + c)^2 - 2*b)*sin(d*x + c) - a)/(d*cos(d*x
+ c)^2 - d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 23.21, size = 60, normalized size = 1.11 \begin {gather*} -\frac {2 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 2 \, b \sin \left (d x + c\right ) - \frac {3 \, a \sin \left (d x + c\right )^{2} - 2 \, b \sin \left (d x + c\right ) - a}{\sin \left (d x + c\right )^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 6.63, size = 146, normalized size = 2.70 \begin {gather*} \frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {10\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{2}}{d\,\left (4\,{\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 {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (b*tan(c/2 + (d*x)/2))/(2*d) - (a/2 + 2*b*tan(c/2 + (d*x)/2) + (a*tan(c/
2 + (d*x)/2)^2)/2 + 10*b*tan(c/2 + (d*x)/2)^3)/(d*(4*tan(c/2 + (d*x)/2)^2 + 4*tan(c/2 + (d*x)/2)^4)) - (a*tan(
c/2 + (d*x)/2)^2)/(8*d) - (a*log(tan(c/2 + (d*x)/2)))/d

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