\(\int \cot (c+d x) (a+a \sin (c+d x))^3 \, dx\) [209]

Optimal result

Integrand size = 19, antiderivative size = 65 \[ \int \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^3 \sin ^3(c+d x)}{3 d} \]

[Out]

a^3*ln(sin(d*x+c))/d+3*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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2786, 45} \[ \int \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

(a^3*Log[Sin[c + d*x]])/d + (3*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a+x)^3}{x} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (3 a^2+\frac {a^3}{x}+3 a x+x^2\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^3 \log (\sin (c+d x))}{d}+\frac {3 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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^3 \sin ^3(c+d x)}{3 d} \]

[In]

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

[Out]

(a^3*Log[Sin[c + d*x]])/d + (3*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)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(-\frac {a^{3} \left (\ln \left (\csc \left (d x +c \right )\right )-\frac {3}{\csc \left (d x +c \right )}-\frac {3}{2 \csc \left (d x +c \right )^{2}}-\frac {1}{3 \csc \left (d x +c \right )^{3}}\right )}{d}\)	\(47\)
default	\(-\frac {a^{3} \left (\ln \left (\csc \left (d x +c \right )\right )-\frac {3}{\csc \left (d x +c \right )}-\frac {3}{2 \csc \left (d x +c \right )^{2}}-\frac {1}{3 \csc \left (d x +c \right )^{3}}\right )}{d}\)	\(47\)
risch	\(-i a^{3} x -\frac {3 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{3} c}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {13 a^{3} \sin \left (d x +c \right )}{4 d}-\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 d}\)	\(103\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.91 \[ \int \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {9 \, a^{3} \cos \left (d x + c\right )^{2} - 6 \, a^{3} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - 10 \, a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

-1/6*(9*a^3*cos(d*x + c)^2 - 6*a^3*log(1/2*sin(d*x + c)) + 2*(a^3*cos(d*x + c)^2 - 10*a^3)*sin(d*x + c))/d

Sympy [F]

\[ \int \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=a^{3} \left (\int \cos {\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 3 \sin {\left (c + d x \right )} \cos {\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**3*(Integral(cos(c + d*x)*csc(c + d*x), x) + Integral(3*sin(c + d*x)*cos(c + d*x)*csc(c + d*x), x) + Integra
l(3*sin(c + d*x)**2*cos(c + d*x)*csc(c + d*x), x) + Integral(sin(c + d*x)**3*cos(c + d*x)*csc(c + d*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} + 6 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 18 \, a^{3} \sin \left (d x + c\right )}{6 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)*(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 + 6*a^3*log(sin(d*x + c)) + 18*a^3*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} + 6 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 18 \, a^{3} \sin \left (d x + c\right )}{6 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)*(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 + 6*a^3*log(abs(sin(d*x + c))) + 18*a^3*sin(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 9.88 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.57 \[ \int \cot (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {10\,a^3\,\sin \left (c+d\,x\right )}{3\,d}-\frac {a^3\,\ln \left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}\right )}{d}+\frac {a^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {3\,a^3\,{\cos \left (c+d\,x\right )}^2}{2\,d}-\frac {a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]

[In]

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

[Out]

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