\(\int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx\) [210]

Optimal result

Integrand size = 25, antiderivative size = 62 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \csc (c+d x)}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2912, 12, 45} \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \sin ^2(c+d x)}{2 d}+\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \csc (c+d x)}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \csc (c+d x)}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(-\frac {a^{3} \left (\csc \left (d x +c \right )+3 \ln \left (\csc \left (d x +c \right )\right )-\frac {3}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\)	\(45\)
default	\(-\frac {a^{3} \left (\csc \left (d x +c \right )+3 \ln \left (\csc \left (d x +c \right )\right )-\frac {3}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\)	\(45\)
parallelrisch	\(-\frac {a^{3} \left (12 \cot \left (\frac {d x}{2}+\frac {c}{2}\right ) \cos \left (d x +c \right )+2 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\cos \left (2 d x +2 c \right )-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}\)	\(94\)
risch	\(-3 i a^{3} x -\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {3 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {6 i a^{3} c}{d}-\frac {2 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(140\)
norman	\(\frac {\frac {2 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3}}{2 d}+\frac {4 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {9 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(192\)

[In]

int(cos(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.26 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {12 \, a^{3} \cos \left (d x + c\right )^{2} - 12 \, a^{3} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 8 \, a^{3} + {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(cos(d*x+c)*csc(d*x+c)**2*(a+a*sin(d*x+c))**3,x)

[Out]

a**3*(Integral(cos(c + d*x)*csc(c + d*x)**2, x) + Integral(3*sin(c + d*x)*cos(c + d*x)*csc(c + d*x)**2, x) + I
ntegral(3*sin(c + d*x)**2*cos(c + d*x)*csc(c + d*x)**2, x) + Integral(sin(c + d*x)**3*cos(c + d*x)*csc(c + d*x
)**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.87 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \sin \left (d x + c\right )^{2} + 6 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 6 \, a^{3} \sin \left (d x + c\right ) - \frac {2 \, a^{3}}{\sin \left (d x + c\right )}}{2 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.89 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \sin \left (d x + c\right )^{2} + 6 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 6 \, a^{3} \sin \left (d x + c\right ) - \frac {2 \, a^{3}}{\sin \left (d x + c\right )}}{2 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.34 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.52 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {3\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {3\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

int((cos(c + d*x)*(a + a*sin(c + d*x))^3)/sin(c + d*x)^2,x)

[Out]

(3*a^3*log(tan(c/2 + (d*x)/2)))/d + (10*a^3*tan(c/2 + (d*x)/2)^2 + 4*a^3*tan(c/2 + (d*x)/2)^3 + 11*a^3*tan(c/2
+ (d*x)/2)^4 - a^3)/(d*(2*tan(c/2 + (d*x)/2) + 4*tan(c/2 + (d*x)/2)^3 + 2*tan(c/2 + (d*x)/2)^5)) - (a^3*tan(c
/2 + (d*x)/2))/(2*d) - (3*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d