\(\int \cot ^2(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\) [290]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 29, antiderivative size = 91 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-3 a^3 x+\frac {a^3 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {a^3 \cos (c+d x)}{d}-\frac {3 a^3 \cot (c+d x)}{d}-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{2 d} \]

[Out]

-3*a^3*x+1/2*a^3*arctanh(cos(d*x+c))/d+a^3*cos(d*x+c)/d-3*a^3*cot(d*x+c)/d-1/3*a^3*cot(d*x+c)^3/d-3/2*a^3*cot(
d*x+c)*csc(d*x+c)/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2951, 3855, 3852, 8, 3853, 2718} \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {a^3 \cos (c+d x)}{d}-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {3 a^3 \cot (c+d x)}{d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}-3 a^3 x \]

[In]

Int[Cot[c + d*x]^2*Csc[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

-3*a^3*x + (a^3*ArcTanh[Cos[c + d*x]])/(2*d) + (a^3*Cos[c + d*x])/d - (3*a^3*Cot[c + d*x])/d - (a^3*Cot[c + d*
x]^3)/(3*d) - (3*a^3*Cot[c + d*x]*Csc[c + d*x])/(2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2951

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
 1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (-3 a^5-2 a^5 \csc (c+d x)+2 a^5 \csc ^2(c+d x)+3 a^5 \csc ^3(c+d x)+a^5 \csc ^4(c+d x)-a^5 \sin (c+d x)\right ) \, dx}{a^2} \\ & = -3 a^3 x+a^3 \int \csc ^4(c+d x) \, dx-a^3 \int \sin (c+d x) \, dx-\left (2 a^3\right ) \int \csc (c+d x) \, dx+\left (2 a^3\right ) \int \csc ^2(c+d x) \, dx+\left (3 a^3\right ) \int \csc ^3(c+d x) \, dx \\ & = -3 a^3 x+\frac {2 a^3 \text {arctanh}(\cos (c+d x))}{d}+\frac {a^3 \cos (c+d x)}{d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {1}{2} \left (3 a^3\right ) \int \csc (c+d x) \, dx-\frac {a^3 \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}-\frac {\left (2 a^3\right ) \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d} \\ & = -3 a^3 x+\frac {a^3 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {a^3 \cos (c+d x)}{d}-\frac {3 a^3 \cot (c+d x)}{d}-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.63 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (-72 c-72 d x+24 \cos (c+d x)-32 \cot \left (\frac {1}{2} (c+d x)\right )-9 \csc ^2\left (\frac {1}{2} (c+d x)\right )+12 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-12 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 \sec ^2\left (\frac {1}{2} (c+d x)\right )+8 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-\frac {1}{2} \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)+32 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{24 d} \]

[In]

Integrate[Cot[c + d*x]^2*Csc[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*(-72*c - 72*d*x + 24*Cos[c + d*x] - 32*Cot[(c + d*x)/2] - 9*Csc[(c + d*x)/2]^2 + 12*Log[Cos[(c + d*x)/2]]
 - 12*Log[Sin[(c + d*x)/2]] + 9*Sec[(c + d*x)/2]^2 + 8*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - (Csc[(c + d*x)/2]^4
*Sin[c + d*x])/2 + 32*Tan[(c + d*x)/2]))/(24*d)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.37

method result size
derivativedivides \(\frac {a^{3} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+3 a^{3} \left (-\frac {\cos ^{3}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )-\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{3 \sin \left (d x +c \right )^{3}}}{d}\) \(125\)
default \(\frac {a^{3} \left (\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+3 a^{3} \left (-\frac {\cos ^{3}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )-\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{3 \sin \left (d x +c \right )^{3}}}{d}\) \(125\)
parallelrisch \(-\frac {\left (\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {\left (\sec ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12}-\frac {5 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right ) \left (\csc ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {3 \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+8 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {19}{4}\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {16 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\left (2 \cos \left (d x +c \right )-\frac {21}{4}\right ) \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 d x \right ) a^{3}}{2 d}\) \(134\)
risch \(-3 a^{3} x +\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a^{3} \left (-12 i {\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{5 i \left (d x +c \right )}+36 i {\mathrm e}^{2 i \left (d x +c \right )}-16 i-9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}\) \(152\)
norman \(\frac {-\frac {a^{3}}{24 d}-\frac {3 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {3 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {23 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {23 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {3 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {3 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-3 a^{3} x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9 a^{3} x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9 a^{3} x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{3} x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {5 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\) \(312\)

[In]

int(cos(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (85) = 170\).

Time = 0.29 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.98 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {32 \, a^{3} \cos \left (d x + c\right )^{3} - 36 \, a^{3} \cos \left (d x + c\right ) - 3 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 6 \, {\left (6 \, a^{3} d x \cos \left (d x + c\right )^{2} - 2 \, a^{3} \cos \left (d x + c\right )^{3} - 6 \, a^{3} d x - a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/12*(32*a^3*cos(d*x + c)^3 - 36*a^3*cos(d*x + c) - 3*(a^3*cos(d*x + c)^2 - a^3)*log(1/2*cos(d*x + c) + 1/2)*
sin(d*x + c) + 3*(a^3*cos(d*x + c)^2 - a^3)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 6*(6*a^3*d*x*cos(d*x +
 c)^2 - 2*a^3*cos(d*x + c)^3 - 6*a^3*d*x - a^3*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c
))

Sympy [F(-1)]

Timed out. \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**4*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.29 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {36 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{3} - 9 \, a^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 6 \, a^{3} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {4 \, a^{3}}{\tan \left (d x + c\right )^{3}}}{12 \, d} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/12*(36*(d*x + c + 1/tan(d*x + c))*a^3 - 9*a^3*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + log(cos(d*x + c) + 1)
- log(cos(d*x + c) - 1)) - 6*a^3*(2*cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) + 4*a^3/tan(
d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.77 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 72 \, {\left (d x + c\right )} a^{3} - 12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 33 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {48 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {22 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 33 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/24*(a^3*tan(1/2*d*x + 1/2*c)^3 + 9*a^3*tan(1/2*d*x + 1/2*c)^2 - 72*(d*x + c)*a^3 - 12*a^3*log(abs(tan(1/2*d*
x + 1/2*c))) + 33*a^3*tan(1/2*d*x + 1/2*c) + 48*a^3/(tan(1/2*d*x + 1/2*c)^2 + 1) + (22*a^3*tan(1/2*d*x + 1/2*c
)^3 - 33*a^3*tan(1/2*d*x + 1/2*c)^2 - 9*a^3*tan(1/2*d*x + 1/2*c) - a^3)/tan(1/2*d*x + 1/2*c)^3)/d

Mupad [B] (verification not implemented)

Time = 9.29 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.74 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {6\,a^3\,\mathrm {atan}\left (\frac {36\,a^6}{6\,a^6-36\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {6\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{6\,a^6-36\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {34\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^3}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}+\frac {11\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d} \]

[In]

int((cos(c + d*x)^2*(a + a*sin(c + d*x))^3)/sin(c + d*x)^4,x)

[Out]

(3*a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^3*tan(c/2 + (d*x)/2)^3)/(24*d) - (a^3*log(tan(c/2 + (d*x)/2)))/(2*d) -
 (6*a^3*atan((36*a^6)/(6*a^6 - 36*a^6*tan(c/2 + (d*x)/2)) + (6*a^6*tan(c/2 + (d*x)/2))/(6*a^6 - 36*a^6*tan(c/2
 + (d*x)/2))))/d - ((34*a^3*tan(c/2 + (d*x)/2)^2)/3 - 13*a^3*tan(c/2 + (d*x)/2)^3 + 11*a^3*tan(c/2 + (d*x)/2)^
4 + a^3/3 + 3*a^3*tan(c/2 + (d*x)/2))/(d*(8*tan(c/2 + (d*x)/2)^3 + 8*tan(c/2 + (d*x)/2)^5)) + (11*a^3*tan(c/2
+ (d*x)/2))/(8*d)