3.292 \(\int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=100 \[ -\frac{a^3 \cot ^5(c+d x)}{5 d}-\frac{4 a^3 \cot ^3(c+d x)}{3 d}+\frac{7 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{3 a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{a^3 \cot (c+d x) \csc (c+d x)}{8 d} \]

[Out]

(7*a^3*ArcTanh[Cos[c + d*x]])/(8*d) - (4*a^3*Cot[c + d*x]^3)/(3*d) - (a^3*Cot[c + d*x]^5)/(5*d) - (a^3*Cot[c +
d*x]*Csc[c + d*x])/(8*d) - (3*a^3*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.240212, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2873, 2611, 3770, 2607, 30, 3768, 14} \[ -\frac{a^3 \cot ^5(c+d x)}{5 d}-\frac{4 a^3 \cot ^3(c+d x)}{3 d}+\frac{7 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{3 a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{a^3 \cot (c+d x) \csc (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*Csc[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

(7*a^3*ArcTanh[Cos[c + d*x]])/(8*d) - (4*a^3*Cot[c + d*x]^3)/(3*d) - (a^3*Cot[c + d*x]^5)/(5*d) - (a^3*Cot[c +
d*x]*Csc[c + d*x])/(8*d) - (3*a^3*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d)

Rule 2873

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

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

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

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3768

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \cot ^2(c+d x) \csc ^4(c+d x) (a+a \sin (c+d x))^3 \, dx &=\int \left (a^3 \cot ^2(c+d x) \csc (c+d x)+3 a^3 \cot ^2(c+d x) \csc ^2(c+d x)+3 a^3 \cot ^2(c+d x) \csc ^3(c+d x)+a^3 \cot ^2(c+d x) \csc ^4(c+d x)\right ) \, dx\\ &=a^3 \int \cot ^2(c+d x) \csc (c+d x) \, dx+a^3 \int \cot ^2(c+d x) \csc ^4(c+d x) \, dx+\left (3 a^3\right ) \int \cot ^2(c+d x) \csc ^2(c+d x) \, dx+\left (3 a^3\right ) \int \cot ^2(c+d x) \csc ^3(c+d x) \, dx\\ &=-\frac{a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{3 a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{1}{2} a^3 \int \csc (c+d x) \, dx-\frac{1}{4} \left (3 a^3\right ) \int \csc ^3(c+d x) \, dx+\frac{a^3 \operatorname{Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{d}+\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,-\cot (c+d x)\right )}{d}\\ &=\frac{a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^3 \cot ^3(c+d x)}{d}-\frac{a^3 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{3 a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{1}{8} \left (3 a^3\right ) \int \csc (c+d x) \, dx+\frac{a^3 \operatorname{Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,-\cot (c+d x)\right )}{d}\\ &=\frac{7 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{4 a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot ^5(c+d x)}{5 d}-\frac{a^3 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{3 a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [B]  time = 0.130221, size = 267, normalized size = 2.67 \[ a^3 \left (-\frac{17 \tan \left (\frac{1}{2} (c+d x)\right )}{30 d}+\frac{17 \cot \left (\frac{1}{2} (c+d x)\right )}{30 d}-\frac{3 \csc ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{3 \sec ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}-\frac{7 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}+\frac{7 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^4\left (\frac{1}{2} (c+d x)\right )}{160 d}-\frac{59 \cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{480 d}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right )}{160 d}+\frac{59 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{480 d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*Csc[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

a^3*((17*Cot[(c + d*x)/2])/(30*d) - Csc[(c + d*x)/2]^2/(32*d) - (59*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(480*
d) - (3*Csc[(c + d*x)/2]^4)/(64*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^4)/(160*d) + (7*Log[Cos[(c + d*x)/2]])
/(8*d) - (7*Log[Sin[(c + d*x)/2]])/(8*d) + Sec[(c + d*x)/2]^2/(32*d) + (3*Sec[(c + d*x)/2]^4)/(64*d) - (17*Tan
[(c + d*x)/2])/(30*d) + (59*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(480*d) + (Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2
])/(160*d))

________________________________________________________________________________________

Maple [A]  time = 0.082, size = 136, normalized size = 1.4 \begin{align*} -{\frac{7\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{7\,{a}^{3}\cos \left ( dx+c \right ) }{8\,d}}-{\frac{7\,{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}}-{\frac{17\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x)

[Out]

-7/8/d*a^3/sin(d*x+c)^2*cos(d*x+c)^3-7/8*a^3*cos(d*x+c)/d-7/8/d*a^3*ln(csc(d*x+c)-cot(d*x+c))-17/15/d*a^3/sin(
d*x+c)^3*cos(d*x+c)^3-3/4/d*a^3/sin(d*x+c)^4*cos(d*x+c)^3-1/5/d*a^3/sin(d*x+c)^5*cos(d*x+c)^3

________________________________________________________________________________________

Maxima [A]  time = 1.13744, size = 209, normalized size = 2.09 \begin{align*} -\frac{45 \, a^{3}{\left (\frac{2 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \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 )} - 60 \, 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 )} + \frac{240 \, a^{3}}{\tan \left (d x + c\right )^{3}} + \frac{16 \,{\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{3}}{\tan \left (d x + c\right )^{5}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/240*(45*a^3*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - log(cos(d*x + c) +
1) + log(cos(d*x + c) - 1)) - 60*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)) + 240*a^3/tan(d*x + c)^3 + 16*(5*tan(d*x + c)^2 + 3)*a^3/tan(d*x + c)^5)/d

________________________________________________________________________________________

Fricas [B]  time = 1.73164, size = 501, normalized size = 5.01 \begin{align*} \frac{272 \, a^{3} \cos \left (d x + c\right )^{5} - 320 \, a^{3} \cos \left (d x + c\right )^{3} + 105 \,{\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, 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 ) - 105 \,{\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, 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 ) + 30 \,{\left (a^{3} \cos \left (d x + c\right )^{3} - 7 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/240*(272*a^3*cos(d*x + c)^5 - 320*a^3*cos(d*x + c)^3 + 105*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^2 + a^3)
*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 105*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^2 + a^3)*log(-1/2*cos
(d*x + c) + 1/2)*sin(d*x + c) + 30*(a^3*cos(d*x + c)^3 - 7*a^3*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^4
- 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**6*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.37693, size = 265, normalized size = 2.65 \begin{align*} \frac{6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 45 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 130 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 840 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 420 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{1918 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 420 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 120 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 130 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 45 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^6*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/960*(6*a^3*tan(1/2*d*x + 1/2*c)^5 + 45*a^3*tan(1/2*d*x + 1/2*c)^4 + 130*a^3*tan(1/2*d*x + 1/2*c)^3 + 120*a^3
*tan(1/2*d*x + 1/2*c)^2 - 840*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 420*a^3*tan(1/2*d*x + 1/2*c) + (1918*a^3*ta
n(1/2*d*x + 1/2*c)^5 + 420*a^3*tan(1/2*d*x + 1/2*c)^4 - 120*a^3*tan(1/2*d*x + 1/2*c)^3 - 130*a^3*tan(1/2*d*x +
1/2*c)^2 - 45*a^3*tan(1/2*d*x + 1/2*c) - 6*a^3)/tan(1/2*d*x + 1/2*c)^5)/d