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

Optimal. Leaf size=118 \[ -\frac{a^2 \cot ^5(c+d x)}{5 d}-\frac{a^2 \cot ^3(c+d x)}{3 d}+\frac{a^2 \cot (c+d x)}{d}-\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{a^2 \cot ^3(c+d x) \csc (c+d x)}{2 d}+\frac{3 a^2 \cot (c+d x) \csc (c+d x)}{4 d}+a^2 x \]

[Out]

a^2*x - (3*a^2*ArcTanh[Cos[c + d*x]])/(4*d) + (a^2*Cot[c + d*x])/d - (a^2*Cot[c + d*x]^3)/(3*d) - (a^2*Cot[c +
d*x]^5)/(5*d) + (3*a^2*Cot[c + d*x]*Csc[c + d*x])/(4*d) - (a^2*Cot[c + d*x]^3*Csc[c + d*x])/(2*d)

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Rubi [A]  time = 0.191195, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2873, 3473, 8, 2611, 3770, 2607, 30} \[ -\frac{a^2 \cot ^5(c+d x)}{5 d}-\frac{a^2 \cot ^3(c+d x)}{3 d}+\frac{a^2 \cot (c+d x)}{d}-\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{a^2 \cot ^3(c+d x) \csc (c+d x)}{2 d}+\frac{3 a^2 \cot (c+d x) \csc (c+d x)}{4 d}+a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*x - (3*a^2*ArcTanh[Cos[c + d*x]])/(4*d) + (a^2*Cot[c + d*x])/d - (a^2*Cot[c + d*x]^3)/(3*d) - (a^2*Cot[c +
d*x]^5)/(5*d) + (3*a^2*Cot[c + d*x]*Csc[c + d*x])/(4*d) - (a^2*Cot[c + d*x]^3*Csc[c + d*x])/(2*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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

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

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]

Rubi steps

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

Mathematica [A]  time = 0.504887, size = 200, normalized size = 1.69 \[ \frac{a^2 \left (-272 \tan \left (\frac{1}{2} (c+d x)\right )+272 \cot \left (\frac{1}{2} (c+d x)\right )+150 \csc ^2\left (\frac{1}{2} (c+d x)\right )+15 \sec ^4\left (\frac{1}{2} (c+d x)\right )-150 \sec ^2\left (\frac{1}{2} (c+d x)\right )+360 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-360 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+96 \sin ^6\left (\frac{1}{2} (c+d x)\right ) \csc ^5(c+d x)-8 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-\frac{3}{2} \sin (c+d x) \csc ^6\left (\frac{1}{2} (c+d x)\right )+\frac{1}{2} (\sin (c+d x)-30) \csc ^4\left (\frac{1}{2} (c+d x)\right )+480 c+480 d x\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(480*c + 480*d*x + 272*Cot[(c + d*x)/2] + 150*Csc[(c + d*x)/2]^2 - 360*Log[Cos[(c + d*x)/2]] + 360*Log[Si
n[(c + d*x)/2]] - 150*Sec[(c + d*x)/2]^2 + 15*Sec[(c + d*x)/2]^4 - 8*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 96*Cs
c[c + d*x]^5*Sin[(c + d*x)/2]^6 + (Csc[(c + d*x)/2]^4*(-30 + Sin[c + d*x]))/2 - (3*Csc[(c + d*x)/2]^6*Sin[c +
d*x])/2 - 272*Tan[(c + d*x)/2]))/(480*d)

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Maple [A]  time = 0.08, size = 170, normalized size = 1.4 \begin{align*} -{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}+{a}^{2}x+{\frac{c{a}^{2}}{d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{a}^{2}\cos \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{4\,d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{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)^4*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x)

[Out]

-1/3*a^2*cot(d*x+c)^3/d+a^2*cot(d*x+c)/d+a^2*x+1/d*c*a^2-1/2/d*a^2/sin(d*x+c)^4*cos(d*x+c)^5+1/4/d*a^2/sin(d*x
+c)^2*cos(d*x+c)^5+1/4*a^2*cos(d*x+c)^3/d+3/4*a^2*cos(d*x+c)/d+3/4/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-1/5/d*a^2/s
in(d*x+c)^5*cos(d*x+c)^5

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Maxima [A]  time = 1.61636, size = 167, normalized size = 1.42 \begin{align*} \frac{40 \,{\left (3 \, d x + 3 \, c + \frac{3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a^{2} - 15 \, a^{2}{\left (\frac{2 \,{\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - \frac{24 \, a^{2}}{\tan \left (d x + c\right )^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/120*(40*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a^2 - 15*a^2*(2*(5*cos(d*x + c)^3 - 3*cos(d*x
+ c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)) - 24*a^2/ta
n(d*x + c)^5)/d

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Fricas [B]  time = 1.53048, size = 621, normalized size = 5.26 \begin{align*} \frac{136 \, a^{2} \cos \left (d x + c\right )^{5} - 280 \, a^{2} \cos \left (d x + c\right )^{3} + 120 \, a^{2} \cos \left (d x + c\right ) - 45 \,{\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 45 \,{\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 30 \,{\left (4 \, a^{2} d x \cos \left (d x + c\right )^{4} - 8 \, a^{2} d x \cos \left (d x + c\right )^{2} - 5 \, a^{2} \cos \left (d x + c\right )^{3} + 4 \, a^{2} d x + 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \,{\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)^4*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/120*(136*a^2*cos(d*x + c)^5 - 280*a^2*cos(d*x + c)^3 + 120*a^2*cos(d*x + c) - 45*(a^2*cos(d*x + c)^4 - 2*a^2
*cos(d*x + c)^2 + a^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 45*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^
2 + a^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 30*(4*a^2*d*x*cos(d*x + c)^4 - 8*a^2*d*x*cos(d*x + c)^2 -
5*a^2*cos(d*x + c)^3 + 4*a^2*d*x + 3*a^2*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))

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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)**4*csc(d*x+c)**6*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.27501, size = 279, normalized size = 2.36 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 5 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 120 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 480 \,{\left (d x + c\right )} a^{2} + 360 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 270 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{822 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 270 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 120 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/480*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 15*a^2*tan(1/2*d*x + 1/2*c)^4 + 5*a^2*tan(1/2*d*x + 1/2*c)^3 - 120*a^2*t
an(1/2*d*x + 1/2*c)^2 + 480*(d*x + c)*a^2 + 360*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 270*a^2*tan(1/2*d*x + 1/2
*c) - (822*a^2*tan(1/2*d*x + 1/2*c)^5 - 270*a^2*tan(1/2*d*x + 1/2*c)^4 - 120*a^2*tan(1/2*d*x + 1/2*c)^3 + 5*a^
2*tan(1/2*d*x + 1/2*c)^2 + 15*a^2*tan(1/2*d*x + 1/2*c) + 3*a^2)/tan(1/2*d*x + 1/2*c)^5)/d