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

Optimal. Leaf size=116 \[ -\frac{4 a^4 \cos ^3(c+d x)}{3 d}+\frac{4 a^4 \cos (c+d x)}{d}-\frac{a^4 \cot (c+d x)}{d}+\frac{a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac{23 a^4 \sin (c+d x) \cos (c+d x)}{8 d}-\frac{4 a^4 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{17 a^4 x}{8} \]

[Out]

(17*a^4*x)/8 - (4*a^4*ArcTanh[Cos[c + d*x]])/d + (4*a^4*Cos[c + d*x])/d - (4*a^4*Cos[c + d*x]^3)/(3*d) - (a^4*
Cot[c + d*x])/d + (23*a^4*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^4*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.168284, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2709, 3770, 3767, 8, 2635, 2633} \[ -\frac{4 a^4 \cos ^3(c+d x)}{3 d}+\frac{4 a^4 \cos (c+d x)}{d}-\frac{a^4 \cot (c+d x)}{d}+\frac{a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac{23 a^4 \sin (c+d x) \cos (c+d x)}{8 d}-\frac{4 a^4 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{17 a^4 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(17*a^4*x)/8 - (4*a^4*ArcTanh[Cos[c + d*x]])/d + (4*a^4*Cos[c + d*x])/d - (4*a^4*Cos[c + d*x]^3)/(3*d) - (a^4*
Cot[c + d*x])/d + (23*a^4*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^4*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d)

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3770

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

Rule 3767

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 8

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

Rule 2635

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

Rule 2633

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

Rubi steps

\begin{align*} \int \cot ^2(c+d x) (a+a \sin (c+d x))^4 \, dx &=\frac{\int \left (5 a^6+4 a^6 \csc (c+d x)+a^6 \csc ^2(c+d x)-5 a^6 \sin ^2(c+d x)-4 a^6 \sin ^3(c+d x)-a^6 \sin ^4(c+d x)\right ) \, dx}{a^2}\\ &=5 a^4 x+a^4 \int \csc ^2(c+d x) \, dx-a^4 \int \sin ^4(c+d x) \, dx+\left (4 a^4\right ) \int \csc (c+d x) \, dx-\left (4 a^4\right ) \int \sin ^3(c+d x) \, dx-\left (5 a^4\right ) \int \sin ^2(c+d x) \, dx\\ &=5 a^4 x-\frac{4 a^4 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{5 a^4 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac{1}{4} \left (3 a^4\right ) \int \sin ^2(c+d x) \, dx-\frac{1}{2} \left (5 a^4\right ) \int 1 \, dx-\frac{a^4 \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}+\frac{\left (4 a^4\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{5 a^4 x}{2}-\frac{4 a^4 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{4 a^4 \cos (c+d x)}{d}-\frac{4 a^4 \cos ^3(c+d x)}{3 d}-\frac{a^4 \cot (c+d x)}{d}+\frac{23 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac{1}{8} \left (3 a^4\right ) \int 1 \, dx\\ &=\frac{17 a^4 x}{8}-\frac{4 a^4 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{4 a^4 \cos (c+d x)}{d}-\frac{4 a^4 \cos ^3(c+d x)}{3 d}-\frac{a^4 \cot (c+d x)}{d}+\frac{23 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.52819, size = 136, normalized size = 1.17 \[ \frac{a^4 \csc \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (408 c \sin (c+d x)+408 d x \sin (c+d x)+320 \sin (2 (c+d x))-32 \sin (4 (c+d x))-48 \cos (c+d x)-147 \cos (3 (c+d x))+3 \cos (5 (c+d x))+768 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-768 \sin (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{384 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(-48*Cos[c + d*x] - 147*Cos[3*(c + d*x)] + 3*Cos[5*(c + d*x)] + 408*c*S
in[c + d*x] + 408*d*x*Sin[c + d*x] - 768*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] + 768*Log[Sin[(c + d*x)/2]]*Sin[c
+ d*x] + 320*Sin[2*(c + d*x)] - 32*Sin[4*(c + d*x)]))/(384*d)

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Maple [A]  time = 0.073, size = 127, normalized size = 1.1 \begin{align*} -{\frac{{a}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{4\,d}}+{\frac{25\,{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+{\frac{17\,{a}^{4}x}{8}}+{\frac{17\,{a}^{4}c}{8\,d}}-{\frac{4\,{a}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+4\,{\frac{{a}^{4}\cos \left ( dx+c \right ) }{d}}+4\,{\frac{{a}^{4}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{4}\cot \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^4*cos(d*x+c)^3*sin(d*x+c)/d+25/8*a^4*cos(d*x+c)*sin(d*x+c)/d+17/8*a^4*x+17/8/d*a^4*c-4/3*a^4*cos(d*x+c)
^3/d+4*a^4*cos(d*x+c)/d+4/d*a^4*ln(csc(d*x+c)-cot(d*x+c))-a^4*cot(d*x+c)/d

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Maxima [A]  time = 1.62824, size = 158, normalized size = 1.36 \begin{align*} -\frac{128 \, a^{4} \cos \left (d x + c\right )^{3} - 3 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{4} - 144 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} + 96 \,{\left (d x + c + \frac{1}{\tan \left (d x + c\right )}\right )} a^{4} - 192 \, a^{4}{\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 )}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(128*a^4*cos(d*x + c)^3 - 3*(4*d*x + 4*c - sin(4*d*x + 4*c))*a^4 - 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*
a^4 + 96*(d*x + c + 1/tan(d*x + c))*a^4 - 192*a^4*(2*cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) -
1)))/d

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Fricas [A]  time = 1.75748, size = 360, normalized size = 3.1 \begin{align*} \frac{6 \, a^{4} \cos \left (d x + c\right )^{5} - 81 \, a^{4} \cos \left (d x + c\right )^{3} - 48 \, a^{4} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 48 \, a^{4} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 51 \, a^{4} \cos \left (d x + c\right ) -{\left (32 \, a^{4} \cos \left (d x + c\right )^{3} - 51 \, a^{4} d x - 96 \, a^{4} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d \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)^2*(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/24*(6*a^4*cos(d*x + c)^5 - 81*a^4*cos(d*x + c)^3 - 48*a^4*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 48*a^4*
log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 51*a^4*cos(d*x + c) - (32*a^4*cos(d*x + c)^3 - 51*a^4*d*x - 96*a^4
*cos(d*x + c))*sin(d*x + c))/(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)**2*csc(d*x+c)**2*(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.40173, size = 262, normalized size = 2.26 \begin{align*} \frac{51 \,{\left (d x + c\right )} a^{4} + 96 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 12 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{12 \,{\left (8 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{4}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} - \frac{2 \,{\left (69 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 93 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 192 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 93 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 256 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 69 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 64 \, a^{4}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(51*(d*x + c)*a^4 + 96*a^4*log(abs(tan(1/2*d*x + 1/2*c))) + 12*a^4*tan(1/2*d*x + 1/2*c) - 12*(8*a^4*tan(1
/2*d*x + 1/2*c) + a^4)/tan(1/2*d*x + 1/2*c) - 2*(69*a^4*tan(1/2*d*x + 1/2*c)^7 + 93*a^4*tan(1/2*d*x + 1/2*c)^5
- 192*a^4*tan(1/2*d*x + 1/2*c)^4 - 93*a^4*tan(1/2*d*x + 1/2*c)^3 - 256*a^4*tan(1/2*d*x + 1/2*c)^2 - 69*a^4*ta
n(1/2*d*x + 1/2*c) - 64*a^4)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d