3.383 \(\int \cos ^2(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=116 \[ \frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}-\frac {a^2 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {9 a^2 x}{8} \]

[Out]

-9/8*a^2*x-2*a^2*arctanh(cos(d*x+c))/d+2*a^2*cos(d*x+c)/d+2/3*a^2*cos(d*x+c)^3/d-a^2*cot(d*x+c)/d+1/8*a^2*cos(
d*x+c)*sin(d*x+c)/d-1/4*a^2*cos(d*x+c)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.20, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2872, 3770, 3767, 8, 2638, 2635, 2633} \[ \frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot (c+d x)}{d}-\frac {a^2 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {9 a^2 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*a^2*x)/8 - (2*a^2*ArcTanh[Cos[c + d*x]])/d + (2*a^2*Cos[c + d*x])/d + (2*a^2*Cos[c + d*x]^3)/(3*d) - (a^2*
Cot[c + d*x])/d + (a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a^2*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d)

Rule 8

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

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]

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 2638

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

Rule 2872

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 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 3770

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

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac {\int \left (-a^6+2 a^6 \csc (c+d x)+a^6 \csc ^2(c+d x)-4 a^6 \sin (c+d x)-a^6 \sin ^2(c+d x)+2 a^6 \sin ^3(c+d x)+a^6 \sin ^4(c+d x)\right ) \, dx}{a^4}\\ &=-a^2 x+a^2 \int \csc ^2(c+d x) \, dx-a^2 \int \sin ^2(c+d x) \, dx+a^2 \int \sin ^4(c+d x) \, dx+\left (2 a^2\right ) \int \csc (c+d x) \, dx+\left (2 a^2\right ) \int \sin ^3(c+d x) \, dx-\left (4 a^2\right ) \int \sin (c+d x) \, dx\\ &=-a^2 x-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {4 a^2 \cos (c+d x)}{d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^2 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac {1}{2} a^2 \int 1 \, dx+\frac {1}{4} \left (3 a^2\right ) \int \sin ^2(c+d x) \, dx-\frac {a^2 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {3 a^2 x}{2}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a^2 \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{8} \left (3 a^2\right ) \int 1 \, dx\\ &=-\frac {9 a^2 x}{8}-\frac {2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a^2 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 83, normalized size = 0.72 \[ \frac {a^2 \left (240 \cos (c+d x)+16 \cos (3 (c+d x))-3 \left (-\sin (4 (c+d x))+32 \cot (c+d x)-64 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+64 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+36 c+36 d x\right )\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(240*Cos[c + d*x] + 16*Cos[3*(c + d*x)] - 3*(36*c + 36*d*x + 32*Cot[c + d*x] + 64*Log[Cos[(c + d*x)/2]] -
64*Log[Sin[(c + d*x)/2]] - Sin[4*(c + d*x)])))/(96*d)

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fricas [A]  time = 0.69, size = 135, normalized size = 1.16 \[ -\frac {6 \, a^{2} \cos \left (d x + c\right )^{5} - 9 \, a^{2} \cos \left (d x + c\right )^{3} + 24 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 24 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 27 \, a^{2} \cos \left (d x + c\right ) - {\left (16 \, a^{2} \cos \left (d x + c\right )^{3} - 27 \, a^{2} d x + 48 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(6*a^2*cos(d*x + c)^5 - 9*a^2*cos(d*x + c)^3 + 24*a^2*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 24*a^2*
log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 27*a^2*cos(d*x + c) - (16*a^2*cos(d*x + c)^3 - 27*a^2*d*x + 48*a^2
*cos(d*x + c))*sin(d*x + c))/(d*sin(d*x + c))

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giac [A]  time = 0.23, size = 210, normalized size = 1.81 \[ -\frac {27 \, {\left (d x + c\right )} a^{2} - 48 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {12 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 21 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 192 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 21 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 64 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(27*(d*x + c)*a^2 - 48*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 12*a^2*tan(1/2*d*x + 1/2*c) + 12*(4*a^2*tan(
1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c) + 2*(3*a^2*tan(1/2*d*x + 1/2*c)^7 - 96*a^2*tan(1/2*d*x + 1/2*c)^6
- 21*a^2*tan(1/2*d*x + 1/2*c)^5 - 192*a^2*tan(1/2*d*x + 1/2*c)^4 + 21*a^2*tan(1/2*d*x + 1/2*c)^3 - 160*a^2*ta
n(1/2*d*x + 1/2*c)^2 - 3*a^2*tan(1/2*d*x + 1/2*c) - 64*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 0.42, size = 137, normalized size = 1.18 \[ -\frac {3 a^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4 d}-\frac {9 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}-\frac {9 a^{2} x}{8}-\frac {9 a^{2} c}{8 d}+\frac {2 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {2 a^{2} \cos \left (d x +c \right )}{d}+\frac {2 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*a^2*cos(d*x+c)^3*sin(d*x+c)/d-9/8*a^2*cos(d*x+c)*sin(d*x+c)/d-9/8*a^2*x-9/8/d*a^2*c+2/3*a^2*cos(d*x+c)^3/
d+2*a^2*cos(d*x+c)/d+2/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-1/d*a^2/sin(d*x+c)*cos(d*x+c)^5

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maxima [A]  time = 0.45, size = 128, normalized size = 1.10 \[ \frac {32 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 48 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(32*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1))*a^2 + 3*(12*d
*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2 - 48*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)/(tan(d*x + c
)^3 + tan(d*x + c)))*a^2)/d

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mupad [B]  time = 8.83, size = 310, normalized size = 2.67 \[ \frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {9\,a^2\,\mathrm {atan}\left (\frac {81\,a^4}{16\,\left (9\,a^4+\frac {81\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}-\frac {9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{9\,a^4+\frac {81\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}}\right )}{4\,d}-\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}-16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}-32\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {19\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}-\frac {80\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {32\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+a^2}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*log(tan(c/2 + (d*x)/2)))/d + (9*a^2*atan((81*a^4)/(16*(9*a^4 + (81*a^4*tan(c/2 + (d*x)/2))/16)) - (9*a^
4*tan(c/2 + (d*x)/2))/(9*a^4 + (81*a^4*tan(c/2 + (d*x)/2))/16)))/(4*d) - ((7*a^2*tan(c/2 + (d*x)/2)^2)/2 - (80
*a^2*tan(c/2 + (d*x)/2)^3)/3 + (19*a^2*tan(c/2 + (d*x)/2)^4)/2 - 32*a^2*tan(c/2 + (d*x)/2)^5 + (a^2*tan(c/2 +
(d*x)/2)^6)/2 - 16*a^2*tan(c/2 + (d*x)/2)^7 + (3*a^2*tan(c/2 + (d*x)/2)^8)/2 + a^2 - (32*a^2*tan(c/2 + (d*x)/2
))/3)/(d*(2*tan(c/2 + (d*x)/2) + 8*tan(c/2 + (d*x)/2)^3 + 12*tan(c/2 + (d*x)/2)^5 + 8*tan(c/2 + (d*x)/2)^7 + 2
*tan(c/2 + (d*x)/2)^9)) + (a^2*tan(c/2 + (d*x)/2))/(2*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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