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

Optimal. Leaf size=92 \[ -\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {3 a^3 \cos (c+d x)}{d}-\frac {a^3 \cot (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a^3 x}{2} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2709, 3770, 3767, 8, 2638, 2635, 2633} \[ -\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {3 a^3 \cos (c+d x)}{d}-\frac {a^3 \cot (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*x)/2 - (3*a^3*ArcTanh[Cos[c + d*x]])/d + (3*a^3*Cos[c + d*x])/d - (a^3*Cos[c + d*x]^3)/(3*d) - (a^3*Cot[c
+ d*x])/d + (3*a^3*Cos[c + d*x]*Sin[c + d*x])/(2*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 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 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 \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac {\int \left (2 a^5+3 a^5 \csc (c+d x)+a^5 \csc ^2(c+d x)-2 a^5 \sin (c+d x)-3 a^5 \sin ^2(c+d x)-a^5 \sin ^3(c+d x)\right ) \, dx}{a^2}\\ &=2 a^3 x+a^3 \int \csc ^2(c+d x) \, dx-a^3 \int \sin ^3(c+d x) \, dx-\left (2 a^3\right ) \int \sin (c+d x) \, dx+\left (3 a^3\right ) \int \csc (c+d x) \, dx-\left (3 a^3\right ) \int \sin ^2(c+d x) \, dx\\ &=2 a^3 x-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^3 \cos (c+d x)}{d}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} \left (3 a^3\right ) \int 1 \, dx-\frac {a^3 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}+\frac {a^3 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a^3 x}{2}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {3 a^3 \cos (c+d x)}{d}-\frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {a^3 \cot (c+d x)}{d}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 1.13, size = 106, normalized size = 1.15 \[ -\frac {a^3 \csc \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left ((15-66 \sin (c+d x)) \cos (c+d x)+(2 \sin (c+d x)+9) \cos (3 (c+d x))-12 \sin (c+d x) \left (6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+c+d x\right )\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/48*(a^3*Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(Cos[c + d*x]*(15 - 66*Sin[c + d*x]) - 12*(c + d*x - 6*Log[Cos[(c
+ d*x)/2]] + 6*Log[Sin[(c + d*x)/2]])*Sin[c + d*x] + Cos[3*(c + d*x)]*(9 + 2*Sin[c + d*x])))/d

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fricas [A]  time = 0.53, size = 121, normalized size = 1.32 \[ -\frac {9 \, a^{3} \cos \left (d x + c\right )^{3} + 9 \, a^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 9 \, a^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, a^{3} \cos \left (d x + c\right ) + {\left (2 \, a^{3} \cos \left (d x + c\right )^{3} - 3 \, a^{3} d x - 18 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \sin \left (d x + c\right )} \]

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))^3,x, algorithm="fricas")

[Out]

-1/6*(9*a^3*cos(d*x + c)^3 + 9*a^3*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 9*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*cos(d*x + c)^3 - 3*a^3*d*x - 18*a^3*cos(d*x + c))*sin(d*x + c))/
(d*sin(d*x + c))

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giac [A]  time = 0.22, size = 162, normalized size = 1.76 \[ \frac {3 \, {\left (d x + c\right )} a^{3} + 18 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, {\left (6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 16 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

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))^3,x, algorithm="giac")

[Out]

1/6*(3*(d*x + c)*a^3 + 18*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 3*a^3*tan(1/2*d*x + 1/2*c) - 3*(6*a^3*tan(1/2*d
*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c) - 2*(9*a^3*tan(1/2*d*x + 1/2*c)^5 - 12*a^3*tan(1/2*d*x + 1/2*c)^4 - 36
*a^3*tan(1/2*d*x + 1/2*c)^2 - 9*a^3*tan(1/2*d*x + 1/2*c) - 16*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 0.32, size = 105, normalized size = 1.14 \[ -\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}+\frac {3 a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {a^{3} x}{2}+\frac {a^{3} c}{2 d}+\frac {3 a^{3} \cos \left (d x +c \right )}{d}+\frac {3 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}-\frac {a^{3} \cot \left (d x +c \right )}{d} \]

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))^3,x)

[Out]

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

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maxima [A]  time = 0.43, size = 93, normalized size = 1.01 \[ -\frac {4 \, a^{3} \cos \left (d x + c\right )^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 12 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{3} - 18 \, a^{3} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

-1/12*(4*a^3*cos(d*x + c)^3 - 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3 + 12*(d*x + c + 1/tan(d*x + c))*a^3 - 18*
a^3*(2*cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 8.77, size = 264, normalized size = 2.87 \[ \frac {3\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {a^3\,\mathrm {atan}\left (\frac {a^6}{6\,a^6-a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {6\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{6\,a^6-a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {-7\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+24\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {32\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-a^3}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a^3*log(tan(c/2 + (d*x)/2)))/d + (a^3*atan(a^6/(6*a^6 - a^6*tan(c/2 + (d*x)/2)) + (6*a^6*tan(c/2 + (d*x)/2)
)/(6*a^6 - a^6*tan(c/2 + (d*x)/2))))/d + (a^3*tan(c/2 + (d*x)/2))/(2*d) + (3*a^3*tan(c/2 + (d*x)/2)^2 + 24*a^3
*tan(c/2 + (d*x)/2)^3 - 3*a^3*tan(c/2 + (d*x)/2)^4 + 8*a^3*tan(c/2 + (d*x)/2)^5 - 7*a^3*tan(c/2 + (d*x)/2)^6 -
a^3 + (32*a^3*tan(c/2 + (d*x)/2))/3)/(d*(2*tan(c/2 + (d*x)/2) + 6*tan(c/2 + (d*x)/2)^3 + 6*tan(c/2 + (d*x)/2)
^5 + 2*tan(c/2 + (d*x)/2)^7))

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

[Out]

Timed out

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