3.398 \(\int \cos (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=137 \[ \frac {a^3 \cos ^3(c+d x)}{d}+\frac {2 a^3 \cos (c+d x)}{d}-\frac {3 a^3 \cot (c+d x)}{d}-\frac {a^3 \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {7 a^3 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {33 a^3 x}{8} \]

[Out]

-33/8*a^3*x-3/2*a^3*arctanh(cos(d*x+c))/d+2*a^3*cos(d*x+c)/d+a^3*cos(d*x+c)^3/d-3*a^3*cot(d*x+c)/d-1/2*a^3*cot
(d*x+c)*csc(d*x+c)/d-7/8*a^3*cos(d*x+c)*sin(d*x+c)/d-1/4*a^3*cos(d*x+c)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.18, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2872, 3770, 3767, 8, 3768, 2638, 2635, 2633} \[ \frac {a^3 \cos ^3(c+d x)}{d}+\frac {2 a^3 \cos (c+d x)}{d}-\frac {3 a^3 \cot (c+d x)}{d}-\frac {a^3 \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {7 a^3 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {33 a^3 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-33*a^3*x)/8 - (3*a^3*ArcTanh[Cos[c + d*x]])/(2*d) + (2*a^3*Cos[c + d*x])/d + (a^3*Cos[c + d*x]^3)/d - (3*a^3
*Cot[c + d*x])/d - (a^3*Cot[c + d*x]*Csc[c + d*x])/(2*d) - (7*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a^3*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 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 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 (c+d x) \cot ^3(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac {\int \left (-5 a^7+a^7 \csc (c+d x)+3 a^7 \csc ^2(c+d x)+a^7 \csc ^3(c+d x)-5 a^7 \sin (c+d x)+a^7 \sin ^2(c+d x)+3 a^7 \sin ^3(c+d x)+a^7 \sin ^4(c+d x)\right ) \, dx}{a^4}\\ &=-5 a^3 x+a^3 \int \csc (c+d x) \, dx+a^3 \int \csc ^3(c+d x) \, dx+a^3 \int \sin ^2(c+d x) \, dx+a^3 \int \sin ^4(c+d x) \, dx+\left (3 a^3\right ) \int \csc ^2(c+d x) \, dx+\left (3 a^3\right ) \int \sin ^3(c+d x) \, dx-\left (5 a^3\right ) \int \sin (c+d x) \, dx\\ &=-5 a^3 x-\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {5 a^3 \cos (c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{2} a^3 \int 1 \, dx+\frac {1}{2} a^3 \int \csc (c+d x) \, dx+\frac {1}{4} \left (3 a^3\right ) \int \sin ^2(c+d x) \, dx-\frac {\left (3 a^3\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {9 a^3 x}{2}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac {2 a^3 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{d}-\frac {3 a^3 \cot (c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {7 a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{8} \left (3 a^3\right ) \int 1 \, dx\\ &=-\frac {33 a^3 x}{8}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac {2 a^3 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{d}-\frac {3 a^3 \cot (c+d x)}{d}-\frac {a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {7 a^3 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 2.86, size = 164, normalized size = 1.20 \[ \frac {(a \sin (c+d x)+a)^3 \left (-132 (c+d x)-16 \sin (2 (c+d x))+\sin (4 (c+d x))+88 \cos (c+d x)+8 \cos (3 (c+d x))+48 \tan \left (\frac {1}{2} (c+d x)\right )-48 \cot \left (\frac {1}{2} (c+d x)\right )-4 \csc ^2\left (\frac {1}{2} (c+d x)\right )+4 \sec ^2\left (\frac {1}{2} (c+d x)\right )+48 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-48 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{32 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + a*Sin[c + d*x])^3*(-132*(c + d*x) + 88*Cos[c + d*x] + 8*Cos[3*(c + d*x)] - 48*Cot[(c + d*x)/2] - 4*Csc[(
c + d*x)/2]^2 - 48*Log[Cos[(c + d*x)/2]] + 48*Log[Sin[(c + d*x)/2]] + 4*Sec[(c + d*x)/2]^2 - 16*Sin[2*(c + d*x
)] + Sin[4*(c + d*x)] + 48*Tan[(c + d*x)/2]))/(32*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)

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fricas [A]  time = 0.49, size = 185, normalized size = 1.35 \[ \frac {8 \, a^{3} \cos \left (d x + c\right )^{5} - 33 \, a^{3} d x \cos \left (d x + c\right )^{2} + 8 \, a^{3} \cos \left (d x + c\right )^{3} + 33 \, a^{3} d x - 12 \, a^{3} \cos \left (d x + c\right ) - 6 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 6 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (2 \, a^{3} \cos \left (d x + c\right )^{5} - 11 \, a^{3} \cos \left (d x + c\right )^{3} + 33 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/8*(8*a^3*cos(d*x + c)^5 - 33*a^3*d*x*cos(d*x + c)^2 + 8*a^3*cos(d*x + c)^3 + 33*a^3*d*x - 12*a^3*cos(d*x + c
) - 6*(a^3*cos(d*x + c)^2 - a^3)*log(1/2*cos(d*x + c) + 1/2) + 6*(a^3*cos(d*x + c)^2 - a^3)*log(-1/2*cos(d*x +
c) + 1/2) + (2*a^3*cos(d*x + c)^5 - 11*a^3*cos(d*x + c)^3 + 33*a^3*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c
)^2 - d)

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giac [A]  time = 0.32, size = 241, normalized size = 1.76 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 33 \, {\left (d x + c\right )} a^{3} + 12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {18 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {2 \, {\left (7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 40 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 56 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 - 33*(d*x + c)*a^3 + 12*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 12*a^3*tan(1/2*d*
x + 1/2*c) - (18*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^3*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c)^2 + 2*(7
*a^3*tan(1/2*d*x + 1/2*c)^7 + 40*a^3*tan(1/2*d*x + 1/2*c)^6 + 15*a^3*tan(1/2*d*x + 1/2*c)^5 + 72*a^3*tan(1/2*d
*x + 1/2*c)^4 - 15*a^3*tan(1/2*d*x + 1/2*c)^3 + 56*a^3*tan(1/2*d*x + 1/2*c)^2 - 7*a^3*tan(1/2*d*x + 1/2*c) + 2
4*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 0.49, size = 161, normalized size = 1.18 \[ -\frac {11 a^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4 d}-\frac {33 a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}-\frac {33 a^{3} x}{8}-\frac {33 a^{3} c}{8 d}+\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} \cos \left (d x +c \right )}{2 d}+\frac {3 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {3 a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c))^3,x)

[Out]

-11/4*a^3*cos(d*x+c)^3*sin(d*x+c)/d-33/8*a^3*cos(d*x+c)*sin(d*x+c)/d-33/8*a^3*x-33/8/d*a^3*c+1/2*a^3*cos(d*x+c
)^3/d+3/2*a^3*cos(d*x+c)/d+3/2/d*a^3*ln(csc(d*x+c)-cot(d*x+c))-3/d*a^3/sin(d*x+c)*cos(d*x+c)^5-1/2/d*a^3/sin(d
*x+c)^2*cos(d*x+c)^5

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maxima [A]  time = 0.44, size = 183, normalized size = 1.34 \[ \frac {16 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{3} + {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 48 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{3} + 8 \, a^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/32*(16*(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^3 + (12*d*x
+ 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3 - 48*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)/(tan(d*x + c)^
3 + tan(d*x + c)))*a^3 + 8*a^3*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) - 4*cos(d*x + c) + 3*log(cos(d*x + c) + 1)
- 3*log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 8.81, size = 347, normalized size = 2.53 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {3\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}+\frac {33\,a^3\,\mathrm {atan}\left (\frac {1089\,a^6}{16\,\left (\frac {99\,a^6}{4}+\frac {1089\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}-\frac {99\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {99\,a^6}{4}+\frac {1089\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}\right )}{4\,d}+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {79\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}-9\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+70\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-51\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+53\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-31\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+22\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-6\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a^3}{2}}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(a + a*sin(c + d*x))^3)/sin(c + d*x)^3,x)

[Out]

(a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (3*a^3*log(tan(c/2 + (d*x)/2)))/(2*d) + (33*a^3*atan((1089*a^6)/(16*((99*a^
6)/4 + (1089*a^6*tan(c/2 + (d*x)/2))/16)) - (99*a^6*tan(c/2 + (d*x)/2))/(4*((99*a^6)/4 + (1089*a^6*tan(c/2 + (
d*x)/2))/16))))/(4*d) + (22*a^3*tan(c/2 + (d*x)/2)^2 - 31*a^3*tan(c/2 + (d*x)/2)^3 + 53*a^3*tan(c/2 + (d*x)/2)
^4 - 51*a^3*tan(c/2 + (d*x)/2)^5 + 70*a^3*tan(c/2 + (d*x)/2)^6 - 9*a^3*tan(c/2 + (d*x)/2)^7 + (79*a^3*tan(c/2
+ (d*x)/2)^8)/2 + a^3*tan(c/2 + (d*x)/2)^9 - a^3/2 - 6*a^3*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 16
*tan(c/2 + (d*x)/2)^4 + 24*tan(c/2 + (d*x)/2)^6 + 16*tan(c/2 + (d*x)/2)^8 + 4*tan(c/2 + (d*x)/2)^10)) + (3*a^3
*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)**3*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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