3.399 \(\int \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=134 \[ \frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {2 a^3 \cos (c+d x)}{d}-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {2 a^3 \cot (c+d x)}{d}-\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {7 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {7 a^3 x}{2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {2709, 3770, 3767, 8, 3768, 2638, 2635, 2633} \[ \frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {2 a^3 \cos (c+d x)}{d}-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {2 a^3 \cot (c+d x)}{d}-\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {7 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {7 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

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

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Mathematica [A]  time = 6.13, size = 201, normalized size = 1.50 \[ \frac {a^3 (\sin (c+d x)+1)^3 \left (-84 (c+d x)-18 \sin (2 (c+d x))-42 \cos (c+d x)+2 \cos (3 (c+d x))+20 \tan \left (\frac {1}{2} (c+d x)\right )-20 \cot \left (\frac {1}{2} (c+d x)\right )-9 \csc ^2\left (\frac {1}{2} (c+d x)\right )+9 \sec ^2\left (\frac {1}{2} (c+d x)\right )-84 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+84 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\frac {1}{2} \sin (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )+8 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)\right )}{24 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*(1 + Sin[c + d*x])^3*(-84*(c + d*x) - 42*Cos[c + d*x] + 2*Cos[3*(c + d*x)] - 20*Cot[(c + d*x)/2] - 9*Csc[
(c + d*x)/2]^2 + 84*Log[Cos[(c + d*x)/2]] - 84*Log[Sin[(c + d*x)/2]] + 9*Sec[(c + d*x)/2]^2 + 8*Csc[c + d*x]^3
*Sin[(c + d*x)/2]^4 - (Csc[(c + d*x)/2]^4*Sin[c + d*x])/2 - 18*Sin[2*(c + d*x)] + 20*Tan[(c + d*x)/2]))/(24*d*
(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)

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fricas [A]  time = 0.48, size = 206, normalized size = 1.54 \[ \frac {18 \, a^{3} \cos \left (d x + c\right )^{5} - 56 \, a^{3} \cos \left (d x + c\right )^{3} + 42 \, a^{3} \cos \left (d x + c\right ) + 21 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 21 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 2 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{5} - 21 \, a^{3} d x \cos \left (d x + c\right )^{2} - 14 \, a^{3} \cos \left (d x + c\right )^{3} + 21 \, a^{3} d x + 21 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(18*a^3*cos(d*x + c)^5 - 56*a^3*cos(d*x + c)^3 + 42*a^3*cos(d*x + c) + 21*(a^3*cos(d*x + c)^2 - a^3)*log(
1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 21*(a^3*cos(d*x + c)^2 - a^3)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c)
+ 2*(2*a^3*cos(d*x + c)^5 - 21*a^3*d*x*cos(d*x + c)^2 - 14*a^3*cos(d*x + c)^3 + 21*a^3*d*x + 21*a^3*cos(d*x +
c))*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))

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giac [B]  time = 0.29, size = 250, normalized size = 1.87 \[ \frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 252 \, {\left (d x + c\right )} a^{3} - 252 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 63 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {154 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 153 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 291 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 192 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 195 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 414 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 167 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 27 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{3}}}{72 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(3*a^3*tan(1/2*d*x + 1/2*c)^3 + 27*a^3*tan(1/2*d*x + 1/2*c)^2 - 252*(d*x + c)*a^3 - 252*a^3*log(abs(tan(1
/2*d*x + 1/2*c))) + 63*a^3*tan(1/2*d*x + 1/2*c) + (154*a^3*tan(1/2*d*x + 1/2*c)^9 + 153*a^3*tan(1/2*d*x + 1/2*
c)^8 + 291*a^3*tan(1/2*d*x + 1/2*c)^7 - 192*a^3*tan(1/2*d*x + 1/2*c)^6 - 195*a^3*tan(1/2*d*x + 1/2*c)^5 - 414*
a^3*tan(1/2*d*x + 1/2*c)^4 - 167*a^3*tan(1/2*d*x + 1/2*c)^3 - 72*a^3*tan(1/2*d*x + 1/2*c)^2 - 27*a^3*tan(1/2*d
*x + 1/2*c) - 3*a^3)/(tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))^3)/d

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maple [A]  time = 0.45, size = 190, normalized size = 1.42 \[ -\frac {7 a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{6 d}-\frac {7 a^{3} \cos \left (d x +c \right )}{2 d}-\frac {7 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-\frac {3 a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}-\frac {3 a^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}-\frac {9 a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}-\frac {7 a^{3} x}{2}-\frac {7 a^{3} c}{2 d}-\frac {3 a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {a^{3} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{3} \cot \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/6*a^3*cos(d*x+c)^3/d-7/2*a^3*cos(d*x+c)/d-7/2/d*a^3*ln(csc(d*x+c)-cot(d*x+c))-3/d*a^3/sin(d*x+c)*cos(d*x+c)
^5-3*a^3*cos(d*x+c)^3*sin(d*x+c)/d-9/2*a^3*cos(d*x+c)*sin(d*x+c)/d-7/2*a^3*x-7/2/d*a^3*c-3/2/d*a^3/sin(d*x+c)^
2*cos(d*x+c)^5-1/3*a^3*cot(d*x+c)^3/d+a^3*cot(d*x+c)/d

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maxima [A]  time = 0.44, size = 185, normalized size = 1.38 \[ \frac {2 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right ) - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{3} - 18 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{3} + 4 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a^{3} + 9 \, a^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*(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 - 18*(3*d*
x + 3*c + (3*tan(d*x + c)^2 + 2)/(tan(d*x + c)^3 + tan(d*x + c)))*a^3 + 4*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1
)/tan(d*x + c)^3)*a^3 + 9*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 = 8.78, size = 339, normalized size = 2.53 \[ \frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {7\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {7\,a^3\,\mathrm {atan}\left (\frac {49\,a^6}{49\,a^6-49\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {49\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{49\,a^6-49\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {-17\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+19\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {64\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+73\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+46\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {107\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^3}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )} \]

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)^4,x)

[Out]

(3*a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^3*tan(c/2 + (d*x)/2)^3)/(24*d) - (7*a^3*log(tan(c/2 + (d*x)/2)))/(2*d)
- (7*a^3*atan((49*a^6)/(49*a^6 - 49*a^6*tan(c/2 + (d*x)/2)) + (49*a^6*tan(c/2 + (d*x)/2))/(49*a^6 - 49*a^6*ta
n(c/2 + (d*x)/2))))/d + (7*a^3*tan(c/2 + (d*x)/2))/(8*d) - (8*a^3*tan(c/2 + (d*x)/2)^2 + (107*a^3*tan(c/2 + (d
*x)/2)^3)/3 + 46*a^3*tan(c/2 + (d*x)/2)^4 + 73*a^3*tan(c/2 + (d*x)/2)^5 + (64*a^3*tan(c/2 + (d*x)/2)^6)/3 + 19
*a^3*tan(c/2 + (d*x)/2)^7 - 17*a^3*tan(c/2 + (d*x)/2)^8 + a^3/3 + 3*a^3*tan(c/2 + (d*x)/2))/(d*(8*tan(c/2 + (d
*x)/2)^3 + 24*tan(c/2 + (d*x)/2)^5 + 24*tan(c/2 + (d*x)/2)^7 + 8*tan(c/2 + (d*x)/2)^9))

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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)**4*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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