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

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

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2872, 3770, 3767, 8, 3768, 2638} \[ -\frac {a^2 \cos (c+d x)}{d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}+\frac {2 a^2 \cot (c+d x)}{d}+\frac {9 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {a^2 \cot (c+d x) \csc (c+d x)}{8 d}+2 a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

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 \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac {\int \left (2 a^6-a^6 \csc (c+d x)-4 a^6 \csc ^2(c+d x)-a^6 \csc ^3(c+d x)+2 a^6 \csc ^4(c+d x)+a^6 \csc ^5(c+d x)+a^6 \sin (c+d x)\right ) \, dx}{a^4}\\ &=2 a^2 x-a^2 \int \csc (c+d x) \, dx-a^2 \int \csc ^3(c+d x) \, dx+a^2 \int \csc ^5(c+d x) \, dx+a^2 \int \sin (c+d x) \, dx+\left (2 a^2\right ) \int \csc ^4(c+d x) \, dx-\left (4 a^2\right ) \int \csc ^2(c+d x) \, dx\\ &=2 a^2 x+\frac {a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a^2 \cos (c+d x)}{d}+\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {1}{2} a^2 \int \csc (c+d x) \, dx+\frac {1}{4} \left (3 a^2\right ) \int \csc ^3(c+d x) \, dx-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}+\frac {\left (4 a^2\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=2 a^2 x+\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cot (c+d x)}{d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {1}{8} \left (3 a^2\right ) \int \csc (c+d x) \, dx\\ &=2 a^2 x+\frac {9 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cot (c+d x)}{d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 1.23, size = 215, normalized size = 1.85 \[ -\frac {a^2 \sin (c+d x) (\sin (c+d x)+1)^2 \left (192 \cot (c+d x)+\csc ^4\left (\frac {1}{2} (c+d x)\right ) (3 \csc (c+d x)+8)-2 \csc ^2\left (\frac {1}{2} (c+d x)\right ) (3 \csc (c+d x)+64)+8 (8 \cos (c+d x)+7) \sec ^4\left (\frac {1}{2} (c+d x)\right )-48 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^5(c+d x)+24 \sin ^2\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)-24 \csc (c+d x) \left (16 (c+d x)-9 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{192 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/192*(a^2*(192*Cot[c + d*x] + Csc[(c + d*x)/2]^4*(8 + 3*Csc[c + d*x]) - 2*Csc[(c + d*x)/2]^2*(64 + 3*Csc[c +
d*x]) - 24*Csc[c + d*x]*(16*(c + d*x) + 9*Log[Cos[(c + d*x)/2]] - 9*Log[Sin[(c + d*x)/2]]) + 8*(7 + 8*Cos[c +
d*x])*Sec[(c + d*x)/2]^4 + 24*Csc[c + d*x]^3*Sin[(c + d*x)/2]^2 - 48*Csc[c + d*x]^5*Sin[(c + d*x)/2]^4)*Sin[c
+ d*x]*(1 + Sin[c + d*x])^2)/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

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fricas [B]  time = 0.53, size = 219, normalized size = 1.89 \[ \frac {96 \, a^{2} d x \cos \left (d x + c\right )^{4} - 48 \, a^{2} \cos \left (d x + c\right )^{5} - 192 \, a^{2} d x \cos \left (d x + c\right )^{2} + 90 \, a^{2} \cos \left (d x + c\right )^{3} + 96 \, a^{2} d x - 54 \, a^{2} \cos \left (d x + c\right ) + 27 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 27 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 32 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/48*(96*a^2*d*x*cos(d*x + c)^4 - 48*a^2*cos(d*x + c)^5 - 192*a^2*d*x*cos(d*x + c)^2 + 90*a^2*cos(d*x + c)^3 +
96*a^2*d*x - 54*a^2*cos(d*x + c) + 27*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(1/2*cos(d*x + c)
+ 1/2) - 27*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(-1/2*cos(d*x + c) + 1/2) - 32*(4*a^2*cos(d*x
+ c)^3 - 3*a^2*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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giac [A]  time = 0.28, size = 162, normalized size = 1.40 \[ \frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 16 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 384 \, {\left (d x + c\right )} a^{2} - 216 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 240 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {384 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {450 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 240 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 + 16*a^2*tan(1/2*d*x + 1/2*c)^3 + 384*(d*x + c)*a^2 - 216*a^2*log(abs(tan(
1/2*d*x + 1/2*c))) - 240*a^2*tan(1/2*d*x + 1/2*c) - 384*a^2/(tan(1/2*d*x + 1/2*c)^2 + 1) + (450*a^2*tan(1/2*d*
x + 1/2*c)^4 + 240*a^2*tan(1/2*d*x + 1/2*c)^3 - 16*a^2*tan(1/2*d*x + 1/2*c) - 3*a^2)/tan(1/2*d*x + 1/2*c)^4)/d

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maple [A]  time = 0.33, size = 149, normalized size = 1.28 \[ -\frac {3 a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}-\frac {3 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{8 d}-\frac {9 a^{2} \cos \left (d x +c \right )}{8 d}-\frac {9 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {2 a^{2} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {2 a^{2} \cot \left (d x +c \right )}{d}+2 a^{2} x +\frac {2 a^{2} c}{d}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.43, size = 167, normalized size = 1.44 \[ \frac {32 \, {\left (3 \, d x + 3 \, c + \frac {3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a^{2} - 3 \, a^{2} {\left (\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/48*(32*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a^2 - 3*a^2*(2*(5*cos(d*x + c)^3 - 3*cos(d*x +
c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)) + 12*a^2*(2*c
os(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 = 8.77, size = 265, normalized size = 2.28 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {9\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}-\frac {4\,a^2\,\mathrm {atan}\left (\frac {16\,a^4}{9\,a^4+16\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{9\,a^4+16\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,d}-\frac {-20\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+32\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {56\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {a^2}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )} \]

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)^5,x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^3)/(12*d) + (a^2*tan(c/2 + (d*x)/2)^4)/(64*d) - (9*a^2*log(tan(c/2 + (d*x)/2)))/(8*d)
- (4*a^2*atan((16*a^4)/(9*a^4 + 16*a^4*tan(c/2 + (d*x)/2)) - (9*a^4*tan(c/2 + (d*x)/2))/(9*a^4 + 16*a^4*tan(c/
2 + (d*x)/2))))/d - (5*a^2*tan(c/2 + (d*x)/2))/(4*d) - ((a^2*tan(c/2 + (d*x)/2)^2)/4 - (56*a^2*tan(c/2 + (d*x)
/2)^3)/3 + 32*a^2*tan(c/2 + (d*x)/2)^4 - 20*a^2*tan(c/2 + (d*x)/2)^5 + a^2/4 + (4*a^2*tan(c/2 + (d*x)/2))/3)/(
d*(16*tan(c/2 + (d*x)/2)^4 + 16*tan(c/2 + (d*x)/2)^6))

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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)**5*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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