3.305 \(\int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=95 \[ \frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot (c+d x)}{a d}-\frac {3 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2839, 3768, 3770, 3767} \[ \frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot (c+d x)}{a d}-\frac {3 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^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 \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac {\int \csc ^4(c+d x) \, dx}{a}+\frac {\int \csc ^5(c+d x) \, dx}{a}\\ &=-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {3 \int \csc ^3(c+d x) \, dx}{4 a}+\frac {\operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a d}\\ &=\frac {\cot (c+d x)}{a d}+\frac {\cot ^3(c+d x)}{3 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {3 \int \csc (c+d x) \, dx}{8 a}\\ &=-\frac {3 \tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac {\cot (c+d x)}{a d}+\frac {\cot ^3(c+d x)}{3 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{8 a d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}\\ \end {align*}

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Mathematica [A]  time = 1.13, size = 125, normalized size = 1.32 \[ -\frac {\csc ^4(c+d x) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (-48 \sin (2 (c+d x))+66 \cos (c+d x)+2 (16 \sin (c+d x)-9) \cos (3 (c+d x))+72 \sin ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{192 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/192*(Csc[c + d*x]^4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(66*Cos[c + d*x] + 72*(Log[Cos[(c + d*x)/2]] -
Log[Sin[(c + d*x)/2]])*Sin[c + d*x]^4 + 2*Cos[3*(c + d*x)]*(-9 + 16*Sin[c + d*x]) - 48*Sin[2*(c + d*x)]))/(a*d
*(1 + Sin[c + d*x]))

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fricas [A]  time = 0.51, size = 143, normalized size = 1.51 \[ \frac {18 \, \cos \left (d x + c\right )^{3} - 9 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 9 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 16 \, {\left (2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 30 \, \cos \left (d x + c\right )}{48 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(18*cos(d*x + c)^3 - 9*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2) + 9*(cos(d*x +
c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2) - 16*(2*cos(d*x + c)^3 - 3*cos(d*x + c))*sin(d*x +
c) - 30*cos(d*x + c))/(a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d)

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giac [A]  time = 0.19, size = 157, normalized size = 1.65 \[ \frac {\frac {72 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} - \frac {150 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 72 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(72*log(abs(tan(1/2*d*x + 1/2*c)))/a + (3*a^3*tan(1/2*d*x + 1/2*c)^4 - 8*a^3*tan(1/2*d*x + 1/2*c)^3 + 24
*a^3*tan(1/2*d*x + 1/2*c)^2 - 72*a^3*tan(1/2*d*x + 1/2*c))/a^4 - (150*tan(1/2*d*x + 1/2*c)^4 - 72*tan(1/2*d*x
+ 1/2*c)^3 + 24*tan(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 3)/(a*tan(1/2*d*x + 1/2*c)^4))/d

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maple [A]  time = 0.43, size = 170, normalized size = 1.79 \[ \frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 a d}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 a d}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {3}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}-\frac {1}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{64 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {1}{24 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64/a/d*tan(1/2*d*x+1/2*c)^4-1/24/a/d*tan(1/2*d*x+1/2*c)^3+1/8/a/d*tan(1/2*d*x+1/2*c)^2-3/8/a/d*tan(1/2*d*x+1
/2*c)+3/8/a/d/tan(1/2*d*x+1/2*c)+3/8/a/d*ln(tan(1/2*d*x+1/2*c))-1/8/a/d/tan(1/2*d*x+1/2*c)^2-1/64/a/d/tan(1/2*
d*x+1/2*c)^4+1/24/a/d/tan(1/2*d*x+1/2*c)^3

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maxima [B]  time = 0.33, size = 195, normalized size = 2.05 \[ -\frac {\frac {\frac {72 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a} - \frac {72 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {{\left (\frac {8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {24 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {72 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 3\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{4}}{a \sin \left (d x + c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*((72*sin(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a - 72*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - (8*sin
(d*x + c)/(cos(d*x + c) + 1) - 24*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 72*sin(d*x + c)^3/(cos(d*x + c) + 1)^3
- 3)*(cos(d*x + c) + 1)^4/(a*sin(d*x + c)^4))/d

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mupad [B]  time = 8.68, size = 151, normalized size = 1.59 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}+\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a\,d}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {1}{4}\right )}{16\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^2/(8*a*d) - tan(c/2 + (d*x)/2)^3/(24*a*d) + tan(c/2 + (d*x)/2)^4/(64*a*d) + (3*log(tan(c/2
+ (d*x)/2)))/(8*a*d) - (3*tan(c/2 + (d*x)/2))/(8*a*d) + (cot(c/2 + (d*x)/2)^4*((2*tan(c/2 + (d*x)/2))/3 - 2*ta
n(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^3 - 1/4))/(16*a*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cos ^{2}{\left (c + d x \right )} \csc ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**5/(a+a*sin(d*x+c)),x)

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)**5/(sin(c + d*x) + 1), x)/a

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