3.429 \(\int \frac {\cot ^4(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=96 \[ \frac {2 \cot ^3(c+d x)}{3 a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {7 \tanh ^{-1}(\cos (c+d x))}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {7 \cot (c+d x) \csc (c+d x)}{8 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2869, 2757, 3768, 3770, 3767} \[ \frac {2 \cot ^3(c+d x)}{3 a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}-\frac {7 \tanh ^{-1}(\cos (c+d x))}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {7 \cot (c+d x) \csc (c+d x)}{8 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*ArcTanh[Cos[c + d*x]])/(8*a^2*d) + (2*Cot[c + d*x])/(a^2*d) + (2*Cot[c + d*x]^3)/(3*a^2*d) - (7*Cot[c + d*
x]*Csc[c + d*x])/(8*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^2*d)

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rule 2869

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[a^(2*m), Int[(d*Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f,
n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 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 ^4(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \csc ^5(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \csc ^3(c+d x)-2 a^2 \csc ^4(c+d x)+a^2 \csc ^5(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \csc ^3(c+d x) \, dx}{a^2}+\frac {\int \csc ^5(c+d x) \, dx}{a^2}-\frac {2 \int \csc ^4(c+d x) \, dx}{a^2}\\ &=-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}+\frac {\int \csc (c+d x) \, dx}{2 a^2}+\frac {3 \int \csc ^3(c+d x) \, dx}{4 a^2}+\frac {2 \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^2 d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}-\frac {7 \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}+\frac {3 \int \csc (c+d x) \, dx}{8 a^2}\\ &=-\frac {7 \tanh ^{-1}(\cos (c+d x))}{8 a^2 d}+\frac {2 \cot (c+d x)}{a^2 d}+\frac {2 \cot ^3(c+d x)}{3 a^2 d}-\frac {7 \cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.52, size = 116, normalized size = 1.21 \[ -\frac {\left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^4 \left (-48 \sin (2 (c+d x))+45 \cos (c+d x)+(32 \sin (c+d x)-21) \cos (3 (c+d x))+84 \sin ^4(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{1536 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1536*((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^4*(45*Cos[c + d*x] + 84*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d
*x)/2]])*Sin[c + d*x]^4 + Cos[3*(c + d*x)]*(-21 + 32*Sin[c + d*x]) - 48*Sin[2*(c + d*x)]))/(a^2*d*(1 + Sin[c +
d*x])^2)

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fricas [A]  time = 0.48, size = 149, normalized size = 1.55 \[ \frac {42 \, \cos \left (d x + c\right )^{3} - 21 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 21 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 32 \, {\left (2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 54 \, \cos \left (d x + c\right )}{48 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}} \]

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*(42*cos(d*x + c)^3 - 21*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2) + 21*(cos(d*x
+ c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2) - 32*(2*cos(d*x + c)^3 - 3*cos(d*x + c))*sin(d*x
+ c) - 54*cos(d*x + c))/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d)

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giac [A]  time = 0.24, size = 157, normalized size = 1.64 \[ \frac {\frac {168 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {350 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 144 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}} + \frac {3 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 48 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 144 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{8}}}{192 \, d} \]

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*(168*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (350*tan(1/2*d*x + 1/2*c)^4 - 144*tan(1/2*d*x + 1/2*c)^3 + 48*
tan(1/2*d*x + 1/2*c)^2 - 16*tan(1/2*d*x + 1/2*c) + 3)/(a^2*tan(1/2*d*x + 1/2*c)^4) + (3*a^6*tan(1/2*d*x + 1/2*
c)^4 - 16*a^6*tan(1/2*d*x + 1/2*c)^3 + 48*a^6*tan(1/2*d*x + 1/2*c)^2 - 144*a^6*tan(1/2*d*x + 1/2*c))/a^8)/d

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maple [A]  time = 0.60, size = 170, normalized size = 1.77 \[ \frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 a^{2} d}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d \,a^{2}}+\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2} d}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{2}}+\frac {3}{4 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{2}}-\frac {1}{4 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{64 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}+\frac {1}{12 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64/a^2/d*tan(1/2*d*x+1/2*c)^4-1/12/a^2/d*tan(1/2*d*x+1/2*c)^3+1/4/a^2/d*tan(1/2*d*x+1/2*c)^2-3/4/d/a^2*tan(1
/2*d*x+1/2*c)+3/4/d/a^2/tan(1/2*d*x+1/2*c)+7/8/d/a^2*ln(tan(1/2*d*x+1/2*c))-1/4/a^2/d/tan(1/2*d*x+1/2*c)^2-1/6
4/a^2/d/tan(1/2*d*x+1/2*c)^4+1/12/a^2/d/tan(1/2*d*x+1/2*c)^3

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maxima [B]  time = 0.35, size = 195, normalized size = 2.03 \[ -\frac {\frac {\frac {144 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {48 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{2}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {{\left (\frac {16 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {48 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {144 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 3\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{4}}{a^{2} \sin \left (d x + c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*((144*sin(d*x + c)/(cos(d*x + c) + 1) - 48*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 16*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 - 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a^2 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 -
(16*sin(d*x + c)/(cos(d*x + c) + 1) - 48*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 144*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 - 3)*(cos(d*x + c) + 1)^4/(a^2*sin(d*x + c)^4))/d

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mupad [B]  time = 8.72, size = 151, normalized size = 1.57 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,a^2\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^2\,d}+\frac {7\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a^2\,d}-\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^2\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {1}{4}\right )}{16\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^2/(4*a^2*d) - tan(c/2 + (d*x)/2)^3/(12*a^2*d) + tan(c/2 + (d*x)/2)^4/(64*a^2*d) + (7*log(ta
n(c/2 + (d*x)/2)))/(8*a^2*d) - (3*tan(c/2 + (d*x)/2))/(4*a^2*d) + (cot(c/2 + (d*x)/2)^4*((4*tan(c/2 + (d*x)/2)
)/3 - 4*tan(c/2 + (d*x)/2)^2 + 12*tan(c/2 + (d*x)/2)^3 - 1/4))/(16*a^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)**5/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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