3.1101 \(\int \cot ^4(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx\)

Optimal. Leaf size=74 \[ -\frac {a \cot ^5(c+d x)}{5 d}-\frac {3 b \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {b \cot ^3(c+d x) \csc (c+d x)}{4 d}+\frac {3 b \cot (c+d x) \csc (c+d x)}{8 d} \]

[Out]

-3/8*b*arctanh(cos(d*x+c))/d-1/5*a*cot(d*x+c)^5/d+3/8*b*cot(d*x+c)*csc(d*x+c)/d-1/4*b*cot(d*x+c)^3*csc(d*x+c)/
d

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Rubi [A]  time = 0.13, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2838, 2607, 30, 2611, 3770} \[ -\frac {a \cot ^5(c+d x)}{5 d}-\frac {3 b \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {b \cot ^3(c+d x) \csc (c+d x)}{4 d}+\frac {3 b \cot (c+d x) \csc (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*ArcTanh[Cos[c + d*x]])/(8*d) - (a*Cot[c + d*x]^5)/(5*d) + (3*b*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (b*Cot
[c + d*x]^3*Csc[c + d*x])/(4*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

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

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Mathematica [A]  time = 0.04, size = 135, normalized size = 1.82 \[ -\frac {a \cot ^5(c+d x)}{5 d}-\frac {b \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}+\frac {5 b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {b \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {5 b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {3 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}-\frac {3 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(a*Cot[c + d*x]^5)/d + (5*b*Csc[(c + d*x)/2]^2)/(32*d) - (b*Csc[(c + d*x)/2]^4)/(64*d) - (3*b*Log[Cos[(c
+ d*x)/2]])/(8*d) + (3*b*Log[Sin[(c + d*x)/2]])/(8*d) - (5*b*Sec[(c + d*x)/2]^2)/(32*d) + (b*Sec[(c + d*x)/2]^
4)/(64*d)

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fricas [B]  time = 0.80, size = 160, normalized size = 2.16 \[ -\frac {16 \, a \cos \left (d x + c\right )^{5} + 15 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 15 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 10 \, {\left (5 \, b \cos \left (d x + c\right )^{3} - 3 \, b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{80 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/80*(16*a*cos(d*x + c)^5 + 15*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(1/2*cos(d*x + c) + 1/2)*sin(d*
x + c) - 15*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 10*(5*b*co
s(d*x + c)^3 - 3*b*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)*sin(d*x + c))

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giac [B]  time = 0.22, size = 173, normalized size = 2.34 \[ \frac {2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 20 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {274 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 40 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{320 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/320*(2*a*tan(1/2*d*x + 1/2*c)^5 + 5*b*tan(1/2*d*x + 1/2*c)^4 - 10*a*tan(1/2*d*x + 1/2*c)^3 - 40*b*tan(1/2*d*
x + 1/2*c)^2 + 120*b*log(abs(tan(1/2*d*x + 1/2*c))) + 20*a*tan(1/2*d*x + 1/2*c) - (274*b*tan(1/2*d*x + 1/2*c)^
5 + 20*a*tan(1/2*d*x + 1/2*c)^4 - 40*b*tan(1/2*d*x + 1/2*c)^3 - 10*a*tan(1/2*d*x + 1/2*c)^2 + 5*b*tan(1/2*d*x
+ 1/2*c) + 2*a)/tan(1/2*d*x + 1/2*c)^5)/d

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maple [A]  time = 0.34, size = 116, normalized size = 1.57 \[ -\frac {a \left (\cos ^{5}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}}-\frac {b \left (\cos ^{5}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}+\frac {b \left (\cos ^{5}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}+\frac {b \left (\cos ^{3}\left (d x +c \right )\right )}{8 d}+\frac {3 b \cos \left (d x +c \right )}{8 d}+\frac {3 b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^6*(a+b*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.32, size = 86, normalized size = 1.16 \[ -\frac {5 \, b {\left (\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {16 \, a}{\tan \left (d x + c\right )^{5}}}{80 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/80*(5*b*(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)) + 16*a/tan(d*x + c)^5)/d

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mupad [B]  time = 9.56, size = 174, normalized size = 2.35 \[ \frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32\,d}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {3\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {a}{5}\right )}{32\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(a + b*sin(c + d*x)))/sin(c + d*x)^6,x)

[Out]

(a*tan(c/2 + (d*x)/2))/(16*d) - (a*tan(c/2 + (d*x)/2)^3)/(32*d) + (a*tan(c/2 + (d*x)/2)^5)/(160*d) - (b*tan(c/
2 + (d*x)/2)^2)/(8*d) + (b*tan(c/2 + (d*x)/2)^4)/(64*d) + (3*b*log(tan(c/2 + (d*x)/2)))/(8*d) - (cot(c/2 + (d*
x)/2)^5*(a/5 + (b*tan(c/2 + (d*x)/2))/2 - a*tan(c/2 + (d*x)/2)^2 + 2*a*tan(c/2 + (d*x)/2)^4 - 4*b*tan(c/2 + (d
*x)/2)^3))/(32*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)**6*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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