3.1057 \(\int \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x)) \, dx\)

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

[Out]

1/8*a*arctanh(cos(d*x+c))/d-1/3*b*cot(d*x+c)^3/d+1/8*a*cot(d*x+c)*csc(d*x+c)/d-1/4*a*cot(d*x+c)*csc(d*x+c)^3/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 = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2838, 2611, 3768, 3770, 2607, 30} \[ \frac {a \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {a \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b \cot ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.74, size = 137, normalized size = 1.85 \[ -\frac {16 \, b \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + 6 \, a \cos \left (d x + c\right )^{3} + 6 \, a \cos \left (d x + c\right ) - 3 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{2} + a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(16*b*cos(d*x + c)^3*sin(d*x + c) + 6*a*cos(d*x + c)^3 + 6*a*cos(d*x + c) - 3*(a*cos(d*x + c)^4 - 2*a*co
s(d*x + c)^2 + a)*log(1/2*cos(d*x + c) + 1/2) + 3*(a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^2 + a)*log(-1/2*cos(d*x
+ c) + 1/2))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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giac [A]  time = 0.19, size = 116, normalized size = 1.57 \[ \frac {3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 8 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 24 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {50 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 24 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*a*tan(1/2*d*x + 1/2*c)^4 + 8*b*tan(1/2*d*x + 1/2*c)^3 - 24*a*log(abs(tan(1/2*d*x + 1/2*c))) - 24*b*ta
n(1/2*d*x + 1/2*c) + (50*a*tan(1/2*d*x + 1/2*c)^4 + 24*b*tan(1/2*d*x + 1/2*c)^3 - 8*b*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 = 0.32, size = 102, normalized size = 1.38 \[ -\frac {a \left (\cos ^{3}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}-\frac {a \left (\cos ^{3}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}-\frac {a \cos \left (d x +c \right )}{8 d}-\frac {a \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {b \left (\cos ^{3}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 80, normalized size = 1.08 \[ -\frac {3 \, a {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {16 \, b}{\tan \left (d x + c\right )^{3}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*a*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - log(cos(d*x + c) + 1)
+ log(cos(d*x + c) - 1)) + 16*b/tan(d*x + c)^3)/d

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mupad [B]  time = 9.33, size = 112, normalized size = 1.51 \[ \frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {a}{4}\right )}{16\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*tan(c/2 + (d*x)/2)^4)/(64*d) - (b*tan(c/2 + (d*x)/2))/(8*d) + (b*tan(c/2 + (d*x)/2)^3)/(24*d) - (a*log(tan(
c/2 + (d*x)/2)))/(8*d) - (cot(c/2 + (d*x)/2)^4*(a/4 + (2*b*tan(c/2 + (d*x)/2))/3 - 2*b*tan(c/2 + (d*x)/2)^3))/
(16*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)**2*csc(d*x+c)**5*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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