\(\int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx\) [1056]

Optimal result

Integrand size = 27, antiderivative size = 52 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {b \text {arctanh}(\cos (c+d x))}{2 d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {b \cot (c+d x) \csc (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2917, 2687, 30, 2691, 3855} \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \cot ^3(c+d x)}{3 d}+\frac {b \text {arctanh}(\cos (c+d x))}{2 d}-\frac {b \cot (c+d x) \csc (c+d x)}{2 d} \]

[In]

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

[Out]

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

Rule 30

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

Rule 2687

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 2691

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 2917

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = a \int \cot ^2(c+d x) \csc ^2(c+d x) \, dx+b \int \cot ^2(c+d x) \csc (c+d x) \, dx \\ & = -\frac {b \cot (c+d x) \csc (c+d x)}{2 d}-\frac {1}{2} b \int \csc (c+d x) \, dx+\frac {a \text {Subst}\left (\int x^2 \, dx,x,-\cot (c+d x)\right )}{d} \\ & = \frac {b \text {arctanh}(\cos (c+d x))}{2 d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {b \cot (c+d x) \csc (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.83 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \cot ^3(c+d x)}{3 d}-\frac {b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(\frac {-\frac {a \left (\cos ^{3}\left (d x +c \right )\right )}{3 \sin \left (d x +c \right )^{3}}+b \left (-\frac {\cos ^{3}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(72\)
default	\(\frac {-\frac {a \left (\cos ^{3}\left (d x +c \right )\right )}{3 \sin \left (d x +c \right )^{3}}+b \left (-\frac {\cos ^{3}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\)	\(72\)
risch	\(\frac {6 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 b \,{\mathrm e}^{5 i \left (d x +c \right )}+2 i a -3 b \,{\mathrm e}^{i \left (d x +c \right )}}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}\)	\(98\)
parallelrisch	\(\frac {-a \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 b \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +3 a \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d}\)	\(99\)
norman	\(\frac {-\frac {a}{24 d}+\frac {a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\)	\(152\)

[In]

int(cos(d*x+c)^2*csc(d*x+c)^4*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (46) = 92\).

Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.29 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {4 \, a \cos \left (d x + c\right )^{3} + 6 \, b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

1/12*(4*a*cos(d*x + c)^3 + 6*b*cos(d*x + c)*sin(d*x + c) + 3*(b*cos(d*x + c)^2 - b)*log(1/2*cos(d*x + c) + 1/2
)*sin(d*x + c) - 3*(b*cos(d*x + c)^2 - b)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*s
in(d*x + c))

Sympy [F]

\[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )} \csc ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**4*(a+b*sin(d*x+c)),x)

[Out]

Integral((a + b*sin(c + d*x))*cos(c + d*x)**2*csc(c + d*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.17 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 \, b {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - \frac {4 \, a}{\tan \left (d x + c\right )^{3}}}{12 \, d} \]

[In]

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

[Out]

1/12*(3*b*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + log(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) - 4*a/tan(d*x
+ c)^3)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (46) = 92\).

Time = 0.34 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.21 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {22 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

[In]

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

[Out]

1/24*(a*tan(1/2*d*x + 1/2*c)^3 + 3*b*tan(1/2*d*x + 1/2*c)^2 - 12*b*log(abs(tan(1/2*d*x + 1/2*c))) - 3*a*tan(1/
2*d*x + 1/2*c) + (22*b*tan(1/2*d*x + 1/2*c)^3 + 3*a*tan(1/2*d*x + 1/2*c)^2 - 3*b*tan(1/2*d*x + 1/2*c) - a)/tan
(1/2*d*x + 1/2*c)^3)/d

Mupad [B] (verification not implemented)

Time = 9.90 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.13 \[ \int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a}{3}\right )}{8\,d} \]

[In]

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

[Out]

(a*tan(c/2 + (d*x)/2)^3)/(24*d) - (a*tan(c/2 + (d*x)/2))/(8*d) + (b*tan(c/2 + (d*x)/2)^2)/(8*d) - (b*log(tan(c
/2 + (d*x)/2)))/(2*d) - (cot(c/2 + (d*x)/2)^3*(a/3 + b*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2)^2))/(8*d)