3.1065 \(\int \cot ^2(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2889, 3048, 3031, 3021, 2735, 3770} \[ \frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{3 d}+\frac {a b \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a b \cot (c+d x) \csc (c+d x)}{3 d}-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}+b^2 (-x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

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 ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=\int \csc ^4(c+d x) (a+b \sin (c+d x))^2 \left (1-\sin ^2(c+d x)\right ) \, dx\\ &=-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}+\frac {1}{3} \int \csc ^3(c+d x) (a+b \sin (c+d x)) \left (2 b-a \sin (c+d x)-3 b \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {a b \cot (c+d x) \csc (c+d x)}{3 d}-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}-\frac {1}{6} \int \csc ^2(c+d x) \left (2 \left (a^2-2 b^2\right )+6 a b \sin (c+d x)+6 b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{3 d}-\frac {a b \cot (c+d x) \csc (c+d x)}{3 d}-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}-\frac {1}{6} \int \csc (c+d x) \left (6 a b+6 b^2 \sin (c+d x)\right ) \, dx\\ &=-b^2 x+\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{3 d}-\frac {a b \cot (c+d x) \csc (c+d x)}{3 d}-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}-(a b) \int \csc (c+d x) \, dx\\ &=-b^2 x+\frac {a b \tanh ^{-1}(\cos (c+d x))}{d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{3 d}-\frac {a b \cot (c+d x) \csc (c+d x)}{3 d}-\frac {\cot (c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2}{3 d}\\ \end {align*}

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Mathematica [B]  time = 6.19, size = 538, normalized size = 5.60 \[ \frac {\sin ^2(c+d x) \csc \left (\frac {1}{2} (c+d x)\right ) \left (a^2 \cos \left (\frac {1}{2} (c+d x)\right )-3 b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^2}{6 d (a+b \sin (c+d x))^2}+\frac {\sin ^2(c+d x) \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 b^2 \sin \left (\frac {1}{2} (c+d x)\right )-a^2 \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^2}{6 d (a+b \sin (c+d x))^2}-\frac {a^2 \sin ^2(c+d x) \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{24 d (a+b \sin (c+d x))^2}+\frac {a^2 \sin ^2(c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{24 d (a+b \sin (c+d x))^2}-\frac {b^2 (c+d x) \sin ^2(c+d x) (a \csc (c+d x)+b)^2}{d (a+b \sin (c+d x))^2}-\frac {a b \sin ^2(c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{4 d (a+b \sin (c+d x))^2}-\frac {a b \sin ^2(c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^2}{d (a+b \sin (c+d x))^2}+\frac {a b \sin ^2(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{4 d (a+b \sin (c+d x))^2}+\frac {a b \sin ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^2}{d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*(c + d*x)*(b + a*Csc[c + d*x])^2*Sin[c + d*x]^2)/(d*(a + b*Sin[c + d*x])^2)) + ((a^2*Cos[(c + d*x)/2] -
3*b^2*Cos[(c + d*x)/2])*Csc[(c + d*x)/2]*(b + a*Csc[c + d*x])^2*Sin[c + d*x]^2)/(6*d*(a + b*Sin[c + d*x])^2)
- (a*b*Csc[(c + d*x)/2]^2*(b + a*Csc[c + d*x])^2*Sin[c + d*x]^2)/(4*d*(a + b*Sin[c + d*x])^2) - (a^2*Cot[(c +
d*x)/2]*Csc[(c + d*x)/2]^2*(b + a*Csc[c + d*x])^2*Sin[c + d*x]^2)/(24*d*(a + b*Sin[c + d*x])^2) + (a*b*(b + a*
Csc[c + d*x])^2*Log[Cos[(c + d*x)/2]]*Sin[c + d*x]^2)/(d*(a + b*Sin[c + d*x])^2) - (a*b*(b + a*Csc[c + d*x])^2
*Log[Sin[(c + d*x)/2]]*Sin[c + d*x]^2)/(d*(a + b*Sin[c + d*x])^2) + (a*b*(b + a*Csc[c + d*x])^2*Sec[(c + d*x)/
2]^2*Sin[c + d*x]^2)/(4*d*(a + b*Sin[c + d*x])^2) + ((b + a*Csc[c + d*x])^2*Sec[(c + d*x)/2]*(-(a^2*Sin[(c + d
*x)/2]) + 3*b^2*Sin[(c + d*x)/2])*Sin[c + d*x]^2)/(6*d*(a + b*Sin[c + d*x])^2) + (a^2*(b + a*Csc[c + d*x])^2*S
ec[(c + d*x)/2]^2*Sin[c + d*x]^2*Tan[(c + d*x)/2])/(24*d*(a + b*Sin[c + d*x])^2)

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fricas [A]  time = 0.75, size = 167, normalized size = 1.74 \[ \frac {2 \, {\left (a^{2} - 3 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, b^{2} \cos \left (d x + c\right ) + 3 \, {\left (a b \cos \left (d x + c\right )^{2} - a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a b \cos \left (d x + c\right )^{2} - a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 6 \, {\left (b^{2} d x \cos \left (d x + c\right )^{2} - b^{2} d x - a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 167, normalized size = 1.74 \[ \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, {\left (d x + c\right )} b^{2} - 24 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {44 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 + 6*a*b*tan(1/2*d*x + 1/2*c)^2 - 24*(d*x + c)*b^2 - 24*a*b*log(abs(tan(1/2*d*
x + 1/2*c))) - 3*a^2*tan(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2*d*x + 1/2*c) + (44*a*b*tan(1/2*d*x + 1/2*c)^3 + 3*a
^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 6*a*b*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/
2*c)^3)/d

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maple [A]  time = 0.42, size = 114, normalized size = 1.19 \[ -\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}-\frac {a b \left (\cos ^{3}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{2}}-\frac {a b \cos \left (d x +c \right )}{d}-\frac {a b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}-b^{2} x -\frac {b^{2} \cot \left (d x +c \right )}{d}-\frac {b^{2} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.52, size = 82, normalized size = 0.85 \[ -\frac {6 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} b^{2} - 3 \, a b {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, a^{2}}{\tan \left (d x + c\right )^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(d*x + c + 1/tan(d*x + c))*b^2 - 3*a*b*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + log(cos(d*x + c) + 1) -
log(cos(d*x + c) - 1)) + 2*a^2/tan(d*x + c)^3)/d

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mupad [B]  time = 9.48, size = 231, normalized size = 2.41 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {a^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}+\frac {a^2\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {b^2\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}+\frac {b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,b\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}+\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}-\frac {a\,b\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*tan(c/2 + (d*x)/2)^3)/(24*d) - (a^2*cot(c/2 + (d*x)/2)^3)/(24*d) + (a^2*cot(c/2 + (d*x)/2))/(8*d) - (b^2*
cot(c/2 + (d*x)/2))/(2*d) - (a^2*tan(c/2 + (d*x)/2))/(8*d) + (b^2*tan(c/2 + (d*x)/2))/(2*d) - (2*b^2*atan((b*c
os(c/2 + (d*x)/2) + a*sin(c/2 + (d*x)/2))/(a*cos(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2))))/d - (a*b*cot(c/2 + (
d*x)/2)^2)/(4*d) + (a*b*tan(c/2 + (d*x)/2)^2)/(4*d) - (a*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/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)**2*csc(d*x+c)**4*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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