3.1064 \(\int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=89 \[ \frac {\left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a b \cot (c+d x)}{d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}-2 a b x+\frac {3 b^2 \cos (c+d x)}{2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 89, 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 = {2889, 3048, 3031, 3023, 2735, 3770} \[ \frac {\left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a b \cot (c+d x)}{d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^2}{2 d}-2 a b x+\frac {3 b^2 \cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*a*b*x + ((a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(2*d) + (3*b^2*Cos[c + d*x])/(2*d) - (a*b*Cot[c + d*x])/d - (
Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + d*x])^2)/(2*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 3023

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.89, size = 155, normalized size = 1.74 \[ \frac {a^2 \left (-\csc ^2\left (\frac {1}{2} (c+d x)\right )\right )+a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )-4 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+8 a b \tan \left (\frac {1}{2} (c+d x)\right )-8 a b \cot \left (\frac {1}{2} (c+d x)\right )-16 a b c-16 a b d x+8 b^2 \cos (c+d x)+8 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*a*b*c - 16*a*b*d*x + 8*b^2*Cos[c + d*x] - 8*a*b*Cot[(c + d*x)/2] - a^2*Csc[(c + d*x)/2]^2 + 4*a^2*Log[Cos
[(c + d*x)/2]] - 8*b^2*Log[Cos[(c + d*x)/2]] - 4*a^2*Log[Sin[(c + d*x)/2]] + 8*b^2*Log[Sin[(c + d*x)/2]] + a^2
*Sec[(c + d*x)/2]^2 + 8*a*b*Tan[(c + d*x)/2])/(8*d)

________________________________________________________________________________________

fricas [B]  time = 0.65, size = 168, normalized size = 1.89 \[ -\frac {8 \, a b d x \cos \left (d x + c\right )^{2} - 4 \, b^{2} \cos \left (d x + c\right )^{3} - 8 \, a b d x - 8 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.20, size = 148, normalized size = 1.66 \[ \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, {\left (d x + c\right )} a b + 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, {\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {16 \, b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 - 16*(d*x + c)*a*b + 8*a*b*tan(1/2*d*x + 1/2*c) - 4*(a^2 - 2*b^2)*log(abs(tan(
1/2*d*x + 1/2*c))) + 16*b^2/(tan(1/2*d*x + 1/2*c)^2 + 1) + (6*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x
+ 1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^2)/d

________________________________________________________________________________________

maple [A]  time = 0.51, size = 126, normalized size = 1.42 \[ -\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {a^{2} \cos \left (d x +c \right )}{2 d}-\frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-2 a b x -\frac {2 a b \cot \left (d x +c \right )}{d}-\frac {2 a b c}{d}+\frac {b^{2} \cos \left (d x +c \right )}{d}+\frac {b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 103, normalized size = 1.16 \[ -\frac {8 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a b - a^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 2 \, b^{2} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/4*(8*(d*x + c + 1/tan(d*x + c))*a*b - a^2*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + log(cos(d*x + c) + 1) - lo
g(cos(d*x + c) - 1)) - 2*b^2*(2*cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)))/d

________________________________________________________________________________________

mupad [B]  time = 10.20, size = 397, normalized size = 4.46 \[ \frac {\cos \left (c+d\,x\right )\,\left (\frac {a^2}{2}-\frac {b^2}{4}\right )-\frac {b^2}{2}+\frac {a^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}-\frac {b^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {b^2\,\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {b^2\,\cos \left (3\,c+3\,d\,x\right )}{4}-2\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{-\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}\right )-\frac {a^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )}{4}+\frac {b^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )}{2}+a\,b\,\sin \left (2\,c+2\,d\,x\right )+2\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b-2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}{-\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2}\right )\,\cos \left (2\,c+2\,d\,x\right )}{d\,\left ({\cos \left (c+d\,x\right )}^2-1\right )} \]

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)^3,x)

[Out]

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

________________________________________________________________________________________

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)**3*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________