3.1073 \(\int \cot ^2(c+d x) \csc (c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=138 \[ \frac {a \left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {1}{2} b x \left (6 a^2-b^2\right )+\frac {15 a b^2 \cos (c+d x)}{2 d}-\frac {3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac {5 b^3 \sin (c+d x) \cos (c+d x)}{2 d} \]

[Out]

-1/2*b*(6*a^2-b^2)*x+1/2*a*(a^2-6*b^2)*arctanh(cos(d*x+c))/d+15/2*a*b^2*cos(d*x+c)/d+5/2*b^3*cos(d*x+c)*sin(d*
x+c)/d-3/2*b*cot(d*x+c)*(a+b*sin(d*x+c))^2/d-1/2*cot(d*x+c)*csc(d*x+c)*(a+b*sin(d*x+c))^3/d

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Rubi [A]  time = 0.47, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2889, 3048, 3047, 3033, 3023, 2735, 3770} \[ \frac {a \left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {1}{2} b x \left (6 a^2-b^2\right )+\frac {15 a b^2 \cos (c+d x)}{2 d}-\frac {3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac {5 b^3 \sin (c+d x) \cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(6*a^2 - b^2)*x)/2 + (a*(a^2 - 6*b^2)*ArcTanh[Cos[c + d*x]])/(2*d) + (15*a*b^2*Cos[c + d*x])/(2*d) + (5*b^
3*Cos[c + d*x]*Sin[c + d*x])/(2*d) - (3*b*Cot[c + d*x]*(a + b*Sin[c + d*x])^2)/(2*d) - (Cot[c + d*x]*Csc[c + d
*x]*(a + b*Sin[c + d*x])^3)/(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 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + 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*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(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, B, 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 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))^3 \, dx &=\int \csc ^3(c+d x) (a+b \sin (c+d x))^3 \left (1-\sin ^2(c+d x)\right ) \, dx\\ &=-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac {1}{2} \int \csc ^2(c+d x) (a+b \sin (c+d x))^2 \left (3 b-a \sin (c+d x)-4 b \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac {1}{2} \int \csc (c+d x) (a+b \sin (c+d x)) \left (-a^2+6 b^2-5 a b \sin (c+d x)-10 b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {5 b^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac {1}{4} \int \csc (c+d x) \left (-2 a \left (a^2-6 b^2\right )-2 b \left (6 a^2-b^2\right ) \sin (c+d x)-30 a b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {15 a b^2 \cos (c+d x)}{2 d}+\frac {5 b^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}+\frac {1}{4} \int \csc (c+d x) \left (-2 a \left (a^2-6 b^2\right )-2 b \left (6 a^2-b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {1}{2} b \left (6 a^2-b^2\right ) x+\frac {15 a b^2 \cos (c+d x)}{2 d}+\frac {5 b^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}-\frac {1}{2} \left (a \left (a^2-6 b^2\right )\right ) \int \csc (c+d x) \, dx\\ &=-\frac {1}{2} b \left (6 a^2-b^2\right ) x+\frac {a \left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac {15 a b^2 \cos (c+d x)}{2 d}+\frac {5 b^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {3 b \cot (c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{2 d}\\ \end {align*}

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Mathematica [A]  time = 1.31, size = 192, normalized size = 1.39 \[ \frac {a^3 \left (-\csc ^2\left (\frac {1}{2} (c+d x)\right )\right )+a^3 \sec ^2\left (\frac {1}{2} (c+d x)\right )-4 a^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+12 a^2 b \tan \left (\frac {1}{2} (c+d x)\right )-12 a^2 b \cot \left (\frac {1}{2} (c+d x)\right )-24 a^2 b c-24 a^2 b d x+24 a b^2 \cos (c+d x)+24 a b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-24 a b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 b^3 \sin (2 (c+d x))+4 b^3 c+4 b^3 d x}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*a^2*b*c + 4*b^3*c - 24*a^2*b*d*x + 4*b^3*d*x + 24*a*b^2*Cos[c + d*x] - 12*a^2*b*Cot[(c + d*x)/2] - a^3*Cs
c[(c + d*x)/2]^2 + 4*a^3*Log[Cos[(c + d*x)/2]] - 24*a*b^2*Log[Cos[(c + d*x)/2]] - 4*a^3*Log[Sin[(c + d*x)/2]]
+ 24*a*b^2*Log[Sin[(c + d*x)/2]] + a^3*Sec[(c + d*x)/2]^2 + 2*b^3*Sin[2*(c + d*x)] + 12*a^2*b*Tan[(c + d*x)/2]
)/(8*d)

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fricas [A]  time = 0.75, size = 216, normalized size = 1.57 \[ \frac {12 \, a b^{2} \cos \left (d x + c\right )^{3} - 2 \, {\left (6 \, a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, a^{2} b - b^{3}\right )} d x + 2 \, {\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right ) - {\left (a^{3} - 6 \, a b^{2} - {\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{3} - 6 \, a b^{2} - {\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (b^{3} \cos \left (d x + c\right )^{3} + {\left (6 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.30, size = 272, normalized size = 1.97 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, {\left (6 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )} - 4 \, {\left (a^{3} - 6 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2}}}{8 \, d} \]

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))^3,x, algorithm="giac")

[Out]

1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^2*b*tan(1/2*d*x + 1/2*c) - 4*(6*a^2*b - b^3)*(d*x + c) - 4*(a^3 - 6*a*b
^2)*log(abs(tan(1/2*d*x + 1/2*c))) + (2*a^3*tan(1/2*d*x + 1/2*c)^6 - 12*a*b^2*tan(1/2*d*x + 1/2*c)^6 - 12*a^2*
b*tan(1/2*d*x + 1/2*c)^5 - 8*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*a^3*tan(1/2*d*x + 1/2*c)^4 + 24*a*b^2*tan(1/2*d*x
+ 1/2*c)^4 - 24*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 8*b^3*tan(1/2*d*x + 1/2*c)^3 + 36*a*b^2*tan(1/2*d*x + 1/2*c)^2
- 12*a^2*b*tan(1/2*d*x + 1/2*c) - a^3)/(tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))^2)/d

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maple [A]  time = 0.50, size = 171, normalized size = 1.24 \[ -\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {a^{3} \cos \left (d x +c \right )}{2 d}-\frac {a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d}-3 a^{2} b x -\frac {3 a^{2} b \cot \left (d x +c \right )}{d}-\frac {3 a^{2} b c}{d}+\frac {3 a \,b^{2} \cos \left (d x +c \right )}{d}+\frac {3 a \,b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {b^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {b^{3} x}{2}+\frac {b^{3} c}{2 d} \]

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))^3,x)

[Out]

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

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maxima [A]  time = 0.50, size = 128, normalized size = 0.93 \[ -\frac {12 \, {\left (d x + c + \frac {1}{\tan \left (d x + c\right )}\right )} a^{2} b - {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{3} - a^{3} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 6 \, a b^{2} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.46, size = 585, normalized size = 4.24 \[ \frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,a\,b^2-\frac {a^3}{2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (6\,a^2\,b+4\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (24\,a\,b^2-\frac {a^3}{2}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (24\,a\,b^2-a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (12\,a^2\,b-4\,b^3\right )+\frac {a^3}{2}+6\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}+\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {b\,\mathrm {atan}\left (\frac {\frac {b\,\left (6\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a\,b^2-a^3\right )-6\,a^2\,b+b^3-b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2-b^2\right )\,3{}\mathrm {i}\right )}{2}+\frac {b\,\left (6\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a\,b^2-a^3\right )-6\,a^2\,b+b^3+b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2-b^2\right )\,3{}\mathrm {i}\right )}{2}}{2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (36\,a^4\,b^2-12\,a^2\,b^4+b^6\right )+6\,a\,b^5+6\,a^5\,b-37\,a^3\,b^3-\frac {b\,\left (6\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a\,b^2-a^3\right )-6\,a^2\,b+b^3-b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2-b^2\right )\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,\left (6\,a^2-b^2\right )\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a\,b^2-a^3\right )-6\,a^2\,b+b^3+b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2-b^2\right )\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}}\right )\,\left (6\,a^2-b^2\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))^3)/sin(c + d*x)^3,x)

[Out]

(a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (log(tan(c/2 + (d*x)/2))*(3*a*b^2 - a^3/2))/d - (tan(c/2 + (d*x)/2)^5*(6*a^
2*b + 4*b^3) - tan(c/2 + (d*x)/2)^4*(24*a*b^2 - a^3/2) - tan(c/2 + (d*x)/2)^2*(24*a*b^2 - a^3) + tan(c/2 + (d*
x)/2)^3*(12*a^2*b - 4*b^3) + a^3/2 + 6*a^2*b*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 8*tan(c/2 + (d*x
)/2)^4 + 4*tan(c/2 + (d*x)/2)^6)) + (3*a^2*b*tan(c/2 + (d*x)/2))/(2*d) + (b*atan(((b*(6*a^2 - b^2)*(tan(c/2 +
(d*x)/2)*(6*a*b^2 - a^3) - 6*a^2*b + b^3 - b*tan(c/2 + (d*x)/2)*(6*a^2 - b^2)*3i))/2 + (b*(6*a^2 - b^2)*(tan(c
/2 + (d*x)/2)*(6*a*b^2 - a^3) - 6*a^2*b + b^3 + b*tan(c/2 + (d*x)/2)*(6*a^2 - b^2)*3i))/2)/(2*tan(c/2 + (d*x)/
2)*(b^6 - 12*a^2*b^4 + 36*a^4*b^2) + 6*a*b^5 + 6*a^5*b - 37*a^3*b^3 - (b*(6*a^2 - b^2)*(tan(c/2 + (d*x)/2)*(6*
a*b^2 - a^3) - 6*a^2*b + b^3 - b*tan(c/2 + (d*x)/2)*(6*a^2 - b^2)*3i)*1i)/2 + (b*(6*a^2 - b^2)*(tan(c/2 + (d*x
)/2)*(6*a*b^2 - a^3) - 6*a^2*b + b^3 + b*tan(c/2 + (d*x)/2)*(6*a^2 - b^2)*3i)*1i)/2))*(6*a^2 - b^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)**3*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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