3.1122 \(\int \cot ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=187 \[ -\frac{b^2 \left (73 a^2-2 b^2\right ) \cos (c+d x)}{8 a d}-\frac{3 a \left (a^2-12 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{3}{2} b x \left (2 a^2-b^2\right )+\frac{17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac{13 b^3 \sin (c+d x) \cos (c+d x)}{4 d} \]

[Out]

(3*b*(2*a^2 - b^2)*x)/2 - (3*a*(a^2 - 12*b^2)*ArcTanh[Cos[c + d*x]])/(8*d) - (b^2*(73*a^2 - 2*b^2)*Cos[c + d*x
])/(8*a*d) - (13*b^3*Cos[c + d*x]*Sin[c + d*x])/(4*d) + (17*b*Cot[c + d*x]*(a + b*Sin[c + d*x])^2)/(8*d) + (5*
Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + d*x])^3)/(8*d) - (Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^4)
/(4*a*d)

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Rubi [A]  time = 0.65629, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2893, 3047, 3033, 3023, 2735, 3770} \[ -\frac{b^2 \left (73 a^2-2 b^2\right ) \cos (c+d x)}{8 a d}-\frac{3 a \left (a^2-12 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac{3}{2} b x \left (2 a^2-b^2\right )+\frac{17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac{13 b^3 \sin (c+d x) \cos (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*Csc[c + d*x]*(a + b*Sin[c + d*x])^3,x]

[Out]

(3*b*(2*a^2 - b^2)*x)/2 - (3*a*(a^2 - 12*b^2)*ArcTanh[Cos[c + d*x]])/(8*d) - (b^2*(73*a^2 - 2*b^2)*Cos[c + d*x
])/(8*a*d) - (13*b^3*Cos[c + d*x]*Sin[c + d*x])/(4*d) + (17*b*Cot[c + d*x]*(a + b*Sin[c + d*x])^2)/(8*d) + (5*
Cot[c + d*x]*Csc[c + d*x]*(a + b*Sin[c + d*x])^3)/(8*d) - (Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^4)
/(4*a*d)

Rule 2893

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*d*f*(n + 1)), x] +
(-Dist[1/(a^2*d^2*(n + 1)*(n + 2)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) -
b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e +
f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/
(a^2*d^2*f*(n + 1)*(n + 2)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Intege
rsQ[2*m, 2*n]) &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])

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 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 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 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 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 ^4(c+d x) \csc (c+d x) (a+b \sin (c+d x))^3 \, dx &=-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}-\frac{\int \csc ^3(c+d x) (a+b \sin (c+d x))^3 \left (15 a^2+3 a b \sin (c+d x)-12 a^2 \sin ^2(c+d x)\right ) \, dx}{12 a^2}\\ &=\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}-\frac{\int \csc ^2(c+d x) (a+b \sin (c+d x))^2 \left (51 a^2 b-3 a \left (3 a^2-2 b^2\right ) \sin (c+d x)-54 a^2 b \sin ^2(c+d x)\right ) \, dx}{24 a^2}\\ &=\frac{17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}-\frac{\int \csc (c+d x) (a+b \sin (c+d x)) \left (-9 a^2 \left (a^2-12 b^2\right )-3 a b \left (21 a^2-2 b^2\right ) \sin (c+d x)-156 a^2 b^2 \sin ^2(c+d x)\right ) \, dx}{24 a^2}\\ &=-\frac{13 b^3 \cos (c+d x) \sin (c+d x)}{4 d}+\frac{17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}-\frac{\int \csc (c+d x) \left (-18 a^3 \left (a^2-12 b^2\right )-72 a^2 b \left (2 a^2-b^2\right ) \sin (c+d x)-6 a b^2 \left (73 a^2-2 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{48 a^2}\\ &=-\frac{b^2 \left (73 a^2-2 b^2\right ) \cos (c+d x)}{8 a d}-\frac{13 b^3 \cos (c+d x) \sin (c+d x)}{4 d}+\frac{17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}-\frac{\int \csc (c+d x) \left (-18 a^3 \left (a^2-12 b^2\right )-72 a^2 b \left (2 a^2-b^2\right ) \sin (c+d x)\right ) \, dx}{48 a^2}\\ &=\frac{3}{2} b \left (2 a^2-b^2\right ) x-\frac{b^2 \left (73 a^2-2 b^2\right ) \cos (c+d x)}{8 a d}-\frac{13 b^3 \cos (c+d x) \sin (c+d x)}{4 d}+\frac{17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}+\frac{1}{8} \left (3 a \left (a^2-12 b^2\right )\right ) \int \csc (c+d x) \, dx\\ &=\frac{3}{2} b \left (2 a^2-b^2\right ) x-\frac{3 a \left (a^2-12 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{b^2 \left (73 a^2-2 b^2\right ) \cos (c+d x)}{8 a d}-\frac{13 b^3 \cos (c+d x) \sin (c+d x)}{4 d}+\frac{17 b \cot (c+d x) (a+b \sin (c+d x))^2}{8 d}+\frac{5 \cot (c+d x) \csc (c+d x) (a+b \sin (c+d x))^3}{8 d}-\frac{\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^4}{4 a d}\\ \end{align*}

Mathematica [B]  time = 6.25432, size = 381, normalized size = 2.04 \[ -\frac{3 b \left (b^2-2 a^2\right ) (c+d x)}{2 d}+\frac{\left (5 a^3-12 a b^2\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{\left (12 a b^2-5 a^3\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{3 \left (a^3-12 a b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}-\frac{3 \left (a^3-12 a b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}+\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \left (4 a^2 b \cos \left (\frac{1}{2} (c+d x)\right )-b^3 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (b^3 \sin \left (\frac{1}{2} (c+d x)\right )-4 a^2 b \sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}-\frac{a^2 b \cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{a^2 b \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{a^3 \csc ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}+\frac{a^3 \sec ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}-\frac{3 a b^2 \cos (c+d x)}{d}-\frac{b^3 \sin (2 (c+d x))}{4 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[c + d*x]^4*Csc[c + d*x]*(a + b*Sin[c + d*x])^3,x]

[Out]

(-3*b*(-2*a^2 + b^2)*(c + d*x))/(2*d) - (3*a*b^2*Cos[c + d*x])/d + ((4*a^2*b*Cos[(c + d*x)/2] - b^3*Cos[(c + d
*x)/2])*Csc[(c + d*x)/2])/(2*d) + ((5*a^3 - 12*a*b^2)*Csc[(c + d*x)/2]^2)/(32*d) - (a^2*b*Cot[(c + d*x)/2]*Csc
[(c + d*x)/2]^2)/(8*d) - (a^3*Csc[(c + d*x)/2]^4)/(64*d) - (3*(a^3 - 12*a*b^2)*Log[Cos[(c + d*x)/2]])/(8*d) +
(3*(a^3 - 12*a*b^2)*Log[Sin[(c + d*x)/2]])/(8*d) + ((-5*a^3 + 12*a*b^2)*Sec[(c + d*x)/2]^2)/(32*d) + (a^3*Sec[
(c + d*x)/2]^4)/(64*d) + (Sec[(c + d*x)/2]*(-4*a^2*b*Sin[(c + d*x)/2] + b^3*Sin[(c + d*x)/2]))/(2*d) - (b^3*Si
n[2*(c + d*x)])/(4*d) + (a^2*b*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(8*d)

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Maple [A]  time = 0.112, size = 316, normalized size = 1.7 \begin{align*} -{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{8\,d}}+{\frac{3\,{a}^{3}\cos \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}}-{\frac{{a}^{2}b \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{d}}+3\,{a}^{2}bx+3\,{\frac{{a}^{2}b\cot \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}bc}{d}}-{\frac{3\,a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d}}-{\frac{9\,a{b}^{2}\cos \left ( dx+c \right ) }{2\,d}}-{\frac{9\,a{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{d\sin \left ( dx+c \right ) }}-{\frac{{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{d}}-{\frac{3\,{b}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,{b}^{3}x}{2}}-{\frac{3\,{b}^{3}c}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^5*(a+b*sin(d*x+c))^3,x)

[Out]

-1/4/d*a^3/sin(d*x+c)^4*cos(d*x+c)^5+1/8/d*a^3/sin(d*x+c)^2*cos(d*x+c)^5+1/8*a^3*cos(d*x+c)^3/d+3/8*a^3*cos(d*
x+c)/d+3/8/d*a^3*ln(csc(d*x+c)-cot(d*x+c))-1/d*a^2*b*cot(d*x+c)^3+3*a^2*b*x+3*a^2*b*cot(d*x+c)/d+3/d*a^2*b*c-3
/2/d*a*b^2/sin(d*x+c)^2*cos(d*x+c)^5-3/2*a*b^2*cos(d*x+c)^3/d-9/2*a*b^2*cos(d*x+c)/d-9/2/d*a*b^2*ln(csc(d*x+c)
-cot(d*x+c))-1/d*b^3/sin(d*x+c)*cos(d*x+c)^5-1/d*b^3*cos(d*x+c)^3*sin(d*x+c)-3/2*b^3*cos(d*x+c)*sin(d*x+c)/d-3
/2*b^3*x-3/2/d*b^3*c

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Maxima [A]  time = 1.67864, size = 286, normalized size = 1.53 \begin{align*} \frac{16 \,{\left (3 \, d x + 3 \, c + \frac{3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a^{2} b - 8 \,{\left (3 \, d x + 3 \, c + \frac{3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} b^{3} - a^{3}{\left (\frac{2 \,{\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 12 \, a b^{2}{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/16*(16*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a^2*b - 8*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)
/(tan(d*x + c)^3 + tan(d*x + c)))*b^3 - a^3*(2*(5*cos(d*x + c)^3 - 3*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x
+ c)^2 + 1) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)) + 12*a*b^2*(2*cos(d*x + c)/(cos(d*x + c)^2 -
1) - 4*cos(d*x + c) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)))/d

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Fricas [A]  time = 1.99008, size = 807, normalized size = 4.32 \begin{align*} -\frac{48 \, a b^{2} \cos \left (d x + c\right )^{5} - 24 \,{\left (2 \, a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right )^{4} + 48 \,{\left (2 \, a^{2} b - b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 10 \,{\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 24 \,{\left (2 \, a^{2} b - b^{3}\right )} d x - 6 \,{\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right ) + 3 \,{\left ({\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + a^{3} - 12 \, a b^{2} - 2 \,{\left (a^{3} - 12 \, 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 ) - 3 \,{\left ({\left (a^{3} - 12 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + a^{3} - 12 \, a b^{2} - 2 \,{\left (a^{3} - 12 \, 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 ) + 8 \,{\left (b^{3} \cos \left (d x + c\right )^{5} + 4 \,{\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{16 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/16*(48*a*b^2*cos(d*x + c)^5 - 24*(2*a^2*b - b^3)*d*x*cos(d*x + c)^4 + 48*(2*a^2*b - b^3)*d*x*cos(d*x + c)^2
+ 10*(a^3 - 12*a*b^2)*cos(d*x + c)^3 - 24*(2*a^2*b - b^3)*d*x - 6*(a^3 - 12*a*b^2)*cos(d*x + c) + 3*((a^3 - 1
2*a*b^2)*cos(d*x + c)^4 + a^3 - 12*a*b^2 - 2*(a^3 - 12*a*b^2)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) - 3*
((a^3 - 12*a*b^2)*cos(d*x + c)^4 + a^3 - 12*a*b^2 - 2*(a^3 - 12*a*b^2)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) +
1/2) + 8*(b^3*cos(d*x + c)^5 + 4*(2*a^2*b - b^3)*cos(d*x + c)^3 - 3*(2*a^2*b - b^3)*cos(d*x + c))*sin(d*x + c
))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**5*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.29662, size = 463, normalized size = 2.48 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 8 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 120 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 32 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 96 \,{\left (2 \, a^{2} b - b^{3}\right )}{\left (d x + c\right )} + 24 \,{\left (a^{3} - 12 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{64 \,{\left (b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, a b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac{50 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 600 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 120 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 32 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/64*(a^3*tan(1/2*d*x + 1/2*c)^4 + 8*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 8*a^3*tan(1/2*d*x + 1/2*c)^2 + 24*a*b^2*ta
n(1/2*d*x + 1/2*c)^2 - 120*a^2*b*tan(1/2*d*x + 1/2*c) + 32*b^3*tan(1/2*d*x + 1/2*c) + 96*(2*a^2*b - b^3)*(d*x
+ c) + 24*(a^3 - 12*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + 64*(b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^2*tan(1/2*d
*x + 1/2*c)^2 - b^3*tan(1/2*d*x + 1/2*c) - 6*a*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - (50*a^3*tan(1/2*d*x + 1/2
*c)^4 - 600*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 120*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 32*b^3*tan(1/2*d*x + 1/2*c)^3 -
8*a^3*tan(1/2*d*x + 1/2*c)^2 + 24*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 8*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d
*x + 1/2*c)^4)/d