3.1121 \(\int \cot ^4(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=194 \[ -\frac{9 a^2 b \cos (c+d x)}{2 d}-\frac{3 a^2 b \cos (c+d x) \cot ^2(c+d x)}{2 d}+\frac{9 a^2 b \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^3 \cot ^3(c+d x)}{3 d}+\frac{a^3 \cot (c+d x)}{d}+a^3 x-\frac{9 a b^2 \cot (c+d x)}{2 d}+\frac{3 a b^2 \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac{9}{2} a b^2 x+\frac{b^3 \cos ^3(c+d x)}{3 d}+\frac{b^3 \cos (c+d x)}{d}-\frac{b^3 \tanh ^{-1}(\cos (c+d x))}{d} \]

[Out]

a^3*x - (9*a*b^2*x)/2 + (9*a^2*b*ArcTanh[Cos[c + d*x]])/(2*d) - (b^3*ArcTanh[Cos[c + d*x]])/d - (9*a^2*b*Cos[c
+ d*x])/(2*d) + (b^3*Cos[c + d*x])/d + (b^3*Cos[c + d*x]^3)/(3*d) + (a^3*Cot[c + d*x])/d - (9*a*b^2*Cot[c + d
*x])/(2*d) + (3*a*b^2*Cos[c + d*x]^2*Cot[c + d*x])/(2*d) - (3*a^2*b*Cos[c + d*x]*Cot[c + d*x]^2)/(2*d) - (a^3*
Cot[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.22253, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {2722, 2592, 302, 206, 2591, 288, 321, 203, 3473, 8} \[ -\frac{9 a^2 b \cos (c+d x)}{2 d}-\frac{3 a^2 b \cos (c+d x) \cot ^2(c+d x)}{2 d}+\frac{9 a^2 b \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^3 \cot ^3(c+d x)}{3 d}+\frac{a^3 \cot (c+d x)}{d}+a^3 x-\frac{9 a b^2 \cot (c+d x)}{2 d}+\frac{3 a b^2 \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac{9}{2} a b^2 x+\frac{b^3 \cos ^3(c+d x)}{3 d}+\frac{b^3 \cos (c+d x)}{d}-\frac{b^3 \tanh ^{-1}(\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*x - (9*a*b^2*x)/2 + (9*a^2*b*ArcTanh[Cos[c + d*x]])/(2*d) - (b^3*ArcTanh[Cos[c + d*x]])/d - (9*a^2*b*Cos[c
+ d*x])/(2*d) + (b^3*Cos[c + d*x])/d + (b^3*Cos[c + d*x]^3)/(3*d) + (a^3*Cot[c + d*x])/d - (9*a*b^2*Cot[c + d
*x])/(2*d) + (3*a*b^2*Cos[c + d*x]^2*Cot[c + d*x])/(2*d) - (3*a^2*b*Cos[c + d*x]*Cot[c + d*x]^2)/(2*d) - (a^3*
Cot[c + d*x]^3)/(3*d)

Rule 2722

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \cot ^4(c+d x) (a+b \sin (c+d x))^3 \, dx &=\int \left (b^3 \cos ^3(c+d x) \cot (c+d x)+3 a b^2 \cos ^2(c+d x) \cot ^2(c+d x)+3 a^2 b \cos (c+d x) \cot ^3(c+d x)+a^3 \cot ^4(c+d x)\right ) \, dx\\ &=a^3 \int \cot ^4(c+d x) \, dx+\left (3 a^2 b\right ) \int \cos (c+d x) \cot ^3(c+d x) \, dx+\left (3 a b^2\right ) \int \cos ^2(c+d x) \cot ^2(c+d x) \, dx+b^3 \int \cos ^3(c+d x) \cot (c+d x) \, dx\\ &=-\frac{a^3 \cot ^3(c+d x)}{3 d}-a^3 \int \cot ^2(c+d x) \, dx-\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{a^3 \cot (c+d x)}{d}+\frac{3 a b^2 \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac{3 a^2 b \cos (c+d x) \cot ^2(c+d x)}{2 d}-\frac{a^3 \cot ^3(c+d x)}{3 d}+a^3 \int 1 \, dx+\frac{\left (9 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}-\frac{\left (9 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}-\frac{b^3 \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=a^3 x-\frac{9 a^2 b \cos (c+d x)}{2 d}+\frac{b^3 \cos (c+d x)}{d}+\frac{b^3 \cos ^3(c+d x)}{3 d}+\frac{a^3 \cot (c+d x)}{d}-\frac{9 a b^2 \cot (c+d x)}{2 d}+\frac{3 a b^2 \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac{3 a^2 b \cos (c+d x) \cot ^2(c+d x)}{2 d}-\frac{a^3 \cot ^3(c+d x)}{3 d}+\frac{\left (9 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{2 d}+\frac{\left (9 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=a^3 x-\frac{9}{2} a b^2 x+\frac{9 a^2 b \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{b^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{9 a^2 b \cos (c+d x)}{2 d}+\frac{b^3 \cos (c+d x)}{d}+\frac{b^3 \cos ^3(c+d x)}{3 d}+\frac{a^3 \cot (c+d x)}{d}-\frac{9 a b^2 \cot (c+d x)}{2 d}+\frac{3 a b^2 \cos ^2(c+d x) \cot (c+d x)}{2 d}-\frac{3 a^2 b \cos (c+d x) \cot ^2(c+d x)}{2 d}-\frac{a^3 \cot ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 6.21857, size = 355, normalized size = 1.83 \[ \frac{a \left (2 a^2-9 b^2\right ) (c+d x)}{2 d}+\frac{b \left (5 b^2-12 a^2\right ) \cos (c+d x)}{4 d}+\frac{\left (2 b^3-9 a^2 b\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{\left (9 a^2 b-2 b^3\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \left (4 a^3 \cos \left (\frac{1}{2} (c+d x)\right )-9 a b^2 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{6 d}+\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (9 a b^2 \sin \left (\frac{1}{2} (c+d x)\right )-4 a^3 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{6 d}-\frac{3 a^2 b \csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{3 a^2 b \sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{a^3 \cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{24 d}+\frac{a^3 \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{24 d}-\frac{3 a b^2 \sin (2 (c+d x))}{4 d}+\frac{b^3 \cos (3 (c+d x))}{12 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*(2*a^2 - 9*b^2)*(c + d*x))/(2*d) + (b*(-12*a^2 + 5*b^2)*Cos[c + d*x])/(4*d) + (b^3*Cos[3*(c + d*x)])/(12*d)
+ ((4*a^3*Cos[(c + d*x)/2] - 9*a*b^2*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(6*d) - (3*a^2*b*Csc[(c + d*x)/2]^2)
/(8*d) - (a^3*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*d) + ((9*a^2*b - 2*b^3)*Log[Cos[(c + d*x)/2]])/(2*d) +
((-9*a^2*b + 2*b^3)*Log[Sin[(c + d*x)/2]])/(2*d) + (3*a^2*b*Sec[(c + d*x)/2]^2)/(8*d) + (Sec[(c + d*x)/2]*(-4*
a^3*Sin[(c + d*x)/2] + 9*a*b^2*Sin[(c + d*x)/2]))/(6*d) - (3*a*b^2*Sin[2*(c + d*x)])/(4*d) + (a^3*Sec[(c + d*x
)/2]^2*Tan[(c + d*x)/2])/(24*d)

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Maple [A]  time = 0.106, size = 264, normalized size = 1.4 \begin{align*} -{\frac{{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{a}^{3}\cot \left ( dx+c \right ) }{d}}+{a}^{3}x+{\frac{{a}^{3}c}{d}}-{\frac{3\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d}}-{\frac{9\,{a}^{2}b\cos \left ( dx+c \right ) }{2\,d}}-{\frac{9\,{a}^{2}b\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-3\,{\frac{a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{d\sin \left ( dx+c \right ) }}-3\,{\frac{a{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{9\,a{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}-{\frac{9\,a{b}^{2}x}{2}}-{\frac{9\,a{b}^{2}c}{2\,d}}+{\frac{{b}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{b}^{3}\cos \left ( dx+c \right ) }{d}}+{\frac{{b}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.66792, size = 252, normalized size = 1.3 \begin{align*} \frac{4 \,{\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^{3} - 18 \,{\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 )} a b^{2} + 2 \,{\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \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 )} b^{3} + 9 \, a^{2} b{\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 )}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a^3 - 18*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)/(
tan(d*x + c)^3 + tan(d*x + c)))*a*b^2 + 2*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3*log(cos(d*x + c) + 1) + 3*log
(cos(d*x + c) - 1))*b^3 + 9*a^2*b*(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 = 1.94293, size = 699, normalized size = 3.6 \begin{align*} \frac{18 \, a b^{2} \cos \left (d x + c\right )^{5} + 8 \,{\left (2 \, a^{3} - 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (9 \, a^{2} b - 2 \, b^{3} -{\left (9 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 3 \,{\left (9 \, a^{2} b - 2 \, b^{3} -{\left (9 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 6 \,{\left (2 \, a^{3} - 9 \, a b^{2}\right )} \cos \left (d x + c\right ) + 2 \,{\left (2 \, b^{3} \cos \left (d x + c\right )^{5} + 3 \,{\left (2 \, a^{3} - 9 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} - 2 \,{\left (9 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (2 \, a^{3} - 9 \, a b^{2}\right )} d x + 3 \,{\left (9 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(18*a*b^2*cos(d*x + c)^5 + 8*(2*a^3 - 9*a*b^2)*cos(d*x + c)^3 - 3*(9*a^2*b - 2*b^3 - (9*a^2*b - 2*b^3)*co
s(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*(9*a^2*b - 2*b^3 - (9*a^2*b - 2*b^3)*cos(d*x + c)^2
)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 6*(2*a^3 - 9*a*b^2)*cos(d*x + c) + 2*(2*b^3*cos(d*x + c)^5 + 3*(
2*a^3 - 9*a*b^2)*d*x*cos(d*x + c)^2 - 2*(9*a^2*b - 2*b^3)*cos(d*x + c)^3 - 3*(2*a^3 - 9*a*b^2)*d*x + 3*(9*a^2*
b - 2*b^3)*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))

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

[Out]

Timed out

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Giac [B]  time = 1.3599, size = 568, normalized size = 2.93 \begin{align*} \frac{3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 27 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 45 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 108 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 36 \,{\left (2 \, a^{3} - 9 \, a b^{2}\right )}{\left (d x + c\right )} - 36 \,{\left (9 \, a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{198 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 44 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 45 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 108 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 135 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 156 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 132 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 324 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 351 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 156 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 126 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 540 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 315 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 148 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 108 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 27 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, 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 )}^{3}}}{72 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(3*a^3*tan(1/2*d*x + 1/2*c)^3 + 27*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 45*a^3*tan(1/2*d*x + 1/2*c) + 108*a*b^2
*tan(1/2*d*x + 1/2*c) + 36*(2*a^3 - 9*a*b^2)*(d*x + c) - 36*(9*a^2*b - 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c))) +
(198*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 44*b^3*tan(1/2*d*x + 1/2*c)^9 + 45*a^3*tan(1/2*d*x + 1/2*c)^8 + 108*a*b^2
*tan(1/2*d*x + 1/2*c)^8 + 135*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 156*b^3*tan(1/2*d*x + 1/2*c)^7 + 132*a^3*tan(1/2*
d*x + 1/2*c)^6 - 324*a*b^2*tan(1/2*d*x + 1/2*c)^6 - 351*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 156*b^3*tan(1/2*d*x + 1
/2*c)^5 + 126*a^3*tan(1/2*d*x + 1/2*c)^4 - 540*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 315*a^2*b*tan(1/2*d*x + 1/2*c)^3
+ 148*b^3*tan(1/2*d*x + 1/2*c)^3 + 36*a^3*tan(1/2*d*x + 1/2*c)^2 - 108*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 27*a^2*
b*tan(1/2*d*x + 1/2*c) - 3*a^3)/(tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))^3)/d