3.159 \(\int \cot ^6(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=202 \[ -\frac{a^2 \cot ^5(c+d x)}{5 d}+\frac{a^2 \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot (c+d x)}{d}-a^2 x+\frac{15 a b \cos (c+d x)}{4 d}-\frac{a b \cos (c+d x) \cot ^4(c+d x)}{2 d}+\frac{5 a b \cos (c+d x) \cot ^2(c+d x)}{4 d}-\frac{15 a b \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{5 b^2 \cot ^3(c+d x)}{6 d}+\frac{5 b^2 \cot (c+d x)}{2 d}+\frac{b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}+\frac{5 b^2 x}{2} \]

[Out]

-(a^2*x) + (5*b^2*x)/2 - (15*a*b*ArcTanh[Cos[c + d*x]])/(4*d) + (15*a*b*Cos[c + d*x])/(4*d) - (a^2*Cot[c + d*x
])/d + (5*b^2*Cot[c + d*x])/(2*d) + (5*a*b*Cos[c + d*x]*Cot[c + d*x]^2)/(4*d) + (a^2*Cot[c + d*x]^3)/(3*d) - (
5*b^2*Cot[c + d*x]^3)/(6*d) + (b^2*Cos[c + d*x]^2*Cot[c + d*x]^3)/(2*d) - (a*b*Cos[c + d*x]*Cot[c + d*x]^4)/(2
*d) - (a^2*Cot[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.169485, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {2722, 2591, 288, 302, 203, 2592, 321, 206, 3473, 8} \[ -\frac{a^2 \cot ^5(c+d x)}{5 d}+\frac{a^2 \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot (c+d x)}{d}-a^2 x+\frac{15 a b \cos (c+d x)}{4 d}-\frac{a b \cos (c+d x) \cot ^4(c+d x)}{2 d}+\frac{5 a b \cos (c+d x) \cot ^2(c+d x)}{4 d}-\frac{15 a b \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac{5 b^2 \cot ^3(c+d x)}{6 d}+\frac{5 b^2 \cot (c+d x)}{2 d}+\frac{b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}+\frac{5 b^2 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + b*Sin[c + d*x])^2,x]

[Out]

-(a^2*x) + (5*b^2*x)/2 - (15*a*b*ArcTanh[Cos[c + d*x]])/(4*d) + (15*a*b*Cos[c + d*x])/(4*d) - (a^2*Cot[c + d*x
])/d + (5*b^2*Cot[c + d*x])/(2*d) + (5*a*b*Cos[c + d*x]*Cot[c + d*x]^2)/(4*d) + (a^2*Cot[c + d*x]^3)/(3*d) - (
5*b^2*Cot[c + d*x]^3)/(6*d) + (b^2*Cos[c + d*x]^2*Cot[c + d*x]^3)/(2*d) - (a*b*Cos[c + d*x]*Cot[c + d*x]^4)/(2
*d) - (a^2*Cot[c + d*x]^5)/(5*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 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 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 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 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 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 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 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 ^6(c+d x) (a+b \sin (c+d x))^2 \, dx &=\int \left (b^2 \cos ^2(c+d x) \cot ^4(c+d x)+2 a b \cos (c+d x) \cot ^5(c+d x)+a^2 \cot ^6(c+d x)\right ) \, dx\\ &=a^2 \int \cot ^6(c+d x) \, dx+(2 a b) \int \cos (c+d x) \cot ^5(c+d x) \, dx+b^2 \int \cos ^2(c+d x) \cot ^4(c+d x) \, dx\\ &=-\frac{a^2 \cot ^5(c+d x)}{5 d}-a^2 \int \cot ^4(c+d x) \, dx-\frac{(2 a b) \operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^3} \, dx,x,\cos (c+d x)\right )}{d}-\frac{b^2 \operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{a^2 \cot ^3(c+d x)}{3 d}+\frac{b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}-\frac{a b \cos (c+d x) \cot ^4(c+d x)}{2 d}-\frac{a^2 \cot ^5(c+d x)}{5 d}+a^2 \int \cot ^2(c+d x) \, dx+\frac{(5 a b) \operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{2 d}-\frac{\left (5 b^2\right ) \operatorname{Subst}\left (\int \frac{x^4}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=-\frac{a^2 \cot (c+d x)}{d}+\frac{5 a b \cos (c+d x) \cot ^2(c+d x)}{4 d}+\frac{a^2 \cot ^3(c+d x)}{3 d}+\frac{b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}-\frac{a b \cos (c+d x) \cot ^4(c+d x)}{2 d}-\frac{a^2 \cot ^5(c+d x)}{5 d}-a^2 \int 1 \, dx-\frac{(15 a b) \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\cos (c+d x)\right )}{4 d}-\frac{\left (5 b^2\right ) \operatorname{Subst}\left (\int \left (-1+x^2+\frac{1}{1+x^2}\right ) \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=-a^2 x+\frac{15 a b \cos (c+d x)}{4 d}-\frac{a^2 \cot (c+d x)}{d}+\frac{5 b^2 \cot (c+d x)}{2 d}+\frac{5 a b \cos (c+d x) \cot ^2(c+d x)}{4 d}+\frac{a^2 \cot ^3(c+d x)}{3 d}-\frac{5 b^2 \cot ^3(c+d x)}{6 d}+\frac{b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}-\frac{a b \cos (c+d x) \cot ^4(c+d x)}{2 d}-\frac{a^2 \cot ^5(c+d x)}{5 d}-\frac{(15 a b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{4 d}-\frac{\left (5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=-a^2 x+\frac{5 b^2 x}{2}-\frac{15 a b \tanh ^{-1}(\cos (c+d x))}{4 d}+\frac{15 a b \cos (c+d x)}{4 d}-\frac{a^2 \cot (c+d x)}{d}+\frac{5 b^2 \cot (c+d x)}{2 d}+\frac{5 a b \cos (c+d x) \cot ^2(c+d x)}{4 d}+\frac{a^2 \cot ^3(c+d x)}{3 d}-\frac{5 b^2 \cot ^3(c+d x)}{6 d}+\frac{b^2 \cos ^2(c+d x) \cot ^3(c+d x)}{2 d}-\frac{a b \cos (c+d x) \cot ^4(c+d x)}{2 d}-\frac{a^2 \cot ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.09302, size = 351, normalized size = 1.74 \[ \frac{\left (560 b^2-368 a^2\right ) \cot \left (\frac{1}{2} (c+d x)\right )+368 a^2 \tan \left (\frac{1}{2} (c+d x)\right )+96 a^2 \sin ^6\left (\frac{1}{2} (c+d x)\right ) \csc ^5(c+d x)-328 a^2 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-\frac{3}{2} a^2 \sin (c+d x) \csc ^6\left (\frac{1}{2} (c+d x)\right )+\frac{41}{2} a^2 \sin (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )-480 a^2 c-480 a^2 d x+960 a b \cos (c+d x)-15 a b \csc ^4\left (\frac{1}{2} (c+d x)\right )+270 a b \csc ^2\left (\frac{1}{2} (c+d x)\right )+15 a b \sec ^4\left (\frac{1}{2} (c+d x)\right )-270 a b \sec ^2\left (\frac{1}{2} (c+d x)\right )+1800 a b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-1800 a b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+120 b^2 \sin (2 (c+d x))-560 b^2 \tan \left (\frac{1}{2} (c+d x)\right )+160 b^2 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-10 b^2 \sin (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )+1200 b^2 c+1200 b^2 d x}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + b*Sin[c + d*x])^2,x]

[Out]

(-480*a^2*c + 1200*b^2*c - 480*a^2*d*x + 1200*b^2*d*x + 960*a*b*Cos[c + d*x] + (-368*a^2 + 560*b^2)*Cot[(c + d
*x)/2] + 270*a*b*Csc[(c + d*x)/2]^2 - 15*a*b*Csc[(c + d*x)/2]^4 - 1800*a*b*Log[Cos[(c + d*x)/2]] + 1800*a*b*Lo
g[Sin[(c + d*x)/2]] - 270*a*b*Sec[(c + d*x)/2]^2 + 15*a*b*Sec[(c + d*x)/2]^4 - 328*a^2*Csc[c + d*x]^3*Sin[(c +
d*x)/2]^4 + 160*b^2*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 96*a^2*Csc[c + d*x]^5*Sin[(c + d*x)/2]^6 + (41*a^2*Cs
c[(c + d*x)/2]^4*Sin[c + d*x])/2 - 10*b^2*Csc[(c + d*x)/2]^4*Sin[c + d*x] - (3*a^2*Csc[(c + d*x)/2]^6*Sin[c +
d*x])/2 + 120*b^2*Sin[2*(c + d*x)] + 368*a^2*Tan[(c + d*x)/2] - 560*b^2*Tan[(c + d*x)/2])/(480*d)

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Maple [A]  time = 0.052, size = 302, normalized size = 1.5 \begin{align*} -{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}-{a}^{2}x-{\frac{{a}^{2}c}{d}}-{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{4\,d}}+{\frac{5\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{15\,ab\cos \left ( dx+c \right ) }{4\,d}}+{\frac{15\,ab\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{4\,d}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{4\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{3\,d\sin \left ( dx+c \right ) }}+{\frac{4\,{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{3\,d}}+{\frac{5\,{b}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{5\,{b}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{5\,{b}^{2}x}{2}}+{\frac{5\,{b}^{2}c}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+b*sin(d*x+c))^2,x)

[Out]

-1/5*a^2*cot(d*x+c)^5/d+1/3*a^2*cot(d*x+c)^3/d-a^2*cot(d*x+c)/d-a^2*x-1/d*a^2*c-1/2/d*a*b/sin(d*x+c)^4*cos(d*x
+c)^7+3/4/d*a*b/sin(d*x+c)^2*cos(d*x+c)^7+3/4/d*a*b*cos(d*x+c)^5+5/4/d*a*b*cos(d*x+c)^3+15/4*a*b*cos(d*x+c)/d+
15/4/d*a*b*ln(csc(d*x+c)-cot(d*x+c))-1/3/d*b^2/sin(d*x+c)^3*cos(d*x+c)^7+4/3/d*b^2/sin(d*x+c)*cos(d*x+c)^7+4/3
/d*b^2*sin(d*x+c)*cos(d*x+c)^5+5/3/d*b^2*sin(d*x+c)*cos(d*x+c)^3+5/2*b^2*cos(d*x+c)*sin(d*x+c)/d+5/2*b^2*x+5/2
/d*b^2*c

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Maxima [A]  time = 1.67638, size = 247, normalized size = 1.22 \begin{align*} -\frac{8 \,{\left (15 \, d x + 15 \, c + \frac{15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a^{2} - 20 \,{\left (15 \, d x + 15 \, c + \frac{15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} b^{2} + 15 \, a b{\left (\frac{2 \,{\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/120*(8*(15*d*x + 15*c + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/tan(d*x + c)^5)*a^2 - 20*(15*d*x + 15*c
+ (15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 - 2)/(tan(d*x + c)^5 + tan(d*x + c)^3))*b^2 + 15*a*b*(2*(9*cos(d*x +
c)^3 - 7*cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - 16*cos(d*x + c) + 15*log(cos(d*x + c) + 1) -
15*log(cos(d*x + c) - 1)))/d

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Fricas [A]  time = 1.63935, size = 786, normalized size = 3.89 \begin{align*} -\frac{60 \, b^{2} \cos \left (d x + c\right )^{7} + 92 \,{\left (2 \, a^{2} - 5 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 140 \,{\left (2 \, a^{2} - 5 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 225 \,{\left (a b \cos \left (d x + c\right )^{4} - 2 \, 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 ) - 225 \,{\left (a b \cos \left (d x + c\right )^{4} - 2 \, 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 ) + 60 \,{\left (2 \, a^{2} - 5 \, b^{2}\right )} \cos \left (d x + c\right ) + 30 \,{\left (2 \,{\left (2 \, a^{2} - 5 \, b^{2}\right )} d x \cos \left (d x + c\right )^{4} - 8 \, a b \cos \left (d x + c\right )^{5} - 4 \,{\left (2 \, a^{2} - 5 \, b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 25 \, a b \cos \left (d x + c\right )^{3} + 2 \,{\left (2 \, a^{2} - 5 \, b^{2}\right )} d x - 15 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, 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(cot(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/120*(60*b^2*cos(d*x + c)^7 + 92*(2*a^2 - 5*b^2)*cos(d*x + c)^5 - 140*(2*a^2 - 5*b^2)*cos(d*x + c)^3 + 225*(
a*b*cos(d*x + c)^4 - 2*a*b*cos(d*x + c)^2 + a*b)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 225*(a*b*cos(d*x +
c)^4 - 2*a*b*cos(d*x + c)^2 + a*b)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 60*(2*a^2 - 5*b^2)*cos(d*x + c
) + 30*(2*(2*a^2 - 5*b^2)*d*x*cos(d*x + c)^4 - 8*a*b*cos(d*x + c)^5 - 4*(2*a^2 - 5*b^2)*d*x*cos(d*x + c)^2 + 2
5*a*b*cos(d*x + c)^3 + 2*(2*a^2 - 5*b^2)*d*x - 15*a*b*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*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(cot(d*x+c)**6*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 2.77852, size = 455, normalized size = 2.25 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 35 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 20 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 240 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1800 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 330 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 540 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 240 \,{\left (2 \, a^{2} - 5 \, b^{2}\right )}{\left (d x + c\right )} - \frac{480 \,{\left (b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a b\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac{4110 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 330 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 540 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 240 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 35 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 20 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 15 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/480*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 15*a*b*tan(1/2*d*x + 1/2*c)^4 - 35*a^2*tan(1/2*d*x + 1/2*c)^3 + 20*b^2*t
an(1/2*d*x + 1/2*c)^3 - 240*a*b*tan(1/2*d*x + 1/2*c)^2 + 1800*a*b*log(abs(tan(1/2*d*x + 1/2*c))) + 330*a^2*tan
(1/2*d*x + 1/2*c) - 540*b^2*tan(1/2*d*x + 1/2*c) - 240*(2*a^2 - 5*b^2)*(d*x + c) - 480*(b^2*tan(1/2*d*x + 1/2*
c)^3 - 4*a*b*tan(1/2*d*x + 1/2*c)^2 - b^2*tan(1/2*d*x + 1/2*c) - 4*a*b)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - (4110
*a*b*tan(1/2*d*x + 1/2*c)^5 + 330*a^2*tan(1/2*d*x + 1/2*c)^4 - 540*b^2*tan(1/2*d*x + 1/2*c)^4 - 240*a*b*tan(1/
2*d*x + 1/2*c)^3 - 35*a^2*tan(1/2*d*x + 1/2*c)^2 + 20*b^2*tan(1/2*d*x + 1/2*c)^2 + 15*a*b*tan(1/2*d*x + 1/2*c)
+ 3*a^2)/tan(1/2*d*x + 1/2*c)^5)/d