3.1252 \(\int \cot ^6(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=159 \[ \frac{5 \left (a^2+8 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{128 d}+\frac{\left (17 a^2-8 b^2\right ) \cot (c+d x) \csc ^5(c+d x)}{48 d}-\frac{\left (59 a^2-104 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{192 d}+\frac{\left (5 a^2-88 b^2\right ) \cot (c+d x) \csc (c+d x)}{128 d}-\frac{a^2 \cot (c+d x) \csc ^7(c+d x)}{8 d}-\frac{2 a b \cot ^7(c+d x)}{7 d} \]

[Out]

(5*(a^2 + 8*b^2)*ArcTanh[Cos[c + d*x]])/(128*d) - (2*a*b*Cot[c + d*x]^7)/(7*d) + ((5*a^2 - 88*b^2)*Cot[c + d*x
]*Csc[c + d*x])/(128*d) - ((59*a^2 - 104*b^2)*Cot[c + d*x]*Csc[c + d*x]^3)/(192*d) + ((17*a^2 - 8*b^2)*Cot[c +
d*x]*Csc[c + d*x]^5)/(48*d) - (a^2*Cot[c + d*x]*Csc[c + d*x]^7)/(8*d)

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Rubi [A]  time = 0.319247, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2911, 2607, 30, 4366, 455, 1814, 1157, 385, 206} \[ \frac{5 \left (a^2+8 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{128 d}+\frac{\left (17 a^2-8 b^2\right ) \cot (c+d x) \csc ^5(c+d x)}{48 d}-\frac{\left (59 a^2-104 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{192 d}+\frac{\left (5 a^2-88 b^2\right ) \cot (c+d x) \csc (c+d x)}{128 d}-\frac{a^2 \cot (c+d x) \csc ^7(c+d x)}{8 d}-\frac{2 a b \cot ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

(5*(a^2 + 8*b^2)*ArcTanh[Cos[c + d*x]])/(128*d) - (2*a*b*Cot[c + d*x]^7)/(7*d) + ((5*a^2 - 88*b^2)*Cot[c + d*x
]*Csc[c + d*x])/(128*d) - ((59*a^2 - 104*b^2)*Cot[c + d*x]*Csc[c + d*x]^3)/(192*d) + ((17*a^2 - 8*b^2)*Cot[c +
d*x]*Csc[c + d*x]^5)/(48*d) - (a^2*Cot[c + d*x]*Csc[c + d*x]^7)/(8*d)

Rule 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4366

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dis
t[d/(b*c), Subst[Int[SubstFor[(1 - d^2*x^2)^((n - 1)/2), Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d]
, x] /; FunctionOfQ[Cos[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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])

Rubi steps

\begin{align*} \int \cot ^6(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \cot ^6(c+d x) \csc ^2(c+d x) \, dx+\int \cot ^6(c+d x) \csc ^3(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^6 \left (a^2+b^2-b^2 x^2\right )}{\left (1-x^2\right )^5} \, dx,x,\cos (c+d x)\right )}{d}+\frac{(2 a b) \operatorname{Subst}\left (\int x^6 \, dx,x,-\cot (c+d x)\right )}{d}\\ &=-\frac{2 a b \cot ^7(c+d x)}{7 d}-\frac{a^2 \cot (c+d x) \csc ^7(c+d x)}{8 d}+\frac{\operatorname{Subst}\left (\int \frac{a^2+8 a^2 x^2+8 a^2 x^4-8 b^2 x^6}{\left (1-x^2\right )^4} \, dx,x,\cos (c+d x)\right )}{8 d}\\ &=-\frac{2 a b \cot ^7(c+d x)}{7 d}+\frac{\left (17 a^2-8 b^2\right ) \cot (c+d x) \csc ^5(c+d x)}{48 d}-\frac{a^2 \cot (c+d x) \csc ^7(c+d x)}{8 d}-\frac{\operatorname{Subst}\left (\int \frac{11 a^2-8 b^2+48 \left (a^2-b^2\right ) x^2-48 b^2 x^4}{\left (1-x^2\right )^3} \, dx,x,\cos (c+d x)\right )}{48 d}\\ &=-\frac{2 a b \cot ^7(c+d x)}{7 d}-\frac{\left (59 a^2-104 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{192 d}+\frac{\left (17 a^2-8 b^2\right ) \cot (c+d x) \csc ^5(c+d x)}{48 d}-\frac{a^2 \cot (c+d x) \csc ^7(c+d x)}{8 d}+\frac{\operatorname{Subst}\left (\int \frac{3 \left (5 a^2-24 b^2\right )-192 b^2 x^2}{\left (1-x^2\right )^2} \, dx,x,\cos (c+d x)\right )}{192 d}\\ &=-\frac{2 a b \cot ^7(c+d x)}{7 d}+\frac{\left (5 a^2-88 b^2\right ) \cot (c+d x) \csc (c+d x)}{128 d}-\frac{\left (59 a^2-104 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{192 d}+\frac{\left (17 a^2-8 b^2\right ) \cot (c+d x) \csc ^5(c+d x)}{48 d}-\frac{a^2 \cot (c+d x) \csc ^7(c+d x)}{8 d}+\frac{\left (5 \left (a^2+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{128 d}\\ &=\frac{5 \left (a^2+8 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{128 d}-\frac{2 a b \cot ^7(c+d x)}{7 d}+\frac{\left (5 a^2-88 b^2\right ) \cot (c+d x) \csc (c+d x)}{128 d}-\frac{\left (59 a^2-104 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{192 d}+\frac{\left (17 a^2-8 b^2\right ) \cot (c+d x) \csc ^5(c+d x)}{48 d}-\frac{a^2 \cot (c+d x) \csc ^7(c+d x)}{8 d}\\ \end{align*}

Mathematica [A]  time = 0.797125, size = 282, normalized size = 1.77 \[ -\frac{7 \left (895 a^2-904 b^2\right ) \cos (3 (c+d x)) \csc ^8(c+d x)+7 \cot (c+d x) \csc ^7(c+d x) \left (1765 a^2+1536 a b \sin (c+d x)+680 b^2\right )+6720 a^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-6720 a^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+2779 a^2 \cos (5 (c+d x)) \csc ^8(c+d x)+105 a^2 \cos (7 (c+d x)) \csc ^8(c+d x)+5376 a b \sin (4 (c+d x)) \csc ^8(c+d x)+2304 a b \sin (6 (c+d x)) \csc ^8(c+d x)+384 a b \sin (8 (c+d x)) \csc ^8(c+d x)+53760 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-53760 b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+3416 b^2 \cos (5 (c+d x)) \csc ^8(c+d x)-1848 b^2 \cos (7 (c+d x)) \csc ^8(c+d x)}{172032 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

-(7*(895*a^2 - 904*b^2)*Cos[3*(c + d*x)]*Csc[c + d*x]^8 + 2779*a^2*Cos[5*(c + d*x)]*Csc[c + d*x]^8 + 3416*b^2*
Cos[5*(c + d*x)]*Csc[c + d*x]^8 + 105*a^2*Cos[7*(c + d*x)]*Csc[c + d*x]^8 - 1848*b^2*Cos[7*(c + d*x)]*Csc[c +
d*x]^8 - 6720*a^2*Log[Cos[(c + d*x)/2]] - 53760*b^2*Log[Cos[(c + d*x)/2]] + 6720*a^2*Log[Sin[(c + d*x)/2]] + 5
3760*b^2*Log[Sin[(c + d*x)/2]] + 7*Cot[c + d*x]*Csc[c + d*x]^7*(1765*a^2 + 680*b^2 + 1536*a*b*Sin[c + d*x]) +
5376*a*b*Csc[c + d*x]^8*Sin[4*(c + d*x)] + 2304*a*b*Csc[c + d*x]^8*Sin[6*(c + d*x)] + 384*a*b*Csc[c + d*x]^8*S
in[8*(c + d*x)])/(172032*d)

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Maple [B]  time = 0.099, size = 333, normalized size = 2.1 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{8}}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{48\,d \left ( \sin \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{192\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{128\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{128\,d}}-{\frac{5\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{384\,d}}-{\frac{5\,{a}^{2}\cos \left ( dx+c \right ) }{128\,d}}-{\frac{5\,{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{128\,d}}-{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{7\,d \left ( \sin \left ( dx+c \right ) \right ) ^{7}}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{6\,d \left ( \sin \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{24\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{16\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{16\,d}}-{\frac{5\,{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{48\,d}}-{\frac{5\,{b}^{2}\cos \left ( dx+c \right ) }{16\,d}}-{\frac{5\,{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{16\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^9*(a+b*sin(d*x+c))^2,x)

[Out]

-1/8/d*a^2/sin(d*x+c)^8*cos(d*x+c)^7-1/48/d*a^2/sin(d*x+c)^6*cos(d*x+c)^7+1/192/d*a^2/sin(d*x+c)^4*cos(d*x+c)^
7-1/128/d*a^2/sin(d*x+c)^2*cos(d*x+c)^7-1/128*a^2*cos(d*x+c)^5/d-5/384*a^2*cos(d*x+c)^3/d-5/128*a^2*cos(d*x+c)
/d-5/128/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-2/7/d*a*b/sin(d*x+c)^7*cos(d*x+c)^7-1/6/d*b^2/sin(d*x+c)^6*cos(d*x+c)
^7+1/24/d*b^2/sin(d*x+c)^4*cos(d*x+c)^7-1/16/d*b^2/sin(d*x+c)^2*cos(d*x+c)^7-1/16*b^2*cos(d*x+c)^5/d-5/48*b^2*
cos(d*x+c)^3/d-5/16*b^2*cos(d*x+c)/d-5/16/d*b^2*ln(csc(d*x+c)-cot(d*x+c))

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Maxima [A]  time = 1.00587, size = 297, normalized size = 1.87 \begin{align*} -\frac{7 \, a^{2}{\left (\frac{2 \,{\left (15 \, \cos \left (d x + c\right )^{7} + 73 \, \cos \left (d x + c\right )^{5} - 55 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{8} - 4 \, \cos \left (d x + c\right )^{6} + 6 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + 1} - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 56 \, b^{2}{\left (\frac{2 \,{\left (33 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac{1536 \, a b}{\tan \left (d x + c\right )^{7}}}{5376 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^9*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/5376*(7*a^2*(2*(15*cos(d*x + c)^7 + 73*cos(d*x + c)^5 - 55*cos(d*x + c)^3 + 15*cos(d*x + c))/(cos(d*x + c)^
8 - 4*cos(d*x + c)^6 + 6*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + 1) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x +
c) - 1)) - 56*b^2*(2*(33*cos(d*x + c)^5 - 40*cos(d*x + c)^3 + 15*cos(d*x + c))/(cos(d*x + c)^6 - 3*cos(d*x + c
)^4 + 3*cos(d*x + c)^2 - 1) + 15*log(cos(d*x + c) + 1) - 15*log(cos(d*x + c) - 1)) + 1536*a*b/tan(d*x + c)^7)/
d

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Fricas [B]  time = 1.80941, size = 853, normalized size = 5.36 \begin{align*} -\frac{1536 \, a b \cos \left (d x + c\right )^{7} \sin \left (d x + c\right ) + 42 \,{\left (5 \, a^{2} - 88 \, b^{2}\right )} \cos \left (d x + c\right )^{7} + 1022 \,{\left (a^{2} + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 770 \,{\left (a^{2} + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 210 \,{\left (a^{2} + 8 \, b^{2}\right )} \cos \left (d x + c\right ) - 105 \,{\left ({\left (a^{2} + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{8} - 4 \,{\left (a^{2} + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{6} + 6 \,{\left (a^{2} + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \,{\left (a^{2} + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 8 \, b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 105 \,{\left ({\left (a^{2} + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{8} - 4 \,{\left (a^{2} + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{6} + 6 \,{\left (a^{2} + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \,{\left (a^{2} + 8 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 8 \, b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{5376 \,{\left (d \cos \left (d x + c\right )^{8} - 4 \, d \cos \left (d x + c\right )^{6} + 6 \, d \cos \left (d x + c\right )^{4} - 4 \, 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)^6*csc(d*x+c)^9*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/5376*(1536*a*b*cos(d*x + c)^7*sin(d*x + c) + 42*(5*a^2 - 88*b^2)*cos(d*x + c)^7 + 1022*(a^2 + 8*b^2)*cos(d*
x + c)^5 - 770*(a^2 + 8*b^2)*cos(d*x + c)^3 + 210*(a^2 + 8*b^2)*cos(d*x + c) - 105*((a^2 + 8*b^2)*cos(d*x + c)
^8 - 4*(a^2 + 8*b^2)*cos(d*x + c)^6 + 6*(a^2 + 8*b^2)*cos(d*x + c)^4 - 4*(a^2 + 8*b^2)*cos(d*x + c)^2 + a^2 +
8*b^2)*log(1/2*cos(d*x + c) + 1/2) + 105*((a^2 + 8*b^2)*cos(d*x + c)^8 - 4*(a^2 + 8*b^2)*cos(d*x + c)^6 + 6*(a
^2 + 8*b^2)*cos(d*x + c)^4 - 4*(a^2 + 8*b^2)*cos(d*x + c)^2 + a^2 + 8*b^2)*log(-1/2*cos(d*x + c) + 1/2))/(d*co
s(d*x + c)^8 - 4*d*cos(d*x + c)^6 + 6*d*cos(d*x + c)^4 - 4*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)**6*csc(d*x+c)**9*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.31952, size = 543, normalized size = 3.42 \begin{align*} \frac{21 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 96 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 112 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 112 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 672 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 168 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 1008 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2016 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 336 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 5040 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3360 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1680 \,{\left (a^{2} + 8 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{4566 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 36528 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 3360 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 336 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 5040 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 2016 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 168 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 1008 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 672 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 112 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 112 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 96 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 21 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8}}}{43008 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^9*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/43008*(21*a^2*tan(1/2*d*x + 1/2*c)^8 + 96*a*b*tan(1/2*d*x + 1/2*c)^7 - 112*a^2*tan(1/2*d*x + 1/2*c)^6 + 112*
b^2*tan(1/2*d*x + 1/2*c)^6 - 672*a*b*tan(1/2*d*x + 1/2*c)^5 + 168*a^2*tan(1/2*d*x + 1/2*c)^4 - 1008*b^2*tan(1/
2*d*x + 1/2*c)^4 + 2016*a*b*tan(1/2*d*x + 1/2*c)^3 + 336*a^2*tan(1/2*d*x + 1/2*c)^2 + 5040*b^2*tan(1/2*d*x + 1
/2*c)^2 - 3360*a*b*tan(1/2*d*x + 1/2*c) - 1680*(a^2 + 8*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + (4566*a^2*tan(1/
2*d*x + 1/2*c)^8 + 36528*b^2*tan(1/2*d*x + 1/2*c)^8 + 3360*a*b*tan(1/2*d*x + 1/2*c)^7 - 336*a^2*tan(1/2*d*x +
1/2*c)^6 - 5040*b^2*tan(1/2*d*x + 1/2*c)^6 - 2016*a*b*tan(1/2*d*x + 1/2*c)^5 - 168*a^2*tan(1/2*d*x + 1/2*c)^4
+ 1008*b^2*tan(1/2*d*x + 1/2*c)^4 + 672*a*b*tan(1/2*d*x + 1/2*c)^3 + 112*a^2*tan(1/2*d*x + 1/2*c)^2 - 112*b^2*
tan(1/2*d*x + 1/2*c)^2 - 96*a*b*tan(1/2*d*x + 1/2*c) - 21*a^2)/tan(1/2*d*x + 1/2*c)^8)/d