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

Optimal. Leaf size=162 \[ -\frac {a^2 \cot ^7(c+d x)}{7 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x)}{d}+\frac {5 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a^2 \cot ^5(c+d x) \csc (c+d x)}{3 d}+\frac {5 a^2 \cot ^3(c+d x) \csc (c+d x)}{12 d}-\frac {5 a^2 \cot (c+d x) \csc (c+d x)}{8 d}-a^2 x \]

[Out]

-a^2*x+5/8*a^2*arctanh(cos(d*x+c))/d-a^2*cot(d*x+c)/d+1/3*a^2*cot(d*x+c)^3/d-1/5*a^2*cot(d*x+c)^5/d-1/7*a^2*co
t(d*x+c)^7/d-5/8*a^2*cot(d*x+c)*csc(d*x+c)/d+5/12*a^2*cot(d*x+c)^3*csc(d*x+c)/d-1/3*a^2*cot(d*x+c)^5*csc(d*x+c
)/d

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Rubi [A]  time = 0.23, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2873, 3473, 8, 2611, 3770, 2607, 30} \[ -\frac {a^2 \cot ^7(c+d x)}{7 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x)}{d}+\frac {5 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a^2 \cot ^5(c+d x) \csc (c+d x)}{3 d}+\frac {5 a^2 \cot ^3(c+d x) \csc (c+d x)}{12 d}-\frac {5 a^2 \cot (c+d x) \csc (c+d x)}{8 d}-a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*x) + (5*a^2*ArcTanh[Cos[c + d*x]])/(8*d) - (a^2*Cot[c + d*x])/d + (a^2*Cot[c + d*x]^3)/(3*d) - (a^2*Cot[
c + d*x]^5)/(5*d) - (a^2*Cot[c + d*x]^7)/(7*d) - (5*a^2*Cot[c + d*x]*Csc[c + d*x])/(8*d) + (5*a^2*Cot[c + d*x]
^3*Csc[c + d*x])/(12*d) - (a^2*Cot[c + d*x]^5*Csc[c + d*x])/(3*d)

Rule 8

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

Rule 30

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

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 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 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 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 ^6(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (a^2 \cot ^6(c+d x)+2 a^2 \cot ^6(c+d x) \csc (c+d x)+a^2 \cot ^6(c+d x) \csc ^2(c+d x)\right ) \, dx\\ &=a^2 \int \cot ^6(c+d x) \, dx+a^2 \int \cot ^6(c+d x) \csc ^2(c+d x) \, dx+\left (2 a^2\right ) \int \cot ^6(c+d x) \csc (c+d x) \, dx\\ &=-\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^5(c+d x) \csc (c+d x)}{3 d}-a^2 \int \cot ^4(c+d x) \, dx-\frac {1}{3} \left (5 a^2\right ) \int \cot ^4(c+d x) \csc (c+d x) \, dx+\frac {a^2 \operatorname {Subst}\left (\int x^6 \, dx,x,-\cot (c+d x)\right )}{d}\\ &=\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^7(c+d x)}{7 d}+\frac {5 a^2 \cot ^3(c+d x) \csc (c+d x)}{12 d}-\frac {a^2 \cot ^5(c+d x) \csc (c+d x)}{3 d}+a^2 \int \cot ^2(c+d x) \, dx+\frac {1}{4} \left (5 a^2\right ) \int \cot ^2(c+d x) \csc (c+d x) \, dx\\ &=-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^7(c+d x)}{7 d}-\frac {5 a^2 \cot (c+d x) \csc (c+d x)}{8 d}+\frac {5 a^2 \cot ^3(c+d x) \csc (c+d x)}{12 d}-\frac {a^2 \cot ^5(c+d x) \csc (c+d x)}{3 d}-\frac {1}{8} \left (5 a^2\right ) \int \csc (c+d x) \, dx-a^2 \int 1 \, dx\\ &=-a^2 x+\frac {5 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^7(c+d x)}{7 d}-\frac {5 a^2 \cot (c+d x) \csc (c+d x)}{8 d}+\frac {5 a^2 \cot ^3(c+d x) \csc (c+d x)}{12 d}-\frac {a^2 \cot ^5(c+d x) \csc (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 1.09, size = 262, normalized size = 1.62 \[ \frac {a^2 \left (9344 \tan \left (\frac {1}{2} (c+d x)\right )-9344 \cot \left (\frac {1}{2} (c+d x)\right )-4620 \csc ^2\left (\frac {1}{2} (c+d x)\right )+70 \sec ^6\left (\frac {1}{2} (c+d x)\right )-840 \sec ^4\left (\frac {1}{2} (c+d x)\right )+4620 \sec ^2\left (\frac {1}{2} (c+d x)\right )-8400 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+8400 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\frac {15}{2} \sin (c+d x) \csc ^8\left (\frac {1}{2} (c+d x)\right )+(33 \sin (c+d x)-70) \csc ^6\left (\frac {1}{2} (c+d x)\right )+(289 \sin (c+d x)+840) \csc ^4\left (\frac {1}{2} (c+d x)\right )-4624 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)+15 \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^6\left (\frac {1}{2} (c+d x)\right )-66 \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right )-13440 c-13440 d x\right )}{13440 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-13440*c - 13440*d*x - 9344*Cot[(c + d*x)/2] - 4620*Csc[(c + d*x)/2]^2 + 8400*Log[Cos[(c + d*x)/2]] - 84
00*Log[Sin[(c + d*x)/2]] + 4620*Sec[(c + d*x)/2]^2 - 840*Sec[(c + d*x)/2]^4 + 70*Sec[(c + d*x)/2]^6 - 4624*Csc
[c + d*x]^3*Sin[(c + d*x)/2]^4 - (15*Csc[(c + d*x)/2]^8*Sin[c + d*x])/2 + Csc[(c + d*x)/2]^6*(-70 + 33*Sin[c +
d*x]) + Csc[(c + d*x)/2]^4*(840 + 289*Sin[c + d*x]) + 9344*Tan[(c + d*x)/2] - 66*Sec[(c + d*x)/2]^4*Tan[(c +
d*x)/2] + 15*Sec[(c + d*x)/2]^6*Tan[(c + d*x)/2]))/(13440*d)

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fricas [B]  time = 0.84, size = 323, normalized size = 1.99 \[ -\frac {2336 \, a^{2} \cos \left (d x + c\right )^{7} - 6496 \, a^{2} \cos \left (d x + c\right )^{5} + 5600 \, a^{2} \cos \left (d x + c\right )^{3} - 1680 \, a^{2} \cos \left (d x + c\right ) - 525 \, {\left (a^{2} \cos \left (d x + c\right )^{6} - 3 \, a^{2} \cos \left (d x + c\right )^{4} + 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 525 \, {\left (a^{2} \cos \left (d x + c\right )^{6} - 3 \, a^{2} \cos \left (d x + c\right )^{4} + 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 70 \, {\left (24 \, a^{2} d x \cos \left (d x + c\right )^{6} - 72 \, a^{2} d x \cos \left (d x + c\right )^{4} - 33 \, a^{2} \cos \left (d x + c\right )^{5} + 72 \, a^{2} d x \cos \left (d x + c\right )^{2} + 40 \, a^{2} \cos \left (d x + c\right )^{3} - 24 \, a^{2} d x - 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{1680 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^8*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/1680*(2336*a^2*cos(d*x + c)^7 - 6496*a^2*cos(d*x + c)^5 + 5600*a^2*cos(d*x + c)^3 - 1680*a^2*cos(d*x + c) -
525*(a^2*cos(d*x + c)^6 - 3*a^2*cos(d*x + c)^4 + 3*a^2*cos(d*x + c)^2 - a^2)*log(1/2*cos(d*x + c) + 1/2)*sin(
d*x + c) + 525*(a^2*cos(d*x + c)^6 - 3*a^2*cos(d*x + c)^4 + 3*a^2*cos(d*x + c)^2 - a^2)*log(-1/2*cos(d*x + c)
+ 1/2)*sin(d*x + c) + 70*(24*a^2*d*x*cos(d*x + c)^6 - 72*a^2*d*x*cos(d*x + c)^4 - 33*a^2*cos(d*x + c)^5 + 72*a
^2*d*x*cos(d*x + c)^2 + 40*a^2*cos(d*x + c)^3 - 24*a^2*d*x - 15*a^2*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x +
c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)*sin(d*x + c))

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giac [A]  time = 0.31, size = 270, normalized size = 1.67 \[ \frac {15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 21 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 630 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 665 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 13440 \, {\left (d x + c\right )} a^{2} - 8400 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 8715 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {21780 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8715 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 3150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 665 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 630 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}}}{13440 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^8*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/13440*(15*a^2*tan(1/2*d*x + 1/2*c)^7 + 70*a^2*tan(1/2*d*x + 1/2*c)^6 - 21*a^2*tan(1/2*d*x + 1/2*c)^5 - 630*a
^2*tan(1/2*d*x + 1/2*c)^4 - 665*a^2*tan(1/2*d*x + 1/2*c)^3 + 3150*a^2*tan(1/2*d*x + 1/2*c)^2 - 13440*(d*x + c)
*a^2 - 8400*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 8715*a^2*tan(1/2*d*x + 1/2*c) + (21780*a^2*tan(1/2*d*x + 1/2*
c)^7 - 8715*a^2*tan(1/2*d*x + 1/2*c)^6 - 3150*a^2*tan(1/2*d*x + 1/2*c)^5 + 665*a^2*tan(1/2*d*x + 1/2*c)^4 + 63
0*a^2*tan(1/2*d*x + 1/2*c)^3 + 21*a^2*tan(1/2*d*x + 1/2*c)^2 - 70*a^2*tan(1/2*d*x + 1/2*c) - 15*a^2)/tan(1/2*d
*x + 1/2*c)^7)/d

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maple [A]  time = 0.38, size = 229, normalized size = 1.41 \[ -\frac {a^{2} \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a^{2} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a^{2} \cot \left (d x +c \right )}{d}-a^{2} x -\frac {a^{2} c}{d}-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{6}}+\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{12 d \sin \left (d x +c \right )^{4}}-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{8 d}-\frac {5 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{24 d}-\frac {5 a^{2} \cos \left (d x +c \right )}{8 d}-\frac {5 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{7 d \sin \left (d x +c \right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^8*(a+a*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-1/3/d*a^2/sin(d*x+c)^6*cos(d*x
+c)^7+1/12/d*a^2/sin(d*x+c)^4*cos(d*x+c)^7-1/8/d*a^2/sin(d*x+c)^2*cos(d*x+c)^7-1/8*a^2*cos(d*x+c)^5/d-5/24*a^2
*cos(d*x+c)^3/d-5/8*a^2*cos(d*x+c)/d-5/8/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-1/7/d*a^2/sin(d*x+c)^7*cos(d*x+c)^7

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maxima [A]  time = 0.43, size = 154, normalized size = 0.95 \[ -\frac {112 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a^{2} - 35 \, a^{2} {\left (\frac {2 \, {\left (33 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {240 \, a^{2}}{\tan \left (d x + c\right )^{7}}}{1680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^8*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/1680*(112*(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 - 35*a^2*(2*(33*c
os(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)) + 240*a^2/tan(d*x + c)^7)/d

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mupad [B]  time = 9.84, size = 351, normalized size = 2.17 \[ \frac {19\,a^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{384\,d}-\frac {15\,a^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{64\,d}+\frac {3\,a^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {a^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{640\,d}-\frac {a^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{192\,d}-\frac {a^2\,{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{896\,d}+\frac {15\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{64\,d}-\frac {19\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{384\,d}-\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{640\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{192\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{896\,d}-\frac {2\,a^2\,\mathrm {atan}\left (\frac {8\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {5\,a^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{8\,d}-\frac {83\,a^2\,\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128\,d}+\frac {83\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^6*(a + a*sin(c + d*x))^2)/sin(c + d*x)^8,x)

[Out]

(19*a^2*cot(c/2 + (d*x)/2)^3)/(384*d) - (15*a^2*cot(c/2 + (d*x)/2)^2)/(64*d) + (3*a^2*cot(c/2 + (d*x)/2)^4)/(6
4*d) + (a^2*cot(c/2 + (d*x)/2)^5)/(640*d) - (a^2*cot(c/2 + (d*x)/2)^6)/(192*d) - (a^2*cot(c/2 + (d*x)/2)^7)/(8
96*d) + (15*a^2*tan(c/2 + (d*x)/2)^2)/(64*d) - (19*a^2*tan(c/2 + (d*x)/2)^3)/(384*d) - (3*a^2*tan(c/2 + (d*x)/
2)^4)/(64*d) - (a^2*tan(c/2 + (d*x)/2)^5)/(640*d) + (a^2*tan(c/2 + (d*x)/2)^6)/(192*d) + (a^2*tan(c/2 + (d*x)/
2)^7)/(896*d) - (2*a^2*atan((8*cos(c/2 + (d*x)/2) + 5*sin(c/2 + (d*x)/2))/(5*cos(c/2 + (d*x)/2) - 8*sin(c/2 +
(d*x)/2))))/d - (5*a^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(8*d) - (83*a^2*cot(c/2 + (d*x)/2))/(128*d)
+ (83*a^2*tan(c/2 + (d*x)/2))/(128*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)**6*csc(d*x+c)**8*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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