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

Optimal. Leaf size=139 \[ \frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {15 a^2 \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {9 a^2 \cot (c+d x) \csc (c+d x)}{4 d}+\frac {3 a^2 x}{2} \]

[Out]

3/2*a^2*x-15/4*a^2*arctanh(cos(d*x+c))/d+2*a^2*cos(d*x+c)/d+a^2*cot(d*x+c)/d-1/5*a^2*cot(d*x+c)^5/d+9/4*a^2*co
t(d*x+c)*csc(d*x+c)/d-1/2*a^2*cot(d*x+c)*csc(d*x+c)^3/d+1/2*a^2*cos(d*x+c)*sin(d*x+c)/d

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Rubi [A]  time = 0.25, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2709, 3770, 3768, 3767, 2638, 2635, 8} \[ \frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {15 a^2 \tanh ^{-1}(\cos (c+d x))}{4 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {9 a^2 \cot (c+d x) \csc (c+d x)}{4 d}+\frac {3 a^2 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*x)/2 - (15*a^2*ArcTanh[Cos[c + d*x]])/(4*d) + (2*a^2*Cos[c + d*x])/d + (a^2*Cot[c + d*x])/d - (a^2*Cot[
c + d*x]^5)/(5*d) + (9*a^2*Cot[c + d*x]*Csc[c + d*x])/(4*d) - (a^2*Cot[c + d*x]*Csc[c + d*x]^3)/(2*d) + (a^2*C
os[c + d*x]*Sin[c + d*x])/(2*d)

Rule 8

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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) (a+a \sin (c+d x))^2 \, dx &=\frac {\int \left (2 a^8+6 a^8 \csc (c+d x)-6 a^8 \csc ^3(c+d x)-2 a^8 \csc ^4(c+d x)+2 a^8 \csc ^5(c+d x)+a^8 \csc ^6(c+d x)-2 a^8 \sin (c+d x)-a^8 \sin ^2(c+d x)\right ) \, dx}{a^6}\\ &=2 a^2 x+a^2 \int \csc ^6(c+d x) \, dx-a^2 \int \sin ^2(c+d x) \, dx-\left (2 a^2\right ) \int \csc ^4(c+d x) \, dx+\left (2 a^2\right ) \int \csc ^5(c+d x) \, dx-\left (2 a^2\right ) \int \sin (c+d x) \, dx+\left (6 a^2\right ) \int \csc (c+d x) \, dx-\left (6 a^2\right ) \int \csc ^3(c+d x) \, dx\\ &=2 a^2 x-\frac {6 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} a^2 \int 1 \, dx+\frac {1}{2} \left (3 a^2\right ) \int \csc ^3(c+d x) \, dx-\left (3 a^2\right ) \int \csc (c+d x) \, dx-\frac {a^2 \operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (c+d x)\right )}{d}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {3 a^2 x}{2}-\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {9 a^2 \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {1}{4} \left (3 a^2\right ) \int \csc (c+d x) \, dx\\ &=\frac {3 a^2 x}{2}-\frac {15 a^2 \tanh ^{-1}(\cos (c+d x))}{4 d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^5(c+d x)}{5 d}+\frac {9 a^2 \cot (c+d x) \csc (c+d x)}{4 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{2 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 1.41, size = 264, normalized size = 1.90 \[ \frac {a^2 (\sin (c+d x)+1)^2 \left (240 (c+d x)+40 \sin (2 (c+d x))+320 \cos (c+d x)-64 \tan \left (\frac {1}{2} (c+d x)\right )+64 \cot \left (\frac {1}{2} (c+d x)\right )-5 \csc ^4\left (\frac {1}{2} (c+d x)\right )+90 \csc ^2\left (\frac {1}{2} (c+d x)\right )+5 \sec ^4\left (\frac {1}{2} (c+d x)\right )-90 \sec ^2\left (\frac {1}{2} (c+d x)\right )+600 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-600 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\frac {1}{2} \sin (c+d x) \csc ^6\left (\frac {1}{2} (c+d x)\right )+\frac {7}{2} \sin (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )-56 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)+\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right )\right )}{160 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Sin[c + d*x])^2*(240*(c + d*x) + 320*Cos[c + d*x] + 64*Cot[(c + d*x)/2] + 90*Csc[(c + d*x)/2]^2 - 5*
Csc[(c + d*x)/2]^4 - 600*Log[Cos[(c + d*x)/2]] + 600*Log[Sin[(c + d*x)/2]] - 90*Sec[(c + d*x)/2]^2 + 5*Sec[(c
+ d*x)/2]^4 - 56*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + (7*Csc[(c + d*x)/2]^4*Sin[c + d*x])/2 - (Csc[(c + d*x)/2]
^6*Sin[c + d*x])/2 + 40*Sin[2*(c + d*x)] - 64*Tan[(c + d*x)/2] + Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2]))/(160*d*
(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

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fricas [B]  time = 0.81, size = 265, normalized size = 1.91 \[ -\frac {20 \, a^{2} \cos \left (d x + c\right )^{7} - 92 \, a^{2} \cos \left (d x + c\right )^{5} + 140 \, a^{2} \cos \left (d x + c\right )^{3} - 60 \, a^{2} \cos \left (d x + c\right ) + 75 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 75 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 10 \, {\left (6 \, a^{2} d x \cos \left (d x + c\right )^{4} + 8 \, a^{2} \cos \left (d x + c\right )^{5} - 12 \, a^{2} d x \cos \left (d x + c\right )^{2} - 25 \, a^{2} \cos \left (d x + c\right )^{3} + 6 \, a^{2} d x + 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40*(20*a^2*cos(d*x + c)^7 - 92*a^2*cos(d*x + c)^5 + 140*a^2*cos(d*x + c)^3 - 60*a^2*cos(d*x + c) + 75*(a^2*
cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 75*(a^2*cos(d*x + c)^4
- 2*a^2*cos(d*x + c)^2 + a^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 10*(6*a^2*d*x*cos(d*x + c)^4 + 8*a^
2*cos(d*x + c)^5 - 12*a^2*d*x*cos(d*x + c)^2 - 25*a^2*cos(d*x + c)^3 + 6*a^2*d*x + 15*a^2*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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giac [B]  time = 0.29, size = 272, normalized size = 1.96 \[ \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 240 \, {\left (d x + c\right )} a^{2} + 600 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {160 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac {1370 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{160 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/160*(a^2*tan(1/2*d*x + 1/2*c)^5 + 5*a^2*tan(1/2*d*x + 1/2*c)^4 - 5*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*a^2*tan(1
/2*d*x + 1/2*c)^2 + 240*(d*x + c)*a^2 + 600*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 70*a^2*tan(1/2*d*x + 1/2*c) -
160*(a^2*tan(1/2*d*x + 1/2*c)^3 - 4*a^2*tan(1/2*d*x + 1/2*c)^2 - a^2*tan(1/2*d*x + 1/2*c) - 4*a^2)/(tan(1/2*d
*x + 1/2*c)^2 + 1)^2 - (1370*a^2*tan(1/2*d*x + 1/2*c)^5 - 70*a^2*tan(1/2*d*x + 1/2*c)^4 - 80*a^2*tan(1/2*d*x +
1/2*c)^3 - 5*a^2*tan(1/2*d*x + 1/2*c)^2 + 5*a^2*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c)^5)/d

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maple [B]  time = 0.45, size = 293, normalized size = 2.11 \[ -\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}+\frac {4 a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )}+\frac {4 a^{2} \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {3 a^{2} x}{2}+\frac {3 a^{2} c}{2 d}-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{4}}+\frac {3 a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{2}}+\frac {3 a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{4 d}+\frac {5 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {15 a^{2} \cos \left (d x +c \right )}{4 d}+\frac {15 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{4 d}-\frac {a^{2} \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a^{2} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a^{2} \cot \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 184, normalized size = 1.32 \[ \frac {20 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a^{2} - 8 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a^{2} - 15 \, a^{2} {\left (\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(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))*a^2
- 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 - 15*a^2*(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) - 1
5*log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 8.90, size = 363, normalized size = 2.61 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}+\frac {15\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}+\frac {-18\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+144\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+61\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+159\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {79\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+14\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}-a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {a^2}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {3\,a^2\,\mathrm {atan}\left (\frac {9\,a^4}{\frac {45\,a^4}{2}-9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {45\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {45\,a^4}{2}-9\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )}{d}-\frac {7\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,d} \]

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)^6,x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^4)/(32*d) - (a^2*tan(c/2 + (d*x)/2)^3)/(32*d) - (a^2*tan(c/2 + (d*x)/2)^2)/(2*d) + (a^
2*tan(c/2 + (d*x)/2)^5)/(160*d) + (15*a^2*log(tan(c/2 + (d*x)/2)))/(4*d) + ((3*a^2*tan(c/2 + (d*x)/2)^2)/5 + 1
4*a^2*tan(c/2 + (d*x)/2)^3 + (79*a^2*tan(c/2 + (d*x)/2)^4)/5 + 159*a^2*tan(c/2 + (d*x)/2)^5 + 61*a^2*tan(c/2 +
(d*x)/2)^6 + 144*a^2*tan(c/2 + (d*x)/2)^7 - 18*a^2*tan(c/2 + (d*x)/2)^8 - a^2/5 - a^2*tan(c/2 + (d*x)/2))/(d*
(32*tan(c/2 + (d*x)/2)^5 + 64*tan(c/2 + (d*x)/2)^7 + 32*tan(c/2 + (d*x)/2)^9)) + (3*a^2*atan((9*a^4)/((45*a^4)
/2 - 9*a^4*tan(c/2 + (d*x)/2)) + (45*a^4*tan(c/2 + (d*x)/2))/(2*((45*a^4)/2 - 9*a^4*tan(c/2 + (d*x)/2)))))/d -
(7*a^2*tan(c/2 + (d*x)/2))/(16*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)**6*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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