3.596 \(\int \cos (c+d x) \cot ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=153 \[ -\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \cos (c+d x)}{d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}+\frac {4 a^2 \cot (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{d}+\frac {5 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {5 a^2 \cot (c+d x) \csc (c+d x)}{8 d}+5 a^2 x \]

[Out]

5*a^2*x+5/8*a^2*arctanh(cos(d*x+c))/d-a^2*cos(d*x+c)/d-1/3*a^2*cos(d*x+c)^3/d+4*a^2*cot(d*x+c)/d-2/3*a^2*cot(d
*x+c)^3/d+5/8*a^2*cot(d*x+c)*csc(d*x+c)/d-1/4*a^2*cot(d*x+c)*csc(d*x+c)^3/d+a^2*cos(d*x+c)*sin(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2872, 3767, 8, 3768, 3770, 2638, 2635, 2633} \[ -\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \cos (c+d x)}{d}-\frac {2 a^2 \cot ^3(c+d x)}{3 d}+\frac {4 a^2 \cot (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{d}+\frac {5 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac {5 a^2 \cot (c+d x) \csc (c+d x)}{8 d}+5 a^2 x \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Cot[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

5*a^2*x + (5*a^2*ArcTanh[Cos[c + d*x]])/(8*d) - (a^2*Cos[c + d*x])/d - (a^2*Cos[c + d*x]^3)/(3*d) + (4*a^2*Cot
[c + d*x])/d - (2*a^2*Cot[c + d*x]^3)/(3*d) + (5*a^2*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (a^2*Cot[c + d*x]*Csc[
c + d*x]^3)/(4*d) + (a^2*Cos[c + d*x]*Sin[c + d*x])/d

Rule 8

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

Rule 2633

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

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 2872

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

________________________________________________________________________________________

Mathematica [A]  time = 1.29, size = 227, normalized size = 1.48 \[ \frac {a^2 (\sin (c+d x)+1)^2 \left (960 (c+d x)+96 \sin (2 (c+d x))-240 \cos (c+d x)-16 \cos (3 (c+d x))-448 \tan \left (\frac {1}{2} (c+d x)\right )+448 \cot \left (\frac {1}{2} (c+d x)\right )-3 \csc ^4\left (\frac {1}{2} (c+d x)\right )+30 \csc ^2\left (\frac {1}{2} (c+d x)\right )+3 \sec ^4\left (\frac {1}{2} (c+d x)\right )-30 \sec ^2\left (\frac {1}{2} (c+d x)\right )-120 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+120 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8 \sin (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )+128 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)\right )}{192 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*Cot[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(1 + Sin[c + d*x])^2*(960*(c + d*x) - 240*Cos[c + d*x] - 16*Cos[3*(c + d*x)] + 448*Cot[(c + d*x)/2] + 30*
Csc[(c + d*x)/2]^2 - 3*Csc[(c + d*x)/2]^4 + 120*Log[Cos[(c + d*x)/2]] - 120*Log[Sin[(c + d*x)/2]] - 30*Sec[(c
+ d*x)/2]^2 + 3*Sec[(c + d*x)/2]^4 + 128*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 8*Csc[(c + d*x)/2]^4*Sin[c + d*x]
+ 96*Sin[2*(c + d*x)] - 448*Tan[(c + d*x)/2]))/(192*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

________________________________________________________________________________________

fricas [A]  time = 0.81, size = 245, normalized size = 1.60 \[ -\frac {16 \, a^{2} \cos \left (d x + c\right )^{7} - 240 \, a^{2} d x \cos \left (d x + c\right )^{4} + 16 \, a^{2} \cos \left (d x + c\right )^{5} + 480 \, a^{2} d x \cos \left (d x + c\right )^{2} - 50 \, a^{2} \cos \left (d x + c\right )^{3} - 240 \, a^{2} d x + 30 \, a^{2} \cos \left (d x + c\right ) - 15 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 16 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{5} - 20 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/48*(16*a^2*cos(d*x + c)^7 - 240*a^2*d*x*cos(d*x + c)^4 + 16*a^2*cos(d*x + c)^5 + 480*a^2*d*x*cos(d*x + c)^2
- 50*a^2*cos(d*x + c)^3 - 240*a^2*d*x + 30*a^2*cos(d*x + c) - 15*(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) + 15*(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) - 16*(3*a^2*cos(d*x + c)^5 - 20*a^2*cos(d*x + c)^3 + 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)

________________________________________________________________________________________

giac [A]  time = 0.30, size = 259, normalized size = 1.69 \[ \frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 16 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 960 \, {\left (d x + c\right )} a^{2} - 120 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 432 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {128 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}} + \frac {250 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 432 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 + 16*a^2*tan(1/2*d*x + 1/2*c)^3 - 24*a^2*tan(1/2*d*x + 1/2*c)^2 + 960*(d*x
+ c)*a^2 - 120*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 432*a^2*tan(1/2*d*x + 1/2*c) - 128*(3*a^2*tan(1/2*d*x + 1
/2*c)^5 + 6*a^2*tan(1/2*d*x + 1/2*c)^4 + 6*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a^2*tan(1/2*d*x + 1/2*c) + 4*a^2)/(t
an(1/2*d*x + 1/2*c)^2 + 1)^3 + (250*a^2*tan(1/2*d*x + 1/2*c)^4 + 432*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*a^2*tan(1
/2*d*x + 1/2*c)^2 - 16*a^2*tan(1/2*d*x + 1/2*c) - 3*a^2)/tan(1/2*d*x + 1/2*c)^4)/d

________________________________________________________________________________________

maple [A]  time = 0.44, size = 247, normalized size = 1.61 \[ -\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{8 d}-\frac {5 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{24 d}-\frac {5 a^{2} \cos \left (d x +c \right )}{8 d}-\frac {5 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {2 a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}+\frac {8 a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )}+\frac {8 a^{2} \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {10 a^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+5 a^{2} x +\frac {5 a^{2} c}{d}-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x)

[Out]

-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))-2/3/d*a^2/sin(d*x+c)^3*cos(d*x+c)^7+8/3/d*a^2/sin(d*x+c)*cos(d*x+c)^7+8/3*a^2*c
os(d*x+c)^5*sin(d*x+c)/d+10/3*a^2*cos(d*x+c)^3*sin(d*x+c)/d+5*a^2*cos(d*x+c)*sin(d*x+c)/d+5*a^2*x+5/d*a^2*c-1/
4/d*a^2/sin(d*x+c)^4*cos(d*x+c)^7

________________________________________________________________________________________

maxima [A]  time = 0.44, size = 206, normalized size = 1.35 \[ -\frac {4 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 16 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a^{2} + 3 \, a^{2} {\left (\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/48*(4*(4*cos(d*x + c)^3 - 6*cos(d*x + c)/(cos(d*x + c)^2 - 1) + 24*cos(d*x + c) - 15*log(cos(d*x + c) + 1)
+ 15*log(cos(d*x + c) - 1))*a^2 - 16*(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 + 3*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) - 15*log(cos(d*x + c) - 1)))/d

________________________________________________________________________________________

mupad [B]  time = 8.94, size = 373, normalized size = 2.44 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-62\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {320\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {233\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{4}+136\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {449\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{12}+32\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-\frac {4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}-\frac {a^2}{4}}{d\,\left (16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {5\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}-\frac {10\,a^2\,\mathrm {atan}\left (\frac {100\,a^4}{\frac {25\,a^4}{2}+100\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {25\,a^4}{2}+100\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )}{d}-\frac {9\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,d} \]

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)^5,x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^3)/(12*d) - (a^2*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^2*tan(c/2 + (d*x)/2)^4)/(64*d) + ((5
*a^2*tan(c/2 + (d*x)/2)^2)/4 + 32*a^2*tan(c/2 + (d*x)/2)^3 - (449*a^2*tan(c/2 + (d*x)/2)^4)/12 + 136*a^2*tan(c
/2 + (d*x)/2)^5 - (233*a^2*tan(c/2 + (d*x)/2)^6)/4 + (320*a^2*tan(c/2 + (d*x)/2)^7)/3 - 62*a^2*tan(c/2 + (d*x)
/2)^8 + 4*a^2*tan(c/2 + (d*x)/2)^9 - a^2/4 - (4*a^2*tan(c/2 + (d*x)/2))/3)/(d*(16*tan(c/2 + (d*x)/2)^4 + 48*ta
n(c/2 + (d*x)/2)^6 + 48*tan(c/2 + (d*x)/2)^8 + 16*tan(c/2 + (d*x)/2)^10)) - (5*a^2*log(tan(c/2 + (d*x)/2)))/(8
*d) - (10*a^2*atan((100*a^4)/((25*a^4)/2 + 100*a^4*tan(c/2 + (d*x)/2)) - (25*a^4*tan(c/2 + (d*x)/2))/(2*((25*a
^4)/2 + 100*a^4*tan(c/2 + (d*x)/2)))))/d - (9*a^2*tan(c/2 + (d*x)/2))/(4*d)

________________________________________________________________________________________

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)**5*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________