3.610 \(\int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=173 \[ -\frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {3 a^3 \cos ^5(c+d x)}{5 d}+\frac {a^3 \cos ^3(c+d x)}{d}+\frac {3 a^3 \cos (c+d x)}{d}-\frac {a^3 \cot (c+d x)}{d}+\frac {a^3 \sin ^5(c+d x) \cos (c+d x)}{2 d}-\frac {11 a^3 \sin ^3(c+d x) \cos (c+d x)}{8 d}+\frac {15 a^3 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {15 a^3 x}{16} \]

[Out]

-15/16*a^3*x-3*a^3*arctanh(cos(d*x+c))/d+3*a^3*cos(d*x+c)/d+a^3*cos(d*x+c)^3/d+3/5*a^3*cos(d*x+c)^5/d-1/7*a^3*
cos(d*x+c)^7/d-a^3*cot(d*x+c)/d+15/16*a^3*cos(d*x+c)*sin(d*x+c)/d-11/8*a^3*cos(d*x+c)*sin(d*x+c)^3/d+1/2*a^3*c
os(d*x+c)*sin(d*x+c)^5/d

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Rubi [A]  time = 0.24, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2872, 3770, 3767, 8, 2638, 2635, 2633} \[ -\frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {3 a^3 \cos ^5(c+d x)}{5 d}+\frac {a^3 \cos ^3(c+d x)}{d}+\frac {3 a^3 \cos (c+d x)}{d}-\frac {a^3 \cot (c+d x)}{d}+\frac {a^3 \sin ^5(c+d x) \cos (c+d x)}{2 d}-\frac {11 a^3 \sin ^3(c+d x) \cos (c+d x)}{8 d}+\frac {15 a^3 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {15 a^3 x}{16} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

(-15*a^3*x)/16 - (3*a^3*ArcTanh[Cos[c + d*x]])/d + (3*a^3*Cos[c + d*x])/d + (a^3*Cos[c + d*x]^3)/d + (3*a^3*Co
s[c + d*x]^5)/(5*d) - (a^3*Cos[c + d*x]^7)/(7*d) - (a^3*Cot[c + d*x])/d + (15*a^3*Cos[c + d*x]*Sin[c + d*x])/(
16*d) - (11*a^3*Cos[c + d*x]*Sin[c + d*x]^3)/(8*d) + (a^3*Cos[c + d*x]*Sin[c + d*x]^5)/(2*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 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 ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac {\int \left (3 a^9 \csc (c+d x)+a^9 \csc ^2(c+d x)-8 a^9 \sin (c+d x)-6 a^9 \sin ^2(c+d x)+6 a^9 \sin ^3(c+d x)+8 a^9 \sin ^4(c+d x)-3 a^9 \sin ^6(c+d x)-a^9 \sin ^7(c+d x)\right ) \, dx}{a^6}\\ &=a^3 \int \csc ^2(c+d x) \, dx-a^3 \int \sin ^7(c+d x) \, dx+\left (3 a^3\right ) \int \csc (c+d x) \, dx-\left (3 a^3\right ) \int \sin ^6(c+d x) \, dx-\left (6 a^3\right ) \int \sin ^2(c+d x) \, dx+\left (6 a^3\right ) \int \sin ^3(c+d x) \, dx-\left (8 a^3\right ) \int \sin (c+d x) \, dx+\left (8 a^3\right ) \int \sin ^4(c+d x) \, dx\\ &=-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {8 a^3 \cos (c+d x)}{d}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{d}-\frac {2 a^3 \cos (c+d x) \sin ^3(c+d x)}{d}+\frac {a^3 \cos (c+d x) \sin ^5(c+d x)}{2 d}-\frac {1}{2} \left (5 a^3\right ) \int \sin ^4(c+d x) \, dx-\left (3 a^3\right ) \int 1 \, dx+\left (6 a^3\right ) \int \sin ^2(c+d x) \, dx-\frac {a^3 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}+\frac {a^3 \operatorname {Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (6 a^3\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-3 a^3 x-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {3 a^3 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{d}+\frac {3 a^3 \cos ^5(c+d x)}{5 d}-\frac {a^3 \cos ^7(c+d x)}{7 d}-\frac {a^3 \cot (c+d x)}{d}-\frac {11 a^3 \cos (c+d x) \sin ^3(c+d x)}{8 d}+\frac {a^3 \cos (c+d x) \sin ^5(c+d x)}{2 d}-\frac {1}{8} \left (15 a^3\right ) \int \sin ^2(c+d x) \, dx+\left (3 a^3\right ) \int 1 \, dx\\ &=-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {3 a^3 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{d}+\frac {3 a^3 \cos ^5(c+d x)}{5 d}-\frac {a^3 \cos ^7(c+d x)}{7 d}-\frac {a^3 \cot (c+d x)}{d}+\frac {15 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {11 a^3 \cos (c+d x) \sin ^3(c+d x)}{8 d}+\frac {a^3 \cos (c+d x) \sin ^5(c+d x)}{2 d}-\frac {1}{16} \left (15 a^3\right ) \int 1 \, dx\\ &=-\frac {15 a^3 x}{16}-\frac {3 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {3 a^3 \cos (c+d x)}{d}+\frac {a^3 \cos ^3(c+d x)}{d}+\frac {3 a^3 \cos ^5(c+d x)}{5 d}-\frac {a^3 \cos ^7(c+d x)}{7 d}-\frac {a^3 \cot (c+d x)}{d}+\frac {15 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {11 a^3 \cos (c+d x) \sin ^3(c+d x)}{8 d}+\frac {a^3 \cos (c+d x) \sin ^5(c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 1.88, size = 168, normalized size = 0.97 \[ \frac {(a \sin (c+d x)+a)^3 \left (-2100 (c+d x)+455 \sin (2 (c+d x))+245 \sin (4 (c+d x))+35 \sin (6 (c+d x))+9065 \cos (c+d x)+875 \cos (3 (c+d x))+49 \cos (5 (c+d x))-5 \cos (7 (c+d x))+1120 \tan \left (\frac {1}{2} (c+d x)\right )-1120 \cot \left (\frac {1}{2} (c+d x)\right )+6720 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-6720 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{2240 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^3,x]

[Out]

((a + a*Sin[c + d*x])^3*(-2100*(c + d*x) + 9065*Cos[c + d*x] + 875*Cos[3*(c + d*x)] + 49*Cos[5*(c + d*x)] - 5*
Cos[7*(c + d*x)] - 1120*Cot[(c + d*x)/2] - 6720*Log[Cos[(c + d*x)/2]] + 6720*Log[Sin[(c + d*x)/2]] + 455*Sin[2
*(c + d*x)] + 245*Sin[4*(c + d*x)] + 35*Sin[6*(c + d*x)] + 1120*Tan[(c + d*x)/2]))/(2240*d*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^6)

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fricas [A]  time = 0.77, size = 173, normalized size = 1.00 \[ -\frac {280 \, a^{3} \cos \left (d x + c\right )^{7} - 70 \, a^{3} \cos \left (d x + c\right )^{5} - 175 \, a^{3} \cos \left (d x + c\right )^{3} + 840 \, a^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 840 \, a^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 525 \, a^{3} \cos \left (d x + c\right ) + {\left (80 \, a^{3} \cos \left (d x + c\right )^{7} - 336 \, a^{3} \cos \left (d x + c\right )^{5} - 560 \, a^{3} \cos \left (d x + c\right )^{3} + 525 \, a^{3} d x - 1680 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{560 \, d \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/560*(280*a^3*cos(d*x + c)^7 - 70*a^3*cos(d*x + c)^5 - 175*a^3*cos(d*x + c)^3 + 840*a^3*log(1/2*cos(d*x + c)
+ 1/2)*sin(d*x + c) - 840*a^3*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 525*a^3*cos(d*x + c) + (80*a^3*cos(
d*x + c)^7 - 336*a^3*cos(d*x + c)^5 - 560*a^3*cos(d*x + c)^3 + 525*a^3*d*x - 1680*a^3*cos(d*x + c))*sin(d*x +
c))/(d*sin(d*x + c))

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giac [A]  time = 0.33, size = 290, normalized size = 1.68 \[ -\frac {525 \, {\left (d x + c\right )} a^{3} - 1680 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 280 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {280 \, {\left (6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (525 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 4480 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 980 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 20160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 945 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 38080 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 49280 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 945 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 32256 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 980 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12992 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 525 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2496 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{7}}}{560 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/560*(525*(d*x + c)*a^3 - 1680*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 280*a^3*tan(1/2*d*x + 1/2*c) + 280*(6*a^
3*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c) + 2*(525*a^3*tan(1/2*d*x + 1/2*c)^13 - 4480*a^3*tan(1/2*d*x
+ 1/2*c)^12 - 980*a^3*tan(1/2*d*x + 1/2*c)^11 - 20160*a^3*tan(1/2*d*x + 1/2*c)^10 + 945*a^3*tan(1/2*d*x + 1/2
*c)^9 - 38080*a^3*tan(1/2*d*x + 1/2*c)^8 - 49280*a^3*tan(1/2*d*x + 1/2*c)^6 - 945*a^3*tan(1/2*d*x + 1/2*c)^5 -
32256*a^3*tan(1/2*d*x + 1/2*c)^4 + 980*a^3*tan(1/2*d*x + 1/2*c)^3 - 12992*a^3*tan(1/2*d*x + 1/2*c)^2 - 525*a^
3*tan(1/2*d*x + 1/2*c) - 2496*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^7)/d

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maple [A]  time = 0.42, size = 190, normalized size = 1.10 \[ -\frac {a^{3} \left (\cos ^{7}\left (d x +c \right )\right )}{7 d}-\frac {a^{3} \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{2 d}-\frac {5 a^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{8 d}-\frac {15 a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{16 d}-\frac {15 a^{3} x}{16}-\frac {15 a^{3} c}{16 d}+\frac {3 a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{d}+\frac {3 a^{3} \cos \left (d x +c \right )}{d}+\frac {3 a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}-\frac {a^{3} \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x)

[Out]

-1/7*a^3*cos(d*x+c)^7/d-1/2*a^3*cos(d*x+c)^5*sin(d*x+c)/d-5/8*a^3*cos(d*x+c)^3*sin(d*x+c)/d-15/16*a^3*cos(d*x+
c)*sin(d*x+c)/d-15/16*a^3*x-15/16/d*a^3*c+3/5*a^3*cos(d*x+c)^5/d+a^3*cos(d*x+c)^3/d+3*a^3*cos(d*x+c)/d+3/d*a^3
*ln(csc(d*x+c)-cot(d*x+c))-1/d*a^3/sin(d*x+c)*cos(d*x+c)^7

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maxima [A]  time = 0.41, size = 186, normalized size = 1.08 \[ -\frac {320 \, a^{3} \cos \left (d x + c\right )^{7} - 224 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{3} + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 280 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{3}}{2240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2240*(320*a^3*cos(d*x + c)^7 - 224*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 30*cos(d*x + c) - 15*log(cos(d*x
+ c) + 1) + 15*log(cos(d*x + c) - 1))*a^3 + 35*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 4
8*sin(2*d*x + 2*c))*a^3 + 280*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 25*tan(d*x + c)^2 + 8)/(tan(d*x + c)^5 + 2
*tan(d*x + c)^3 + tan(d*x + c)))*a^3)/d

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mupad [B]  time = 9.38, size = 429, normalized size = 2.48 \[ \frac {3\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {15\,a^3\,\mathrm {atan}\left (\frac {225\,a^6}{64\,\left (\frac {45\,a^6}{4}+\frac {225\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}-\frac {45\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {45\,a^6}{4}+\frac {225\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}\right )}{8\,d}+\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {\frac {19\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{4}-32\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-144\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\frac {111\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{4}-272\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+35\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-352\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {113\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{4}-\frac {1152\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {464\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{5}+\frac {13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}-\frac {624\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}+a^3}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}+14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+42\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+70\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+70\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+42\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+14\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^6*(a + a*sin(c + d*x))^3)/sin(c + d*x)^2,x)

[Out]

(3*a^3*log(tan(c/2 + (d*x)/2)))/d + (15*a^3*atan((225*a^6)/(64*((45*a^6)/4 + (225*a^6*tan(c/2 + (d*x)/2))/64))
- (45*a^6*tan(c/2 + (d*x)/2))/(4*((45*a^6)/4 + (225*a^6*tan(c/2 + (d*x)/2))/64))))/(8*d) + (a^3*tan(c/2 + (d*
x)/2))/(2*d) - ((13*a^3*tan(c/2 + (d*x)/2)^2)/4 - (464*a^3*tan(c/2 + (d*x)/2)^3)/5 + 28*a^3*tan(c/2 + (d*x)/2)
^4 - (1152*a^3*tan(c/2 + (d*x)/2)^5)/5 + (113*a^3*tan(c/2 + (d*x)/2)^6)/4 - 352*a^3*tan(c/2 + (d*x)/2)^7 + 35*
a^3*tan(c/2 + (d*x)/2)^8 - 272*a^3*tan(c/2 + (d*x)/2)^9 + (111*a^3*tan(c/2 + (d*x)/2)^10)/4 - 144*a^3*tan(c/2
+ (d*x)/2)^11 - 32*a^3*tan(c/2 + (d*x)/2)^13 + (19*a^3*tan(c/2 + (d*x)/2)^14)/4 + a^3 - (624*a^3*tan(c/2 + (d*
x)/2))/35)/(d*(2*tan(c/2 + (d*x)/2) + 14*tan(c/2 + (d*x)/2)^3 + 42*tan(c/2 + (d*x)/2)^5 + 70*tan(c/2 + (d*x)/2
)^7 + 70*tan(c/2 + (d*x)/2)^9 + 42*tan(c/2 + (d*x)/2)^11 + 14*tan(c/2 + (d*x)/2)^13 + 2*tan(c/2 + (d*x)/2)^15)
)

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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)**2*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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