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

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

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2872, 3770, 3767, 8, 3768, 2638, 2635, 2633} \[ -\frac {2 a^2 \cos ^3(c+d x)}{3 d}-\frac {4 a^2 \cos (c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x)}{d}+\frac {a^2 \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {5 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{d}+\frac {5 a^2 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 3.37, size = 209, normalized size = 1.37 \[ \frac {a^2 (\sin (c+d x)+1)^2 \left (60 (c+d x)-24 \sin (2 (c+d x))-3 \sin (4 (c+d x))-432 \cos (c+d x)-16 \cos (3 (c+d x))-64 \tan \left (\frac {1}{2} (c+d x)\right )+64 \cot \left (\frac {1}{2} (c+d x)\right )-24 \csc ^2\left (\frac {1}{2} (c+d x)\right )+24 \sec ^2\left (\frac {1}{2} (c+d x)\right )-480 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+480 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-2 \sin (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )+32 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)\right )}{96 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Sin[c + d*x])^2*(60*(c + d*x) - 432*Cos[c + d*x] - 16*Cos[3*(c + d*x)] + 64*Cot[(c + d*x)/2] - 24*Cs
c[(c + d*x)/2]^2 + 480*Log[Cos[(c + d*x)/2]] - 480*Log[Sin[(c + d*x)/2]] + 24*Sec[(c + d*x)/2]^2 + 32*Csc[c +
d*x]^3*Sin[(c + d*x)/2]^4 - 2*Csc[(c + d*x)/2]^4*Sin[c + d*x] - 24*Sin[2*(c + d*x)] - 3*Sin[4*(c + d*x)] - 64*
Tan[(c + d*x)/2]))/(96*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

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fricas [A]  time = 0.75, size = 219, normalized size = 1.43 \[ \frac {6 \, a^{2} \cos \left (d x + c\right )^{7} - 3 \, a^{2} \cos \left (d x + c\right )^{5} + 20 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} \cos \left (d x + c\right ) + 60 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 60 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - {\left (16 \, a^{2} \cos \left (d x + c\right )^{5} - 15 \, a^{2} d x \cos \left (d x + c\right )^{2} + 80 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} d x - 120 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(6*a^2*cos(d*x + c)^7 - 3*a^2*cos(d*x + c)^5 + 20*a^2*cos(d*x + c)^3 - 15*a^2*cos(d*x + c) + 60*(a^2*cos(
d*x + c)^2 - a^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 60*(a^2*cos(d*x + c)^2 - a^2)*log(-1/2*cos(d*x +
c) + 1/2)*sin(d*x + c) - (16*a^2*cos(d*x + c)^5 - 15*a^2*d*x*cos(d*x + c)^2 + 80*a^2*cos(d*x + c)^3 + 15*a^2*d
*x - 120*a^2*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))

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giac [A]  time = 0.28, size = 274, normalized size = 1.79 \[ \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, {\left (d x + c\right )} a^{2} - 120 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {220 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} + \frac {2 \, {\left (15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 144 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 336 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 304 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 112 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*tan(1/2*d*x + 1/2*c)^2 + 15*(d*x + c)*a^2 - 120*a^2*log(abs(tan(1/2*d
*x + 1/2*c))) - 15*a^2*tan(1/2*d*x + 1/2*c) + (220*a^2*tan(1/2*d*x + 1/2*c)^3 + 15*a^2*tan(1/2*d*x + 1/2*c)^2
- 6*a^2*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^3 + 2*(15*a^2*tan(1/2*d*x + 1/2*c)^7 - 144*a^2*tan(1/
2*d*x + 1/2*c)^6 - 9*a^2*tan(1/2*d*x + 1/2*c)^5 - 336*a^2*tan(1/2*d*x + 1/2*c)^4 + 9*a^2*tan(1/2*d*x + 1/2*c)^
3 - 304*a^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2*tan(1/2*d*x + 1/2*c) - 112*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 0.43, size = 223, normalized size = 1.46 \[ \frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )}+\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{12 d}+\frac {5 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {5 a^{2} x}{8}+\frac {5 a^{2} c}{8 d}-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{2}}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{d}-\frac {5 a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3 d}-\frac {5 a^{2} \cos \left (d x +c \right )}{d}-\frac {5 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}-\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.76, size = 190, normalized size = 1.24 \[ -\frac {4 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} + 3 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2} - 4 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a^{2}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/24*(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 + 3*(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^2 - 4*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 - 2)/(t
an(d*x + c)^5 + tan(d*x + c)^3))*a^2)/d

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mupad [B]  time = 8.93, size = 384, normalized size = 2.51 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {5\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {5\,a^2\,\mathrm {atan}\left (\frac {25\,a^4}{16\,\left (\frac {25\,a^4}{2}+\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}-\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {25\,a^4}{2}+\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}\right )}{4\,d}-\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {-15\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+98\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-\frac {41\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+232\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {104\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {644\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {248\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^2}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )} \]

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)^4,x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^2)/(4*d) + (a^2*tan(c/2 + (d*x)/2)^3)/(24*d) - (5*a^2*log(tan(c/2 + (d*x)/2)))/d - (5*
a^2*atan((25*a^4)/(16*((25*a^4)/2 + (25*a^4*tan(c/2 + (d*x)/2))/16)) - (25*a^4*tan(c/2 + (d*x)/2))/(2*((25*a^4
)/2 + (25*a^4*tan(c/2 + (d*x)/2))/16))))/(4*d) - (5*a^2*tan(c/2 + (d*x)/2))/(8*d) - ((248*a^2*tan(c/2 + (d*x)/
2)^3)/3 - (11*a^2*tan(c/2 + (d*x)/2)^2)/3 - 8*a^2*tan(c/2 + (d*x)/2)^4 + (644*a^2*tan(c/2 + (d*x)/2)^5)/3 - (1
04*a^2*tan(c/2 + (d*x)/2)^6)/3 + 232*a^2*tan(c/2 + (d*x)/2)^7 - (41*a^2*tan(c/2 + (d*x)/2)^8)/3 + 98*a^2*tan(c
/2 + (d*x)/2)^9 - 15*a^2*tan(c/2 + (d*x)/2)^10 + a^2/3 + 2*a^2*tan(c/2 + (d*x)/2))/(d*(8*tan(c/2 + (d*x)/2)^3
+ 32*tan(c/2 + (d*x)/2)^5 + 48*tan(c/2 + (d*x)/2)^7 + 32*tan(c/2 + (d*x)/2)^9 + 8*tan(c/2 + (d*x)/2)^11))

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

[Out]

Timed out

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