3.43 \(\int \cot ^6(c+d x) (a+a \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=198 \[ -\frac {4 a^4 \cos ^3(c+d x)}{3 d}-\frac {4 a^4 \cos (c+d x)}{d}-\frac {a^4 \cot ^5(c+d x)}{5 d}-\frac {5 a^4 \cot ^3(c+d x)}{3 d}+\frac {10 a^4 \cot (c+d x)}{d}+\frac {a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {15 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 a^4 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^4 \cot (c+d x) \csc ^3(c+d x)}{d}+\frac {5 a^4 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {97 a^4 x}{8} \]

[Out]

97/8*a^4*x+5/2*a^4*arctanh(cos(d*x+c))/d-4*a^4*cos(d*x+c)/d-4/3*a^4*cos(d*x+c)^3/d+10*a^4*cot(d*x+c)/d-5/3*a^4
*cot(d*x+c)^3/d-1/5*a^4*cot(d*x+c)^5/d+5/2*a^4*cot(d*x+c)*csc(d*x+c)/d-a^4*cot(d*x+c)*csc(d*x+c)^3/d+15/8*a^4*
cos(d*x+c)*sin(d*x+c)/d+1/4*a^4*cos(d*x+c)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.43, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {2709, 3767, 8, 3768, 3770, 2638, 2635, 2633} \[ -\frac {4 a^4 \cos ^3(c+d x)}{3 d}-\frac {4 a^4 \cos (c+d x)}{d}-\frac {a^4 \cot ^5(c+d x)}{5 d}-\frac {5 a^4 \cot ^3(c+d x)}{3 d}+\frac {10 a^4 \cot (c+d x)}{d}+\frac {a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {15 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 a^4 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {a^4 \cot (c+d x) \csc ^3(c+d x)}{d}+\frac {5 a^4 \cot (c+d x) \csc (c+d x)}{2 d}+\frac {97 a^4 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + a*Sin[c + d*x])^4,x]

[Out]

(97*a^4*x)/8 + (5*a^4*ArcTanh[Cos[c + d*x]])/(2*d) - (4*a^4*Cos[c + d*x])/d - (4*a^4*Cos[c + d*x]^3)/(3*d) + (
10*a^4*Cot[c + d*x])/d - (5*a^4*Cot[c + d*x]^3)/(3*d) - (a^4*Cot[c + d*x]^5)/(5*d) + (5*a^4*Cot[c + d*x]*Csc[c
+ d*x])/(2*d) - (a^4*Cot[c + d*x]*Csc[c + d*x]^3)/d + (15*a^4*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^4*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 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))^4 \, dx &=\frac {\int \left (14 a^{10}-14 a^{10} \csc ^2(c+d x)-8 a^{10} \csc ^3(c+d x)+3 a^{10} \csc ^4(c+d x)+4 a^{10} \csc ^5(c+d x)+a^{10} \csc ^6(c+d x)+8 a^{10} \sin (c+d x)-3 a^{10} \sin ^2(c+d x)-4 a^{10} \sin ^3(c+d x)-a^{10} \sin ^4(c+d x)\right ) \, dx}{a^6}\\ &=14 a^4 x+a^4 \int \csc ^6(c+d x) \, dx-a^4 \int \sin ^4(c+d x) \, dx+\left (3 a^4\right ) \int \csc ^4(c+d x) \, dx-\left (3 a^4\right ) \int \sin ^2(c+d x) \, dx+\left (4 a^4\right ) \int \csc ^5(c+d x) \, dx-\left (4 a^4\right ) \int \sin ^3(c+d x) \, dx-\left (8 a^4\right ) \int \csc ^3(c+d x) \, dx+\left (8 a^4\right ) \int \sin (c+d x) \, dx-\left (14 a^4\right ) \int \csc ^2(c+d x) \, dx\\ &=14 a^4 x-\frac {8 a^4 \cos (c+d x)}{d}+\frac {4 a^4 \cot (c+d x) \csc (c+d x)}{d}-\frac {a^4 \cot (c+d x) \csc ^3(c+d x)}{d}+\frac {3 a^4 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac {1}{4} \left (3 a^4\right ) \int \sin ^2(c+d x) \, dx-\frac {1}{2} \left (3 a^4\right ) \int 1 \, dx+\left (3 a^4\right ) \int \csc ^3(c+d x) \, dx-\left (4 a^4\right ) \int \csc (c+d x) \, dx-\frac {a^4 \operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (c+d x)\right )}{d}-\frac {\left (3 a^4\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}+\frac {\left (4 a^4\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (14 a^4\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=\frac {25 a^4 x}{2}+\frac {4 a^4 \tanh ^{-1}(\cos (c+d x))}{d}-\frac {4 a^4 \cos (c+d x)}{d}-\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {10 a^4 \cot (c+d x)}{d}-\frac {5 a^4 \cot ^3(c+d x)}{3 d}-\frac {a^4 \cot ^5(c+d x)}{5 d}+\frac {5 a^4 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {a^4 \cot (c+d x) \csc ^3(c+d x)}{d}+\frac {15 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac {1}{8} \left (3 a^4\right ) \int 1 \, dx+\frac {1}{2} \left (3 a^4\right ) \int \csc (c+d x) \, dx\\ &=\frac {97 a^4 x}{8}+\frac {5 a^4 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac {4 a^4 \cos (c+d x)}{d}-\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {10 a^4 \cot (c+d x)}{d}-\frac {5 a^4 \cot ^3(c+d x)}{3 d}-\frac {a^4 \cot ^5(c+d x)}{5 d}+\frac {5 a^4 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {a^4 \cot (c+d x) \csc ^3(c+d x)}{d}+\frac {15 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 1.58, size = 283, normalized size = 1.43 \[ \frac {a^4 (\sin (c+d x)+1)^4 \left (5820 (c+d x)+480 \sin (2 (c+d x))-15 \sin (4 (c+d x))-2400 \cos (c+d x)-160 \cos (3 (c+d x))-2752 \tan \left (\frac {1}{2} (c+d x)\right )+2752 \cot \left (\frac {1}{2} (c+d x)\right )-30 \csc ^4\left (\frac {1}{2} (c+d x)\right )+300 \csc ^2\left (\frac {1}{2} (c+d x)\right )+30 \sec ^4\left (\frac {1}{2} (c+d x)\right )-300 \sec ^2\left (\frac {1}{2} (c+d x)\right )-1200 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+1200 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\frac {3}{2} \sin (c+d x) \csc ^6\left (\frac {1}{2} (c+d x)\right )+96 \sin ^6\left (\frac {1}{2} (c+d x)\right ) \csc ^5(c+d x)-\frac {79}{2} \sin (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )+632 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)\right )}{480 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + a*Sin[c + d*x])^4,x]

[Out]

(a^4*(1 + Sin[c + d*x])^4*(5820*(c + d*x) - 2400*Cos[c + d*x] - 160*Cos[3*(c + d*x)] + 2752*Cot[(c + d*x)/2] +
300*Csc[(c + d*x)/2]^2 - 30*Csc[(c + d*x)/2]^4 + 1200*Log[Cos[(c + d*x)/2]] - 1200*Log[Sin[(c + d*x)/2]] - 30
0*Sec[(c + d*x)/2]^2 + 30*Sec[(c + d*x)/2]^4 + 632*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 96*Csc[c + d*x]^5*Sin[(
c + d*x)/2]^6 - (79*Csc[(c + d*x)/2]^4*Sin[c + d*x])/2 - (3*Csc[(c + d*x)/2]^6*Sin[c + d*x])/2 + 480*Sin[2*(c
+ d*x)] - 15*Sin[4*(c + d*x)] - 2752*Tan[(c + d*x)/2]))/(480*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8)

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fricas [A]  time = 0.47, size = 291, normalized size = 1.47 \[ \frac {30 \, a^{4} \cos \left (d x + c\right )^{9} - 345 \, a^{4} \cos \left (d x + c\right )^{7} + 2231 \, a^{4} \cos \left (d x + c\right )^{5} - 3395 \, a^{4} \cos \left (d x + c\right )^{3} + 1455 \, a^{4} \cos \left (d x + c\right ) + 150 \, {\left (a^{4} \cos \left (d x + c\right )^{4} - 2 \, a^{4} \cos \left (d x + c\right )^{2} + a^{4}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 150 \, {\left (a^{4} \cos \left (d x + c\right )^{4} - 2 \, a^{4} \cos \left (d x + c\right )^{2} + a^{4}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 5 \, {\left (32 \, a^{4} \cos \left (d x + c\right )^{7} - 291 \, a^{4} d x \cos \left (d x + c\right )^{4} + 32 \, a^{4} \cos \left (d x + c\right )^{5} + 582 \, a^{4} d x \cos \left (d x + c\right )^{2} - 100 \, a^{4} \cos \left (d x + c\right )^{3} - 291 \, a^{4} d x + 60 \, a^{4} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(30*a^4*cos(d*x + c)^9 - 345*a^4*cos(d*x + c)^7 + 2231*a^4*cos(d*x + c)^5 - 3395*a^4*cos(d*x + c)^3 + 14
55*a^4*cos(d*x + c) + 150*(a^4*cos(d*x + c)^4 - 2*a^4*cos(d*x + c)^2 + a^4)*log(1/2*cos(d*x + c) + 1/2)*sin(d*
x + c) - 150*(a^4*cos(d*x + c)^4 - 2*a^4*cos(d*x + c)^2 + a^4)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 5*(
32*a^4*cos(d*x + c)^7 - 291*a^4*d*x*cos(d*x + c)^4 + 32*a^4*cos(d*x + c)^5 + 582*a^4*d*x*cos(d*x + c)^2 - 100*
a^4*cos(d*x + c)^3 - 291*a^4*d*x + 60*a^4*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 [A]  time = 1.39, size = 339, normalized size = 1.71 \[ \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 85 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 240 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5820 \, {\left (d x + c\right )} a^{4} - 1200 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 2670 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {40 \, {\left (45 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 192 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 69 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 384 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 69 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 128 \, a^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}} + \frac {2740 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2670 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 240 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 85 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 30 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{4}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(3*a^4*tan(1/2*d*x + 1/2*c)^5 + 30*a^4*tan(1/2*d*x + 1/2*c)^4 + 85*a^4*tan(1/2*d*x + 1/2*c)^3 - 240*a^4*
tan(1/2*d*x + 1/2*c)^2 + 5820*(d*x + c)*a^4 - 1200*a^4*log(abs(tan(1/2*d*x + 1/2*c))) - 2670*a^4*tan(1/2*d*x +
1/2*c) - 40*(45*a^4*tan(1/2*d*x + 1/2*c)^7 + 192*a^4*tan(1/2*d*x + 1/2*c)^6 + 69*a^4*tan(1/2*d*x + 1/2*c)^5 +
384*a^4*tan(1/2*d*x + 1/2*c)^4 - 69*a^4*tan(1/2*d*x + 1/2*c)^3 + 320*a^4*tan(1/2*d*x + 1/2*c)^2 - 45*a^4*tan(
1/2*d*x + 1/2*c) + 128*a^4)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4 + (2740*a^4*tan(1/2*d*x + 1/2*c)^5 + 2670*a^4*tan(1
/2*d*x + 1/2*c)^4 + 240*a^4*tan(1/2*d*x + 1/2*c)^3 - 85*a^4*tan(1/2*d*x + 1/2*c)^2 - 30*a^4*tan(1/2*d*x + 1/2*
c) - 3*a^4)/tan(1/2*d*x + 1/2*c)^5)/d

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maple [A]  time = 0.23, size = 293, normalized size = 1.48 \[ \frac {97 a^{4} x}{8}+\frac {7 a^{4} \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{d}+\frac {97 a^{4} c}{8 d}-\frac {a^{4} \left (\cos ^{5}\left (d x +c \right )\right )}{2 d}-\frac {5 a^{4} \left (\cos ^{3}\left (d x +c \right )\right )}{6 d}-\frac {a^{4} \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{4}}-\frac {2 a^{4} \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )^{3}}+\frac {7 a^{4} \left (\cos ^{7}\left (d x +c \right )\right )}{d \sin \left (d x +c \right )}+\frac {35 a^{4} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4 d}+\frac {105 a^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}-\frac {a^{4} \left (\cos ^{7}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {a^{4} \cot \left (d x +c \right )}{d}-\frac {5 a^{4} \cos \left (d x +c \right )}{2 d}-\frac {a^{4} \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a^{4} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {5 a^{4} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+a*sin(d*x+c))^4,x)

[Out]

97/8*a^4*x+7/d*a^4*sin(d*x+c)*cos(d*x+c)^5+97/8/d*a^4*c-1/2/d*a^4*cos(d*x+c)^5-5/6*a^4*cos(d*x+c)^3/d-1/d*a^4/
sin(d*x+c)^4*cos(d*x+c)^7-2/d*a^4/sin(d*x+c)^3*cos(d*x+c)^7+7/d*a^4/sin(d*x+c)*cos(d*x+c)^7+35/4/d*a^4*cos(d*x
+c)^3*sin(d*x+c)+105/8*a^4*cos(d*x+c)*sin(d*x+c)/d-1/2/d*a^4/sin(d*x+c)^2*cos(d*x+c)^7-a^4*cot(d*x+c)/d-5/2*a^
4*cos(d*x+c)/d-1/5*a^4*cot(d*x+c)^5/d+1/3*a^4*cot(d*x+c)^3/d-5/2/d*a^4*ln(csc(d*x+c)-cot(d*x+c))

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maxima [A]  time = 0.41, size = 313, normalized size = 1.58 \[ -\frac {40 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{4} + 15 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{4} - 120 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a^{4} + 8 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} a^{4} + 30 \, a^{4} {\left (\frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - 16 \, \cos \left (d x + c\right ) + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(40*(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^4 + 15*(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^4 - 120*(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^4 + 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^4 + 30*a^4*(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*c
os(d*x + c) + 15*log(cos(d*x + c) + 1) - 15*log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 6.85, size = 454, normalized size = 2.29 \[ \frac {17\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,d}-\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2\,d}+\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{16\,d}+\frac {a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,d}-\frac {5\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {97\,a^4\,\mathrm {atan}\left (\frac {9409\,a^8}{16\,\left (\frac {485\,a^8}{4}+\frac {9409\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}-\frac {485\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {485\,a^8}{4}+\frac {9409\,a^8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}\right )}{4\,d}-\frac {-58\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+496\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-\frac {1567\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{3}+962\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-\frac {18437\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{15}+\frac {2296\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {3986\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+\frac {868\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {2312\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}-8\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {97\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^4}{5}}{d\,\left (32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+128\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+192\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+128\,{\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 {89\,a^4\,\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(cot(c + d*x)^6*(a + a*sin(c + d*x))^4,x)

[Out]

(17*a^4*tan(c/2 + (d*x)/2)^3)/(96*d) - (a^4*tan(c/2 + (d*x)/2)^2)/(2*d) + (a^4*tan(c/2 + (d*x)/2)^4)/(16*d) +
(a^4*tan(c/2 + (d*x)/2)^5)/(160*d) - (5*a^4*log(tan(c/2 + (d*x)/2)))/(2*d) - (97*a^4*atan((9409*a^8)/(16*((485
*a^8)/4 + (9409*a^8*tan(c/2 + (d*x)/2))/16)) - (485*a^8*tan(c/2 + (d*x)/2))/(4*((485*a^8)/4 + (9409*a^8*tan(c/
2 + (d*x)/2))/16))))/(4*d) - ((97*a^4*tan(c/2 + (d*x)/2)^2)/15 - 8*a^4*tan(c/2 + (d*x)/2)^3 - (2312*a^4*tan(c/
2 + (d*x)/2)^4)/15 + (868*a^4*tan(c/2 + (d*x)/2)^5)/3 - (3986*a^4*tan(c/2 + (d*x)/2)^6)/5 + (2296*a^4*tan(c/2
+ (d*x)/2)^7)/3 - (18437*a^4*tan(c/2 + (d*x)/2)^8)/15 + 962*a^4*tan(c/2 + (d*x)/2)^9 - (1567*a^4*tan(c/2 + (d*
x)/2)^10)/3 + 496*a^4*tan(c/2 + (d*x)/2)^11 - 58*a^4*tan(c/2 + (d*x)/2)^12 + a^4/5 + 2*a^4*tan(c/2 + (d*x)/2))
/(d*(32*tan(c/2 + (d*x)/2)^5 + 128*tan(c/2 + (d*x)/2)^7 + 192*tan(c/2 + (d*x)/2)^9 + 128*tan(c/2 + (d*x)/2)^11
+ 32*tan(c/2 + (d*x)/2)^13)) - (89*a^4*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(cot(d*x+c)**6*(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

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