3.1251 \(\int \cot ^6(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=158 \[ -\frac {a^2 \cot ^7(c+d x)}{7 d}+\frac {5 a b \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a b \cot ^5(c+d x) \csc (c+d x)}{3 d}+\frac {5 a b \cot ^3(c+d x) \csc (c+d x)}{12 d}-\frac {5 a b \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b^2 \cot ^5(c+d x)}{5 d}+\frac {b^2 \cot ^3(c+d x)}{3 d}-\frac {b^2 \cot (c+d x)}{d}-b^2 x \]

[Out]

-b^2*x+5/8*a*b*arctanh(cos(d*x+c))/d-b^2*cot(d*x+c)/d+1/3*b^2*cot(d*x+c)^3/d-1/5*b^2*cot(d*x+c)^5/d-1/7*a^2*co
t(d*x+c)^7/d-5/8*a*b*cot(d*x+c)*csc(d*x+c)/d+5/12*a*b*cot(d*x+c)^3*csc(d*x+c)/d-1/3*a*b*cot(d*x+c)^5*csc(d*x+c
)/d

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Rubi [A]  time = 0.41, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2911, 2611, 3770, 14, 203} \[ -\frac {a^2 \cot ^7(c+d x)}{7 d}+\frac {5 a b \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {a b \cot ^5(c+d x) \csc (c+d x)}{3 d}+\frac {5 a b \cot ^3(c+d x) \csc (c+d x)}{12 d}-\frac {5 a b \cot (c+d x) \csc (c+d x)}{8 d}-\frac {b^2 \cot ^5(c+d x)}{5 d}+\frac {b^2 \cot ^3(c+d x)}{3 d}-\frac {b^2 \cot (c+d x)}{d}-b^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*x) + (5*a*b*ArcTanh[Cos[c + d*x]])/(8*d) - (b^2*Cot[c + d*x])/d + (b^2*Cot[c + d*x]^3)/(3*d) - (b^2*Cot[
c + d*x]^5)/(5*d) - (a^2*Cot[c + d*x]^7)/(7*d) - (5*a*b*Cot[c + d*x]*Csc[c + d*x])/(8*d) + (5*a*b*Cot[c + d*x]
^3*Csc[c + d*x])/(12*d) - (a*b*Cot[c + d*x]^5*Csc[c + d*x])/(3*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^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 \cot ^6(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x))^2 \, dx &=(2 a b) \int \cot ^6(c+d x) \csc (c+d x) \, dx+\int \cot ^6(c+d x) \csc ^2(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {a b \cot ^5(c+d x) \csc (c+d x)}{3 d}-\frac {1}{3} (5 a b) \int \cot ^4(c+d x) \csc (c+d x) \, dx+\frac {\operatorname {Subst}\left (\int \frac {a^2+\frac {b^2 x^2}{1+x^2}}{x^8} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {5 a b \cot ^3(c+d x) \csc (c+d x)}{12 d}-\frac {a b \cot ^5(c+d x) \csc (c+d x)}{3 d}+\frac {1}{4} (5 a b) \int \cot ^2(c+d x) \csc (c+d x) \, dx+\frac {\operatorname {Subst}\left (\int \left (\frac {a^2}{x^8}+\frac {b^2}{x^6}-\frac {b^2}{x^4}+\frac {b^2}{x^2}-\frac {b^2}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {b^2 \cot (c+d x)}{d}+\frac {b^2 \cot ^3(c+d x)}{3 d}-\frac {b^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^7(c+d x)}{7 d}-\frac {5 a b \cot (c+d x) \csc (c+d x)}{8 d}+\frac {5 a b \cot ^3(c+d x) \csc (c+d x)}{12 d}-\frac {a b \cot ^5(c+d x) \csc (c+d x)}{3 d}-\frac {1}{8} (5 a b) \int \csc (c+d x) \, dx-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-b^2 x+\frac {5 a b \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac {b^2 \cot (c+d x)}{d}+\frac {b^2 \cot ^3(c+d x)}{3 d}-\frac {b^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^7(c+d x)}{7 d}-\frac {5 a b \cot (c+d x) \csc (c+d x)}{8 d}+\frac {5 a b \cot ^3(c+d x) \csc (c+d x)}{12 d}-\frac {a b \cot ^5(c+d x) \csc (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 1.40, size = 280, normalized size = 1.77 \[ \frac {\csc ^7(c+d x) \left (-84 \left (15 a^2-41 b^2\right ) \cos (3 (c+d x))-28 \left (15 a^2+71 b^2\right ) \cos (5 (c+d x))-60 a^2 \cos (7 (c+d x))+980 a b \sin (4 (c+d x))-1155 a b \sin (6 (c+d x))+8820 b^2 c \sin (3 (c+d x))+8820 b^2 d x \sin (3 (c+d x))-2940 b^2 c \sin (5 (c+d x))-2940 b^2 d x \sin (5 (c+d x))+420 b^2 c \sin (7 (c+d x))+420 b^2 d x \sin (7 (c+d x))+644 b^2 \cos (7 (c+d x))\right )-350 \cot (c+d x) \csc ^6(c+d x) \left (6 \left (a^2+b^2\right )+17 a b \sin (c+d x)\right )+16800 a b \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-14700 b^2 (c+d x) \csc ^6(c+d x)}{26880 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14700*b^2*(c + d*x)*Csc[c + d*x]^6 + 16800*a*b*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]]) - 350*Cot[c +
d*x]*Csc[c + d*x]^6*(6*(a^2 + b^2) + 17*a*b*Sin[c + d*x]) + Csc[c + d*x]^7*(-84*(15*a^2 - 41*b^2)*Cos[3*(c +
d*x)] - 28*(15*a^2 + 71*b^2)*Cos[5*(c + d*x)] - 60*a^2*Cos[7*(c + d*x)] + 644*b^2*Cos[7*(c + d*x)] + 8820*b^2*
c*Sin[3*(c + d*x)] + 8820*b^2*d*x*Sin[3*(c + d*x)] + 980*a*b*Sin[4*(c + d*x)] - 2940*b^2*c*Sin[5*(c + d*x)] -
2940*b^2*d*x*Sin[5*(c + d*x)] - 1155*a*b*Sin[6*(c + d*x)] + 420*b^2*c*Sin[7*(c + d*x)] + 420*b^2*d*x*Sin[7*(c
+ d*x)]))/(26880*d)

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fricas [B]  time = 1.03, size = 320, normalized size = 2.03 \[ \frac {16 \, {\left (15 \, a^{2} - 161 \, b^{2}\right )} \cos \left (d x + c\right )^{7} + 6496 \, b^{2} \cos \left (d x + c\right )^{5} - 5600 \, b^{2} \cos \left (d x + c\right )^{3} + 1680 \, b^{2} \cos \left (d x + c\right ) + 525 \, {\left (a b \cos \left (d x + c\right )^{6} - 3 \, a b \cos \left (d x + c\right )^{4} + 3 \, a b \cos \left (d x + c\right )^{2} - a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 525 \, {\left (a b \cos \left (d x + c\right )^{6} - 3 \, a b \cos \left (d x + c\right )^{4} + 3 \, a b \cos \left (d x + c\right )^{2} - a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 70 \, {\left (24 \, b^{2} d x \cos \left (d x + c\right )^{6} - 72 \, b^{2} d x \cos \left (d x + c\right )^{4} - 33 \, a b \cos \left (d x + c\right )^{5} + 72 \, b^{2} d x \cos \left (d x + c\right )^{2} + 40 \, a b \cos \left (d x + c\right )^{3} - 24 \, b^{2} d x - 15 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{1680 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^8*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/1680*(16*(15*a^2 - 161*b^2)*cos(d*x + c)^7 + 6496*b^2*cos(d*x + c)^5 - 5600*b^2*cos(d*x + c)^3 + 1680*b^2*co
s(d*x + c) + 525*(a*b*cos(d*x + c)^6 - 3*a*b*cos(d*x + c)^4 + 3*a*b*cos(d*x + c)^2 - a*b)*log(1/2*cos(d*x + c)
+ 1/2)*sin(d*x + c) - 525*(a*b*cos(d*x + c)^6 - 3*a*b*cos(d*x + c)^4 + 3*a*b*cos(d*x + c)^2 - a*b)*log(-1/2*c
os(d*x + c) + 1/2)*sin(d*x + c) - 70*(24*b^2*d*x*cos(d*x + c)^6 - 72*b^2*d*x*cos(d*x + c)^4 - 33*a*b*cos(d*x +
c)^5 + 72*b^2*d*x*cos(d*x + c)^2 + 40*a*b*cos(d*x + c)^3 - 24*b^2*d*x - 15*a*b*cos(d*x + c))*sin(d*x + c))/((
d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)*sin(d*x + c))

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giac [B]  time = 0.34, size = 356, normalized size = 2.25 \[ \frac {15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 70 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 105 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 84 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 630 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 315 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 980 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3150 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 13440 \, {\left (d x + c\right )} b^{2} - 8400 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 525 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9240 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {21780 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 525 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 9240 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 3150 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 315 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 980 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 630 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 84 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 70 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}}}{13440 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^8*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/13440*(15*a^2*tan(1/2*d*x + 1/2*c)^7 + 70*a*b*tan(1/2*d*x + 1/2*c)^6 - 105*a^2*tan(1/2*d*x + 1/2*c)^5 + 84*b
^2*tan(1/2*d*x + 1/2*c)^5 - 630*a*b*tan(1/2*d*x + 1/2*c)^4 + 315*a^2*tan(1/2*d*x + 1/2*c)^3 - 980*b^2*tan(1/2*
d*x + 1/2*c)^3 + 3150*a*b*tan(1/2*d*x + 1/2*c)^2 - 13440*(d*x + c)*b^2 - 8400*a*b*log(abs(tan(1/2*d*x + 1/2*c)
)) - 525*a^2*tan(1/2*d*x + 1/2*c) + 9240*b^2*tan(1/2*d*x + 1/2*c) + (21780*a*b*tan(1/2*d*x + 1/2*c)^7 + 525*a^
2*tan(1/2*d*x + 1/2*c)^6 - 9240*b^2*tan(1/2*d*x + 1/2*c)^6 - 3150*a*b*tan(1/2*d*x + 1/2*c)^5 - 315*a^2*tan(1/2
*d*x + 1/2*c)^4 + 980*b^2*tan(1/2*d*x + 1/2*c)^4 + 630*a*b*tan(1/2*d*x + 1/2*c)^3 + 105*a^2*tan(1/2*d*x + 1/2*
c)^2 - 84*b^2*tan(1/2*d*x + 1/2*c)^2 - 70*a*b*tan(1/2*d*x + 1/2*c) - 15*a^2)/tan(1/2*d*x + 1/2*c)^7)/d

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maple [A]  time = 0.46, size = 222, normalized size = 1.41 \[ -\frac {a^{2} \left (\cos ^{7}\left (d x +c \right )\right )}{7 d \sin \left (d x +c \right )^{7}}-\frac {a b \left (\cos ^{7}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{6}}+\frac {a b \left (\cos ^{7}\left (d x +c \right )\right )}{12 d \sin \left (d x +c \right )^{4}}-\frac {a b \left (\cos ^{7}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}-\frac {a b \left (\cos ^{5}\left (d x +c \right )\right )}{8 d}-\frac {5 a b \left (\cos ^{3}\left (d x +c \right )\right )}{24 d}-\frac {5 a b \cos \left (d x +c \right )}{8 d}-\frac {5 a b \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {b^{2} \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}+\frac {b^{2} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {b^{2} \cot \left (d x +c \right )}{d}-b^{2} x -\frac {b^{2} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^8*(a+b*sin(d*x+c))^2,x)

[Out]

-1/7/d*a^2/sin(d*x+c)^7*cos(d*x+c)^7-1/3/d*a*b/sin(d*x+c)^6*cos(d*x+c)^7+1/12/d*a*b/sin(d*x+c)^4*cos(d*x+c)^7-
1/8/d*a*b/sin(d*x+c)^2*cos(d*x+c)^7-1/8*a*b*cos(d*x+c)^5/d-5/24*a*b*cos(d*x+c)^3/d-5/8*a*b*cos(d*x+c)/d-5/8/d*
a*b*ln(csc(d*x+c)-cot(d*x+c))-1/5*b^2*cot(d*x+c)^5/d+1/3*b^2*cot(d*x+c)^3/d-b^2*cot(d*x+c)/d-b^2*x-1/d*b^2*c

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maxima [A]  time = 0.42, size = 153, normalized size = 0.97 \[ -\frac {112 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{\tan \left (d x + c\right )^{5}}\right )} b^{2} - 35 \, a b {\left (\frac {2 \, {\left (33 \, \cos \left (d x + c\right )^{5} - 40 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} + 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {240 \, a^{2}}{\tan \left (d x + c\right )^{7}}}{1680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^8*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/1680*(112*(15*d*x + 15*c + (15*tan(d*x + c)^4 - 5*tan(d*x + c)^2 + 3)/tan(d*x + c)^5)*b^2 - 35*a*b*(2*(33*c
os(d*x + c)^5 - 40*cos(d*x + c)^3 + 15*cos(d*x + c))/(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 - 1
) + 15*log(cos(d*x + c) + 1) - 15*log(cos(d*x + c) - 1)) + 240*a^2/tan(d*x + c)^7)/d

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mupad [B]  time = 11.87, size = 379, normalized size = 2.40 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{896\,d}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {4\,b^4}{\frac {5\,a\,b^3}{2}-4\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {5\,a\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {5\,a\,b^3}{2}-4\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,a^2}{128}-\frac {11\,b^2}{16}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^2-\frac {28\,b^2}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (5\,a^2-88\,b^2\right )+\frac {a^2}{7}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2-\frac {4\,b^2}{5}\right )-6\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+30\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {2\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}}{128\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {3\,a^2}{128}-\frac {7\,b^2}{96}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {a^2}{128}-\frac {b^2}{160}\right )}{d}+\frac {15\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{64\,d}-\frac {3\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}+\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{192\,d}-\frac {5\,a\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^6*(a + b*sin(c + d*x))^2)/sin(c + d*x)^8,x)

[Out]

(a^2*tan(c/2 + (d*x)/2)^7)/(896*d) - (2*b^2*atan((4*b^4)/((5*a*b^3)/2 - 4*b^4*tan(c/2 + (d*x)/2)) + (5*a*b^3*t
an(c/2 + (d*x)/2))/(2*((5*a*b^3)/2 - 4*b^4*tan(c/2 + (d*x)/2)))))/d - (tan(c/2 + (d*x)/2)*((5*a^2)/128 - (11*b
^2)/16))/d - (tan(c/2 + (d*x)/2)^4*(3*a^2 - (28*b^2)/3) - tan(c/2 + (d*x)/2)^6*(5*a^2 - 88*b^2) + a^2/7 - tan(
c/2 + (d*x)/2)^2*(a^2 - (4*b^2)/5) - 6*a*b*tan(c/2 + (d*x)/2)^3 + 30*a*b*tan(c/2 + (d*x)/2)^5 + (2*a*b*tan(c/2
+ (d*x)/2))/3)/(128*d*tan(c/2 + (d*x)/2)^7) + (tan(c/2 + (d*x)/2)^3*((3*a^2)/128 - (7*b^2)/96))/d - (tan(c/2
+ (d*x)/2)^5*(a^2/128 - b^2/160))/d + (15*a*b*tan(c/2 + (d*x)/2)^2)/(64*d) - (3*a*b*tan(c/2 + (d*x)/2)^4)/(64*
d) + (a*b*tan(c/2 + (d*x)/2)^6)/(192*d) - (5*a*b*log(tan(c/2 + (d*x)/2)))/(8*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)**8*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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