3.1325 \(\int \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=174 \[ -\frac {x \left (2 a^2-5 b^2\right )}{2 b^3}+\frac {b \cot (c+d x)}{a^2 d}+\frac {\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac {2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 b^3 d}-\frac {a \cos (c+d x)}{b^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\sin (c+d x) \cos (c+d x)}{2 b d} \]

[Out]

-1/2*(2*a^2-5*b^2)*x/b^3+2*(a^2-b^2)^(5/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^3/b^3/d+1/2*(5*a
^2-2*b^2)*arctanh(cos(d*x+c))/a^3/d-a*cos(d*x+c)/b^2/d+b*cot(d*x+c)/a^2/d-1/2*cot(d*x+c)*csc(d*x+c)/a/d+1/2*co
s(d*x+c)*sin(d*x+c)/b/d

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Rubi [A]  time = 0.38, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2896, 3057, 2660, 618, 204, 3770} \[ \frac {2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 b^3 d}+\frac {\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {x \left (2 a^2-5 b^2\right )}{2 b^3}+\frac {b \cot (c+d x)}{a^2 d}-\frac {a \cos (c+d x)}{b^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\sin (c+d x) \cos (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a^2 - 5*b^2)*x)/(2*b^3) + (2*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^3
*d) + ((5*a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(2*a^3*d) - (a*Cos[c + d*x])/(b^2*d) + (b*Cot[c + d*x])/(a^2*d)
- (Cot[c + d*x]*Csc[c + d*x])/(2*a*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*b*d)

Rule 204

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2896

Int[cos[(e_.) + (f_.)*(x_)]^6*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(a*d*f*(n + 1)), x] +
(Dist[1/(a^2*b^2*d^2*(n + 1)*(n + 2)*(m + n + 5)*(m + n + 6)), Int[(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e + f*
x])^m*Simp[a^4*(n + 1)*(n + 2)*(n + 3)*(n + 5) - a^2*b^2*(n + 2)*(2*n + 1)*(m + n + 5)*(m + n + 6) + b^4*(m +
n + 2)*(m + n + 3)*(m + n + 5)*(m + n + 6) + a*b*m*(a^2*(n + 1)*(n + 2) - b^2*(m + n + 5)*(m + n + 6))*Sin[e +
f*x] - (a^4*(n + 1)*(n + 2)*(4 + n)*(n + 5) + b^4*(m + n + 2)*(m + n + 4)*(m + n + 5)*(m + n + 6) - a^2*b^2*(
n + 1)*(n + 2)*(m + n + 5)*(2*n + 2*m + 13))*Sin[e + f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(d*
Sin[e + f*x])^(n + 2)*(a + b*Sin[e + f*x])^(m + 1))/(a^2*d^2*f*(n + 1)*(n + 2)), x] - Simp[(a*(n + 5)*Cos[e +
f*x]*(d*Sin[e + f*x])^(n + 3)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d^3*f*(m + n + 5)*(m + n + 6)), x] + Simp[(Co
s[e + f*x]*(d*Sin[e + f*x])^(n + 4)*(a + b*Sin[e + f*x])^(m + 1))/(b*d^4*f*(m + n + 6)), x]) /; FreeQ[{a, b, d
, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*m, 2*n] && NeQ[n, -1] && NeQ[n, -2] && NeQ[m + n + 5, 0]
&& NeQ[m + n + 6, 0] &&  !IGtQ[m, 0]

Rule 3057

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(C*x)/(b*d), x] + (Dist[(A*b^2 - a*b*B + a
^2*C)/(b*(b*c - a*d)), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)), Int[
1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^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 \frac {\cos ^3(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac {a \cos (c+d x)}{b^2 d}+\frac {b \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\int \frac {\csc (c+d x) \left (-2 b^2 \left (5 a^2-2 b^2\right )-2 a b \left (a^2-b^2\right ) \sin (c+d x)-2 a^2 \left (2 a^2-5 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 a^2 b^2}\\ &=-\frac {\left (2 a^2-5 b^2\right ) x}{2 b^3}-\frac {a \cos (c+d x)}{b^2 d}+\frac {b \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\left (5 a^2-2 b^2\right ) \int \csc (c+d x) \, dx}{2 a^3}+\frac {\left (a^2-b^2\right )^3 \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^3 b^3}\\ &=-\frac {\left (2 a^2-5 b^2\right ) x}{2 b^3}+\frac {\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {a \cos (c+d x)}{b^2 d}+\frac {b \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\left (2 \left (a^2-b^2\right )^3\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 b^3 d}\\ &=-\frac {\left (2 a^2-5 b^2\right ) x}{2 b^3}+\frac {\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {a \cos (c+d x)}{b^2 d}+\frac {b \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\left (4 \left (a^2-b^2\right )^3\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 b^3 d}\\ &=-\frac {\left (2 a^2-5 b^2\right ) x}{2 b^3}+\frac {2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b^3 d}+\frac {\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {a \cos (c+d x)}{b^2 d}+\frac {b \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\cos (c+d x) \sin (c+d x)}{2 b d}\\ \end {align*}

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Mathematica [A]  time = 5.26, size = 259, normalized size = 1.49 \[ \frac {-8 a^5 c-8 a^5 d x-8 a^4 b \cos (c+d x)+2 a^3 b^2 \sin (2 (c+d x))+20 a^3 b^2 c+20 a^3 b^2 d x-a^2 b^3 \csc ^2\left (\frac {1}{2} (c+d x)\right )+a^2 b^3 \sec ^2\left (\frac {1}{2} (c+d x)\right )-20 a^2 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+20 a^2 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+16 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )-4 a b^4 \tan \left (\frac {1}{2} (c+d x)\right )+4 a b^4 \cot \left (\frac {1}{2} (c+d x)\right )+8 b^5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 b^5 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 a^3 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^5*c + 20*a^3*b^2*c - 8*a^5*d*x + 20*a^3*b^2*d*x + 16*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/S
qrt[a^2 - b^2]] - 8*a^4*b*Cos[c + d*x] + 4*a*b^4*Cot[(c + d*x)/2] - a^2*b^3*Csc[(c + d*x)/2]^2 + 20*a^2*b^3*Lo
g[Cos[(c + d*x)/2]] - 8*b^5*Log[Cos[(c + d*x)/2]] - 20*a^2*b^3*Log[Sin[(c + d*x)/2]] + 8*b^5*Log[Sin[(c + d*x)
/2]] + a^2*b^3*Sec[(c + d*x)/2]^2 + 2*a^3*b^2*Sin[2*(c + d*x)] - 4*a*b^4*Tan[(c + d*x)/2])/(8*a^3*b^3*d)

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fricas [B]  time = 1.82, size = 770, normalized size = 4.43 \[ \left [-\frac {4 \, a^{4} b \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, a^{5} - 5 \, a^{3} b^{2}\right )} d x \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} - 5 \, a^{3} b^{2}\right )} d x + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (2 \, a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right ) + {\left (5 \, a^{2} b^{3} - 2 \, b^{5} - {\left (5 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (5 \, a^{2} b^{3} - 2 \, b^{5} - {\left (5 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (a^{3} b^{2} \cos \left (d x + c\right )^{3} - {\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{3} b^{3} d \cos \left (d x + c\right )^{2} - a^{3} b^{3} d\right )}}, -\frac {4 \, a^{4} b \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, a^{5} - 5 \, a^{3} b^{2}\right )} d x \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} - 5 \, a^{3} b^{2}\right )} d x - 4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 2 \, {\left (2 \, a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right ) + {\left (5 \, a^{2} b^{3} - 2 \, b^{5} - {\left (5 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (5 \, a^{2} b^{3} - 2 \, b^{5} - {\left (5 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (a^{3} b^{2} \cos \left (d x + c\right )^{3} - {\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{3} b^{3} d \cos \left (d x + c\right )^{2} - a^{3} b^{3} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*a^4*b*cos(d*x + c)^3 + 2*(2*a^5 - 5*a^3*b^2)*d*x*cos(d*x + c)^2 - 2*(2*a^5 - 5*a^3*b^2)*d*x + 2*(a^4
- 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^
2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*c
os(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 2*(2*a^4*b + a^2*b^3)*cos(d*x + c) + (5*a^2*b^3 - 2*b^5 - (
5*a^2*b^3 - 2*b^5)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) - (5*a^2*b^3 - 2*b^5 - (5*a^2*b^3 - 2*b^5)*cos(
d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^3*b^2*cos(d*x + c)^3 - (a^3*b^2 + 2*a*b^4)*cos(d*x + c))*sin(d
*x + c))/(a^3*b^3*d*cos(d*x + c)^2 - a^3*b^3*d), -1/4*(4*a^4*b*cos(d*x + c)^3 + 2*(2*a^5 - 5*a^3*b^2)*d*x*cos(
d*x + c)^2 - 2*(2*a^5 - 5*a^3*b^2)*d*x - 4*(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2)*sq
rt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 2*(2*a^4*b + a^2*b^3)*cos(d*x + c
) + (5*a^2*b^3 - 2*b^5 - (5*a^2*b^3 - 2*b^5)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) - (5*a^2*b^3 - 2*b^5
- (5*a^2*b^3 - 2*b^5)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^3*b^2*cos(d*x + c)^3 - (a^3*b^2 + 2*
a*b^4)*cos(d*x + c))*sin(d*x + c))/(a^3*b^3*d*cos(d*x + c)^2 - a^3*b^3*d)]

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giac [B]  time = 0.21, size = 431, normalized size = 2.48 \[ \frac {\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {4 \, {\left (2 \, a^{2} - 5 \, b^{2}\right )} {\left (d x + c\right )}}{b^{3}} - \frac {4 \, {\left (5 \, a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3} b^{3}} + \frac {10 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 8 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 19 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 8 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2} b^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2} a^{3} b^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c))/a^2 - 4*(2*a^2 - 5*b^2)*(d*x + c)/b^3 - 4*(5*a^2 -
2*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 16*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(pi*floor(1/2*(d*x + c)/pi
+ 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^3*b^3) + (10*a^2*b^2*
tan(1/2*d*x + 1/2*c)^6 - 4*b^4*tan(1/2*d*x + 1/2*c)^6 - 8*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 4*a*b^3*tan(1/2*d*x +
1/2*c)^5 - 16*a^4*tan(1/2*d*x + 1/2*c)^4 + 19*a^2*b^2*tan(1/2*d*x + 1/2*c)^4 - 8*b^4*tan(1/2*d*x + 1/2*c)^4 +
8*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 8*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 16*a^4*tan(1/2*d*x + 1/2*c)^2 + 8*a^2*b^2*t
an(1/2*d*x + 1/2*c)^2 - 4*b^4*tan(1/2*d*x + 1/2*c)^2 + 4*a*b^3*tan(1/2*d*x + 1/2*c) - a^2*b^2)/((tan(1/2*d*x +
1/2*c)^3 + tan(1/2*d*x + 1/2*c))^2*a^3*b^2))/d

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maple [B]  time = 0.48, size = 483, normalized size = 2.78 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{2 d \,a^{2}}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3}}+\frac {5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b}-\frac {1}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{3}}+\frac {b}{2 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{3} \sqrt {a^{2}-b^{2}}}-\frac {6 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d b \sqrt {a^{2}-b^{2}}}+\frac {6 b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d a \sqrt {a^{2}-b^{2}}}-\frac {2 b^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{3} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/a/d*tan(1/2*d*x+1/2*c)^2-1/2/d/a^2*tan(1/2*d*x+1/2*c)*b-1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c
)^3-2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^2*a+1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2
*c)-2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*a-2/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a^2+5/d/b*arctan(tan(1/2*d*x+1/2*c
))-1/8/a/d/tan(1/2*d*x+1/2*c)^2-5/2/a/d*ln(tan(1/2*d*x+1/2*c))+1/d/a^3*ln(tan(1/2*d*x+1/2*c))*b^2+1/2/d/a^2*b/
tan(1/2*d*x+1/2*c)+2/d*a^3/b^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-6/d*a/
b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+6/d*b/a/(a^2-b^2)^(1/2)*arctan(1/2*
(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-2/d/a^3*b^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2
*b)/(a^2-b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 13.09, size = 4898, normalized size = 28.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^2/(8*a*d) - (a/2 - 2*b*tan(c/2 + (d*x)/2) - (4*tan(c/2 + (d*x)/2)^3*(a^2 + b^2))/b + (tan(c
/2 + (d*x)/2)^2*(a*b^2 + 8*a^3))/b^2 + (tan(c/2 + (d*x)/2)^4*(a*b^2 + 16*a^3))/(2*b^2) + (2*tan(c/2 + (d*x)/2)
^5*(2*a^2 - b^2))/b)/(d*(4*a^2*tan(c/2 + (d*x)/2)^2 + 8*a^2*tan(c/2 + (d*x)/2)^4 + 4*a^2*tan(c/2 + (d*x)/2)^6)
) - (b*tan(c/2 + (d*x)/2))/(2*a^2*d) + (atan((((a^2*1i - (b^2*5i)/2)*((4*(20*a^3*b^12 - 28*a^15 + 272*a^5*b^10
- 1272*a^7*b^8 + 1709*a^9*b^6 - 998*a^11*b^4 + 270*a^13*b^2))/(a^6*b^5) + ((a^2*1i - (b^2*5i)/2)*(((a^2*1i -
(b^2*5i)/2)*((4*(64*a^5*b^12 - 208*a^7*b^10 + 160*a^9*b^8 - 28*a^11*b^6))/(a^6*b^5) + (((4*(32*a^8*b^10 - 24*a
^10*b^8))/(a^6*b^5) + (4*tan(c/2 + (d*x)/2)*(128*a^7*b^14 - 136*a^9*b^12 + 16*a^11*b^10))/(a^6*b^8))*(a^2*1i -
(b^2*5i)/2))/b^3 + (4*tan(c/2 + (d*x)/2)*(128*a^4*b^16 - 456*a^6*b^14 + 484*a^8*b^12 - 200*a^10*b^10 + 32*a^1
2*b^8))/(a^6*b^8)))/b^3 - (4*(184*a^4*b^12 - 32*a^2*b^14 - 360*a^6*b^10 + 78*a^8*b^8 + 240*a^10*b^6 - 152*a^12
*b^4 + 24*a^14*b^2))/(a^6*b^5) + (4*tan(c/2 + (d*x)/2)*(8*a^3*b^16 - 160*a^5*b^14 + 1320*a^7*b^12 - 2128*a^9*b
^10 + 1242*a^11*b^8 - 280*a^13*b^6 + 16*a^15*b^4))/(a^6*b^8)))/b^3 + (4*tan(c/2 + (d*x)/2)*(8*b^18 - 68*a^2*b^
16 + 1040*a^4*b^14 - 3900*a^6*b^12 + 5738*a^8*b^10 - 4201*a^10*b^8 + 1684*a^12*b^6 - 360*a^14*b^4 + 32*a^16*b^
2))/(a^6*b^8))*1i)/b^3 + ((a^2*1i - (b^2*5i)/2)*((4*(20*a^3*b^12 - 28*a^15 + 272*a^5*b^10 - 1272*a^7*b^8 + 170
9*a^9*b^6 - 998*a^11*b^4 + 270*a^13*b^2))/(a^6*b^5) + ((a^2*1i - (b^2*5i)/2)*(((a^2*1i - (b^2*5i)/2)*((4*(64*a
^5*b^12 - 208*a^7*b^10 + 160*a^9*b^8 - 28*a^11*b^6))/(a^6*b^5) - (((4*(32*a^8*b^10 - 24*a^10*b^8))/(a^6*b^5) +
(4*tan(c/2 + (d*x)/2)*(128*a^7*b^14 - 136*a^9*b^12 + 16*a^11*b^10))/(a^6*b^8))*(a^2*1i - (b^2*5i)/2))/b^3 + (
4*tan(c/2 + (d*x)/2)*(128*a^4*b^16 - 456*a^6*b^14 + 484*a^8*b^12 - 200*a^10*b^10 + 32*a^12*b^8))/(a^6*b^8)))/b
^3 + (4*(184*a^4*b^12 - 32*a^2*b^14 - 360*a^6*b^10 + 78*a^8*b^8 + 240*a^10*b^6 - 152*a^12*b^4 + 24*a^14*b^2))/
(a^6*b^5) - (4*tan(c/2 + (d*x)/2)*(8*a^3*b^16 - 160*a^5*b^14 + 1320*a^7*b^12 - 2128*a^9*b^10 + 1242*a^11*b^8 -
280*a^13*b^6 + 16*a^15*b^4))/(a^6*b^8)))/b^3 + (4*tan(c/2 + (d*x)/2)*(8*b^18 - 68*a^2*b^16 + 1040*a^4*b^14 -
3900*a^6*b^12 + 5738*a^8*b^10 - 4201*a^10*b^8 + 1684*a^12*b^6 - 360*a^14*b^4 + 32*a^16*b^2))/(a^6*b^8))*1i)/b^
3)/(((a^2*1i - (b^2*5i)/2)*((4*(20*a^3*b^12 - 28*a^15 + 272*a^5*b^10 - 1272*a^7*b^8 + 1709*a^9*b^6 - 998*a^11*
b^4 + 270*a^13*b^2))/(a^6*b^5) + ((a^2*1i - (b^2*5i)/2)*(((a^2*1i - (b^2*5i)/2)*((4*(64*a^5*b^12 - 208*a^7*b^1
0 + 160*a^9*b^8 - 28*a^11*b^6))/(a^6*b^5) + (((4*(32*a^8*b^10 - 24*a^10*b^8))/(a^6*b^5) + (4*tan(c/2 + (d*x)/2
)*(128*a^7*b^14 - 136*a^9*b^12 + 16*a^11*b^10))/(a^6*b^8))*(a^2*1i - (b^2*5i)/2))/b^3 + (4*tan(c/2 + (d*x)/2)*
(128*a^4*b^16 - 456*a^6*b^14 + 484*a^8*b^12 - 200*a^10*b^10 + 32*a^12*b^8))/(a^6*b^8)))/b^3 - (4*(184*a^4*b^12
- 32*a^2*b^14 - 360*a^6*b^10 + 78*a^8*b^8 + 240*a^10*b^6 - 152*a^12*b^4 + 24*a^14*b^2))/(a^6*b^5) + (4*tan(c/
2 + (d*x)/2)*(8*a^3*b^16 - 160*a^5*b^14 + 1320*a^7*b^12 - 2128*a^9*b^10 + 1242*a^11*b^8 - 280*a^13*b^6 + 16*a^
15*b^4))/(a^6*b^8)))/b^3 + (4*tan(c/2 + (d*x)/2)*(8*b^18 - 68*a^2*b^16 + 1040*a^4*b^14 - 3900*a^6*b^12 + 5738*
a^8*b^10 - 4201*a^10*b^8 + 1684*a^12*b^6 - 360*a^14*b^4 + 32*a^16*b^2))/(a^6*b^8)))/b^3 - ((a^2*1i - (b^2*5i)/
2)*((4*(20*a^3*b^12 - 28*a^15 + 272*a^5*b^10 - 1272*a^7*b^8 + 1709*a^9*b^6 - 998*a^11*b^4 + 270*a^13*b^2))/(a^
6*b^5) + ((a^2*1i - (b^2*5i)/2)*(((a^2*1i - (b^2*5i)/2)*((4*(64*a^5*b^12 - 208*a^7*b^10 + 160*a^9*b^8 - 28*a^1
1*b^6))/(a^6*b^5) - (((4*(32*a^8*b^10 - 24*a^10*b^8))/(a^6*b^5) + (4*tan(c/2 + (d*x)/2)*(128*a^7*b^14 - 136*a^
9*b^12 + 16*a^11*b^10))/(a^6*b^8))*(a^2*1i - (b^2*5i)/2))/b^3 + (4*tan(c/2 + (d*x)/2)*(128*a^4*b^16 - 456*a^6*
b^14 + 484*a^8*b^12 - 200*a^10*b^10 + 32*a^12*b^8))/(a^6*b^8)))/b^3 + (4*(184*a^4*b^12 - 32*a^2*b^14 - 360*a^6
*b^10 + 78*a^8*b^8 + 240*a^10*b^6 - 152*a^12*b^4 + 24*a^14*b^2))/(a^6*b^5) - (4*tan(c/2 + (d*x)/2)*(8*a^3*b^16
- 160*a^5*b^14 + 1320*a^7*b^12 - 2128*a^9*b^10 + 1242*a^11*b^8 - 280*a^13*b^6 + 16*a^15*b^4))/(a^6*b^8)))/b^3
+ (4*tan(c/2 + (d*x)/2)*(8*b^18 - 68*a^2*b^16 + 1040*a^4*b^14 - 3900*a^6*b^12 + 5738*a^8*b^10 - 4201*a^10*b^8
+ 1684*a^12*b^6 - 360*a^14*b^4 + 32*a^16*b^2))/(a^6*b^8)))/b^3 + (8*(70*a^14 + 20*b^14 + 22*a^2*b^12 - 642*a^
4*b^10 + 1937*a^6*b^8 - 2549*a^8*b^6 + 1695*a^10*b^4 - 553*a^12*b^2))/(a^6*b^5) - (8*tan(c/2 + (d*x)/2)*(64*a^
17 + 700*a^5*b^12 - 3060*a^7*b^10 + 5412*a^9*b^8 - 4940*a^11*b^6 + 2448*a^13*b^4 - 624*a^15*b^2))/(a^6*b^8)))*
(a^2*1i - (b^2*5i)/2)*2i)/(b^3*d) - (log(tan(c/2 + (d*x)/2))*((5*a^2)/2 - b^2))/(a^3*d) + (atan((((-(a + b)^5*
(a - b)^5)^(1/2)*((4*(20*a^3*b^12 - 28*a^15 + 272*a^5*b^10 - 1272*a^7*b^8 + 1709*a^9*b^6 - 998*a^11*b^4 + 270*
a^13*b^2))/(a^6*b^5) + (4*tan(c/2 + (d*x)/2)*(8*b^18 - 68*a^2*b^16 + 1040*a^4*b^14 - 3900*a^6*b^12 + 5738*a^8*
b^10 - 4201*a^10*b^8 + 1684*a^12*b^6 - 360*a^14*b^4 + 32*a^16*b^2))/(a^6*b^8) + ((-(a + b)^5*(a - b)^5)^(1/2)*
((4*tan(c/2 + (d*x)/2)*(8*a^3*b^16 - 160*a^5*b^14 + 1320*a^7*b^12 - 2128*a^9*b^10 + 1242*a^11*b^8 - 280*a^13*b
^6 + 16*a^15*b^4))/(a^6*b^8) - (4*(184*a^4*b^12 - 32*a^2*b^14 - 360*a^6*b^10 + 78*a^8*b^8 + 240*a^10*b^6 - 152
*a^12*b^4 + 24*a^14*b^2))/(a^6*b^5) + ((-(a + b)^5*(a - b)^5)^(1/2)*((4*(64*a^5*b^12 - 208*a^7*b^10 + 160*a^9*
b^8 - 28*a^11*b^6))/(a^6*b^5) + (((4*(32*a^8*b^10 - 24*a^10*b^8))/(a^6*b^5) + (4*tan(c/2 + (d*x)/2)*(128*a^7*b
^14 - 136*a^9*b^12 + 16*a^11*b^10))/(a^6*b^8))*(-(a + b)^5*(a - b)^5)^(1/2))/(a^3*b^3) + (4*tan(c/2 + (d*x)/2)
*(128*a^4*b^16 - 456*a^6*b^14 + 484*a^8*b^12 - 200*a^10*b^10 + 32*a^12*b^8))/(a^6*b^8)))/(a^3*b^3)))/(a^3*b^3)
)*1i)/(a^3*b^3) + ((-(a + b)^5*(a - b)^5)^(1/2)*((4*(20*a^3*b^12 - 28*a^15 + 272*a^5*b^10 - 1272*a^7*b^8 + 170
9*a^9*b^6 - 998*a^11*b^4 + 270*a^13*b^2))/(a^6*b^5) + (4*tan(c/2 + (d*x)/2)*(8*b^18 - 68*a^2*b^16 + 1040*a^4*b
^14 - 3900*a^6*b^12 + 5738*a^8*b^10 - 4201*a^10*b^8 + 1684*a^12*b^6 - 360*a^14*b^4 + 32*a^16*b^2))/(a^6*b^8) +
((-(a + b)^5*(a - b)^5)^(1/2)*((4*(184*a^4*b^12 - 32*a^2*b^14 - 360*a^6*b^10 + 78*a^8*b^8 + 240*a^10*b^6 - 15
2*a^12*b^4 + 24*a^14*b^2))/(a^6*b^5) - (4*tan(c/2 + (d*x)/2)*(8*a^3*b^16 - 160*a^5*b^14 + 1320*a^7*b^12 - 2128
*a^9*b^10 + 1242*a^11*b^8 - 280*a^13*b^6 + 16*a^15*b^4))/(a^6*b^8) + ((-(a + b)^5*(a - b)^5)^(1/2)*((4*(64*a^5
*b^12 - 208*a^7*b^10 + 160*a^9*b^8 - 28*a^11*b^6))/(a^6*b^5) - (((4*(32*a^8*b^10 - 24*a^10*b^8))/(a^6*b^5) + (
4*tan(c/2 + (d*x)/2)*(128*a^7*b^14 - 136*a^9*b^12 + 16*a^11*b^10))/(a^6*b^8))*(-(a + b)^5*(a - b)^5)^(1/2))/(a
^3*b^3) + (4*tan(c/2 + (d*x)/2)*(128*a^4*b^16 - 456*a^6*b^14 + 484*a^8*b^12 - 200*a^10*b^10 + 32*a^12*b^8))/(a
^6*b^8)))/(a^3*b^3)))/(a^3*b^3))*1i)/(a^3*b^3))/((8*(70*a^14 + 20*b^14 + 22*a^2*b^12 - 642*a^4*b^10 + 1937*a^6
*b^8 - 2549*a^8*b^6 + 1695*a^10*b^4 - 553*a^12*b^2))/(a^6*b^5) + ((-(a + b)^5*(a - b)^5)^(1/2)*((4*(20*a^3*b^1
2 - 28*a^15 + 272*a^5*b^10 - 1272*a^7*b^8 + 1709*a^9*b^6 - 998*a^11*b^4 + 270*a^13*b^2))/(a^6*b^5) + (4*tan(c/
2 + (d*x)/2)*(8*b^18 - 68*a^2*b^16 + 1040*a^4*b^14 - 3900*a^6*b^12 + 5738*a^8*b^10 - 4201*a^10*b^8 + 1684*a^12
*b^6 - 360*a^14*b^4 + 32*a^16*b^2))/(a^6*b^8) + ((-(a + b)^5*(a - b)^5)^(1/2)*((4*tan(c/2 + (d*x)/2)*(8*a^3*b^
16 - 160*a^5*b^14 + 1320*a^7*b^12 - 2128*a^9*b^10 + 1242*a^11*b^8 - 280*a^13*b^6 + 16*a^15*b^4))/(a^6*b^8) - (
4*(184*a^4*b^12 - 32*a^2*b^14 - 360*a^6*b^10 + 78*a^8*b^8 + 240*a^10*b^6 - 152*a^12*b^4 + 24*a^14*b^2))/(a^6*b
^5) + ((-(a + b)^5*(a - b)^5)^(1/2)*((4*(64*a^5*b^12 - 208*a^7*b^10 + 160*a^9*b^8 - 28*a^11*b^6))/(a^6*b^5) +
(((4*(32*a^8*b^10 - 24*a^10*b^8))/(a^6*b^5) + (4*tan(c/2 + (d*x)/2)*(128*a^7*b^14 - 136*a^9*b^12 + 16*a^11*b^1
0))/(a^6*b^8))*(-(a + b)^5*(a - b)^5)^(1/2))/(a^3*b^3) + (4*tan(c/2 + (d*x)/2)*(128*a^4*b^16 - 456*a^6*b^14 +
484*a^8*b^12 - 200*a^10*b^10 + 32*a^12*b^8))/(a^6*b^8)))/(a^3*b^3)))/(a^3*b^3)))/(a^3*b^3) - ((-(a + b)^5*(a -
b)^5)^(1/2)*((4*(20*a^3*b^12 - 28*a^15 + 272*a^5*b^10 - 1272*a^7*b^8 + 1709*a^9*b^6 - 998*a^11*b^4 + 270*a^13
*b^2))/(a^6*b^5) + (4*tan(c/2 + (d*x)/2)*(8*b^18 - 68*a^2*b^16 + 1040*a^4*b^14 - 3900*a^6*b^12 + 5738*a^8*b^10
- 4201*a^10*b^8 + 1684*a^12*b^6 - 360*a^14*b^4 + 32*a^16*b^2))/(a^6*b^8) + ((-(a + b)^5*(a - b)^5)^(1/2)*((4*
(184*a^4*b^12 - 32*a^2*b^14 - 360*a^6*b^10 + 78*a^8*b^8 + 240*a^10*b^6 - 152*a^12*b^4 + 24*a^14*b^2))/(a^6*b^5
) - (4*tan(c/2 + (d*x)/2)*(8*a^3*b^16 - 160*a^5*b^14 + 1320*a^7*b^12 - 2128*a^9*b^10 + 1242*a^11*b^8 - 280*a^1
3*b^6 + 16*a^15*b^4))/(a^6*b^8) + ((-(a + b)^5*(a - b)^5)^(1/2)*((4*(64*a^5*b^12 - 208*a^7*b^10 + 160*a^9*b^8
- 28*a^11*b^6))/(a^6*b^5) - (((4*(32*a^8*b^10 - 24*a^10*b^8))/(a^6*b^5) + (4*tan(c/2 + (d*x)/2)*(128*a^7*b^14
- 136*a^9*b^12 + 16*a^11*b^10))/(a^6*b^8))*(-(a + b)^5*(a - b)^5)^(1/2))/(a^3*b^3) + (4*tan(c/2 + (d*x)/2)*(12
8*a^4*b^16 - 456*a^6*b^14 + 484*a^8*b^12 - 200*a^10*b^10 + 32*a^12*b^8))/(a^6*b^8)))/(a^3*b^3)))/(a^3*b^3)))/(
a^3*b^3) - (8*tan(c/2 + (d*x)/2)*(64*a^17 + 700*a^5*b^12 - 3060*a^7*b^10 + 5412*a^9*b^8 - 4940*a^11*b^6 + 2448
*a^13*b^4 - 624*a^15*b^2))/(a^6*b^8)))*(-(a + b)^5*(a - b)^5)^(1/2)*2i)/(a^3*b^3*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)**3/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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