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

Optimal. Leaf size=124 \[ -\frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a b^3 d}+\frac {x \left (2 a^2-3 b^2\right )}{2 b^3}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2895, 3057, 2660, 618, 204, 3770} \[ -\frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a b^3 d}+\frac {x \left (2 a^2-3 b^2\right )}{2 b^3}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2 - 3*b^2)*x)/(2*b^3) - (2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*b^3*d)
- ArcTanh[Cos[c + d*x]]/(a*d) + (a*Cos[c + d*x])/(b^2*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 2895

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(a*(n + 3)*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d*f*(m
+ n + 3)*(m + n + 4)), x] + (-Dist[1/(b^2*(m + n + 3)*(m + n + 4)), Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x
])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*
(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x] - Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e
+ f*x])^(m + 1))/(b*d^2*f*(m + n + 4)), x]) /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[
m, 0] || IntegersQ[2*m, 2*n]) &&  !m < -1 &&  !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[m + n + 4, 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 (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {a \cos (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\int \frac {\csc (c+d x) \left (-2 b^2-a b \sin (c+d x)-\left (2 a^2-3 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 b^2}\\ &=\frac {\left (2 a^2-3 b^2\right ) x}{2 b^3}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\int \csc (c+d x) \, dx}{a}-\frac {\left (a^2-b^2\right )^2 \int \frac {1}{a+b \sin (c+d x)} \, dx}{a b^3}\\ &=\frac {\left (2 a^2-3 b^2\right ) x}{2 b^3}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\left (2 \left (a^2-b^2\right )^2\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 b^3 d}\\ &=\frac {\left (2 a^2-3 b^2\right ) x}{2 b^3}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\left (4 \left (a^2-b^2\right )^2\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 b^3 d}\\ &=\frac {\left (2 a^2-3 b^2\right ) x}{2 b^3}-\frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a d}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 143, normalized size = 1.15 \[ -\frac {-4 a^3 c-4 a^3 d x+8 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )-4 a^2 b \cos (c+d x)+a b^2 \sin (2 (c+d x))+6 a b^2 c+6 a b^2 d x-4 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 a b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(-4*a^3*c + 6*a*b^2*c - 4*a^3*d*x + 6*a*b^2*d*x + 8*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqr
t[a^2 - b^2]] - 4*a^2*b*Cos[c + d*x] + 4*b^3*Log[Cos[(c + d*x)/2]] - 4*b^3*Log[Sin[(c + d*x)/2]] + a*b^2*Sin[2
*(c + d*x)])/(a*b^3*d)

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fricas [A]  time = 0.65, size = 350, normalized size = 2.82 \[ \left [-\frac {a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} b \cos \left (d x + c\right ) + b^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - b^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x - {\left (-a^{2} + b^{2}\right )}^{\frac {3}{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right )}{2 \, a b^{3} d}, -\frac {a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} b \cos \left (d x + c\right ) + b^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - b^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x - 2 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right )}{2 \, a b^{3} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(a*b^2*cos(d*x + c)*sin(d*x + c) - 2*a^2*b*cos(d*x + c) + b^3*log(1/2*cos(d*x + c) + 1/2) - b^3*log(-1/2
*cos(d*x + c) + 1/2) - (2*a^3 - 3*a*b^2)*d*x - (-a^2 + b^2)^(3/2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*s
in(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*cos(d*x + c)
^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)))/(a*b^3*d), -1/2*(a*b^2*cos(d*x + c)*sin(d*x + c) - 2*a^2*b*cos(d*x + c)
+ b^3*log(1/2*cos(d*x + c) + 1/2) - b^3*log(-1/2*cos(d*x + c) + 1/2) - (2*a^3 - 3*a*b^2)*d*x - 2*(a^2 - b^2)^
(3/2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))))/(a*b^3*d)]

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giac [A]  time = 0.65, size = 183, normalized size = 1.48 \[ \frac {\frac {2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {{\left (2 \, a^{2} - 3 \, b^{2}\right )} {\left (d x + c\right )}}{b^{3}} - \frac {4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a b^{3}} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*log(abs(tan(1/2*d*x + 1/2*c)))/a + (2*a^2 - 3*b^2)*(d*x + c)/b^3 - 4*(a^4 - 2*a^2*b^2 + b^4)*(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*b^3)
+ 2*(b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c) + 2*a)/((tan(1/2*d*x + 1/
2*c)^2 + 1)^2*b^2))/d

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maple [B]  time = 0.20, size = 334, normalized size = 2.69 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{3} \sqrt {a^{2}-b^{2}}}+\frac {4 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d b \sqrt {a^{2}-b^{2}}}-\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d a \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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^2-3/d/b*arctan(tan(1/2*d*x+1/2*c))+1/a/d*ln(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))+4/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*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))

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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)^3*cot(d*x+c)/(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 = 6.64, size = 1320, normalized size = 10.65 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/(a*d) - sin(2*c + 2*d*x)/(4*b*d) - (3*atan((2*a^3*cos(c/2 + (d*x)/2
) + 2*b^3*sin(c/2 + (d*x)/2) - 3*a*b^2*cos(c/2 + (d*x)/2))/(2*b^3*cos(c/2 + (d*x)/2) - 2*a^3*sin(c/2 + (d*x)/2
) + 3*a*b^2*sin(c/2 + (d*x)/2))))/(b*d) + (a*cos(c + d*x))/(b^2*d) + (2*a^2*atan((2*a^3*cos(c/2 + (d*x)/2) + 2
*b^3*sin(c/2 + (d*x)/2) - 3*a*b^2*cos(c/2 + (d*x)/2))/(2*b^3*cos(c/2 + (d*x)/2) - 2*a^3*sin(c/2 + (d*x)/2) + 3
*a*b^2*sin(c/2 + (d*x)/2))))/(b^3*d) + (atan((b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2)
*64i - a^12*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*16i - a^6*sin(c/2 + (d*x)/2)*(b^6 - a
^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2)*16i - a^3*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2)*4
2i + a^3*b^9*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*66i - a^5*b^7*cos(c/2 + (d*x)/2)*(b^
6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*176i + a^7*b^5*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(
1/2)*178i - a^9*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*81i - a^2*b^4*sin(c/2 + (d*x)
/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2)*116i + a^4*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4
*b^2)^(3/2)*72i + a^2*b^10*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*148i - a^4*b^8*sin(c/2
+ (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*460i + a^6*b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4
+ 3*a^4*b^2)^(1/2)*577i - a^8*b^4*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*368i + a^10*b^
2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*120i + a*b^5*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*
a^2*b^4 + 3*a^4*b^2)^(3/2)*32i + a^5*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2)*14i + a^11
*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*14i)/(64*b^15*sin(c/2 + (d*x)/2) + 32*a*b^14*c
os(c/2 + (d*x)/2) - 120*a^3*b^12*cos(c/2 + (d*x)/2) + 180*a^5*b^10*cos(c/2 + (d*x)/2) - 137*a^7*b^8*cos(c/2 +
(d*x)/2) + 54*a^9*b^6*cos(c/2 + (d*x)/2) - 9*a^11*b^4*cos(c/2 + (d*x)/2) - 256*a^2*b^13*sin(c/2 + (d*x)/2) + 4
16*a^4*b^11*sin(c/2 + (d*x)/2) - 351*a^6*b^9*sin(c/2 + (d*x)/2) + 161*a^8*b^7*sin(c/2 + (d*x)/2) - 37*a^10*b^5
*sin(c/2 + (d*x)/2) + 3*a^12*b^3*sin(c/2 + (d*x)/2)))*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*2i)/(a*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)**3*cot(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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