3.1306 \(\int \frac {\cos (c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=123 \[ \frac {b \cot (c+d x)}{a^2 d}-\frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 b d}+\frac {\left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {x}{b} \]

[Out]

x/b-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^3/b/d+1/2*(3*a^2-2*b^2)*arctanh(cos(d
*x+c))/a^3/d+b*cot(d*x+c)/a^2/d-1/2*cot(d*x+c)*csc(d*x+c)/a/d

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Rubi [A]  time = 0.30, antiderivative size = 123, 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 = {2893, 3057, 2660, 618, 204, 3770} \[ -\frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 b d}+\frac {\left (3 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac {b \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/b - (2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b*d) + ((3*a^2 - 2*b^2)*ArcT
anh[Cos[c + d*x]])/(2*a^3*d) + (b*Cot[c + d*x])/(a^2*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*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 2893

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

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Mathematica [A]  time = 1.69, size = 204, normalized size = 1.66 \[ \frac {8 a^3 c+8 a^3 d x-16 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )-a^2 b \csc ^2\left (\frac {1}{2} (c+d x)\right )+a^2 b \sec ^2\left (\frac {1}{2} (c+d x)\right )-12 a^2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 a b^2 \tan \left (\frac {1}{2} (c+d x)\right )+4 a b^2 \cot \left (\frac {1}{2} (c+d x)\right )+8 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 a^3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^3*c + 8*a^3*d*x - 16*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 4*a*b^2*Cot[(c
+ d*x)/2] - a^2*b*Csc[(c + d*x)/2]^2 + 12*a^2*b*Log[Cos[(c + d*x)/2]] - 8*b^3*Log[Cos[(c + d*x)/2]] - 12*a^2*b
*Log[Sin[(c + d*x)/2]] + 8*b^3*Log[Sin[(c + d*x)/2]] + a^2*b*Sec[(c + d*x)/2]^2 - 4*a*b^2*Tan[(c + d*x)/2])/(8
*a^3*b*d)

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fricas [B]  time = 0.96, size = 572, normalized size = 4.65 \[ \left [\frac {4 \, a^{3} d x \cos \left (d x + c\right )^{2} - 4 \, a^{3} d x - 4 \, 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 ) - 2 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{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 ) - {\left (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\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 (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{3} b d \cos \left (d x + c\right )^{2} - a^{3} b d\right )}}, \frac {4 \, a^{3} d x \cos \left (d x + c\right )^{2} - 4 \, a^{3} d x - 4 \, 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 ) + 4 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{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 ) - {\left (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\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 (3 \, a^{2} b - 2 \, b^{3} - {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{3} b d \cos \left (d x + c\right )^{2} - a^{3} b d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[1/4*(4*a^3*d*x*cos(d*x + c)^2 - 4*a^3*d*x - 4*a*b^2*cos(d*x + c)*sin(d*x + c) + 2*a^2*b*cos(d*x + c) - 2*((a^
2 - b^2)*cos(d*x + c)^2 - a^2 + b^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*cos(d*x + c)^2 - 2*a*b*s
in(d*x + c) - a^2 - b^2)) - (3*a^2*b - 2*b^3 - (3*a^2*b - 2*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) +
(3*a^2*b - 2*b^3 - (3*a^2*b - 2*b^3)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/(a^3*b*d*cos(d*x + c)^2 -
a^3*b*d), 1/4*(4*a^3*d*x*cos(d*x + c)^2 - 4*a^3*d*x - 4*a*b^2*cos(d*x + c)*sin(d*x + c) + 2*a^2*b*cos(d*x + c)
+ 4*((a^2 - b^2)*cos(d*x + c)^2 - a^2 + b^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*co
s(d*x + c))) - (3*a^2*b - 2*b^3 - (3*a^2*b - 2*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) + (3*a^2*b - 2
*b^3 - (3*a^2*b - 2*b^3)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/(a^3*b*d*cos(d*x + c)^2 - a^3*b*d)]

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giac [A]  time = 0.20, size = 217, normalized size = 1.76 \[ \frac {\frac {8 \, {\left (d x + c\right )}}{b} + \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {4 \, {\left (3 \, a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {16 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3} b} + \frac {18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/8*(8*(d*x + c)/b + (a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c))/a^2 - 4*(3*a^2 - 2*b^2)*log(abs(tan
(1/2*d*x + 1/2*c)))/a^3 - 16*(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^3*b) + (18*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(
1/2*d*x + 1/2*c)^2 + 4*a*b*tan(1/2*d*x + 1/2*c) - a^2)/(a^3*tan(1/2*d*x + 1/2*c)^2))/d

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maple [B]  time = 0.48, size = 286, normalized size = 2.33 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{2 d \,a^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b}-\frac {1}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{3}}+\frac {b}{2 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d b \sqrt {a^{2}-b^{2}}}+\frac {4 b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d a \sqrt {a^{2}-b^{2}}}-\frac {2 b^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{3} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*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+2/d/b*arctan(tan(1/2*d*x+1/2*c))-1/8/a/d/tan(1/2*d
*x+1/2*c)^2-3/2/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/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))+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/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)^4*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 = 12.68, size = 2718, normalized size = 22.10 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^3*(a + b*sin(c + d*x))),x)

[Out]

(b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*1i)/(a^3*b*d*1i - a^3*b*d*cos(2*c + 2*d*x)*1i) - (a^3*atan((2*
a^3*cos(c/2 + (d*x)/2) + 2*b^3*sin(c/2 + (d*x)/2) - 3*a^2*b*sin(c/2 + (d*x)/2))/(2*a^3*sin(c/2 + (d*x)/2) - 2*
b^3*cos(c/2 + (d*x)/2) + 3*a^2*b*cos(c/2 + (d*x)/2)))*2i)/(a^3*b*d*1i - a^3*b*d*cos(2*c + 2*d*x)*1i) + (2*atan
((32*b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 14*a^12*sin(c/2 + (d*x)/2)*(b^6 - a^6
- 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 14*a^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 36*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) + 2*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) + 8*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) - 24*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) - 82*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) + 63*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) + 2*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) - 11*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) + 56*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) - 106
*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) + 72*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) + 16*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) + 1
9*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) + 13*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))/(b^15*sin(c/2 + (d*x)/2)*32i + a*b^14*cos(c/2 + (d*x)/2)*16i - a^14*b*sin(c/2
+ (d*x)/2)*6i - a^3*b^12*cos(c/2 + (d*x)/2)*108i + a^5*b^10*cos(c/2 + (d*x)/2)*309i - a^7*b^8*cos(c/2 + (d*x)
/2)*469i + a^9*b^6*cos(c/2 + (d*x)/2)*390i - a^11*b^4*cos(c/2 + (d*x)/2)*165i + a^13*b^2*cos(c/2 + (d*x)/2)*27
i - a^2*b^13*sin(c/2 + (d*x)/2)*224i + a^4*b^11*sin(c/2 + (d*x)/2)*670i - a^6*b^9*sin(c/2 + (d*x)/2)*1080i + a
^8*b^7*sin(c/2 + (d*x)/2)*982i - a^10*b^5*sin(c/2 + (d*x)/2)*482i + a^12*b^3*sin(c/2 + (d*x)/2)*108i))*(b^6 -
a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(a^3*b*d*1i - a^3*b*d*cos(2*c + 2*d*x)*1i) - (a^2*b*cos(c + d*x)*1i)/(a^3*
b*d*1i - a^3*b*d*cos(2*c + 2*d*x)*1i) + (a^3*atan((2*a^3*cos(c/2 + (d*x)/2) + 2*b^3*sin(c/2 + (d*x)/2) - 3*a^2
*b*sin(c/2 + (d*x)/2))/(2*a^3*sin(c/2 + (d*x)/2) - 2*b^3*cos(c/2 + (d*x)/2) + 3*a^2*b*cos(c/2 + (d*x)/2)))*cos
(2*c + 2*d*x)*2i)/(a^3*b*d*1i - a^3*b*d*cos(2*c + 2*d*x)*1i) - (a^2*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2
))*3i)/(2*(a^3*b*d*1i - a^3*b*d*cos(2*c + 2*d*x)*1i)) + (a*b^2*sin(2*c + 2*d*x)*1i)/(a^3*b*d*1i - a^3*b*d*cos(
2*c + 2*d*x)*1i) - (2*cos(2*c + 2*d*x)*atan((32*b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/
2) - 14*a^12*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 14*a^6*sin(c/2 + (d*x)/2)*(b^6 - a
^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 36*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) +
2*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) + 8*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) - 24*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) -
82*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) + 63*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) + 2*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)
- 11*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) + 56*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) - 106*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) + 72*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) + 16*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) + 19*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
) + 13*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))/(b^15*sin(c/2 + (d*x)/2)*32i + a*b
^14*cos(c/2 + (d*x)/2)*16i - a^14*b*sin(c/2 + (d*x)/2)*6i - a^3*b^12*cos(c/2 + (d*x)/2)*108i + a^5*b^10*cos(c/
2 + (d*x)/2)*309i - a^7*b^8*cos(c/2 + (d*x)/2)*469i + a^9*b^6*cos(c/2 + (d*x)/2)*390i - a^11*b^4*cos(c/2 + (d*
x)/2)*165i + a^13*b^2*cos(c/2 + (d*x)/2)*27i - a^2*b^13*sin(c/2 + (d*x)/2)*224i + a^4*b^11*sin(c/2 + (d*x)/2)*
670i - a^6*b^9*sin(c/2 + (d*x)/2)*1080i + a^8*b^7*sin(c/2 + (d*x)/2)*982i - a^10*b^5*sin(c/2 + (d*x)/2)*482i +
a^12*b^3*sin(c/2 + (d*x)/2)*108i))*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(a^3*b*d*1i - a^3*b*d*cos(2*c +
2*d*x)*1i) - (b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*1i)/(a^3*b*d*1i - a^3*b*d*cos(2
*c + 2*d*x)*1i) + (a^2*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*3i)/(2*(a^3*b*d*1i - a^3*
b*d*cos(2*c + 2*d*x)*1i))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{4}{\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Integral(cos(c + d*x)**4*csc(c + d*x)**3/(a + b*sin(c + d*x)), x)

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