3.1293 \(\int \frac {\cot ^2(c+d x) \csc ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=194 \[ \frac {b \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}+\frac {2 b^3 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^5 d}-\frac {b \left (a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac {\left (a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {\left (a^4+4 a^2 b^2-8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^5 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d} \]

[Out]

1/8*(a^4+4*a^2*b^2-8*b^4)*arctanh(cos(d*x+c))/a^5/d-1/3*b*(a^2-3*b^2)*cot(d*x+c)/a^4/d+1/8*(a^2-4*b^2)*cot(d*x
+c)*csc(d*x+c)/a^3/d+1/3*b*cot(d*x+c)*csc(d*x+c)^2/a^2/d-1/4*cot(d*x+c)*csc(d*x+c)^3/a/d+2*b^3*arctan((b+a*tan
(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))*(a^2-b^2)^(1/2)/a^5/d

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Rubi [A]  time = 0.95, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2889, 3056, 3055, 3001, 3770, 2660, 618, 204} \[ \frac {2 b^3 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^5 d}-\frac {b \left (a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac {\left (4 a^2 b^2+a^4-8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^5 d}+\frac {\left (a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {b \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^2*Csc[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

(2*b^3*Sqrt[a^2 - b^2]*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^5*d) + ((a^4 + 4*a^2*b^2 - 8*b^4)*
ArcTanh[Cos[c + d*x]])/(8*a^5*d) - (b*(a^2 - 3*b^2)*Cot[c + d*x])/(3*a^4*d) + ((a^2 - 4*b^2)*Cot[c + d*x]*Csc[
c + d*x])/(8*a^3*d) + (b*Cot[c + d*x]*Csc[c + d*x]^2)/(3*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*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 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 {\cot ^2(c+d x) \csc ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\int \frac {\csc ^5(c+d x) \left (1-\sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx\\ &=-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {\int \frac {\csc ^4(c+d x) \left (-4 b-a \sin (c+d x)+3 b \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 a}\\ &=\frac {b \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {\int \frac {\csc ^3(c+d x) \left (-3 \left (a^2-4 b^2\right )+a b \sin (c+d x)-8 b^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{12 a^2}\\ &=\frac {\left (a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {b \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {\int \frac {\csc ^2(c+d x) \left (8 b \left (a^2-3 b^2\right )-a \left (3 a^2+4 b^2\right ) \sin (c+d x)-3 b \left (a^2-4 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{24 a^3}\\ &=-\frac {b \left (a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac {\left (a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {b \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {\int \frac {\csc (c+d x) \left (-3 \left (a^4+4 a^2 b^2-8 b^4\right )-3 a b \left (a^2-4 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{24 a^4}\\ &=-\frac {b \left (a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac {\left (a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {b \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {\left (b^3 \left (a^2-b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^5}-\frac {\left (a^4+4 a^2 b^2-8 b^4\right ) \int \csc (c+d x) \, dx}{8 a^5}\\ &=\frac {\left (a^4+4 a^2 b^2-8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^5 d}-\frac {b \left (a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac {\left (a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {b \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}+\frac {\left (2 b^3 \left (a^2-b^2\right )\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^5 d}\\ &=\frac {\left (a^4+4 a^2 b^2-8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^5 d}-\frac {b \left (a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac {\left (a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {b \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}-\frac {\left (4 b^3 \left (a^2-b^2\right )\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^5 d}\\ &=\frac {2 b^3 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^5 d}+\frac {\left (a^4+4 a^2 b^2-8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^5 d}-\frac {b \left (a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac {\left (a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac {b \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}-\frac {\cot (c+d x) \csc ^3(c+d x)}{4 a d}\\ \end {align*}

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Mathematica [B]  time = 6.27, size = 430, normalized size = 2.22 \[ \frac {b \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 a^2 d}-\frac {b \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{24 a^2 d}+\frac {2 b^3 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (a \sin \left (\frac {1}{2} (c+d x)\right )+b \cos \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^5 d}+\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \left (3 b^3 \cos \left (\frac {1}{2} (c+d x)\right )-a^2 b \cos \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^4 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (a^2 b \sin \left (\frac {1}{2} (c+d x)\right )-3 b^3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^4 d}+\frac {\left (a^2-4 b^2\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 a^3 d}+\frac {\left (4 b^2-a^2\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 a^3 d}+\frac {\left (-a^4-4 a^2 b^2+8 b^4\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 a^5 d}+\frac {\left (a^4+4 a^2 b^2-8 b^4\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 a^5 d}-\frac {\csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 a d}+\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 a d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cot[c + d*x]^2*Csc[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

(2*b^3*Sqrt[a^2 - b^2]*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^2]])/(
a^5*d) + ((-(a^2*b*Cos[(c + d*x)/2]) + 3*b^3*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(6*a^4*d) + ((a^2 - 4*b^2)*Cs
c[(c + d*x)/2]^2)/(32*a^3*d) + (b*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*a^2*d) - Csc[(c + d*x)/2]^4/(64*a*d
) + ((a^4 + 4*a^2*b^2 - 8*b^4)*Log[Cos[(c + d*x)/2]])/(8*a^5*d) + ((-a^4 - 4*a^2*b^2 + 8*b^4)*Log[Sin[(c + d*x
)/2]])/(8*a^5*d) + ((-a^2 + 4*b^2)*Sec[(c + d*x)/2]^2)/(32*a^3*d) + Sec[(c + d*x)/2]^4/(64*a*d) + (Sec[(c + d*
x)/2]*(a^2*b*Sin[(c + d*x)/2] - 3*b^3*Sin[(c + d*x)/2]))/(6*a^4*d) - (b*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(
24*a^2*d)

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fricas [B]  time = 1.02, size = 808, normalized size = 4.16 \[ \left [-\frac {6 \, {\left (a^{4} - 4 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} - 24 \, {\left (b^{3} \cos \left (d x + c\right )^{4} - 2 \, b^{3} \cos \left (d x + c\right )^{2} + b^{3}\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 ) + 6 \, {\left (a^{4} + 4 \, a^{2} b^{2}\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4} - 2 \, {\left (a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4} - 2 \, {\left (a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 16 \, {\left (3 \, a b^{3} \cos \left (d x + c\right ) + {\left (a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{48 \, {\left (a^{5} d \cos \left (d x + c\right )^{4} - 2 \, a^{5} d \cos \left (d x + c\right )^{2} + a^{5} d\right )}}, -\frac {6 \, {\left (a^{4} - 4 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 48 \, {\left (b^{3} \cos \left (d x + c\right )^{4} - 2 \, b^{3} \cos \left (d x + c\right )^{2} + b^{3}\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 ) + 6 \, {\left (a^{4} + 4 \, a^{2} b^{2}\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4} - 2 \, {\left (a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4} - 2 \, {\left (a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 16 \, {\left (3 \, a b^{3} \cos \left (d x + c\right ) + {\left (a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{48 \, {\left (a^{5} d \cos \left (d x + c\right )^{4} - 2 \, a^{5} d \cos \left (d x + c\right )^{2} + a^{5} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[-1/48*(6*(a^4 - 4*a^2*b^2)*cos(d*x + c)^3 - 24*(b^3*cos(d*x + c)^4 - 2*b^3*cos(d*x + c)^2 + b^3)*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*sin(d*x + c) - a^2 - b^2)) + 6*(a^4 + 4*a^2*b^2)*c
os(d*x + c) - 3*((a^4 + 4*a^2*b^2 - 8*b^4)*cos(d*x + c)^4 + a^4 + 4*a^2*b^2 - 8*b^4 - 2*(a^4 + 4*a^2*b^2 - 8*b
^4)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) + 3*((a^4 + 4*a^2*b^2 - 8*b^4)*cos(d*x + c)^4 + a^4 + 4*a^2*b^
2 - 8*b^4 - 2*(a^4 + 4*a^2*b^2 - 8*b^4)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2) - 16*(3*a*b^3*cos(d*x + c
) + (a^3*b - 3*a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/(a^5*d*cos(d*x + c)^4 - 2*a^5*d*cos(d*x + c)^2 + a^5*d), -
1/48*(6*(a^4 - 4*a^2*b^2)*cos(d*x + c)^3 + 48*(b^3*cos(d*x + c)^4 - 2*b^3*cos(d*x + c)^2 + b^3)*sqrt(a^2 - b^2
)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + 6*(a^4 + 4*a^2*b^2)*cos(d*x + c) - 3*((a^4 +
4*a^2*b^2 - 8*b^4)*cos(d*x + c)^4 + a^4 + 4*a^2*b^2 - 8*b^4 - 2*(a^4 + 4*a^2*b^2 - 8*b^4)*cos(d*x + c)^2)*log(
1/2*cos(d*x + c) + 1/2) + 3*((a^4 + 4*a^2*b^2 - 8*b^4)*cos(d*x + c)^4 + a^4 + 4*a^2*b^2 - 8*b^4 - 2*(a^4 + 4*a
^2*b^2 - 8*b^4)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2) - 16*(3*a*b^3*cos(d*x + c) + (a^3*b - 3*a*b^3)*co
s(d*x + c)^3)*sin(d*x + c))/(a^5*d*cos(d*x + c)^4 - 2*a^5*d*cos(d*x + c)^2 + a^5*d)]

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giac [A]  time = 0.20, size = 336, normalized size = 1.73 \[ \frac {\frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} - \frac {24 \, {\left (a^{4} + 4 \, a^{2} b^{2} - 8 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{5}} + \frac {384 \, {\left (a^{2} b^{3} - b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{5}} + \frac {50 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 200 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 400 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 96 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{4}}{a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/192*((3*a^3*tan(1/2*d*x + 1/2*c)^4 - 8*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 24*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 24*a
^2*b*tan(1/2*d*x + 1/2*c) - 96*b^3*tan(1/2*d*x + 1/2*c))/a^4 - 24*(a^4 + 4*a^2*b^2 - 8*b^4)*log(abs(tan(1/2*d*
x + 1/2*c)))/a^5 + 384*(a^2*b^3 - b^5)*(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^5) + (50*a^4*tan(1/2*d*x + 1/2*c)^4 + 200*a^2*b^2*tan(1/2*d*x + 1
/2*c)^4 - 400*b^4*tan(1/2*d*x + 1/2*c)^4 - 24*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 96*a*b^3*tan(1/2*d*x + 1/2*c)^3 -
24*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 + 8*a^3*b*tan(1/2*d*x + 1/2*c) - 3*a^4)/(a^5*tan(1/2*d*x + 1/2*c)^4))/d

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maple [A]  time = 0.49, size = 315, normalized size = 1.62 \[ \frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{64 d a}-\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{2}}+\frac {b^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{8 d \,a^{2}}-\frac {b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{4}}-\frac {1}{64 d a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{2 d \,a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{4}}{d \,a^{5}}+\frac {b}{24 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {b^{2}}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {b}{8 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b^{3}}{2 d \,a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 \sqrt {a^{2}-b^{2}}\, b^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^5/(a+b*sin(d*x+c)),x)

[Out]

1/64/d/a*tan(1/2*d*x+1/2*c)^4-1/24/d/a^2*b*tan(1/2*d*x+1/2*c)^3+1/8/d/a^3*b^2*tan(1/2*d*x+1/2*c)^2+1/8/d/a^2*t
an(1/2*d*x+1/2*c)*b-1/2/d/a^4*b^3*tan(1/2*d*x+1/2*c)-1/64/d/a/tan(1/2*d*x+1/2*c)^4-1/8/a/d*ln(tan(1/2*d*x+1/2*
c))-1/2/d/a^3*ln(tan(1/2*d*x+1/2*c))*b^2+1/d/a^5*ln(tan(1/2*d*x+1/2*c))*b^4+1/24/d/a^2*b/tan(1/2*d*x+1/2*c)^3-
1/8/d*b^2/a^3/tan(1/2*d*x+1/2*c)^2-1/8/d/a^2*b/tan(1/2*d*x+1/2*c)+1/2/d*b^3/a^4/tan(1/2*d*x+1/2*c)+2/d*(a^2-b^
2)^(1/2)/a^5*b^3*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)^2*csc(d*x+c)^5/(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 = 12.07, size = 873, normalized size = 4.50 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {b}{8\,a^2}-\frac {b^3}{2\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^2\,b-8\,b^3\right )+\frac {a^3}{4}-\frac {2\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16\,a^4\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^4+4\,a^2\,b^2-8\,b^4\right )}{8\,a^5\,d}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}+\frac {b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}+\frac {b^3\,\mathrm {atan}\left (\frac {\frac {b^3\,\sqrt {b^2-a^2}\,\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^9+2\,a^7\,b^2-32\,a^5\,b^4+32\,a^3\,b^6\right )}{4\,a^7}-\frac {a^9\,b+12\,a^7\,b^3-16\,a^5\,b^5}{4\,a^8}+\frac {b^3\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (24\,a^{10}-32\,a^8\,b^2\right )}{4\,a^7}\right )\,\sqrt {b^2-a^2}}{a^5}\right )\,1{}\mathrm {i}}{a^5}-\frac {b^3\,\sqrt {b^2-a^2}\,\left (\frac {a^9\,b+12\,a^7\,b^3-16\,a^5\,b^5}{4\,a^8}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^9+2\,a^7\,b^2-32\,a^5\,b^4+32\,a^3\,b^6\right )}{4\,a^7}+\frac {b^3\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (24\,a^{10}-32\,a^8\,b^2\right )}{4\,a^7}\right )\,\sqrt {b^2-a^2}}{a^5}\right )\,1{}\mathrm {i}}{a^5}}{\frac {a^6\,b^3+3\,a^4\,b^5-12\,a^2\,b^7+8\,b^9}{2\,a^8}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^4\,b^4-10\,a^2\,b^6+8\,b^8\right )}{2\,a^7}+\frac {b^3\,\sqrt {b^2-a^2}\,\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^9+2\,a^7\,b^2-32\,a^5\,b^4+32\,a^3\,b^6\right )}{4\,a^7}-\frac {a^9\,b+12\,a^7\,b^3-16\,a^5\,b^5}{4\,a^8}+\frac {b^3\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (24\,a^{10}-32\,a^8\,b^2\right )}{4\,a^7}\right )\,\sqrt {b^2-a^2}}{a^5}\right )}{a^5}+\frac {b^3\,\sqrt {b^2-a^2}\,\left (\frac {a^9\,b+12\,a^7\,b^3-16\,a^5\,b^5}{4\,a^8}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^9+2\,a^7\,b^2-32\,a^5\,b^4+32\,a^3\,b^6\right )}{4\,a^7}+\frac {b^3\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (24\,a^{10}-32\,a^8\,b^2\right )}{4\,a^7}\right )\,\sqrt {b^2-a^2}}{a^5}\right )}{a^5}}\right )\,\sqrt {b^2-a^2}\,2{}\mathrm {i}}{a^5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(sin(c + d*x)^5*(a + b*sin(c + d*x))),x)

[Out]

tan(c/2 + (d*x)/2)^4/(64*a*d) + (tan(c/2 + (d*x)/2)*(b/(8*a^2) - b^3/(2*a^4)))/d - (tan(c/2 + (d*x)/2)^3*(2*a^
2*b - 8*b^3) + a^3/4 - (2*a^2*b*tan(c/2 + (d*x)/2))/3 + 2*a*b^2*tan(c/2 + (d*x)/2)^2)/(16*a^4*d*tan(c/2 + (d*x
)/2)^4) - (log(tan(c/2 + (d*x)/2))*(a^4 - 8*b^4 + 4*a^2*b^2))/(8*a^5*d) - (b*tan(c/2 + (d*x)/2)^3)/(24*a^2*d)
+ (b^2*tan(c/2 + (d*x)/2)^2)/(8*a^3*d) + (b^3*atan(((b^3*(b^2 - a^2)^(1/2)*((tan(c/2 + (d*x)/2)*(a^9 + 32*a^3*
b^6 - 32*a^5*b^4 + 2*a^7*b^2))/(4*a^7) - (a^9*b - 16*a^5*b^5 + 12*a^7*b^3)/(4*a^8) + (b^3*(2*a^2*b - (tan(c/2
+ (d*x)/2)*(24*a^10 - 32*a^8*b^2))/(4*a^7))*(b^2 - a^2)^(1/2))/a^5)*1i)/a^5 - (b^3*(b^2 - a^2)^(1/2)*((a^9*b -
16*a^5*b^5 + 12*a^7*b^3)/(4*a^8) - (tan(c/2 + (d*x)/2)*(a^9 + 32*a^3*b^6 - 32*a^5*b^4 + 2*a^7*b^2))/(4*a^7) +
(b^3*(2*a^2*b - (tan(c/2 + (d*x)/2)*(24*a^10 - 32*a^8*b^2))/(4*a^7))*(b^2 - a^2)^(1/2))/a^5)*1i)/a^5)/((8*b^9
- 12*a^2*b^7 + 3*a^4*b^5 + a^6*b^3)/(2*a^8) + (tan(c/2 + (d*x)/2)*(8*b^8 - 10*a^2*b^6 + 2*a^4*b^4))/(2*a^7) +
(b^3*(b^2 - a^2)^(1/2)*((tan(c/2 + (d*x)/2)*(a^9 + 32*a^3*b^6 - 32*a^5*b^4 + 2*a^7*b^2))/(4*a^7) - (a^9*b - 1
6*a^5*b^5 + 12*a^7*b^3)/(4*a^8) + (b^3*(2*a^2*b - (tan(c/2 + (d*x)/2)*(24*a^10 - 32*a^8*b^2))/(4*a^7))*(b^2 -
a^2)^(1/2))/a^5))/a^5 + (b^3*(b^2 - a^2)^(1/2)*((a^9*b - 16*a^5*b^5 + 12*a^7*b^3)/(4*a^8) - (tan(c/2 + (d*x)/2
)*(a^9 + 32*a^3*b^6 - 32*a^5*b^4 + 2*a^7*b^2))/(4*a^7) + (b^3*(2*a^2*b - (tan(c/2 + (d*x)/2)*(24*a^10 - 32*a^8
*b^2))/(4*a^7))*(b^2 - a^2)^(1/2))/a^5))/a^5))*(b^2 - a^2)^(1/2)*2i)/(a^5*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}{\left (c + d x \right )} \csc ^{5}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)**5/(a+b*sin(d*x+c)),x)

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)**5/(a + b*sin(c + d*x)), x)

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