3.1309 \(\int \frac {\cot ^4(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=244 \[ \frac {b \cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}+\frac {2 b^2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^6 d}-\frac {b \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}+\frac {\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 a^3 d}+\frac {b \left (3 a^4-12 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^6 d}-\frac {\left (3 a^4-20 a^2 b^2+15 b^4\right ) \cot (c+d x)}{15 a^5 d}-\frac {\cot (c+d x) \csc ^4(c+d x)}{5 a d} \]

[Out]

2*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^6/d+1/8*b*(3*a^4-12*a^2*b^2+8*b^4)*ar
ctanh(cos(d*x+c))/a^6/d-1/15*(3*a^4-20*a^2*b^2+15*b^4)*cot(d*x+c)/a^5/d-1/8*b*(5*a^2-4*b^2)*cot(d*x+c)*csc(d*x
+c)/a^4/d+1/15*(6*a^2-5*b^2)*cot(d*x+c)*csc(d*x+c)^2/a^3/d+1/4*b*cot(d*x+c)*csc(d*x+c)^3/a^2/d-1/5*cot(d*x+c)*
csc(d*x+c)^4/a/d

________________________________________________________________________________________

Rubi [A]  time = 1.05, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2893, 3055, 3001, 3770, 2660, 618, 204} \[ \frac {2 b^2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^6 d}-\frac {\left (-20 a^2 b^2+3 a^4+15 b^4\right ) \cot (c+d x)}{15 a^5 d}+\frac {b \left (-12 a^2 b^2+3 a^4+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^6 d}+\frac {\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 a^3 d}-\frac {b \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}+\frac {b \cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x)}{5 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^4*Csc[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

(2*b^2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^6*d) + (b*(3*a^4 - 12*a^2*b^2 +
8*b^4)*ArcTanh[Cos[c + d*x]])/(8*a^6*d) - ((3*a^4 - 20*a^2*b^2 + 15*b^4)*Cot[c + d*x])/(15*a^5*d) - (b*(5*a^2
- 4*b^2)*Cot[c + d*x]*Csc[c + d*x])/(8*a^4*d) + ((6*a^2 - 5*b^2)*Cot[c + d*x]*Csc[c + d*x]^2)/(15*a^3*d) + (b*
Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^4)/(5*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 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 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 ^4(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b \cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x)}{5 a d}-\frac {\int \frac {\csc ^4(c+d x) \left (4 \left (6 a^2-5 b^2\right )-a b \sin (c+d x)-5 \left (4 a^2-3 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{20 a^2}\\ &=\frac {\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 a^3 d}+\frac {b \cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x)}{5 a d}-\frac {\int \frac {\csc ^3(c+d x) \left (-15 b \left (5 a^2-4 b^2\right )-a \left (12 a^2-5 b^2\right ) \sin (c+d x)+8 b \left (6 a^2-5 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{60 a^3}\\ &=-\frac {b \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}+\frac {\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 a^3 d}+\frac {b \cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x)}{5 a d}-\frac {\int \frac {\csc ^2(c+d x) \left (-8 \left (3 a^4-20 a^2 b^2+15 b^4\right )+a b \left (21 a^2-20 b^2\right ) \sin (c+d x)-15 b^2 \left (5 a^2-4 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{120 a^4}\\ &=-\frac {\left (3 a^4-20 a^2 b^2+15 b^4\right ) \cot (c+d x)}{15 a^5 d}-\frac {b \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}+\frac {\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 a^3 d}+\frac {b \cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x)}{5 a d}-\frac {\int \frac {\csc (c+d x) \left (15 b \left (3 a^4-12 a^2 b^2+8 b^4\right )-15 a b^2 \left (5 a^2-4 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{120 a^5}\\ &=-\frac {\left (3 a^4-20 a^2 b^2+15 b^4\right ) \cot (c+d x)}{15 a^5 d}-\frac {b \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}+\frac {\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 a^3 d}+\frac {b \cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x)}{5 a d}+\frac {\left (b^2 \left (a^2-b^2\right )^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^6}-\frac {\left (b \left (3 a^4-12 a^2 b^2+8 b^4\right )\right ) \int \csc (c+d x) \, dx}{8 a^6}\\ &=\frac {b \left (3 a^4-12 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^6 d}-\frac {\left (3 a^4-20 a^2 b^2+15 b^4\right ) \cot (c+d x)}{15 a^5 d}-\frac {b \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}+\frac {\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 a^3 d}+\frac {b \cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x)}{5 a d}+\frac {\left (2 b^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^6 d}\\ &=\frac {b \left (3 a^4-12 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^6 d}-\frac {\left (3 a^4-20 a^2 b^2+15 b^4\right ) \cot (c+d x)}{15 a^5 d}-\frac {b \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}+\frac {\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 a^3 d}+\frac {b \cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x)}{5 a d}-\frac {\left (4 b^2 \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^6 d}\\ &=\frac {2 b^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^6 d}+\frac {b \left (3 a^4-12 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^6 d}-\frac {\left (3 a^4-20 a^2 b^2+15 b^4\right ) \cot (c+d x)}{15 a^5 d}-\frac {b \left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^4 d}+\frac {\left (6 a^2-5 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{15 a^3 d}+\frac {b \cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac {\cot (c+d x) \csc ^4(c+d x)}{5 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 1.82, size = 507, normalized size = 2.08 \[ \frac {96 a^5 \tan \left (\frac {1}{2} (c+d x)\right )-3 a^5 \sin (c+d x) \csc ^6\left (\frac {1}{2} (c+d x)\right )+21 a^5 \sin (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )-336 a^5 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)+6 a^5 \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right )+15 a^4 b \csc ^4\left (\frac {1}{2} (c+d x)\right )-150 a^4 b \csc ^2\left (\frac {1}{2} (c+d x)\right )-15 a^4 b \sec ^4\left (\frac {1}{2} (c+d x)\right )+150 a^4 b \sec ^2\left (\frac {1}{2} (c+d x)\right )-360 a^4 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+360 a^4 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-640 a^3 b^2 \tan \left (\frac {1}{2} (c+d x)\right )-20 a^3 b^2 \sin (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )+320 a^3 b^2 \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)+120 a^2 b^3 \csc ^2\left (\frac {1}{2} (c+d x)\right )-120 a^2 b^3 \sec ^2\left (\frac {1}{2} (c+d x)\right )+1440 a^2 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-1440 a^2 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+1920 b^2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )-32 \left (3 a^5-20 a^3 b^2+15 a b^4\right ) \cot \left (\frac {1}{2} (c+d x)\right )+480 a b^4 \tan \left (\frac {1}{2} (c+d x)\right )-960 b^5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+960 b^5 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{960 a^6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^4*Csc[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

(1920*b^2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] - 32*(3*a^5 - 20*a^3*b^2 + 15*a*b
^4)*Cot[(c + d*x)/2] - 150*a^4*b*Csc[(c + d*x)/2]^2 + 120*a^2*b^3*Csc[(c + d*x)/2]^2 + 15*a^4*b*Csc[(c + d*x)/
2]^4 + 360*a^4*b*Log[Cos[(c + d*x)/2]] - 1440*a^2*b^3*Log[Cos[(c + d*x)/2]] + 960*b^5*Log[Cos[(c + d*x)/2]] -
360*a^4*b*Log[Sin[(c + d*x)/2]] + 1440*a^2*b^3*Log[Sin[(c + d*x)/2]] - 960*b^5*Log[Sin[(c + d*x)/2]] + 150*a^4
*b*Sec[(c + d*x)/2]^2 - 120*a^2*b^3*Sec[(c + d*x)/2]^2 - 15*a^4*b*Sec[(c + d*x)/2]^4 - 336*a^5*Csc[c + d*x]^3*
Sin[(c + d*x)/2]^4 + 320*a^3*b^2*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 21*a^5*Csc[(c + d*x)/2]^4*Sin[c + d*x] -
20*a^3*b^2*Csc[(c + d*x)/2]^4*Sin[c + d*x] - 3*a^5*Csc[(c + d*x)/2]^6*Sin[c + d*x] + 96*a^5*Tan[(c + d*x)/2] -
640*a^3*b^2*Tan[(c + d*x)/2] + 480*a*b^4*Tan[(c + d*x)/2] + 6*a^5*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2])/(960*a
^6*d)

________________________________________________________________________________________

fricas [B]  time = 1.83, size = 1051, normalized size = 4.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[-1/240*(16*(3*a^5 - 20*a^3*b^2 + 15*a*b^4)*cos(d*x + c)^5 + 80*(7*a^3*b^2 - 6*a*b^4)*cos(d*x + c)^3 + 120*((a
^2*b^2 - b^4)*cos(d*x + c)^4 + a^2*b^2 - b^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*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2))*sin(d*x + c) - 15*(3*a^4*b - 12*a^2*b^3 +
8*b^5 + (3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d*x + c)^4 - 2*(3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d*x + c)^2)*log(
1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 15*(3*a^4*b - 12*a^2*b^3 + 8*b^5 + (3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d
*x + c)^4 - 2*(3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 240*(
a^3*b^2 - a*b^4)*cos(d*x + c) - 30*((5*a^4*b - 4*a^2*b^3)*cos(d*x + c)^3 - (3*a^4*b - 4*a^2*b^3)*cos(d*x + c))
*sin(d*x + c))/((a^6*d*cos(d*x + c)^4 - 2*a^6*d*cos(d*x + c)^2 + a^6*d)*sin(d*x + c)), -1/240*(16*(3*a^5 - 20*
a^3*b^2 + 15*a*b^4)*cos(d*x + c)^5 + 80*(7*a^3*b^2 - 6*a*b^4)*cos(d*x + c)^3 + 240*((a^2*b^2 - b^4)*cos(d*x +
c)^4 + a^2*b^2 - b^4 - 2*(a^2*b^2 - b^4)*cos(d*x + c)^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^
2 - b^2)*cos(d*x + c)))*sin(d*x + c) - 15*(3*a^4*b - 12*a^2*b^3 + 8*b^5 + (3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d
*x + c)^4 - 2*(3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 15*(3*
a^4*b - 12*a^2*b^3 + 8*b^5 + (3*a^4*b - 12*a^2*b^3 + 8*b^5)*cos(d*x + c)^4 - 2*(3*a^4*b - 12*a^2*b^3 + 8*b^5)*
cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 240*(a^3*b^2 - a*b^4)*cos(d*x + c) - 30*((5*a^4*b
- 4*a^2*b^3)*cos(d*x + c)^3 - (3*a^4*b - 4*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6*d*cos(d*x + c)^4 - 2*a^6
*d*cos(d*x + c)^2 + a^6*d)*sin(d*x + c))]

________________________________________________________________________________________

giac [B]  time = 0.22, size = 484, normalized size = 1.98 \[ \frac {\frac {6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 60 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 600 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 480 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{5}} - \frac {120 \, {\left (3 \, a^{4} b - 12 \, a^{2} b^{3} + 8 \, b^{5}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{6}} + \frac {1920 \, {\left (a^{4} b^{2} - 2 \, 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^{6}} + \frac {822 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3288 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2192 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 60 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 600 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 480 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 40 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{5}}{a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^6/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/960*((6*a^4*tan(1/2*d*x + 1/2*c)^5 - 15*a^3*b*tan(1/2*d*x + 1/2*c)^4 - 30*a^4*tan(1/2*d*x + 1/2*c)^3 + 40*a^
2*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*a^3*b*tan(1/2*d*x + 1/2*c)^2 - 120*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 60*a^4*ta
n(1/2*d*x + 1/2*c) - 600*a^2*b^2*tan(1/2*d*x + 1/2*c) + 480*b^4*tan(1/2*d*x + 1/2*c))/a^5 - 120*(3*a^4*b - 12*
a^2*b^3 + 8*b^5)*log(abs(tan(1/2*d*x + 1/2*c)))/a^6 + 1920*(a^4*b^2 - 2*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^6) + (822*a^4*b*t
an(1/2*d*x + 1/2*c)^5 - 3288*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 2192*b^5*tan(1/2*d*x + 1/2*c)^5 - 60*a^5*tan(1/2
*d*x + 1/2*c)^4 + 600*a^3*b^2*tan(1/2*d*x + 1/2*c)^4 - 480*a*b^4*tan(1/2*d*x + 1/2*c)^4 - 120*a^4*b*tan(1/2*d*
x + 1/2*c)^3 + 120*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 30*a^5*tan(1/2*d*x + 1/2*c)^2 - 40*a^3*b^2*tan(1/2*d*x + 1
/2*c)^2 + 15*a^4*b*tan(1/2*d*x + 1/2*c) - 6*a^5)/(a^6*tan(1/2*d*x + 1/2*c)^5))/d

________________________________________________________________________________________

maple [B]  time = 0.49, size = 583, normalized size = 2.39 \[ \frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d a}-\frac {\left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{64 d \,a^{2}}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{32 d a}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{24 d \,a^{3}}+\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{8 d \,a^{2}}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{8 d \,a^{4}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}-\frac {5 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{3}}+\frac {b^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{5}}-\frac {1}{160 d a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {1}{32 d a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {b^{2}}{24 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {1}{16 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {5 b^{2}}{8 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b^{4}}{2 d \,a^{5} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{64 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}-\frac {b}{8 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b^{3}}{8 d \,a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{2}}+\frac {3 b^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{4}}-\frac {b^{5} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{6}}+\frac {2 b^{2} \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^{2} \sqrt {a^{2}-b^{2}}}-\frac {4 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) b^{4}}{d \,a^{4} \sqrt {a^{2}-b^{2}}}+\frac {2 b^{6} \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^{6} \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)^6/(a+b*sin(d*x+c)),x)

[Out]

1/160/d/a*tan(1/2*d*x+1/2*c)^5-1/64/d/a^2*tan(1/2*d*x+1/2*c)^4*b-1/32/d/a*tan(1/2*d*x+1/2*c)^3+1/24/d/a^3*tan(
1/2*d*x+1/2*c)^3*b^2+1/8/d/a^2*tan(1/2*d*x+1/2*c)^2*b-1/8/d/a^4*tan(1/2*d*x+1/2*c)^2*b^3+1/16/a/d*tan(1/2*d*x+
1/2*c)-5/8/d/a^3*b^2*tan(1/2*d*x+1/2*c)+1/2/d/a^5*b^4*tan(1/2*d*x+1/2*c)-1/160/d/a/tan(1/2*d*x+1/2*c)^5+1/32/d
/a/tan(1/2*d*x+1/2*c)^3-1/24/d/a^3/tan(1/2*d*x+1/2*c)^3*b^2-1/16/a/d/tan(1/2*d*x+1/2*c)+5/8/d/a^3/tan(1/2*d*x+
1/2*c)*b^2-1/2/d/a^5/tan(1/2*d*x+1/2*c)*b^4+1/64/d/a^2*b/tan(1/2*d*x+1/2*c)^4-1/8/d/a^2*b/tan(1/2*d*x+1/2*c)^2
+1/8/d*b^3/a^4/tan(1/2*d*x+1/2*c)^2-3/8/d/a^2*b*ln(tan(1/2*d*x+1/2*c))+3/2/d/a^4*b^3*ln(tan(1/2*d*x+1/2*c))-1/
d/a^6*b^5*ln(tan(1/2*d*x+1/2*c))+2/d*b^2/a^2/(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/a^4/(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))*b^4+2/d/a^6*b^6/(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))

________________________________________________________________________________________

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)^6/(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?

________________________________________________________________________________________

mupad [B]  time = 12.18, size = 1082, normalized size = 4.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^6*(a + b*sin(c + d*x))),x)

[Out]

tan(c/2 + (d*x)/2)^5/(160*a*d) + (tan(c/2 + (d*x)/2)^2*(b/(32*a^2) + (b*(3/(32*a) - b^2/(8*a^3)))/a))/d - (tan
(c/2 + (d*x)/2)*(b^2/(8*a^3) - 1/(16*a) + (2*b*(b/(16*a^2) + (2*b*(3/(32*a) - b^2/(8*a^3)))/a))/a))/d - (tan(c
/2 + (d*x)/2)^3*(1/(32*a) - b^2/(24*a^3)))/d - (b*tan(c/2 + (d*x)/2)^4)/(64*a^2*d) + (tan(c/2 + (d*x)/2)^2*(a^
4 - (4*a^2*b^2)/3) - tan(c/2 + (d*x)/2)^4*(2*a^4 + 16*b^4 - 20*a^2*b^2) - a^4/5 + tan(c/2 + (d*x)/2)^3*(4*a*b^
3 - 4*a^3*b) + (a^3*b*tan(c/2 + (d*x)/2))/2)/(32*a^5*d*tan(c/2 + (d*x)/2)^5) - (log(tan(c/2 + (d*x)/2))*((3*a^
4*b)/8 + b^5 - (3*a^2*b^3)/2))/(a^6*d) + (b^2*atan(((b^2*(-(a + b)^3*(a - b)^3)^(1/2)*((tan(c/2 + (d*x)/2)*(3*
a^10*b - 32*a^4*b^7 + 64*a^6*b^5 - 34*a^8*b^3))/(4*a^9) - (16*a^6*b^6 - 28*a^8*b^4 + 11*a^10*b^2)/(4*a^10) + (
b^2*(2*a^2*b - (tan(c/2 + (d*x)/2)*(24*a^12 - 32*a^10*b^2))/(4*a^9))*(-(a + b)^3*(a - b)^3)^(1/2))/a^6)*1i)/a^
6 - (b^2*(-(a + b)^3*(a - b)^3)^(1/2)*((16*a^6*b^6 - 28*a^8*b^4 + 11*a^10*b^2)/(4*a^10) - (tan(c/2 + (d*x)/2)*
(3*a^10*b - 32*a^4*b^7 + 64*a^6*b^5 - 34*a^8*b^3))/(4*a^9) + (b^2*(2*a^2*b - (tan(c/2 + (d*x)/2)*(24*a^12 - 32
*a^10*b^2))/(4*a^9))*(-(a + b)^3*(a - b)^3)^(1/2))/a^6)*1i)/a^6)/((8*b^11 - 28*a^2*b^9 + 35*a^4*b^7 - 18*a^6*b
^5 + 3*a^8*b^3)/(2*a^10) + (tan(c/2 + (d*x)/2)*(8*b^10 - 26*a^2*b^8 + 28*a^4*b^6 - 10*a^6*b^4))/(2*a^9) + (b^2
*(-(a + b)^3*(a - b)^3)^(1/2)*((tan(c/2 + (d*x)/2)*(3*a^10*b - 32*a^4*b^7 + 64*a^6*b^5 - 34*a^8*b^3))/(4*a^9)
- (16*a^6*b^6 - 28*a^8*b^4 + 11*a^10*b^2)/(4*a^10) + (b^2*(2*a^2*b - (tan(c/2 + (d*x)/2)*(24*a^12 - 32*a^10*b^
2))/(4*a^9))*(-(a + b)^3*(a - b)^3)^(1/2))/a^6))/a^6 + (b^2*(-(a + b)^3*(a - b)^3)^(1/2)*((16*a^6*b^6 - 28*a^8
*b^4 + 11*a^10*b^2)/(4*a^10) - (tan(c/2 + (d*x)/2)*(3*a^10*b - 32*a^4*b^7 + 64*a^6*b^5 - 34*a^8*b^3))/(4*a^9)
+ (b^2*(2*a^2*b - (tan(c/2 + (d*x)/2)*(24*a^12 - 32*a^10*b^2))/(4*a^9))*(-(a + b)^3*(a - b)^3)^(1/2))/a^6))/a^
6))*(-(a + b)^3*(a - b)^3)^(1/2)*2i)/(a^6*d)

________________________________________________________________________________________

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)**4*csc(d*x+c)**6/(a+b*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________