3.1308 \(\int \frac{\cot ^4(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=198 \[ -\frac{2 b \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^5 d}-\frac{b \left (4 a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac{\left (-12 a^2 b^2+3 a^4+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^5 d}+\frac{\left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac{b \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d} \]

[Out]

(-2*b*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^5*d) - ((3*a^4 - 12*a^2*b^2 + 8*b
^4)*ArcTanh[Cos[c + d*x]])/(8*a^5*d) - (b*(4*a^2 - 3*b^2)*Cot[c + d*x])/(3*a^4*d) + ((5*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
)

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Rubi [A]  time = 0.760678, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2893, 3055, 3001, 3770, 2660, 618, 204} \[ -\frac{2 b \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^5 d}-\frac{b \left (4 a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac{\left (-12 a^2 b^2+3 a^4+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^5 d}+\frac{\left (5 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac{b \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^5*d) - ((3*a^4 - 12*a^2*b^2 + 8*b
^4)*ArcTanh[Cos[c + d*x]])/(8*a^5*d) - (b*(4*a^2 - 3*b^2)*Cot[c + d*x])/(3*a^4*d) + ((5*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 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 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 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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 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 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])

Rubi steps

\begin{align*} \int \frac{\cot ^4(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx &=\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 (5 a^2-4 b^2\right )-a b \sin (c+d x)-4 \left (3 a^2-2 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{12 a^2}\\ &=\frac{\left (5 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 (4 a^2-3 b^2\right )-a \left (9 a^2-4 b^2\right ) \sin (c+d x)+3 b \left (5 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 (4 a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac{\left (5 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 (3 a^4-12 a^2 b^2+8 b^4\right )+3 a b \left (5 a^2-4 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{24 a^4}\\ &=-\frac{b \left (4 a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac{\left (5 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 \left (a^2-b^2\right )^2\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^5}+\frac{\left (3 a^4-12 a^2 b^2+8 b^4\right ) \int \csc (c+d x) \, dx}{8 a^5}\\ &=-\frac{\left (3 a^4-12 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^5 d}-\frac{b \left (4 a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac{\left (5 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 \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^5 d}\\ &=-\frac{\left (3 a^4-12 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^5 d}-\frac{b \left (4 a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac{\left (5 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 \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^5 d}\\ &=-\frac{2 b \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^5 d}-\frac{\left (3 a^4-12 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\cos (c+d x))}{8 a^5 d}-\frac{b \left (4 a^2-3 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac{\left (5 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*}

Mathematica [B]  time = 6.21264, size = 433, normalized size = 2.19 \[ \frac{\left (5 a^2-4 b^2\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{32 a^3 d}+\frac{\left (4 b^2-5 a^2\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{32 a^3 d}+\frac{\left (-12 a^2 b^2+3 a^4+8 b^4\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{8 a^5 d}+\frac{\left (12 a^2 b^2-3 a^4-8 b^4\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 a^5 d}+\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \left (3 b^3 \cos \left (\frac{1}{2} (c+d x)\right )-4 a^2 b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{6 a^4 d}+\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (4 a^2 b \sin \left (\frac{1}{2} (c+d x)\right )-3 b^3 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{6 a^4 d}-\frac{2 b \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (a \sin \left (\frac{1}{2} (c+d x)\right )+b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^5 d}+\frac{b \cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{24 a^2 d}-\frac{b \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{24 a^2 d}-\frac{\csc ^4\left (\frac{1}{2} (c+d x)\right )}{64 a d}+\frac{\sec ^4\left (\frac{1}{2} (c+d x)\right )}{64 a d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*b*(a^2 - b^2)^(3/2)*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) + ((-4*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) + ((5*a^2 - 4*b^2)
*Csc[(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) + ((-3*a^4 + 12*a^2*b^2 - 8*b^4)*Log[Cos[(c + d*x)/2]])/(8*a^5*d) + ((3*a^4 - 12*a^2*b^2 + 8*b^4)*Log[Sin
[(c + d*x)/2]])/(8*a^5*d) + ((-5*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]*(4*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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Maple [B]  time = 0.128, size = 455, normalized size = 2.3 \begin{align*}{\frac{1}{64\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}}-{\frac{b}{24\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{{b}^{2}}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{5\,b}{8\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{{b}^{3}}{2\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{b}{da\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+4\,{\frac{{b}^{3}}{d{a}^{3}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-2\,{\frac{{b}^{5}}{d{a}^{5}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{64\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}}+{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{{b}^{2}}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{3}{8\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{3\,{b}^{2}}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{{b}^{4}}{d{a}^{5}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{b}{24\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-{\frac{5\,b}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{{b}^{3}}{2\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*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*tan(1/2*d*x+1/2*c)^3*b-1/8/d/a*tan(1/2*d*x+1/2*c)^2+1/8/d/a^3*tan(1/2
*d*x+1/2*c)^2*b^2+5/8/d/a^2*tan(1/2*d*x+1/2*c)*b-1/2/d/a^4*b^3*tan(1/2*d*x+1/2*c)-2/d/a*b/(a^2-b^2)^(1/2)*arct
an(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+4/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))-2/d*b^5/a^5/(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
))-1/64/d/a/tan(1/2*d*x+1/2*c)^4+1/8/d/a/tan(1/2*d*x+1/2*c)^2-1/8/d/a^3*b^2/tan(1/2*d*x+1/2*c)^2+3/8/d/a*ln(ta
n(1/2*d*x+1/2*c))-3/2/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)^3-5/8/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)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.22323, size = 2068, normalized size = 10.44 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[-1/48*(6*(5*a^4 - 4*a^2*b^2)*cos(d*x + c)^3 + 24*((a^2*b - b^3)*cos(d*x + c)^4 + a^2*b - b^3 - 2*(a^2*b - b^3
)*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)) - 6*(3*a^4 - 4*a^2*b^2)*cos(d*x + c) + 3*((3*a^4 - 12*a^2*b^2 + 8*b^4)*cos(d*x + c)^4 + 3*a^4 - 12*a^2*
b^2 + 8*b^4 - 2*(3*a^4 - 12*a^2*b^2 + 8*b^4)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) - 3*((3*a^4 - 12*a^2*
b^2 + 8*b^4)*cos(d*x + c)^4 + 3*a^4 - 12*a^2*b^2 + 8*b^4 - 2*(3*a^4 - 12*a^2*b^2 + 8*b^4)*cos(d*x + c)^2)*log(
-1/2*cos(d*x + c) + 1/2) - 16*((4*a^3*b - 3*a*b^3)*cos(d*x + c)^3 - 3*(a^3*b - a*b^3)*cos(d*x + c))*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*(5*a^4 - 4*a^2*b^2)*cos(d*x + c)^3 - 48*
((a^2*b - b^3)*cos(d*x + c)^4 + a^2*b - b^3 - 2*(a^2*b - b^3)*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))) - 6*(3*a^4 - 4*a^2*b^2)*cos(d*x + c) + 3*((3*a^4 - 12*a^2*b^2 + 8
*b^4)*cos(d*x + c)^4 + 3*a^4 - 12*a^2*b^2 + 8*b^4 - 2*(3*a^4 - 12*a^2*b^2 + 8*b^4)*cos(d*x + c)^2)*log(1/2*cos
(d*x + c) + 1/2) - 3*((3*a^4 - 12*a^2*b^2 + 8*b^4)*cos(d*x + c)^4 + 3*a^4 - 12*a^2*b^2 + 8*b^4 - 2*(3*a^4 - 12
*a^2*b^2 + 8*b^4)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2) - 16*((4*a^3*b - 3*a*b^3)*cos(d*x + c)^3 - 3*(a
^3*b - a*b^3)*cos(d*x + c))*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**5/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.26145, size = 506, normalized size = 2.56 \begin{align*} \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^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 120 \, 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 (3 \, a^{4} - 12 \, 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^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \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{150 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 600 \, 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} + 120 \, 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^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*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^3*tan(1/2*d*x + 1/2*c)^2 + 24*a*b
^2*tan(1/2*d*x + 1/2*c)^2 + 120*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*(3*a^4 - 12
*a^2*b^2 + 8*b^4)*log(abs(tan(1/2*d*x + 1/2*c)))/a^5 - 384*(a^4*b - 2*a^2*b^3 + b^5)*(pi*floor(1/2*(d*x + c)/p
i + 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) - (150*a^4*tan(1
/2*d*x + 1/2*c)^4 - 600*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 + 120*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^4*tan(1/2*d*x + 1/2*c)^2 + 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