3.1140 \(\int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=218 \[ -\frac {3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}+\frac {3 b \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^5 d \sqrt {a^2-b^2}}+\frac {3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^5 d}-\frac {\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2} \]

[Out]

3/2*(a^2-4*b^2)*arctanh(cos(d*x+c))/a^5/d-1/2*(a^2-12*b^2)*cot(d*x+c)/a^4/b/d+1/2*(a^2-2*b^2)*cot(d*x+c)/a^2/b
/d/(a+b*sin(d*x+c))^2-1/2*cot(d*x+c)*csc(d*x+c)/a/d/(a+b*sin(d*x+c))^2-3*b*cot(d*x+c)/a^3/d/(a+b*sin(d*x+c))+3
*b*(3*a^2-4*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^5/d/(a^2-b^2)^(1/2)

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Rubi [A]  time = 0.77, antiderivative size = 218, normalized size of antiderivative = 1.00, 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 = {2890, 3055, 3001, 3770, 2660, 618, 204} \[ \frac {3 b \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^5 d \sqrt {a^2-b^2}}-\frac {\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac {3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^5 d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2890

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]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(a*d*f*(n + 1)), x] +
(Dist[1/(a^2*b*d*(n + 1)*(m + 1)), Int[(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1)*Simp[a^2*(n + 1)
*(n + 2) - b^2*(m + n + 2)*(m + n + 3) + a*b*(m + 1)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m
+ n + 4))*Sin[e + f*x]^2, x], x], x] - Simp[((a^2*(n + 1) - b^2*(m + n + 2))*Cos[e + f*x]*(d*Sin[e + f*x])^(n
+ 2)*(a + b*Sin[e + f*x])^(m + 1))/(a^2*b*d^2*f*(n + 1)*(m + 1)), x]) /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2
- b^2, 0] && IntegersQ[2*m, 2*n] && LtQ[m, -1] && LtQ[n, -1]

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 {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac {\int \frac {\csc ^2(c+d x) \left (2 \left (a^2-6 b^2\right )-2 a b \sin (c+d x)+8 b^2 \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx}{4 a^2 b}\\ &=\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac {3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}+\frac {\int \frac {\csc ^2(c+d x) \left (2 \left (a^4-13 a^2 b^2+12 b^4\right )-4 a b \left (a^2-b^2\right ) \sin (c+d x)+12 b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 a^3 b \left (a^2-b^2\right )}\\ &=-\frac {\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac {3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}+\frac {\int \frac {\csc (c+d x) \left (-6 b \left (a^4-5 a^2 b^2+4 b^4\right )+12 a b^2 \left (a^2-b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 a^4 b \left (a^2-b^2\right )}\\ &=-\frac {\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac {3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}-\frac {\left (3 \left (a^2-4 b^2\right )\right ) \int \csc (c+d x) \, dx}{2 a^5}+\frac {\left (12 a^2 b^2 \left (a^2-b^2\right )+6 b^2 \left (a^4-5 a^2 b^2+4 b^4\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{4 a^5 b \left (a^2-b^2\right )}\\ &=\frac {3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^5 d}-\frac {\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac {3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}+\frac {\left (3 b \left (3 a^2-4 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 {3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^5 d}-\frac {\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac {3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}-\frac {\left (6 b \left (3 a^2-4 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 {3 b \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^5 \sqrt {a^2-b^2} d}+\frac {3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^5 d}-\frac {\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac {3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 6.18, size = 319, normalized size = 1.46 \[ -\frac {3 b \tan \left (\frac {1}{2} (c+d x)\right )}{2 a^4 d}+\frac {3 b \cot \left (\frac {1}{2} (c+d x)\right )}{2 a^4 d}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^3 d}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^3 d}-\frac {3 \left (a^2-4 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^5 d}+\frac {3 \left (a^2-4 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^5 d}+\frac {3 b \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (a \sin \left (\frac {1}{2} (c+d x)\right )+b \cos \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^5 d \sqrt {a^2-b^2}}+\frac {6 b^2 \cos (c+d x)-a^2 \cos (c+d x)}{2 a^4 d (a+b \sin (c+d x))}+\frac {b^2 \cos (c+d x)-a^2 \cos (c+d x)}{2 a^3 d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.20, size = 1560, normalized size = 7.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*(a^6 - 10*a^4*b^2 + 9*a^2*b^4)*cos(d*x + c)^3 + 3*(3*a^4*b - a^2*b^3 - 4*b^5 + (3*a^2*b^3 - 4*b^5)*cos
(d*x + c)^4 - (3*a^4*b + 2*a^2*b^3 - 8*b^5)*cos(d*x + c)^2 + 2*(3*a^3*b^2 - 4*a*b^4 - (3*a^3*b^2 - 4*a*b^4)*co
s(d*x + c)^2)*sin(d*x + c))*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^6 - 7*a^4*b^2 + 6*a^2*b^4)*cos(d*x + c) + 3*(a^6 - 4*a^4*b^2 - a^2*b^4 + 4*b^6 + (a^4*
b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^4 - (a^6 - 3*a^4*b^2 - 6*a^2*b^4 + 8*b^6)*cos(d*x + c)^2 + 2*(a^5*b - 5*
a^3*b^3 + 4*a*b^5 - (a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) -
3*(a^6 - 4*a^4*b^2 - a^2*b^4 + 4*b^6 + (a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^4 - (a^6 - 3*a^4*b^2 - 6*a^2
*b^4 + 8*b^6)*cos(d*x + c)^2 + 2*(a^5*b - 5*a^3*b^3 + 4*a*b^5 - (a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c)^2)*
sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*((a^5*b - 13*a^3*b^3 + 12*a*b^5)*cos(d*x + c)^3 + 3*(a^5*b + 3*
a^3*b^3 - 4*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^7*b^2 - a^5*b^4)*d*cos(d*x + c)^4 - (a^9 + a^7*b^2 - 2*a^5*
b^4)*d*cos(d*x + c)^2 + (a^9 - a^5*b^4)*d - 2*((a^8*b - a^6*b^3)*d*cos(d*x + c)^2 - (a^8*b - a^6*b^3)*d)*sin(d
*x + c)), 1/4*(4*(a^6 - 10*a^4*b^2 + 9*a^2*b^4)*cos(d*x + c)^3 - 6*(3*a^4*b - a^2*b^3 - 4*b^5 + (3*a^2*b^3 - 4
*b^5)*cos(d*x + c)^4 - (3*a^4*b + 2*a^2*b^3 - 8*b^5)*cos(d*x + c)^2 + 2*(3*a^3*b^2 - 4*a*b^4 - (3*a^3*b^2 - 4*
a*b^4)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c
))) - 6*(a^6 - 7*a^4*b^2 + 6*a^2*b^4)*cos(d*x + c) + 3*(a^6 - 4*a^4*b^2 - a^2*b^4 + 4*b^6 + (a^4*b^2 - 5*a^2*b
^4 + 4*b^6)*cos(d*x + c)^4 - (a^6 - 3*a^4*b^2 - 6*a^2*b^4 + 8*b^6)*cos(d*x + c)^2 + 2*(a^5*b - 5*a^3*b^3 + 4*a
*b^5 - (a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*(a^6 - 4*a^
4*b^2 - a^2*b^4 + 4*b^6 + (a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^4 - (a^6 - 3*a^4*b^2 - 6*a^2*b^4 + 8*b^6)
*cos(d*x + c)^2 + 2*(a^5*b - 5*a^3*b^3 + 4*a*b^5 - (a^5*b - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))
*log(-1/2*cos(d*x + c) + 1/2) + 2*((a^5*b - 13*a^3*b^3 + 12*a*b^5)*cos(d*x + c)^3 + 3*(a^5*b + 3*a^3*b^3 - 4*a
*b^5)*cos(d*x + c))*sin(d*x + c))/((a^7*b^2 - a^5*b^4)*d*cos(d*x + c)^4 - (a^9 + a^7*b^2 - 2*a^5*b^4)*d*cos(d*
x + c)^2 + (a^9 - a^5*b^4)*d - 2*((a^8*b - a^6*b^3)*d*cos(d*x + c)^2 - (a^8*b - a^6*b^3)*d)*sin(d*x + c))]

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giac [A]  time = 0.29, size = 395, normalized size = 1.81 \[ -\frac {\frac {12 \, {\left (a^{2} - 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{5}} - \frac {24 \, {\left (3 \, a^{2} b - 4 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{5}} - \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}} - \frac {6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 24 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 32 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 48 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 16 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 112 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 76 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{4}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2} a^{5}}}{8 \, d} \]

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))^3,x, algorithm="giac")

[Out]

-1/8*(12*(a^2 - 4*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^5 - 24*(3*a^2*b - 4*b^3)*(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) - (a^3*tan(1/2*d*x +
1/2*c)^2 - 12*a^2*b*tan(1/2*d*x + 1/2*c))/a^6 - (6*a^4*tan(1/2*d*x + 1/2*c)^6 - 24*a^2*b^2*tan(1/2*d*x + 1/2*
c)^6 + 12*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 32*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 5*a^4*tan(1/2*d*x + 1/2*c)^4 + 48*a
^2*b^2*tan(1/2*d*x + 1/2*c)^4 + 16*b^4*tan(1/2*d*x + 1/2*c)^4 + 4*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 112*a*b^3*tan
(1/2*d*x + 1/2*c)^3 - 12*a^4*tan(1/2*d*x + 1/2*c)^2 + 76*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) - a^4)/((a*tan(1/2*d*x + 1/2*c)^3 + 2*b*tan(1/2*d*x + 1/2*c)^2 + a*tan(1/2*d*x + 1/2*c))^2*a^5))/d

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maple [B]  time = 0.91, size = 642, normalized size = 2.94 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{3}}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{2 d \,a^{4}}-\frac {1}{8 d \,a^{3} \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 d \,a^{3}}+\frac {6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{5}}+\frac {3 b}{2 d \,a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}+\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{d \,a^{4} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} a}+\frac {3 b^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}+\frac {14 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{4}}{d \,a^{5} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {5 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}+\frac {20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{3}}{d \,a^{4} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {2}{d a \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}+\frac {7 b^{2}}{d \,a^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}+\frac {9 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^{3} \sqrt {a^{2}-b^{2}}}-\frac {12 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} \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))^3,x)

[Out]

1/8/d/a^3*tan(1/2*d*x+1/2*c)^2-3/2/d/a^4*tan(1/2*d*x+1/2*c)*b-1/8/d/a^3/tan(1/2*d*x+1/2*c)^2-3/2/d/a^3*ln(tan(
1/2*d*x+1/2*c))+6/d/a^5*ln(tan(1/2*d*x+1/2*c))*b^2+3/2/d*b/a^4/tan(1/2*d*x+1/2*c)-3/d/a^2*b/(tan(1/2*d*x+1/2*c
)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^3+8/d/a^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+
a)^2*tan(1/2*d*x+1/2*c)^3*b^3-2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a*tan(1/2*d*x+1/2*c)^2+3
/d/a^3*b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2+14/d/a^5/(tan(1/2*d*x+1/2*
c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2*b^4-5/d/a^2*b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/
2*c)*b+a)^2*tan(1/2*d*x+1/2*c)+20/d/a^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)
*b^3-2/d/a/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2+7/d/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1
/2*c)*b+a)^2*b^2+9/d/a^3*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))-12/d/a^5*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))^3,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 = 9.90, size = 1100, normalized size = 5.05 \[ \text {result too large to display} \]

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))^3),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^3*d) + (tan(c/2 + (d*x)/2)^2*(50*a*b^2 - 9*a^3) - tan(c/2 + (d*x)/2)^5*(6*a^2*b - 32
*b^3) - tan(c/2 + (d*x)/2)^3*(10*a^2*b - 104*b^3) - a^3/2 + 4*a^2*b*tan(c/2 + (d*x)/2) + (tan(c/2 + (d*x)/2)^4
*(112*b^4 - 17*a^4 + 72*a^2*b^2))/(2*a))/(d*(4*a^6*tan(c/2 + (d*x)/2)^2 + 4*a^6*tan(c/2 + (d*x)/2)^6 + tan(c/2
+ (d*x)/2)^4*(8*a^6 + 16*a^4*b^2) + 16*a^5*b*tan(c/2 + (d*x)/2)^3 + 16*a^5*b*tan(c/2 + (d*x)/2)^5)) - (3*b*ta
n(c/2 + (d*x)/2))/(2*a^4*d) - (log(tan(c/2 + (d*x)/2))*(3*a^2 - 12*b^2))/(2*a^5*d) - (b*atan(((b*(-(a + b)*(a
- b))^(1/2)*(3*a^2 - 4*b^2)*((12*a^7*b - 24*a^5*b^3)/a^8 - (tan(c/2 + (d*x)/2)*(3*a^7 + 48*a^3*b^4 - 36*a^5*b^
2))/a^7 + (3*b*(-(a + b)*(a - b))^(1/2)*(2*a^2*b - (tan(c/2 + (d*x)/2)*(6*a^10 - 8*a^8*b^2))/a^7)*(3*a^2 - 4*b
^2))/(2*(a^7 - a^5*b^2)))*3i)/(2*(a^7 - a^5*b^2)) - (b*(-(a + b)*(a - b))^(1/2)*(3*a^2 - 4*b^2)*((tan(c/2 + (d
*x)/2)*(3*a^7 + 48*a^3*b^4 - 36*a^5*b^2))/a^7 - (12*a^7*b - 24*a^5*b^3)/a^8 + (3*b*(-(a + b)*(a - b))^(1/2)*(2
*a^2*b - (tan(c/2 + (d*x)/2)*(6*a^10 - 8*a^8*b^2))/a^7)*(3*a^2 - 4*b^2))/(2*(a^7 - a^5*b^2)))*3i)/(2*(a^7 - a^
5*b^2)))/((27*a^4*b + 144*b^5 - 144*a^2*b^3)/a^8 + (2*tan(c/2 + (d*x)/2)*(72*b^4 - 54*a^2*b^2))/a^7 + (3*b*(-(
a + b)*(a - b))^(1/2)*(3*a^2 - 4*b^2)*((12*a^7*b - 24*a^5*b^3)/a^8 - (tan(c/2 + (d*x)/2)*(3*a^7 + 48*a^3*b^4 -
36*a^5*b^2))/a^7 + (3*b*(-(a + b)*(a - b))^(1/2)*(2*a^2*b - (tan(c/2 + (d*x)/2)*(6*a^10 - 8*a^8*b^2))/a^7)*(3
*a^2 - 4*b^2))/(2*(a^7 - a^5*b^2))))/(2*(a^7 - a^5*b^2)) + (3*b*(-(a + b)*(a - b))^(1/2)*(3*a^2 - 4*b^2)*((tan
(c/2 + (d*x)/2)*(3*a^7 + 48*a^3*b^4 - 36*a^5*b^2))/a^7 - (12*a^7*b - 24*a^5*b^3)/a^8 + (3*b*(-(a + b)*(a - b))
^(1/2)*(2*a^2*b - (tan(c/2 + (d*x)/2)*(6*a^10 - 8*a^8*b^2))/a^7)*(3*a^2 - 4*b^2))/(2*(a^7 - a^5*b^2))))/(2*(a^
7 - a^5*b^2))))*(-(a + b)*(a - b))^(1/2)*(3*a^2 - 4*b^2)*3i)/(d*(a^7 - a^5*b^2))

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

[Out]

Timed out

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