3.1139 \(\int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=182 \[ \frac {3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d \sqrt {a^2-b^2}}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2} \]

[Out]

3*b*arctanh(cos(d*x+c))/a^4/d+1/2*(a^2-3*b^2)*cos(d*x+c)/a^2/b/d/(a+b*sin(d*x+c))^2-cot(d*x+c)/a/d/(a+b*sin(d*
x+c))^2-1/2*(a^2+6*b^2)*cos(d*x+c)/a^3/b/d/(a+b*sin(d*x+c))-3*(a^2-2*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2
-b^2)^(1/2))/a^4/d/(a^2-b^2)^(1/2)

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Rubi [A]  time = 0.47, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2890, 3055, 3001, 3770, 2660, 618, 204} \[ -\frac {3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d \sqrt {a^2-b^2}}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}+\frac {3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(a^2 - 2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*Sqrt[a^2 - b^2]*d) + (3*b*ArcTanh[Cos
[c + d*x]])/(a^4*d) + ((a^2 - 3*b^2)*Cos[c + d*x])/(2*a^2*b*d*(a + b*Sin[c + d*x])^2) - Cot[c + d*x]/(a*d*(a +
b*Sin[c + d*x])^2) - ((a^2 + 6*b^2)*Cos[c + d*x])/(2*a^3*b*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 ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}+\frac {\int \frac {\csc (c+d x) \left (-6 b^2-2 a b \sin (c+d x)+\left (a^2+3 b^2\right ) \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx}{2 a^2 b}\\ &=\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}+\frac {\int \frac {\csc (c+d x) \left (-6 b^2 \left (a^2-b^2\right )-3 a b \left (a^2-b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^3 b \left (a^2-b^2\right )}\\ &=\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}-\frac {(3 b) \int \csc (c+d x) \, dx}{a^4}-\frac {\left (3 \left (a^2-2 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 a^4}\\ &=\frac {3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}-\frac {\left (3 \left (a^2-2 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^4 d}\\ &=\frac {3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}+\frac {\left (6 \left (a^2-2 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^4 d}\\ &=-\frac {3 \left (a^2-2 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^4 \sqrt {a^2-b^2} d}+\frac {3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 2.55, size = 184, normalized size = 1.01 \[ \frac {-\frac {6 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {a^2 \left (a^2-b^2\right ) \cos (c+d x)}{b (a+b \sin (c+d x))^2}-\frac {a \left (a^2+4 b^2\right ) \cos (c+d x)}{b (a+b \sin (c+d x))}+a \tan \left (\frac {1}{2} (c+d x)\right )-a \cot \left (\frac {1}{2} (c+d x)\right )-6 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*(a^2 - 2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - a*Cot[(c + d*x)/2] + 6*
b*Log[Cos[(c + d*x)/2]] - 6*b*Log[Sin[(c + d*x)/2]] + (a^2*(a^2 - b^2)*Cos[c + d*x])/(b*(a + b*Sin[c + d*x])^2
) - (a*(a^2 + 4*b^2)*Cos[c + d*x])/(b*(a + b*Sin[c + d*x])) + a*Tan[(c + d*x)/2])/(2*a^4*d)

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fricas [B]  time = 0.94, size = 1064, normalized size = 5.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(2*(a^5 + 5*a^3*b^2 - 6*a*b^4)*cos(d*x + c)^3 - 18*(a^4*b - a^2*b^3)*cos(d*x + c)*sin(d*x + c) + 3*(2*a^
3*b - 4*a*b^3 - 2*(a^3*b - 2*a*b^3)*cos(d*x + c)^2 + (a^4 - a^2*b^2 - 2*b^4 - (a^2*b^2 - 2*b^4)*cos(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^5 + a^3*b^2 - 2*a*b^4)*cos(d*x + c) + 6*(2*a^3*b^2 - 2*a*b^4 - 2*(a^3*b^2 - a*b^4)*cos(d*x + c)^2 +
(a^4*b - b^5 - (a^2*b^3 - b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 6*(2*a^3*b^2 - 2*a
*b^4 - 2*(a^3*b^2 - a*b^4)*cos(d*x + c)^2 + (a^4*b - b^5 - (a^2*b^3 - b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(-
1/2*cos(d*x + c) + 1/2))/(2*(a^7*b - a^5*b^3)*d*cos(d*x + c)^2 - 2*(a^7*b - a^5*b^3)*d + ((a^6*b^2 - a^4*b^4)*
d*cos(d*x + c)^2 - (a^8 - a^4*b^4)*d)*sin(d*x + c)), -1/2*((a^5 + 5*a^3*b^2 - 6*a*b^4)*cos(d*x + c)^3 - 9*(a^4
*b - a^2*b^3)*cos(d*x + c)*sin(d*x + c) + 3*(2*a^3*b - 4*a*b^3 - 2*(a^3*b - 2*a*b^3)*cos(d*x + c)^2 + (a^4 - a
^2*b^2 - 2*b^4 - (a^2*b^2 - 2*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))) - 3*(a^5 + a^3*b^2 - 2*a*b^4)*cos(d*x + c) + 3*(2*a^3*b^2 - 2*a*b^4 - 2*(a^3*b
^2 - a*b^4)*cos(d*x + c)^2 + (a^4*b - b^5 - (a^2*b^3 - b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c)
+ 1/2) - 3*(2*a^3*b^2 - 2*a*b^4 - 2*(a^3*b^2 - a*b^4)*cos(d*x + c)^2 + (a^4*b - b^5 - (a^2*b^3 - b^5)*cos(d*x
+ c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/(2*(a^7*b - a^5*b^3)*d*cos(d*x + c)^2 - 2*(a^7*b - a^5*b^
3)*d + ((a^6*b^2 - a^4*b^4)*d*cos(d*x + c)^2 - (a^8 - a^4*b^4)*d)*sin(d*x + c))]

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giac [A]  time = 0.26, size = 273, normalized size = 1.50 \[ -\frac {\frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} + \frac {6 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (a^{2} - 2 \, b^{2}\right )}}{\sqrt {a^{2} - b^{2}} a^{4}} - \frac {6 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 14 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{2} b\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2} a^{4}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 - tan(1/2*d*x + 1/2*c)/a^3 + 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)))*(a^2 - 2*b^2)/(sqrt(a^2 - b^2)*a^4) - (6*b*tan(
1/2*d*x + 1/2*c) - a)/(a^4*tan(1/2*d*x + 1/2*c)) - 2*(a^3*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^2*tan(1/2*d*x + 1/2*c
)^3 - 5*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 10*b^3*tan(1/2*d*x + 1/2*c)^2 - a^3*tan(1/2*d*x + 1/2*c) - 14*a*b^2*tan
(1/2*d*x + 1/2*c) - 5*a^2*b)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2*a^4))/d

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maple [B]  time = 0.79, size = 489, normalized size = 2.69 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{3}}-\frac {1}{2 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{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 {6 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) 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 {5 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{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 {10 \left (\tan ^{2}\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 {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{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 {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) 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 {5 b}{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 {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^{2} \sqrt {a^{2}-b^{2}}}+\frac {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 ) b^{2}}{d \,a^{4} \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)^2/(a+b*sin(d*x+c))^3,x)

[Out]

1/2/d/a^3*tan(1/2*d*x+1/2*c)-1/2/d/a^3/tan(1/2*d*x+1/2*c)-3/d/a^4*b*ln(tan(1/2*d*x+1/2*c))+1/d/a/(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)^3-6/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*tan(1/2*d*x+1/2*c)^3*b^2-5/d/a^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*b-10/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)^2*b^3-1/d/a/(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)-14/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*tan(1/2*d*x+1/2*c)*b^2-5/d/a^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b-3/d/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))+6/d/a^4/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*ta
n(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))*b^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)^2/(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 = 9.88, size = 956, normalized size = 5.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^2*(a + b*sin(c + d*x))^3),x)

[Out]

tan(c/2 + (d*x)/2)/(2*a^3*d) - (tan(c/2 + (d*x)/2)^2*(4*a^2 + 32*b^2) + a^2 - tan(c/2 + (d*x)/2)^4*(a^2 - 12*b
^2) + (2*tan(c/2 + (d*x)/2)^3*(7*a^2*b + 10*b^3))/a + 14*a*b*tan(c/2 + (d*x)/2))/(d*(2*a^5*tan(c/2 + (d*x)/2)^
5 + tan(c/2 + (d*x)/2)^3*(4*a^5 + 8*a^3*b^2) + 2*a^5*tan(c/2 + (d*x)/2) + 8*a^4*b*tan(c/2 + (d*x)/2)^2 + 8*a^4
*b*tan(c/2 + (d*x)/2)^4)) - (3*b*log(tan(c/2 + (d*x)/2)))/(a^4*d) - (atan((((-(a + b)*(a - b))^(1/2)*(a^2 - 2*
b^2)*((3*a^6 - 12*a^4*b^2)/a^6 + (tan(c/2 + (d*x)/2)*(12*a^4*b - 24*a^2*b^3))/a^5 - (3*(-(a + b)*(a - b))^(1/2
)*(2*a^2*b - (tan(c/2 + (d*x)/2)*(6*a^8 - 8*a^6*b^2))/a^5)*(a^2 - 2*b^2))/(2*(a^6 - a^4*b^2)))*3i)/(2*(a^6 - a
^4*b^2)) + ((-(a + b)*(a - b))^(1/2)*(a^2 - 2*b^2)*((3*a^6 - 12*a^4*b^2)/a^6 + (tan(c/2 + (d*x)/2)*(12*a^4*b -
24*a^2*b^3))/a^5 + (3*(-(a + b)*(a - b))^(1/2)*(2*a^2*b - (tan(c/2 + (d*x)/2)*(6*a^8 - 8*a^6*b^2))/a^5)*(a^2
- 2*b^2))/(2*(a^6 - a^4*b^2)))*3i)/(2*(a^6 - a^4*b^2)))/((2*(9*a^2*b - 18*b^3))/a^6 + (2*tan(c/2 + (d*x)/2)*(9
*a^2 - 18*b^2))/a^5 + (3*(-(a + b)*(a - b))^(1/2)*(a^2 - 2*b^2)*((3*a^6 - 12*a^4*b^2)/a^6 + (tan(c/2 + (d*x)/2
)*(12*a^4*b - 24*a^2*b^3))/a^5 - (3*(-(a + b)*(a - b))^(1/2)*(2*a^2*b - (tan(c/2 + (d*x)/2)*(6*a^8 - 8*a^6*b^2
))/a^5)*(a^2 - 2*b^2))/(2*(a^6 - a^4*b^2))))/(2*(a^6 - a^4*b^2)) - (3*(-(a + b)*(a - b))^(1/2)*(a^2 - 2*b^2)*(
(3*a^6 - 12*a^4*b^2)/a^6 + (tan(c/2 + (d*x)/2)*(12*a^4*b - 24*a^2*b^3))/a^5 + (3*(-(a + b)*(a - b))^(1/2)*(2*a
^2*b - (tan(c/2 + (d*x)/2)*(6*a^8 - 8*a^6*b^2))/a^5)*(a^2 - 2*b^2))/(2*(a^6 - a^4*b^2))))/(2*(a^6 - a^4*b^2)))
)*(-(a + b)*(a - b))^(1/2)*(a^2 - 2*b^2)*3i)/(d*(a^6 - a^4*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{4}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**2/(a+b*sin(d*x+c))**3,x)

[Out]

Integral(cos(c + d*x)**4*csc(c + d*x)**2/(a + b*sin(c + d*x))**3, x)

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