3.1084 \(\int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=193 \[ \frac {2 b \cot (c+d x) \csc (c+d x)}{a^3 d}-\frac {4 \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}-\frac {2 b^2 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^5 d \sqrt {a^2-b^2}}-\frac {b \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac {\left (a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}+\frac {\cot (c+d x) \csc ^2(c+d x)}{a d (a+b \sin (c+d x))} \]

[Out]

-b*(a^2-4*b^2)*arctanh(cos(d*x+c))/a^5/d+1/3*(a^2-12*b^2)*cot(d*x+c)/a^4/d+2*b*cot(d*x+c)*csc(d*x+c)/a^3/d-4/3
*cot(d*x+c)*csc(d*x+c)^2/a^2/d+cot(d*x+c)*csc(d*x+c)^2/a/d/(a+b*sin(d*x+c))-2*b^2*(3*a^2-4*b^2)*arctan((b+a*ta
n(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 = 1.04, antiderivative size = 193, 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} \[ -\frac {2 b^2 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^5 d \sqrt {a^2-b^2}}+\frac {\left (a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac {b \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac {2 b \cot (c+d x) \csc (c+d x)}{a^3 d}-\frac {4 \cot (c+d x) \csc ^2(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc ^2(c+d x)}{a d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 6.34, size = 385, normalized size = 1.99 \[ -\frac {b^3 \cos (c+d x)}{a^4 d (a+b \sin (c+d x))}+\frac {b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{4 a^3 d}-\frac {b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{4 a^3 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 a^2 d}+\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{24 a^2 d}+\frac {\left (a^2 b-4 b^3\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}+\frac {\left (4 b^3-a^2 b\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^5 d}-\frac {2 b^2 \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (a \sin \left (\frac {1}{2} (c+d x)\right )+b \cos \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^5 d \sqrt {a^2-b^2}}+\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \left (a^2 \cos \left (\frac {1}{2} (c+d x)\right )-9 b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^4 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (9 b^2 \sin \left (\frac {1}{2} (c+d x)\right )-a^2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^4 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*b^2*(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) + ((a^2*Cos[(c + d*x)/2] - 9*b^2*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(6*a^4*d) + (b*Cs
c[(c + d*x)/2]^2)/(4*a^3*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*a^2*d) + ((-(a^2*b) + 4*b^3)*Log[Cos[(
c + d*x)/2]])/(a^5*d) + ((a^2*b - 4*b^3)*Log[Sin[(c + d*x)/2]])/(a^5*d) - (b*Sec[(c + d*x)/2]^2)/(4*a^3*d) + (
Sec[(c + d*x)/2]*(-(a^2*Sin[(c + d*x)/2]) + 9*b^2*Sin[(c + d*x)/2]))/(6*a^4*d) - (b^3*Cos[c + d*x])/(a^4*d*(a
+ b*Sin[c + d*x])) + (Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(24*a^2*d)

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fricas [B]  time = 0.84, size = 1471, normalized size = 7.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(2*(a^6 - 7*a^4*b^2 + 6*a^2*b^4)*cos(d*x + c)^3 - 3*(3*a^2*b^3 - 4*b^5 + (3*a^2*b^3 - 4*b^5)*cos(d*x + c
)^4 - 2*(3*a^2*b^3 - 4*b^5)*cos(d*x + c)^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)*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)) +
12*(a^4*b^2 - a^2*b^4)*cos(d*x + c) + 3*(a^4*b^2 - 5*a^2*b^4 + 4*b^6 + (a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x +
c)^4 - 2*(a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^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^4*b^2 - 5*a^2*b^4 + 4*b^6 + (a^4*b^2
- 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^4 - 2*(a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^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 - 5*a^3*b^3 + 4*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^7*b -
a^5*b^3)*d*cos(d*x + c)^4 - 2*(a^7*b - a^5*b^3)*d*cos(d*x + c)^2 + (a^7*b - a^5*b^3)*d - ((a^8 - a^6*b^2)*d*c
os(d*x + c)^2 - (a^8 - a^6*b^2)*d)*sin(d*x + c)), -1/6*(2*(a^6 - 7*a^4*b^2 + 6*a^2*b^4)*cos(d*x + c)^3 - 6*(3*
a^2*b^3 - 4*b^5 + (3*a^2*b^3 - 4*b^5)*cos(d*x + c)^4 - 2*(3*a^2*b^3 - 4*b^5)*cos(d*x + c)^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))) + 12*(a^4*b^2 - a^2*b^4)*cos(d*x + c) + 3*(a^4*b^2 - 5*a^2*b^4 + 4*b^6 + (a^4*b^2 -
5*a^2*b^4 + 4*b^6)*cos(d*x + c)^4 - 2*(a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^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^4*b^2 - 5
*a^2*b^4 + 4*b^6 + (a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)^4 - 2*(a^4*b^2 - 5*a^2*b^4 + 4*b^6)*cos(d*x + c)
^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 - 5*a^3*b^3 + 4*a*b^5)*cos(d*x +
c))*sin(d*x + c))/((a^7*b - a^5*b^3)*d*cos(d*x + c)^4 - 2*(a^7*b - a^5*b^3)*d*cos(d*x + c)^2 + (a^7*b - a^5*b
^3)*d - ((a^8 - a^6*b^2)*d*cos(d*x + c)^2 - (a^8 - a^6*b^2)*d)*sin(d*x + c))]

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giac [A]  time = 0.22, size = 329, normalized size = 1.70 \[ \frac {\frac {24 \, {\left (a^{2} b - 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{5}} - \frac {48 \, {\left (3 \, a^{2} b^{2} - 4 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{5}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}} - \frac {48 \, {\left (b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a b^{3}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a^{5}} - \frac {44 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 176 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(24*(a^2*b - 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c)))/a^5 - 48*(3*a^2*b^2 - 4*b^4)*(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) + (a^4*tan(1/2*d
*x + 1/2*c)^3 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^2 - 3*a^4*tan(1/2*d*x + 1/2*c) + 36*a^2*b^2*tan(1/2*d*x + 1/2*c))
/a^6 - 48*(b^4*tan(1/2*d*x + 1/2*c) + a*b^3)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*a^5) -
(44*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 176*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*a^3*tan(1/2*d*x + 1/2*c)^2 + 36*a*b^2*t
an(1/2*d*x + 1/2*c)^2 - 6*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/(a^5*tan(1/2*d*x + 1/2*c)^3))/d

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maple [B]  time = 0.82, size = 390, normalized size = 2.02 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{2}}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{4 d \,a^{3}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{2}}+\frac {3 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{4}}-\frac {1}{24 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{8 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 b^{2}}{2 d \,a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{4 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}-\frac {4 b^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{5}}-\frac {2 b^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{5} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {2 b^{3}}{d \,a^{4} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {6 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) b^{2}}{d \,a^{3} \sqrt {a^{2}-b^{2}}}+\frac {8 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{5} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24/d/a^2*tan(1/2*d*x+1/2*c)^3-1/4/d/a^3*tan(1/2*d*x+1/2*c)^2*b-1/8/d/a^2*tan(1/2*d*x+1/2*c)+3/2/d/a^4*b^2*ta
n(1/2*d*x+1/2*c)-1/24/a^2/d/tan(1/2*d*x+1/2*c)^3+1/8/d/a^2/tan(1/2*d*x+1/2*c)-3/2/d/a^4/tan(1/2*d*x+1/2*c)*b^2
+1/4/d/a^3*b/tan(1/2*d*x+1/2*c)^2+1/d/a^3*b*ln(tan(1/2*d*x+1/2*c))-4/d/a^5*b^3*ln(tan(1/2*d*x+1/2*c))-2/d*b^4/
a^5/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*tan(1/2*d*x+1/2*c)-2/d*b^3/a^4/(tan(1/2*d*x+1/2*c)^2*a+2
*tan(1/2*d*x+1/2*c)*b+a)-6/d/a^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))*b^2+
8/d*b^4/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))

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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)^4/(a+b*sin(d*x+c))^2,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 = 9.99, size = 1089, normalized size = 5.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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