3.1292 \(\int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=153 \[ \frac {b \cot (c+d x) \csc (c+d x)}{2 a^2 d}-\frac {2 b^2 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d}-\frac {b \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac {\left (a^2-3 b^2\right ) \cot (c+d x)}{3 a^3 d}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.67, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ -\frac {2 b^2 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \cot (c+d x)}{3 a^3 d}-\frac {b \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac {b \cot (c+d x) \csc (c+d x)}{2 a^2 d}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [B]  time = 6.23, size = 351, normalized size = 2.29 \[ \frac {b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}-\frac {b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}+\frac {\left (a^2 b-2 b^3\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac {\left (2 b^3-a^2 b\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}-\frac {2 b^2 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (a \sin \left (\frac {1}{2} (c+d x)\right )+b \cos \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^4 d}+\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \left (a^2 \cos \left (\frac {1}{2} (c+d x)\right )-3 b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^3 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (3 b^2 \sin \left (\frac {1}{2} (c+d x)\right )-a^2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^3 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 a d}+\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{24 a d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*b^2*Sqrt[a^2 - 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^4*d) + ((a^2*Cos[(c + d*x)/2] - 3*b^2*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(6*a^3*d) + (b*Csc[(c + d*x)/2]^2
)/(8*a^2*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*a*d) + ((-(a^2*b) + 2*b^3)*Log[Cos[(c + d*x)/2]])/(2*a
^4*d) + ((a^2*b - 2*b^3)*Log[Sin[(c + d*x)/2]])/(2*a^4*d) - (b*Sec[(c + d*x)/2]^2)/(8*a^2*d) + (Sec[(c + d*x)/
2]*(-(a^2*Sin[(c + d*x)/2]) + 3*b^2*Sin[(c + d*x)/2]))/(6*a^3*d) + (Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(24*a
*d)

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fricas [A]  time = 1.13, size = 591, normalized size = 3.86 \[ \left [-\frac {6 \, a^{2} b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 12 \, a b^{2} \cos \left (d x + c\right ) - 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (b^{2} \cos \left (d x + c\right )^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a^{2} b - 2 \, b^{3} - {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{2} b - 2 \, b^{3} - {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right )}{12 \, {\left (a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d\right )} \sin \left (d x + c\right )}, -\frac {6 \, a^{2} b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 12 \, a b^{2} \cos \left (d x + c\right ) - 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 12 \, {\left (b^{2} \cos \left (d x + c\right )^{2} - b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a^{2} b - 2 \, b^{3} - {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{2} b - 2 \, b^{3} - {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right )}{12 \, {\left (a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d\right )} \sin \left (d x + c\right )}\right ] \]

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)),x, algorithm="fricas")

[Out]

[-1/12*(6*a^2*b*cos(d*x + c)*sin(d*x + c) - 12*a*b^2*cos(d*x + c) - 4*(a^3 - 3*a*b^2)*cos(d*x + c)^3 - 6*(b^2*
cos(d*x + c)^2 - b^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) - 3*(a^2*b - 2*b^3 - (a^2*b - 2*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*
x + c) + 3*(a^2*b - 2*b^3 - (a^2*b - 2*b^3)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c))/((a^4*d
*cos(d*x + c)^2 - a^4*d)*sin(d*x + c)), -1/12*(6*a^2*b*cos(d*x + c)*sin(d*x + c) - 12*a*b^2*cos(d*x + c) - 4*(
a^3 - 3*a*b^2)*cos(d*x + c)^3 - 12*(b^2*cos(d*x + c)^2 - b^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sq
rt(a^2 - b^2)*cos(d*x + c)))*sin(d*x + c) - 3*(a^2*b - 2*b^3 - (a^2*b - 2*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x
+ c) + 1/2)*sin(d*x + c) + 3*(a^2*b - 2*b^3 - (a^2*b - 2*b^3)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2)*si
n(d*x + c))/((a^4*d*cos(d*x + c)^2 - a^4*d)*sin(d*x + c))]

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giac [A]  time = 0.19, size = 270, normalized size = 1.76 \[ \frac {\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} + \frac {12 \, {\left (a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {48 \, {\left (a^{2} b^{2} - b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{4}} - \frac {22 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

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)),x, algorithm="giac")

[Out]

1/24*((a^2*tan(1/2*d*x + 1/2*c)^3 - 3*a*b*tan(1/2*d*x + 1/2*c)^2 - 3*a^2*tan(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2
*d*x + 1/2*c))/a^3 + 12*(a^2*b - 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 - 48*(a^2*b^2 - b^4)*(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^4) - (22
*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 44*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a*b^2*tan(1/
2*d*x + 1/2*c)^2 - 3*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/(a^4*tan(1/2*d*x + 1/2*c)^3))/d

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maple [A]  time = 0.46, size = 250, normalized size = 1.63 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{8 d \,a^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{3}}-\frac {1}{24 d a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b^{2}}{2 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{8 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}-\frac {b^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}}-\frac {2 \sqrt {a^{2}-b^{2}}\, b^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{4}} \]

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)),x)

[Out]

1/24/d/a*tan(1/2*d*x+1/2*c)^3-1/8/d/a^2*tan(1/2*d*x+1/2*c)^2*b-1/8/a/d*tan(1/2*d*x+1/2*c)+1/2/d/a^3*b^2*tan(1/
2*d*x+1/2*c)-1/24/d/a/tan(1/2*d*x+1/2*c)^3+1/8/a/d/tan(1/2*d*x+1/2*c)-1/2/d/a^3/tan(1/2*d*x+1/2*c)*b^2+1/8/d/a
^2*b/tan(1/2*d*x+1/2*c)^2+1/2/d/a^2*b*ln(tan(1/2*d*x+1/2*c))-1/d/a^4*b^3*ln(tan(1/2*d*x+1/2*c))-2/d*(a^2-b^2)^
(1/2)/a^4*b^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)),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 = 13.51, size = 749, normalized size = 4.90 \[ \frac {a^3\,\left (\frac {\cos \left (c+d\,x\right )}{8}+\frac {\cos \left (3\,c+3\,d\,x\right )}{24}\right )-a^2\,\left (\frac {b\,\sin \left (2\,c+2\,d\,x\right )}{8}-\frac {b\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sin \left (3\,c+3\,d\,x\right )}{16}+\frac {3\,b\,\sin \left (c+d\,x\right )\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{16}\right )+a\,\left (\frac {b^2\,\cos \left (c+d\,x\right )}{8}-\frac {b^2\,\cos \left (3\,c+3\,d\,x\right )}{8}\right )+\frac {3\,b^3\,\sin \left (c+d\,x\right )\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{8}-\frac {b^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {b^2\,\sin \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {a^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}-a^2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}+a\,b^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}-a^3\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,3{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2-12\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {b^2-a^2}\,3{}\mathrm {i}}{4}-\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )\,\mathrm {atan}\left (\frac {a^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}-a^2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}+a\,b^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}-a^3\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,3{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2-12\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{4}}{\frac {a^4\,d\,\sin \left (3\,c+3\,d\,x\right )}{8}-\frac {3\,a^4\,d\,\sin \left (c+d\,x\right )}{8}} \]

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))),x)

[Out]

(a^3*(cos(c + d*x)/8 + cos(3*c + 3*d*x)/24) - a^2*((b*sin(2*c + 2*d*x))/8 - (b*log(sin(c/2 + (d*x)/2)/cos(c/2
+ (d*x)/2))*sin(3*c + 3*d*x))/16 + (3*b*sin(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/16) + a*((b^2
*cos(c + d*x))/8 - (b^2*cos(3*c + 3*d*x))/8) + (3*b^3*sin(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))
/8 - (b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x))/8 + (b^2*sin(c + d*x)*atan((a^4*sin(c/2
+ (d*x)/2)*(b^2 - a^2)^(1/2)*1i + b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i - a^2*b^2*sin(c/2 + (d*x)/2)*(b
^2 - a^2)^(1/2)*8i + a*b^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i - a^3*b*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2
)*3i)/(a^5*cos(c/2 + (d*x)/2) + 8*b^5*sin(c/2 + (d*x)/2) + 4*a*b^4*cos(c/2 + (d*x)/2) + 4*a^4*b*sin(c/2 + (d*x
)/2) - 5*a^3*b^2*cos(c/2 + (d*x)/2) - 12*a^2*b^3*sin(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2)*3i)/4 - (b^2*sin(3*c +
3*d*x)*atan((a^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i + b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i - a^2*
b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i + a*b^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i - a^3*b*cos(c/2 +
(d*x)/2)*(b^2 - a^2)^(1/2)*3i)/(a^5*cos(c/2 + (d*x)/2) + 8*b^5*sin(c/2 + (d*x)/2) + 4*a*b^4*cos(c/2 + (d*x)/2)
+ 4*a^4*b*sin(c/2 + (d*x)/2) - 5*a^3*b^2*cos(c/2 + (d*x)/2) - 12*a^2*b^3*sin(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/
2)*1i)/4)/((a^4*d*sin(3*c + 3*d*x))/8 - (3*a^4*d*sin(c + d*x))/8)

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

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)),x)

[Out]

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

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