3.1083 \(\int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=157 \[ \frac {3 b \cot (c+d x)}{a^3 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {2 b \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d \sqrt {a^2-b^2}}+\frac {\left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac {\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))} \]

[Out]

1/2*(a^2-6*b^2)*arctanh(cos(d*x+c))/a^4/d+3*b*cot(d*x+c)/a^3/d-3/2*cot(d*x+c)*csc(d*x+c)/a^2/d+cot(d*x+c)*csc(
d*x+c)/a/d/(a+b*sin(d*x+c))+2*b*(2*a^2-3*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 = 0.77, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2889, 3056, 3055, 3001, 3770, 2660, 618, 204} \[ \frac {2 b \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d \sqrt {a^2-b^2}}+\frac {\left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac {3 b \cot (c+d x)}{a^3 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(2*a^2 - 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*Sqrt[a^2 - b^2]*d) + ((a^2 - 6*b^2
)*ArcTanh[Cos[c + d*x]])/(2*a^4*d) + (3*b*Cot[c + d*x])/(a^3*d) - (3*Cot[c + d*x]*Csc[c + d*x])/(2*a^2*d) + (C
ot[c + d*x]*Csc[c + d*x])/(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 (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \frac {\csc ^3(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 (c+d x)}{a d (a+b \sin (c+d x))}+\frac {\int \frac {\csc ^3(c+d x) \left (3 \left (a^2-b^2\right )-2 \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 {3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}+\frac {\int \frac {\csc ^2(c+d x) \left (-6 b \left (a^2-b^2\right )-a \left (a^2-b^2\right ) \sin (c+d x)+3 b \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac {3 b \cot (c+d x)}{a^3 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}+\frac {\int \frac {\csc (c+d x) \left (-a^4+7 a^2 b^2-6 b^4+3 a b \left (a^2-b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {3 b \cot (c+d x)}{a^3 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}-\frac {\left (a^2-6 b^2\right ) \int \csc (c+d x) \, dx}{2 a^4}+\frac {\left (b \left (2 a^2-3 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^4}\\ &=\frac {\left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac {3 b \cot (c+d x)}{a^3 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}+\frac {\left (2 b \left (2 a^2-3 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 {\left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac {3 b \cot (c+d x)}{a^3 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}-\frac {\left (4 b \left (2 a^2-3 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 \left (2 a^2-3 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 {\left (a^2-6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac {3 b \cot (c+d x)}{a^3 d}-\frac {3 \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 3.09, size = 196, normalized size = 1.25 \[ \frac {-\frac {16 b \left (3 b^2-2 a^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-4 \left (a^2-6 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 \left (a^2-6 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-a^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )+a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+\frac {8 a b^2 \cos (c+d x)}{a+b \sin (c+d x)}-8 a b \tan \left (\frac {1}{2} (c+d x)\right )+8 a b \cot \left (\frac {1}{2} (c+d x)\right )}{8 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-16*b*(-2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 8*a*b*Cot[(c + d*
x)/2] - a^2*Csc[(c + d*x)/2]^2 + 4*(a^2 - 6*b^2)*Log[Cos[(c + d*x)/2]] - 4*(a^2 - 6*b^2)*Log[Sin[(c + d*x)/2]]
+ a^2*Sec[(c + d*x)/2]^2 + (8*a*b^2*Cos[c + d*x])/(a + b*Sin[c + d*x]) - 8*a*b*Tan[(c + d*x)/2])/(8*a^4*d)

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fricas [B]  time = 1.29, size = 1130, normalized size = 7.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 257, normalized size = 1.64 \[ -\frac {\frac {4 \, {\left (a^{2} - 6 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {16 \, {\left (2 \, a^{2} b - 3 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{4}} - \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} - \frac {16 \, {\left (b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a^{4}} - \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*(a^2 - 6*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 - 16*(2*a^2*b - 3*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^4) - (a^2*tan(1/2*d*x +
1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2*c))/a^4 - 16*(b^3*tan(1/2*d*x + 1/2*c) + a*b^2)/((a*tan(1/2*d*x + 1/2*c)^2
+ 2*b*tan(1/2*d*x + 1/2*c) + a)*a^4) - (6*a^2*tan(1/2*d*x + 1/2*c)^2 - 36*b^2*tan(1/2*d*x + 1/2*c)^2 + 8*a*b*t
an(1/2*d*x + 1/2*c) - a^2)/(a^4*tan(1/2*d*x + 1/2*c)^2))/d

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maple [B]  time = 0.81, size = 307, normalized size = 1.96 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{2} d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{d \,a^{3}}-\frac {1}{8 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{4}}+\frac {b}{d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2 b^{2}}{d \,a^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {4 b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{2} \sqrt {a^{2}-b^{2}}}-\frac {6 b^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{4} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d/a^2*tan(1/2*d*x+1/2*c)^2-1/d/a^3*tan(1/2*d*x+1/2*c)*b-1/8/a^2/d/tan(1/2*d*x+1/2*c)^2-1/2/d/a^2*ln(tan(1/
2*d*x+1/2*c))+3/d/a^4*ln(tan(1/2*d*x+1/2*c))*b^2+1/d*b/a^3/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)*tan(1/2*d*x+1/2*c)+2/d*b^2/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)
+4/d/a^2*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))-6/d*b^3/a^4/(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)^3/(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 = 9.89, size = 966, normalized size = 6.15 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{2}-16\,b^2\right )+\frac {a^2}{2}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^2\,b+2\,b^3\right )}{a}-3\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+8\,b\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-6\,b^2\right )}{2\,a^4\,d}-\frac {b\,\mathrm {atan}\left (\frac {\frac {b\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {5\,a^6\,b-12\,a^4\,b^3}{a^6}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^6-16\,a^4\,b^2+24\,a^2\,b^4\right )}{a^5}+\frac {b\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^8-8\,a^6\,b^2\right )}{a^5}\right )\,\left (2\,a^2-3\,b^2\right )}{a^6-a^4\,b^2}\right )\,1{}\mathrm {i}}{a^6-a^4\,b^2}-\frac {b\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^6-16\,a^4\,b^2+24\,a^2\,b^4\right )}{a^5}-\frac {5\,a^6\,b-12\,a^4\,b^3}{a^6}+\frac {b\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^8-8\,a^6\,b^2\right )}{a^5}\right )\,\left (2\,a^2-3\,b^2\right )}{a^6-a^4\,b^2}\right )\,1{}\mathrm {i}}{a^6-a^4\,b^2}}{\frac {2\,\left (2\,a^4\,b-15\,a^2\,b^3+18\,b^5\right )}{a^6}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (18\,b^4-12\,a^2\,b^2\right )}{a^5}+\frac {b\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {5\,a^6\,b-12\,a^4\,b^3}{a^6}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^6-16\,a^4\,b^2+24\,a^2\,b^4\right )}{a^5}+\frac {b\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^8-8\,a^6\,b^2\right )}{a^5}\right )\,\left (2\,a^2-3\,b^2\right )}{a^6-a^4\,b^2}\right )}{a^6-a^4\,b^2}+\frac {b\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^6-16\,a^4\,b^2+24\,a^2\,b^4\right )}{a^5}-\frac {5\,a^6\,b-12\,a^4\,b^3}{a^6}+\frac {b\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^8-8\,a^6\,b^2\right )}{a^5}\right )\,\left (2\,a^2-3\,b^2\right )}{a^6-a^4\,b^2}\right )}{a^6-a^4\,b^2}}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2-3\,b^2\right )\,2{}\mathrm {i}}{d\,\left (a^6-a^4\,b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (tan(c/2 + (d*x)/2)^2*(a^2/2 - 16*b^2) + a^2/2 - (4*tan(c/2 + (d*x)/2)^3*(a^2
*b + 2*b^3))/a - 3*a*b*tan(c/2 + (d*x)/2))/(d*(4*a^4*tan(c/2 + (d*x)/2)^2 + 4*a^4*tan(c/2 + (d*x)/2)^4 + 8*a^3
*b*tan(c/2 + (d*x)/2)^3)) - (b*tan(c/2 + (d*x)/2))/(a^3*d) - (log(tan(c/2 + (d*x)/2))*(a^2 - 6*b^2))/(2*a^4*d)
- (b*atan(((b*(-(a + b)*(a - b))^(1/2)*(2*a^2 - 3*b^2)*((5*a^6*b - 12*a^4*b^3)/a^6 - (tan(c/2 + (d*x)/2)*(a^6
+ 24*a^2*b^4 - 16*a^4*b^2))/a^5 + (b*(-(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)*(2*a^2 - 3*b^2))/(a^6 - a^4*b^2))*1i)/(a^6 - a^4*b^2) - (b*(-(a + b)*(a - b))^(1/2)*(2*a^2 - 3*b^2)*
((tan(c/2 + (d*x)/2)*(a^6 + 24*a^2*b^4 - 16*a^4*b^2))/a^5 - (5*a^6*b - 12*a^4*b^3)/a^6 + (b*(-(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)*(2*a^2 - 3*b^2))/(a^6 - a^4*b^2))*1i)/(a^6 - a
^4*b^2))/((2*(2*a^4*b + 18*b^5 - 15*a^2*b^3))/a^6 + (2*tan(c/2 + (d*x)/2)*(18*b^4 - 12*a^2*b^2))/a^5 + (b*(-(a
+ b)*(a - b))^(1/2)*(2*a^2 - 3*b^2)*((5*a^6*b - 12*a^4*b^3)/a^6 - (tan(c/2 + (d*x)/2)*(a^6 + 24*a^2*b^4 - 16*
a^4*b^2))/a^5 + (b*(-(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)*(2*a^2 -
3*b^2))/(a^6 - a^4*b^2)))/(a^6 - a^4*b^2) + (b*(-(a + b)*(a - b))^(1/2)*(2*a^2 - 3*b^2)*((tan(c/2 + (d*x)/2)*(
a^6 + 24*a^2*b^4 - 16*a^4*b^2))/a^5 - (5*a^6*b - 12*a^4*b^3)/a^6 + (b*(-(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)*(2*a^2 - 3*b^2))/(a^6 - a^4*b^2)))/(a^6 - a^4*b^2)))*(-(a + b)*(a -
b))^(1/2)*(2*a^2 - 3*b^2)*2i)/(d*(a^6 - a^4*b^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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