3.1132 \(\int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=158 \[ \frac {2 b \cot (c+d x)}{a^3 d}-\frac {\cos (c+d x)}{2 a^2 d \left (1-\cos ^2(c+d x)\right )}+\frac {6 b \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d}+\frac {3 \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a^3 d (a+b \sin (c+d x))} \]

[Out]

3/2*(a^2-2*b^2)*arctanh(cos(d*x+c))/a^4/d-1/2*cos(d*x+c)/a^2/d/(1-cos(d*x+c)^2)+2*b*cot(d*x+c)/a^3/d-(a^2-b^2)
*cos(d*x+c)/a^3/d/(a+b*sin(d*x+c))+6*b*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 = 0.45, antiderivative size = 180, normalized size of antiderivative = 1.14, number of steps used = 7, 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} \[ \frac {6 b \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d}-\frac {\left (a^2-3 b^2\right ) \cot (c+d x)}{a^3 b d}+\frac {3 \left (a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^4 d}+\frac {\left (2 a^2-3 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 4.30, size = 191, normalized size = 1.21 \[ \frac {48 b \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )-12 \left (a^2-2 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 \left (a^2-2 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {8 a \left (b^2-a^2\right ) \cos (c+d x)}{a+b \sin (c+d x)}-a^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )+a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )-8 a b \tan \left (\frac {1}{2} (c+d x)\right )+8 a b \cot \left (\frac {1}{2} (c+d x)\right )}{8 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*b*Sqrt[a^2 - b^2]*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 8*a*b*Cot[(c + d*x)/2] - a^2*Csc[(c +
d*x)/2]^2 + 12*(a^2 - 2*b^2)*Log[Cos[(c + d*x)/2]] - 12*(a^2 - 2*b^2)*Log[Sin[(c + d*x)/2]] + a^2*Sec[(c + d*
x)/2]^2 + (8*a*(-a^2 + 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 = 0.99, size = 804, normalized size = 5.09 \[ \left [-\frac {6 \, a^{2} b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (a b \cos \left (d x + c\right )^{2} - a b + {\left (b^{2} \cos \left (d x + c\right )^{2} - b^{2}\right )} \sin \left (d x + c\right )\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 ) - 6 \, {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right ) + 3 \, {\left (a^{3} - 2 \, a b^{2} - {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{2} b - 2 \, b^{3} - {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (a^{3} - 2 \, a b^{2} - {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{2} b - 2 \, b^{3} - {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{5} d \cos \left (d x + c\right )^{2} - a^{5} d + {\left (a^{4} b d \cos \left (d x + c\right )^{2} - a^{4} b d\right )} \sin \left (d x + c\right )\right )}}, -\frac {6 \, a^{2} b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (a b \cos \left (d x + c\right )^{2} - a b + {\left (b^{2} \cos \left (d x + c\right )^{2} - b^{2}\right )} \sin \left (d x + c\right )\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 ) - 6 \, {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right ) + 3 \, {\left (a^{3} - 2 \, a b^{2} - {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{2} b - 2 \, b^{3} - {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (a^{3} - 2 \, a b^{2} - {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{2} b - 2 \, b^{3} - {\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{5} d \cos \left (d x + c\right )^{2} - a^{5} d + {\left (a^{4} b d \cos \left (d x + c\right )^{2} - a^{4} b d\right )} \sin \left (d x + c\right )\right )}}\right ] \]

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))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 275, normalized size = 1.74 \[ -\frac {\frac {12 \, {\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {48 \, {\left (a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{4}} - \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} + \frac {16 \, {\left (a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} - a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a^{4}} - \frac {18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(12*(a^2 - 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 - 48*(a^2*b - 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*(a^2*b*tan(1/2*d*x + 1/2*c) - b^3*tan(1/2*d*x + 1/2*c) + a^3 - 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) - (18*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*tan(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 = 0.87, size = 339, normalized size = 2.15 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{2} d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{d \,a^{3}}-\frac {1}{8 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{4}}+\frac {b}{d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {2}{d a \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2 b^{2}}{d \,a^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {6 b \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*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)^2-3/2/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/a^2*b/(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)*t
an(1/2*d*x+1/2*c)-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/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)+6/d/a^4*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))

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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))^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 = 9.73, size = 675, normalized size = 4.27 \[ \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 {17\,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 (3\,a^2-6\,b^2\right )}{2\,a^4\,d}-\frac {6\,b\,\mathrm {atanh}\left (\frac {72\,b^4\,\sqrt {b^2-a^2}}{18\,a^4\,b+72\,b^5-90\,a^2\,b^3-216\,a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+72\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {144\,b^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}}-\frac {54\,b^2\,\sqrt {b^2-a^2}}{18\,a^2\,b-90\,b^3+\frac {72\,b^5}{a^2}+72\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {216\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}+\frac {144\,b^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3}}+\frac {18\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{18\,a\,b-\frac {90\,b^3}{a}+\frac {72\,b^5}{a^3}+72\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {216\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2}+\frac {144\,b^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4}}-\frac {144\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{18\,a^3\,b-90\,a\,b^3+\frac {72\,b^5}{a}-216\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+72\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {144\,b^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2}}+\frac {144\,b^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{18\,a^5\,b+72\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^2-90\,a^3\,b^3-216\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^4+72\,a\,b^5+144\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^6}\right )\,\sqrt {b^2-a^2}}{a^4\,d} \]

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))^2),x)

[Out]

tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (tan(c/2 + (d*x)/2)^2*((17*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))*(3*a^2 - 6*b^2))/(2
*a^4*d) - (6*b*atanh((72*b^4*(b^2 - a^2)^(1/2))/(18*a^4*b + 72*b^5 - 90*a^2*b^3 - 216*a*b^4*tan(c/2 + (d*x)/2)
+ 72*a^3*b^2*tan(c/2 + (d*x)/2) + (144*b^6*tan(c/2 + (d*x)/2))/a) - (54*b^2*(b^2 - a^2)^(1/2))/(18*a^2*b - 90
*b^3 + (72*b^5)/a^2 + 72*a*b^2*tan(c/2 + (d*x)/2) - (216*b^4*tan(c/2 + (d*x)/2))/a + (144*b^6*tan(c/2 + (d*x)/
2))/a^3) + (18*b*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(18*a*b - (90*b^3)/a + (72*b^5)/a^3 + 72*b^2*tan(c/2 +
(d*x)/2) - (216*b^4*tan(c/2 + (d*x)/2))/a^2 + (144*b^6*tan(c/2 + (d*x)/2))/a^4) - (144*b^3*tan(c/2 + (d*x)/2)*
(b^2 - a^2)^(1/2))/(18*a^3*b - 90*a*b^3 + (72*b^5)/a - 216*b^4*tan(c/2 + (d*x)/2) + 72*a^2*b^2*tan(c/2 + (d*x)
/2) + (144*b^6*tan(c/2 + (d*x)/2))/a^2) + (144*b^5*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(72*a*b^5 + 18*a^5*b
- 90*a^3*b^3 + 144*b^6*tan(c/2 + (d*x)/2) - 216*a^2*b^4*tan(c/2 + (d*x)/2) + 72*a^4*b^2*tan(c/2 + (d*x)/2)))*(
b^2 - a^2)^(1/2))/(a^4*d)

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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))**2,x)

[Out]

Timed out

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