\(\int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [1138]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 175 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {x}{b^3}-\frac {\left (2 a^4-a^2 b^2+2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b^3 \sqrt {a^2-b^2} d}-\frac {\text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2970, 3136, 2739, 632, 210, 3855} \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}-\frac {\left (2 a^4-a^2 b^2+2 b^4\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 b^3 d \sqrt {a^2-b^2}}+\frac {x}{b^3} \]

[In]

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

[Out]

x/b^3 - ((2*a^4 - a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^3*Sqrt[a^2 - b^2]*
d) - ArcTanh[Cos[c + d*x]]/(a^3*d) - ((a^2 - b^2)*Cos[c + d*x])/(2*a*b^2*d*(a + b*Sin[c + d*x])^2) + ((3*a^2 +
 2*b^2)*Cos[c + d*x])/(2*a^2*b^2*d*(a + b*Sin[c + d*x]))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

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 2739

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 2970

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(a^2 - b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 1)/(a*b^2*d*f
*(m + 1))), x] + (-Dist[1/(a^2*b^2*(m + 1)*(m + 2)), Int[(a + b*Sin[e + f*x])^(m + 2)*(d*Sin[e + f*x])^n*Simp[
a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 3) + a*b*(m + 2)*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*(m +
 n + 2)*(m + n + 4))*Sin[e + f*x]^2, x], x], x] + Simp[(a^2*(n - m + 1) - b^2*(m + n + 2))*Cos[e + f*x]*(a + b
*Sin[e + f*x])^(m + 2)*((d*Sin[e + f*x])^(n + 1)/(a^2*b^2*d*f*(m + 1)*(m + 2))), x]) /; FreeQ[{a, b, d, e, f,
n}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*m, 2*n] && LtQ[m, -1] &&  !LtQ[n, -1] && (LtQ[m, -2] || EqQ[m + n +
 4, 0])

Rule 3136

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C*(x/(b*d)), x] + (Dist[(A*b^2 - a*b*B + a
^2*C)/(b*(b*c - a*d)), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)), Int[
1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac {\int \frac {\csc (c+d x) \left (-2 b^2-a b \sin (c+d x)-2 a^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^2 b^2} \\ & = \frac {x}{b^3}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}+\frac {\int \csc (c+d x) \, dx}{a^3}+\frac {\left (-2 a^4+a^2 b^2-2 b^4\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 a^3 b^3} \\ & = \frac {x}{b^3}-\frac {\text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac {\left (2 a^4-a^2 b^2+2 b^4\right ) \text {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^3 b^3 d} \\ & = \frac {x}{b^3}-\frac {\text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}+\frac {\left (2 \left (2 a^4-a^2 b^2+2 b^4\right )\right ) \text {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^3 b^3 d} \\ & = \frac {x}{b^3}-\frac {\left (2 a^4-a^2 b^2+2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b^3 \sqrt {a^2-b^2} d}-\frac {\text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (3 a^2+2 b^2\right ) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.66 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.01 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {-\frac {2 \left (2 a^4-a^2 b^2+2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b^3 \sqrt {a^2-b^2}}+2 \left (\frac {c+d x}{b^3}-\frac {\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}\right )+\frac {\cos (c+d x) \left (2 a^3+3 a b^2+b \left (3 a^2+2 b^2\right ) \sin (c+d x)\right )}{a^2 b^2 (a+b \sin (c+d x))^2}}{2 d} \]

[In]

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

[Out]

((-2*(2*a^4 - a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^3*Sqrt[a^2 - b^2]) + 2
*((c + d*x)/b^3 - Log[Cos[(c + d*x)/2]]/a^3 + Log[Sin[(c + d*x)/2]]/a^3) + (Cos[c + d*x]*(2*a^3 + 3*a*b^2 + b*
(3*a^2 + 2*b^2)*Sin[c + d*x]))/(a^2*b^2*(a + b*Sin[c + d*x])^2))/(2*d)

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.37

method result size
derivativedivides \(\frac {-\frac {2 \left (\frac {\left (-\frac {1}{2} a^{3} b^{2}-2 a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {b \left (2 a^{4}+7 a^{2} b^{2}+6 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {a \,b^{2} \left (7 a^{2}+8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a^{2} b \left (2 a^{2}+3 b^{2}\right )}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {\left (2 a^{4}-a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3} b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\) \(240\)
default \(\frac {-\frac {2 \left (\frac {\left (-\frac {1}{2} a^{3} b^{2}-2 a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {b \left (2 a^{4}+7 a^{2} b^{2}+6 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {a \,b^{2} \left (7 a^{2}+8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a^{2} b \left (2 a^{2}+3 b^{2}\right )}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {\left (2 a^{4}-a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3} b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}}{d}\) \(240\)
risch \(\frac {x}{b^{3}}-\frac {i \left (-4 i a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-i a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+8 i a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}+7 i a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+6 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+7 b^{2} a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-3 a^{2} b^{2}-2 b^{4}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} a^{2} d \,b^{3}}+\frac {i a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, d b a}+\frac {i b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}-\frac {i a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}+\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, d b a}-\frac {i b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}\) \(643\)

[In]

int(cos(d*x+c)^4*csc(d*x+c)/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-2/a^3/b^3*(((-1/2*a^3*b^2-2*a*b^4)*tan(1/2*d*x+1/2*c)^3-1/2*b*(2*a^4+7*a^2*b^2+6*b^4)*tan(1/2*d*x+1/2*c)
^2-1/2*a*b^2*(7*a^2+8*b^2)*tan(1/2*d*x+1/2*c)-1/2*a^2*b*(2*a^2+3*b^2))/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x
+1/2*c)+a)^2+1/2*(2*a^4-a^2*b^2+2*b^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)
))+1/a^3*ln(tan(1/2*d*x+1/2*c))+2/b^3*arctan(tan(1/2*d*x+1/2*c)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 479 vs. \(2 (166) = 332\).

Time = 0.52 (sec) , antiderivative size = 1042, normalized size of antiderivative = 5.95 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\cos ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)/(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 de

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.57 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {d x + c}{b^{3}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {{\left (2 \, a^{4} - a^{2} b^{2} + 2 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3} b^{3}} + \frac {a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 7 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 7 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a^{4} + 3 \, a^{2} b^{2}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2} a^{3} b^{2}}}{d} \]

[In]

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

[Out]

((d*x + c)/b^3 + log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (2*a^4 - a^2*b^2 + 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^3*b^3) + (a^3*b*tan(1/2*
d*x + 1/2*c)^3 + 4*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 2*a^4*tan(1/2*d*x + 1/2*c)^2 + 7*a^2*b^2*tan(1/2*d*x + 1/2*c
)^2 + 6*b^4*tan(1/2*d*x + 1/2*c)^2 + 7*a^3*b*tan(1/2*d*x + 1/2*c) + 8*a*b^3*tan(1/2*d*x + 1/2*c) + 2*a^4 + 3*a
^2*b^2)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2*a^3*b^2))/d

Mupad [B] (verification not implemented)

Time = 16.36 (sec) , antiderivative size = 3001, normalized size of antiderivative = 17.15 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

log(tan(c/2 + (d*x)/2))/(a^3*d) + ((2*a^2 + 3*b^2)/(a*b^2) + (tan(c/2 + (d*x)/2)*(7*a^2 + 8*b^2))/(a^2*b) + (t
an(c/2 + (d*x)/2)^3*(a^2 + 4*b^2))/(a^2*b) + (tan(c/2 + (d*x)/2)^2*(2*a^4 + 6*b^4 + 7*a^2*b^2))/(a^3*b^2))/(d*
(tan(c/2 + (d*x)/2)^2*(2*a^2 + 4*b^2) + a^2*tan(c/2 + (d*x)/2)^4 + a^2 + 4*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b*ta
n(c/2 + (d*x)/2))) - (2*atan((144*tan(c/2 + (d*x)/2))/((64*b^2)/a^2 - (64*b^4)/a^4 - (64*a*tan(c/2 + (d*x)/2))
/b + (64*b*tan(c/2 + (d*x)/2))/a + (144*a^3*tan(c/2 + (d*x)/2))/b^3 - 144) + (144*a)/(144*a*tan(c/2 + (d*x)/2)
 - (144*b^3)/a^2 + (64*b^5)/a^4 - (64*b^7)/a^6 - (64*b^2*tan(c/2 + (d*x)/2))/a + (64*b^4*tan(c/2 + (d*x)/2))/a
^3) + 64/(64*tan(c/2 + (d*x)/2) - (144*a)/b + (64*b)/a - (64*b^3)/a^3 - (64*a^2*tan(c/2 + (d*x)/2))/b^2 + (144
*a^4*tan(c/2 + (d*x)/2))/b^4) + (64*tan(c/2 + (d*x)/2))/((64*b^2)/a^2 + (144*a^2)/b^2 - (64*a*tan(c/2 + (d*x)/
2))/b + (64*a^3*tan(c/2 + (d*x)/2))/b^3 - (144*a^5*tan(c/2 + (d*x)/2))/b^5 - 64) + 64/(64*tan(c/2 + (d*x)/2) +
 (144*b)/a - (64*b^3)/a^3 + (64*b^5)/a^5 - (64*b^2*tan(c/2 + (d*x)/2))/a^2 - (144*a^2*tan(c/2 + (d*x)/2))/b^2)
 - (64*b*tan(c/2 + (d*x)/2))/(64*b - (64*a^2)/b + (144*a^4)/b^3 - (64*a^3*tan(c/2 + (d*x)/2))/b^2 + (64*a^5*ta
n(c/2 + (d*x)/2))/b^4 - (144*a^7*tan(c/2 + (d*x)/2))/b^6)))/(b^3*d) + (atan((((-(a + b)*(a - b))^(1/2)*((8*(14
*a^9 + 4*a^3*b^6 + 28*a^5*b^4 - 15*a^7*b^2))/(a^6*b^5) + ((-(a + b)*(a - b))^(1/2)*((8*(16*a^2*b^10 - 12*a^4*b
^8 + 14*a^6*b^6 + 16*a^8*b^4 - 12*a^10*b^2))/(a^6*b^5) + ((-(a + b)*(a - b))^(1/2)*(a^4 + b^4 - (a^2*b^2)/2)*(
(8*(32*a^5*b^10 - 24*a^7*b^8 + 14*a^9*b^6))/(a^6*b^5) + ((-(a + b)*(a - b))^(1/2)*((8*(16*a^8*b^10 - 12*a^10*b
^8))/(a^6*b^5) + (8*tan(c/2 + (d*x)/2)*(64*a^7*b^14 - 68*a^9*b^12 + 8*a^11*b^10))/(a^6*b^8))*(a^4 + b^4 - (a^2
*b^2)/2))/(a^3*b^5 - a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(64*a^4*b^14 - 68*a^6*b^12 + 48*a^8*b^10 - 16*a^10*b^8))
/(a^6*b^8)))/(a^3*b^5 - a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(4*a^3*b^12 - 36*a^5*b^10 + 89*a^7*b^8 - 100*a^9*b^6
+ 8*a^11*b^4))/(a^6*b^8))*(a^4 + b^4 - (a^2*b^2)/2))/(a^3*b^5 - a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(4*b^12 - 4*a
^2*b^10 + 73*a^4*b^8 - 68*a^6*b^6 + 48*a^8*b^4 - 16*a^10*b^2))/(a^6*b^8))*(a^4 + b^4 - (a^2*b^2)/2)*1i)/(a^3*b
^5 - a^5*b^3) + ((-(a + b)*(a - b))^(1/2)*((8*(14*a^9 + 4*a^3*b^6 + 28*a^5*b^4 - 15*a^7*b^2))/(a^6*b^5) - ((-(
a + b)*(a - b))^(1/2)*((8*(16*a^2*b^10 - 12*a^4*b^8 + 14*a^6*b^6 + 16*a^8*b^4 - 12*a^10*b^2))/(a^6*b^5) - ((-(
a + b)*(a - b))^(1/2)*(a^4 + b^4 - (a^2*b^2)/2)*((8*(32*a^5*b^10 - 24*a^7*b^8 + 14*a^9*b^6))/(a^6*b^5) - ((-(a
 + b)*(a - b))^(1/2)*((8*(16*a^8*b^10 - 12*a^10*b^8))/(a^6*b^5) + (8*tan(c/2 + (d*x)/2)*(64*a^7*b^14 - 68*a^9*
b^12 + 8*a^11*b^10))/(a^6*b^8))*(a^4 + b^4 - (a^2*b^2)/2))/(a^3*b^5 - a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(64*a^4
*b^14 - 68*a^6*b^12 + 48*a^8*b^10 - 16*a^10*b^8))/(a^6*b^8)))/(a^3*b^5 - a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(4*a
^3*b^12 - 36*a^5*b^10 + 89*a^7*b^8 - 100*a^9*b^6 + 8*a^11*b^4))/(a^6*b^8))*(a^4 + b^4 - (a^2*b^2)/2))/(a^3*b^5
 - a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(4*b^12 - 4*a^2*b^10 + 73*a^4*b^8 - 68*a^6*b^6 + 48*a^8*b^4 - 16*a^10*b^2)
)/(a^6*b^8))*(a^4 + b^4 - (a^2*b^2)/2)*1i)/(a^3*b^5 - a^5*b^3))/((16*(14*a^6 + 4*b^6 + 12*a^2*b^4 - 3*a^4*b^2)
)/(a^6*b^5) + ((-(a + b)*(a - b))^(1/2)*((8*(14*a^9 + 4*a^3*b^6 + 28*a^5*b^4 - 15*a^7*b^2))/(a^6*b^5) + ((-(a
+ b)*(a - b))^(1/2)*((8*(16*a^2*b^10 - 12*a^4*b^8 + 14*a^6*b^6 + 16*a^8*b^4 - 12*a^10*b^2))/(a^6*b^5) + ((-(a
+ b)*(a - b))^(1/2)*(a^4 + b^4 - (a^2*b^2)/2)*((8*(32*a^5*b^10 - 24*a^7*b^8 + 14*a^9*b^6))/(a^6*b^5) + ((-(a +
 b)*(a - b))^(1/2)*((8*(16*a^8*b^10 - 12*a^10*b^8))/(a^6*b^5) + (8*tan(c/2 + (d*x)/2)*(64*a^7*b^14 - 68*a^9*b^
12 + 8*a^11*b^10))/(a^6*b^8))*(a^4 + b^4 - (a^2*b^2)/2))/(a^3*b^5 - a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(64*a^4*b
^14 - 68*a^6*b^12 + 48*a^8*b^10 - 16*a^10*b^8))/(a^6*b^8)))/(a^3*b^5 - a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(4*a^3
*b^12 - 36*a^5*b^10 + 89*a^7*b^8 - 100*a^9*b^6 + 8*a^11*b^4))/(a^6*b^8))*(a^4 + b^4 - (a^2*b^2)/2))/(a^3*b^5 -
 a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(4*b^12 - 4*a^2*b^10 + 73*a^4*b^8 - 68*a^6*b^6 + 48*a^8*b^4 - 16*a^10*b^2))/
(a^6*b^8))*(a^4 + b^4 - (a^2*b^2)/2))/(a^3*b^5 - a^5*b^3) - ((-(a + b)*(a - b))^(1/2)*((8*(14*a^9 + 4*a^3*b^6
+ 28*a^5*b^4 - 15*a^7*b^2))/(a^6*b^5) - ((-(a + b)*(a - b))^(1/2)*((8*(16*a^2*b^10 - 12*a^4*b^8 + 14*a^6*b^6 +
 16*a^8*b^4 - 12*a^10*b^2))/(a^6*b^5) - ((-(a + b)*(a - b))^(1/2)*(a^4 + b^4 - (a^2*b^2)/2)*((8*(32*a^5*b^10 -
 24*a^7*b^8 + 14*a^9*b^6))/(a^6*b^5) - ((-(a + b)*(a - b))^(1/2)*((8*(16*a^8*b^10 - 12*a^10*b^8))/(a^6*b^5) +
(8*tan(c/2 + (d*x)/2)*(64*a^7*b^14 - 68*a^9*b^12 + 8*a^11*b^10))/(a^6*b^8))*(a^4 + b^4 - (a^2*b^2)/2))/(a^3*b^
5 - a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(64*a^4*b^14 - 68*a^6*b^12 + 48*a^8*b^10 - 16*a^10*b^8))/(a^6*b^8)))/(a^3
*b^5 - a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(4*a^3*b^12 - 36*a^5*b^10 + 89*a^7*b^8 - 100*a^9*b^6 + 8*a^11*b^4))/(a
^6*b^8))*(a^4 + b^4 - (a^2*b^2)/2))/(a^3*b^5 - a^5*b^3) + (8*tan(c/2 + (d*x)/2)*(4*b^12 - 4*a^2*b^10 + 73*a^4*
b^8 - 68*a^6*b^6 + 48*a^8*b^4 - 16*a^10*b^2))/(a^6*b^8))*(a^4 + b^4 - (a^2*b^2)/2))/(a^3*b^5 - a^5*b^3) - (16*
tan(c/2 + (d*x)/2)*(32*a^9 + 32*a^5*b^4 - 16*a^7*b^2))/(a^6*b^8)))*(-(a + b)*(a - b))^(1/2)*(a^4 + b^4 - (a^2*
b^2)/2)*2i)/(d*(a^3*b^5 - a^5*b^3))