\(\int \frac {\cot ^4(c+d x)}{a+b \sec (c+d x)} \, dx\) [298]

Optimal result

Integrand size = 21, antiderivative size = 177 \[ \int \frac {\cot ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {x}{a}-\frac {2 b^5 \text {arctanh}\left (\frac {\sqrt {a^2-b^2} \tan \left (\frac {1}{2} (c+d x)\right )}{a+b}\right )}{a \left (a^2-b^2\right )^{5/2} d}+\frac {a \left (a^2-2 b^2\right ) \cot (c+d x)}{\left (a^2-b^2\right )^2 d}-\frac {a \cot ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {b \left (a^2-2 b^2\right ) \csc (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {b \csc ^3(c+d x)}{3 \left (a^2-b^2\right ) d} \]

[Out]

x/a-2*b^5*arctanh((a^2-b^2)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b))/a/(a^2-b^2)^(5/2)/d+a*(a^2-2*b^2)*cot(d*x+c)/(a^2-
b^2)^2/d-1/3*a*cot(d*x+c)^3/(a^2-b^2)/d-b*(a^2-2*b^2)*csc(d*x+c)/(a^2-b^2)^2/d+1/3*b*csc(d*x+c)^3/(a^2-b^2)/d

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.45, number of steps used = 15, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3983, 2981, 2686, 3554, 8, 2814, 2738, 214} \[ \int \frac {\cot ^4(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {a \cot ^3(c+d x)}{3 d \left (a^2-b^2\right )}-\frac {a b^2 \cot (c+d x)}{d \left (a^2-b^2\right )^2}+\frac {a \cot (c+d x)}{d \left (a^2-b^2\right )}+\frac {b \csc ^3(c+d x)}{3 d \left (a^2-b^2\right )}-\frac {b \csc (c+d x)}{d \left (a^2-b^2\right )}-\frac {a b^2 x}{\left (a^2-b^2\right )^2}+\frac {a x}{a^2-b^2}+\frac {b^4 x}{a \left (a^2-b^2\right )^2}+\frac {b^3 \csc (c+d x)}{d \left (a^2-b^2\right )^2}-\frac {2 b^5 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d (a-b)^{5/2} (a+b)^{5/2}} \]

[In]

Int[Cot[c + d*x]^4/(a + b*Sec[c + d*x]),x]

[Out]

-((a*b^2*x)/(a^2 - b^2)^2) + (b^4*x)/(a*(a^2 - b^2)^2) + (a*x)/(a^2 - b^2) - (2*b^5*ArcTanh[(Sqrt[a - b]*Tan[(
c + d*x)/2])/Sqrt[a + b]])/(a*(a - b)^(5/2)*(a + b)^(5/2)*d) - (a*b^2*Cot[c + d*x])/((a^2 - b^2)^2*d) + (a*Cot
[c + d*x])/((a^2 - b^2)*d) - (a*Cot[c + d*x]^3)/(3*(a^2 - b^2)*d) + (b^3*Csc[c + d*x])/((a^2 - b^2)^2*d) - (b*
Csc[c + d*x])/((a^2 - b^2)*d) + (b*Csc[c + d*x]^3)/(3*(a^2 - b^2)*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 214

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

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*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 2814

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

Rule 2981

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[a*(d^2/(a^2 - b^2)), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[b*(d/(a^2 - b^2)), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[a^2*(d^2/(g^2*(a^2 - b^2
))), Int[(g*Cos[e + f*x])^(p + 2)*((d*Sin[e + f*x])^(n - 2)/(a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3983

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[Cos[c + d*x]^m
*((b + a*Sin[c + d*x])^n/Sin[c + d*x]^(m + n)), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[
n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos (c+d x) \cot ^4(c+d x)}{b+a \cos (c+d x)} \, dx \\ & = \frac {a \int \cot ^4(c+d x) \, dx}{a^2-b^2}-\frac {b \int \cot ^3(c+d x) \csc (c+d x) \, dx}{a^2-b^2}+\frac {b^2 \int \frac {\cos (c+d x) \cot ^2(c+d x)}{b+a \cos (c+d x)} \, dx}{a^2-b^2} \\ & = -\frac {a \cot ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {\left (a b^2\right ) \int \cot ^2(c+d x) \, dx}{\left (a^2-b^2\right )^2}-\frac {b^3 \int \cot (c+d x) \csc (c+d x) \, dx}{\left (a^2-b^2\right )^2}+\frac {b^4 \int \frac {\cos (c+d x)}{b+a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2}-\frac {a \int \cot ^2(c+d x) \, dx}{a^2-b^2}+\frac {b \text {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\csc (c+d x)\right )}{\left (a^2-b^2\right ) d} \\ & = \frac {b^4 x}{a \left (a^2-b^2\right )^2}-\frac {a b^2 \cot (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {a \cot (c+d x)}{\left (a^2-b^2\right ) d}-\frac {a \cot ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {b \csc (c+d x)}{\left (a^2-b^2\right ) d}+\frac {b \csc ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {\left (a b^2\right ) \int 1 \, dx}{\left (a^2-b^2\right )^2}-\frac {b^5 \int \frac {1}{b+a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )^2}+\frac {a \int 1 \, dx}{a^2-b^2}+\frac {b^3 \text {Subst}(\int 1 \, dx,x,\csc (c+d x))}{\left (a^2-b^2\right )^2 d} \\ & = -\frac {a b^2 x}{\left (a^2-b^2\right )^2}+\frac {b^4 x}{a \left (a^2-b^2\right )^2}+\frac {a x}{a^2-b^2}-\frac {a b^2 \cot (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {a \cot (c+d x)}{\left (a^2-b^2\right ) d}-\frac {a \cot ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {b^3 \csc (c+d x)}{\left (a^2-b^2\right )^2 d}-\frac {b \csc (c+d x)}{\left (a^2-b^2\right ) d}+\frac {b \csc ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {\left (2 b^5\right ) \text {Subst}\left (\int \frac {1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a \left (a^2-b^2\right )^2 d} \\ & = -\frac {a b^2 x}{\left (a^2-b^2\right )^2}+\frac {b^4 x}{a \left (a^2-b^2\right )^2}+\frac {a x}{a^2-b^2}-\frac {2 b^5 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a (a-b)^{5/2} (a+b)^{5/2} d}-\frac {a b^2 \cot (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {a \cot (c+d x)}{\left (a^2-b^2\right ) d}-\frac {a \cot ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {b^3 \csc (c+d x)}{\left (a^2-b^2\right )^2 d}-\frac {b \csc (c+d x)}{\left (a^2-b^2\right ) d}+\frac {b \csc ^3(c+d x)}{3 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(416\) vs. \(2(177)=354\).

Time = 6.44 (sec) , antiderivative size = 416, normalized size of antiderivative = 2.35 \[ \int \frac {\cot ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {(c+d x) (b+a \cos (c+d x)) \sec (c+d x)}{a d (a+b \sec (c+d x))}+\frac {2 b^5 \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x)) \sec (c+d x)}{a \sqrt {a^2-b^2} \left (-a^2+b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\left (8 a \cos \left (\frac {1}{2} (c+d x)\right )+11 b \cos \left (\frac {1}{2} (c+d x)\right )\right ) (b+a \cos (c+d x)) \csc \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)}{12 (a+b)^2 d (a+b \sec (c+d x))}-\frac {(b+a \cos (c+d x)) \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)}{24 (a+b) d (a+b \sec (c+d x))}+\frac {(b+a \cos (c+d x)) \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (-8 a \sin \left (\frac {1}{2} (c+d x)\right )+11 b \sin \left (\frac {1}{2} (c+d x)\right )\right )}{12 (-a+b)^2 d (a+b \sec (c+d x))}-\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{24 (-a+b) d (a+b \sec (c+d x))} \]

[In]

Integrate[Cot[c + d*x]^4/(a + b*Sec[c + d*x]),x]

[Out]

((c + d*x)*(b + a*Cos[c + d*x])*Sec[c + d*x])/(a*d*(a + b*Sec[c + d*x])) + (2*b^5*ArcTanh[((-a + b)*Tan[(c + d
*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x])*Sec[c + d*x])/(a*Sqrt[a^2 - b^2]*(-a^2 + b^2)^2*d*(a + b*Sec[c +
d*x])) + ((8*a*Cos[(c + d*x)/2] + 11*b*Cos[(c + d*x)/2])*(b + a*Cos[c + d*x])*Csc[(c + d*x)/2]*Sec[c + d*x])/
(12*(a + b)^2*d*(a + b*Sec[c + d*x])) - ((b + a*Cos[c + d*x])*Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2*Sec[c + d*x]
)/(24*(a + b)*d*(a + b*Sec[c + d*x])) + ((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]*Sec[c + d*x]*(-8*a*Sin[(c + d*x
)/2] + 11*b*Sin[(c + d*x)/2]))/(12*(-a + b)^2*d*(a + b*Sec[c + d*x])) - ((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]
^2*Sec[c + d*x]*Tan[(c + d*x)/2])/(24*(-a + b)*d*(a + b*Sec[c + d*x]))

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b}{3}-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{8 \left (a -b \right )^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {1}{24 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-5 a -7 b}{8 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b^{5} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(184\)
default	\(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b}{3}-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{8 \left (a -b \right )^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {1}{24 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-5 a -7 b}{8 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b^{5} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(184\)
risch	\(\frac {x}{a}-\frac {2 i \left (3 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-6 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-6 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+9 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-2 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+8 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+6 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-12 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-6 b^{3} {\mathrm e}^{i \left (d x +c \right )}-4 a^{3}+7 a \,b^{2}\right )}{3 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} d}+\frac {b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}-\frac {b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}\)	\(371\)

[In]

int(cot(d*x+c)^4/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/8/(a-b)^2*(1/3*tan(1/2*d*x+1/2*c)^3*a-1/3*tan(1/2*d*x+1/2*c)^3*b-5*tan(1/2*d*x+1/2*c)*a+7*tan(1/2*d*x+1
/2*c)*b)+2/a*arctan(tan(1/2*d*x+1/2*c))-1/24/(a+b)/tan(1/2*d*x+1/2*c)^3-1/8/(a+b)^2*(-5*a-7*b)/tan(1/2*d*x+1/2
*c)-2/(a+b)^2/(a-b)^2*b^5/a/((a-b)*(a+b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (168) = 336\).

Time = 0.31 (sec) , antiderivative size = 742, normalized size of antiderivative = 4.19 \[ \int \frac {\cot ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\left [\frac {4 \, a^{5} b - 14 \, a^{3} b^{3} + 10 \, a b^{5} + 2 \, {\left (4 \, a^{6} - 11 \, a^{4} b^{2} + 7 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (b^{5} \cos \left (d x + c\right )^{2} - b^{5}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 6 \, {\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 2 \, a^{2} b^{4}\right )} \cos \left (d x + c\right ) + 6 \, {\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d x \cos \left (d x + c\right )^{2} - {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d x\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d\right )} \sin \left (d x + c\right )}, \frac {2 \, a^{5} b - 7 \, a^{3} b^{3} + 5 \, a b^{5} + {\left (4 \, a^{6} - 11 \, a^{4} b^{2} + 7 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (b^{5} \cos \left (d x + c\right )^{2} - b^{5}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 2 \, a^{2} b^{4}\right )} \cos \left (d x + c\right ) + 3 \, {\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d x \cos \left (d x + c\right )^{2} - {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d x\right )} \sin \left (d x + c\right )}{3 \, {\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]

[In]

integrate(cot(d*x+c)^4/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/6*(4*a^5*b - 14*a^3*b^3 + 10*a*b^5 + 2*(4*a^6 - 11*a^4*b^2 + 7*a^2*b^4)*cos(d*x + c)^3 + 3*(b^5*cos(d*x + c
)^2 - b^5)*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d
*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2))*sin(d*x + c) - 6*(a^
5*b - 3*a^3*b^3 + 2*a*b^5)*cos(d*x + c)^2 - 6*(a^6 - 3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c) + 6*((a^6 - 3*a^4*b^2
+ 3*a^2*b^4 - b^6)*d*x*cos(d*x + c)^2 - (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*x)*sin(d*x + c))/(((a^7 - 3*a^5
*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2 - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d)*sin(d*x + c)), 1/3*(2*a^
5*b - 7*a^3*b^3 + 5*a*b^5 + (4*a^6 - 11*a^4*b^2 + 7*a^2*b^4)*cos(d*x + c)^3 - 3*(b^5*cos(d*x + c)^2 - b^5)*sqr
t(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*sin(d*x + c) - 3*(a^5*
b - 3*a^3*b^3 + 2*a*b^5)*cos(d*x + c)^2 - 3*(a^6 - 3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c) + 3*((a^6 - 3*a^4*b^2 +
3*a^2*b^4 - b^6)*d*x*cos(d*x + c)^2 - (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*x)*sin(d*x + c))/(((a^7 - 3*a^5*b
^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2 - (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d)*sin(d*x + c))]

Sympy [F]

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

[In]

integrate(cot(d*x+c)**4/(a+b*sec(d*x+c)),x)

[Out]

Integral(cot(c + d*x)**4/(a + b*sec(c + d*x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(cot(d*x+c)^4/(a+b*sec(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*a^2-4*b^2>0)', see `assume?`
for more de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1073 vs. \(2 (168) = 336\).

Time = 0.48 (sec) , antiderivative size = 1073, normalized size of antiderivative = 6.06 \[ \int \frac {\cot ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate(cot(d*x+c)^4/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/24*(24*((a^4 - a^3*b - 2*a^2*b^2 + 2*a*b^3 + b^4)*sqrt(-a^2 + b^2)*abs(a^5 - 2*a^3*b^2 + a*b^4)*abs(-a + b)
- (a^9 - a^8*b - 4*a^7*b^2 + 4*a^6*b^3 + 6*a^5*b^4 - 7*a^4*b^5 - 4*a^3*b^6 + 6*a^2*b^7 + a*b^8 - 2*b^9)*sqrt(
-a^2 + b^2)*abs(-a + b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(a^4*b - 2*a^2*
b^3 + b^5 + sqrt((a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*
b^4 - b^5) + (a^4*b - 2*a^2*b^3 + b^5)^2))/(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5))))/((a^5 - 2*a^
3*b^2 + a*b^4)^2*(a^2 - 2*a*b + b^2) + (a^6*b - 2*a^5*b^2 - a^4*b^3 + 4*a^3*b^4 - a^2*b^5 - 2*a*b^6 + b^7)*abs
(a^5 - 2*a^3*b^2 + a*b^4)) + 24*(a^9 - a^8*b - 4*a^7*b^2 + 4*a^6*b^3 + 6*a^5*b^4 - 7*a^4*b^5 - 4*a^3*b^6 + 6*a
^2*b^7 + a*b^8 - 2*b^9 + a^4*abs(a^5 - 2*a^3*b^2 + a*b^4) - a^3*b*abs(a^5 - 2*a^3*b^2 + a*b^4) - 2*a^2*b^2*abs
(a^5 - 2*a^3*b^2 + a*b^4) + 2*a*b^3*abs(a^5 - 2*a^3*b^2 + a*b^4) + b^4*abs(a^5 - 2*a^3*b^2 + a*b^4))*(pi*floor
(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(a^4*b - 2*a^2*b^3 + b^5 - sqrt((a^5 + a^4*b - 2*
a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5) + (a^4*b - 2*a^2*b^3 +
b^5)^2))/(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5))))/(a^4*b*abs(a^5 - 2*a^3*b^2 + a*b^4) - 2*a^2*b^
3*abs(a^5 - 2*a^3*b^2 + a*b^4) + b^5*abs(a^5 - 2*a^3*b^2 + a*b^4) - (a^5 - 2*a^3*b^2 + a*b^4)^2) - (a^2*tan(1/
2*d*x + 1/2*c)^3 - 2*a*b*tan(1/2*d*x + 1/2*c)^3 + b^2*tan(1/2*d*x + 1/2*c)^3 - 15*a^2*tan(1/2*d*x + 1/2*c) + 3
6*a*b*tan(1/2*d*x + 1/2*c) - 21*b^2*tan(1/2*d*x + 1/2*c))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (15*a*tan(1/2*d*x
+ 1/2*c)^2 + 21*b*tan(1/2*d*x + 1/2*c)^2 - a - b)/((a^2 + 2*a*b + b^2)*tan(1/2*d*x + 1/2*c)^3))/d

Mupad [B] (verification not implemented)

Time = 23.38 (sec) , antiderivative size = 3859, normalized size of antiderivative = 21.80 \[ \int \frac {\cot ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

(a^10*((cos(3*c + 3*d*x)*4i)/3 - sin(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*6i + atan(sin(c/2 +
(d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x)*2i) + a*((b^9*8i)/3 - b^9*cos(2*c + 2*d*x)*4i) - a^7*((b^3*14i)/
3 - b^3*cos(2*c + 2*d*x)*10i) + a^5*(b^5*10i - b^5*cos(2*c + 2*d*x)*18i) - a^3*((b^7*26i)/3 - b^7*cos(2*c + 2*
d*x)*14i) + a^9*((b*2i)/3 - b*cos(2*c + 2*d*x)*2i) + a^8*(b^2*cos(c + d*x)*1i - (b^2*cos(3*c + 3*d*x)*19i)/3 +
b^2*sin(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*30i - b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x
)/2))*sin(3*c + 3*d*x)*10i) - a^2*(b^8*cos(c + d*x)*1i - (b^8*cos(3*c + 3*d*x)*7i)/3 + b^8*sin(c + d*x)*atan(s
in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*30i - b^8*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x)*1
0i) - a^6*(b^4*cos(c + d*x)*3i - b^4*cos(3*c + 3*d*x)*11i + b^4*sin(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 +
(d*x)/2))*60i - b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x)*20i) + a^4*(b^6*cos(c + d*x)
*3i - (b^6*cos(3*c + 3*d*x)*25i)/3 + b^6*sin(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*60i - b^6*at
an(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x)*20i) + b^10*sin(c + d*x)*atan(sin(c/2 + (d*x)/2)/co
s(c/2 + (d*x)/2))*6i - b^10*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x)*2i + b^5*atanh((2*b^1
1*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(3/2) - a^21*sin(c/2 + (d
*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 2*b^21*sin(c/2 + (d*x)/2)*(a^10
- b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + a^20*b*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*
a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 15*a^2*b^19*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8
- 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 5*a^3*b^18*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^
4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 55*a^4*b^17*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 +
10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 35*a^5*b^16*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6
*b^4 - 5*a^8*b^2)^(1/2) - 130*a^6*b^15*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 -
5*a^8*b^2)^(1/2) + 110*a^7*b^14*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8
*b^2)^(1/2) + 215*a^8*b^13*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^
(1/2) - 205*a^9*b^12*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2)
- 253*a^10*b^11*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 251
*a^11*b^10*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 210*a^12
*b^9*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 210*a^13*b^8*s
in(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 120*a^14*b^7*sin(c/2
+ (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 120*a^15*b^6*sin(c/2 + (d*
x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 45*a^16*b^5*sin(c/2 + (d*x)/2)*(
a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 45*a^17*b^4*sin(c/2 + (d*x)/2)*(a^10 -
b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 10*a^18*b^3*sin(c/2 + (d*x)/2)*(a^10 - b^10 +
5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 10*a^19*b^2*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b
^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a^26 + 5*a^2*b^24 - 50*a^4*b^22 + 230*a^
6*b^20 - 645*a^8*b^18 + 1231*a^10*b^16 - 1688*a^12*b^14 + 1708*a^14*b^12 - 1286*a^16*b^10 + 715*a^18*b^8 - 286
*a^20*b^6 + 78*a^22*b^4 - 13*a^24*b^2)))*sin(3*c + 3*d*x)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 -
5*a^8*b^2)^(1/2)*2i - b^5*atanh((2*b^11*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4
- 5*a^8*b^2)^(3/2) - a^21*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^
(1/2) + 2*b^21*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + a^20
*b*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 15*a^2*b^19*sin(
c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 5*a^3*b^18*sin(c/2 + (d
*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 55*a^4*b^17*sin(c/2 + (d*x)/2)*
(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 35*a^5*b^16*sin(c/2 + (d*x)/2)*(a^10 -
b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 130*a^6*b^15*sin(c/2 + (d*x)/2)*(a^10 - b^10
+ 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 110*a^7*b^14*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^
2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 215*a^8*b^13*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8
- 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 205*a^9*b^12*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a
^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 253*a^10*b^11*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^
6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 251*a^11*b^10*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 1
0*a^6*b^4 - 5*a^8*b^2)^(1/2) + 210*a^12*b^9*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*
b^4 - 5*a^8*b^2)^(1/2) - 210*a^13*b^8*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 -
5*a^8*b^2)^(1/2) - 120*a^14*b^7*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*
b^2)^(1/2) + 120*a^15*b^6*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(
1/2) + 45*a^16*b^5*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) -
45*a^17*b^4*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) - 10*a^18
*b^3*sin(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2) + 10*a^19*b^2*si
n(c/2 + (d*x)/2)*(a^10 - b^10 + 5*a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a
^26 + 5*a^2*b^24 - 50*a^4*b^22 + 230*a^6*b^20 - 645*a^8*b^18 + 1231*a^10*b^16 - 1688*a^12*b^14 + 1708*a^14*b^1
2 - 1286*a^16*b^10 + 715*a^18*b^8 - 286*a^20*b^6 + 78*a^22*b^4 - 13*a^24*b^2)))*sin(c + d*x)*(a^10 - b^10 + 5*
a^2*b^8 - 10*a^4*b^6 + 10*a^6*b^4 - 5*a^8*b^2)^(1/2)*6i)/(a^11*d*sin(3*c + 3*d*x)*1i - a^11*d*sin(c + d*x)*3i
+ a^3*b^8*d*sin(3*c + 3*d*x)*5i - a^5*b^6*d*sin(3*c + 3*d*x)*10i + a^7*b^4*d*sin(3*c + 3*d*x)*10i - a^9*b^2*d*
sin(3*c + 3*d*x)*5i + a*b^10*d*sin(c + d*x)*3i - a*b^10*d*sin(3*c + 3*d*x)*1i - a^3*b^8*d*sin(c + d*x)*15i + a
^5*b^6*d*sin(c + d*x)*30i - a^7*b^4*d*sin(c + d*x)*30i + a^9*b^2*d*sin(c + d*x)*15i)