\(\int \cot ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx\) [139]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 23, antiderivative size = 131 \[ \int \cot ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {7 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} d}+\frac {a}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}} \]

[Out]

-2*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d+7/8*arctanh(1/2*(a+a*sec(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2
^(1/2)*a^(1/2)/d+1/4*a/d/(a+a*sec(d*x+c))^(1/2)+1/2*a/d/(1-sec(d*x+c))/(a+a*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3965, 105, 157, 162, 65, 213} \[ \int \cot ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {7 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} d}+\frac {a}{4 d \sqrt {a \sec (c+d x)+a}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a \sec (c+d x)+a}} \]

[In]

Int[Cot[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(-2*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d + (7*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2
]*Sqrt[a])])/(4*Sqrt[2]*d) + a/(4*d*Sqrt[a + a*Sec[c + d*x]]) + a/(2*d*(1 - Sec[c + d*x])*Sqrt[a + a*Sec[c + d
*x]])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
 c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 157

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
 Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 213

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

Rule 3965

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

Rubi steps \begin{align*} \text {integral}& = \frac {a^4 \text {Subst}\left (\int \frac {1}{x (-a+a x)^2 (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {a}{2 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}-\frac {a \text {Subst}\left (\int \frac {2 a^2+\frac {3 a^2 x}{2}}{x (-a+a x) (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{2 d} \\ & = \frac {a}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}+\frac {\text {Subst}\left (\int \frac {-2 a^4+\frac {a^4 x}{4}}{x (-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{2 a^2 d} \\ & = \frac {a}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}+\frac {a \text {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d}-\frac {\left (7 a^2\right ) \text {Subst}\left (\int \frac {1}{(-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{8 d} \\ & = \frac {a}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{d}-\frac {(7 a) \text {Subst}\left (\int \frac {1}{-2 a+x^2} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{4 d} \\ & = -\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {7 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} d}+\frac {a}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.55 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.66 \[ \int \cot ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\frac {\cot ^2(c+d x) \left (-2-7 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1+\sec (c+d x))\right ) (-1+\sec (c+d x))+8 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\sec (c+d x)\right ) (-1+\sec (c+d x))\right ) \sqrt {a (1+\sec (c+d x))}}{4 d} \]

[In]

Integrate[Cot[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(Cot[c + d*x]^2*(-2 - 7*Hypergeometric2F1[-1/2, 1, 1/2, (1 + Sec[c + d*x])/2]*(-1 + Sec[c + d*x]) + 8*Hypergeo
metric2F1[-1/2, 1, 1/2, 1 + Sec[c + d*x]]*(-1 + Sec[c + d*x]))*Sqrt[a*(1 + Sec[c + d*x])])/(4*d)

Maple [A] (verified)

Time = 2.62 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05

method result size
default \(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-7 \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-16 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+6 \cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )\right )}{8 d}\) \(137\)

[In]

int(cot(d*x+c)^3*(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/8/d*(a*(1+sec(d*x+c)))^(1/2)*(-7*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)/(-cos(d*x+c)
/(cos(d*x+c)+1))^(1/2))-16*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+6*cot
(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 426, normalized size of antiderivative = 3.25 \[ \int \cot ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\left [\frac {8 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sqrt {a} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 7 \, {\left (\sqrt {2} \cos \left (d x + c\right )^{2} - \sqrt {2}\right )} \sqrt {a} \log \left (\frac {2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{16 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}}, -\frac {7 \, {\left (\sqrt {2} \cos \left (d x + c\right )^{2} - \sqrt {2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - 8 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) - 2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{8 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}}\right ] \]

[In]

integrate(cot(d*x+c)^3*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/16*(8*(cos(d*x + c)^2 - 1)*sqrt(a)*log(-8*a*cos(d*x + c)^2 + 4*(2*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*sq
rt((a*cos(d*x + c) + a)/cos(d*x + c)) - 8*a*cos(d*x + c) - a) + 7*(sqrt(2)*cos(d*x + c)^2 - sqrt(2))*sqrt(a)*l
og((2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) + 3*a*cos(d*x + c) + a)/(cos(d*x +
c) - 1)) + 4*(3*cos(d*x + c)^2 - cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^2 - d)
, -1/8*(7*(sqrt(2)*cos(d*x + c)^2 - sqrt(2))*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*
x + c))*cos(d*x + c)/(a*cos(d*x + c) + a)) - 8*(cos(d*x + c)^2 - 1)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x
 + c) + a)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + a)) - 2*(3*cos(d*x + c)^2 - cos(d*x + c))*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^2 - d)]

Sympy [F]

\[ \int \cot ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \cot ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(cot(d*x+c)**3*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*cot(c + d*x)**3, x)

Maxima [F]

\[ \int \cot ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cot(d*x+c)^3*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sec(d*x + c) + a)*cot(d*x + c)^3, x)

Giac [F]

\[ \int \cot ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cot(d*x+c)^3*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \cot ^3(c+d x) \sqrt {a+a \sec (c+d x)} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^3\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]

[In]

int(cot(c + d*x)^3*(a + a/cos(c + d*x))^(1/2),x)

[Out]

int(cot(c + d*x)^3*(a + a/cos(c + d*x))^(1/2), x)