\(\int \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2} \, dx\) [151]

Optimal result

Integrand size = 23, antiderivative size = 109 \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {5 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} d}+\frac {a \sqrt {a+a \sec (c+d x)}}{2 d (1-\sec (c+d x))} \]

[Out]

-2*a^(3/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d+5/4*a^(3/2)*arctanh(1/2*(a+a*sec(d*x+c))^(1/2)*2^(1/2)/a^
(1/2))*2^(1/2)/d+1/2*a*(a+a*sec(d*x+c))^(1/2)/d/(1-sec(d*x+c))

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3965, 105, 162, 65, 213} \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {5 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} d}+\frac {a \sqrt {a \sec (c+d x)+a}}{2 d (1-\sec (c+d x))} \]

[In]

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

[Out]

(-2*a^(3/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d + (5*a^(3/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2
]*Sqrt[a])])/(2*Sqrt[2]*d) + (a*Sqrt[a + a*Sec[c + d*x]])/(2*d*(1 - 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 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 \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d} \\ & = \frac {a \sqrt {a+a \sec (c+d x)}}{2 d (1-\sec (c+d x))}-\frac {a \text {Subst}\left (\int \frac {2 a^2+\frac {a^2 x}{2}}{x (-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{2 d} \\ & = \frac {a \sqrt {a+a \sec (c+d x)}}{2 d (1-\sec (c+d x))}+\frac {a^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d}-\frac {\left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{(-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{4 d} \\ & = \frac {a \sqrt {a+a \sec (c+d x)}}{2 d (1-\sec (c+d x))}+\frac {(2 a) \text {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{d}-\frac {\left (5 a^2\right ) \text {Subst}\left (\int \frac {1}{-2 a+x^2} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{2 d} \\ & = -\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {5 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} d}+\frac {a \sqrt {a+a \sec (c+d x)}}{2 d (1-\sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.92 \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\frac {(a (1+\sec (c+d x)))^{3/2} \left (-2 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+\frac {5 \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\sqrt {1+\sec (c+d x)}}{2 (-1+\sec (c+d x))}\right )}{d (1+\sec (c+d x))^{3/2}} \]

[In]

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

[Out]

((a*(1 + Sec[c + d*x]))^(3/2)*(-2*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + (5*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]]
)/(2*Sqrt[2]) - Sqrt[1 + Sec[c + d*x]]/(2*(-1 + Sec[c + d*x]))))/(d*(1 + Sec[c + d*x])^(3/2))

Maple [A] (verified)

Time = 1.94 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.27

method	result	size
default	\(\frac {a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (5 \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+8 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-2 \cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )\right )}{4 d}\)	\(138\)

[In]

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

[Out]

1/4/d*a*(a*(1+sec(d*x+c)))^(1/2)*(5*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))+8*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-2*cot(
d*x+c)^2-2*cot(d*x+c)*csc(d*x+c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (86) = 172\).

Time = 0.33 (sec) , antiderivative size = 378, normalized size of antiderivative = 3.47 \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\left [\frac {4 \, a \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + 8 \, {\left (a \cos \left (d x + c\right ) - a\right )} \sqrt {a} \log \left (-2 \, a \cos \left (d x + c\right ) + 2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - a\right ) + 5 \, {\left (\sqrt {2} a \cos \left (d x + c\right ) - \sqrt {2} a\right )} \sqrt {a} \log \left (\frac {2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) - 1}\right )}{8 \, {\left (d \cos \left (d x + c\right ) - d\right )}}, -\frac {5 \, {\left (\sqrt {2} a \cos \left (d x + c\right ) - \sqrt {2} a\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - 8 \, {\left (a \cos \left (d x + c\right ) - a\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - 2 \, a \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right ) - d\right )}}\right ] \]

[In]

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

[Out]

[1/8*(4*a*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) + 8*(a*cos(d*x + c) - a)*sqrt(a)*log(-2*a*cos(d
*x + c) + 2*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) - a) + 5*(sqrt(2)*a*cos(d*x + c) - sq
rt(2)*a)*sqrt(a)*log((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)))/(d*cos(d*x + c) - d), -1/4*(5*(sqrt(2)*a*cos(d*x + c) - sqrt(2)*a)*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*(a*cos(d*x +
c) - a)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c) + a)) -
2*a*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c))/(d*cos(d*x + c) - d)]

Sympy [F]

\[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \cot ^{3}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

[In]

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

[Out]

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