\(\int \cot (c+d x) (a+a \sec (c+d x))^{3/2} \, dx\) [150]

Optimal result

Integrand size = 21, antiderivative size = 73 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3965, 85, 65, 213} \[ \int \cot (c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{d} \]

[In]

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

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 85

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\frac {\left (2 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )-2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )\right ) (a (1+\sec (c+d x)))^{3/2}}{d (1+\sec (c+d x))^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25

method	result	size
default	\(-\frac {2 a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\sqrt {2}\, \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )\right )}{d}\)	\(91\)

[In]

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

[Out]

-2/d*a*(a*(1+sec(d*x+c)))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))
+2^(1/2)*arctan(1/2*2^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (58) = 116\).

Time = 0.33 (sec) , antiderivative size = 243, normalized size of antiderivative = 3.33 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{3/2} \, dx=\left [\frac {\sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) - 1}\right ) + a^{\frac {3}{2}} \log \left (-2 \, a \cos \left (d x + c\right ) - 2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - a\right )}{d}, \frac {2 \, {\left (\sqrt {2} \sqrt {-a} a \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - \sqrt {-a} a \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right )\right )}}{d}\right ] \]

[In]

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

[Out]

[(sqrt(2)*a^(3/2)*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)) + a^(3/2)*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))/d, 2*(sqrt(2)*sqrt(-a)*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)) - sqrt(-a)*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)))/d]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(cot(d*x+c)*(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), x)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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