\(\int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\) [168]

Optimal result

Integrand size = 23, antiderivative size = 66 \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}-\frac {4 a^2 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3972, 464, 209} \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^2 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{d} \]

[In]

Int[Cot[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(-2*a^(5/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (4*a^2*Cot[c + d*x]*Sqrt[a + a*Sec[c
+ d*x]])/d

Rule 209

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

Rule 464

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

Rule 3972

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

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

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.88 \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (2 \cos (c+d x)-\frac {\text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) (-1+\cos (c+d x))}{\sqrt {1-\sec (c+d x)}}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{3/2} (a (1+\sec (c+d x)))^{5/2}}{d \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \]

[In]

Integrate[Cot[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 6.62 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.35

method	result	size
default	\(-\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+2 \cot \left (d x +c \right )\right )}{d}\)	\(89\)

[In]

int(cot(d*x+c)^2*(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/d*a^2*(a*(1+sec(d*x+c)))^(1/2)*((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(
d*x+c)/(cos(d*x+c)+1))^(1/2))+2*cot(d*x+c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (58) = 116\).

Time = 0.35 (sec) , antiderivative size = 270, normalized size of antiderivative = 4.09 \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {\sqrt {-a} a^{2} \log \left (-\frac {8 \, a \cos \left (d x + c\right )^{3} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) + 1}\right ) \sin \left (d x + c\right ) - 8 \, a^{2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, d \sin \left (d x + c\right )}, -\frac {a^{\frac {5}{2}} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) - a}\right ) \sin \left (d x + c\right ) + 4 \, a^{2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{d \sin \left (d x + c\right )}\right ] \]

[In]

integrate(cot(d*x+c)^2*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/2*(sqrt(-a)*a^2*log(-(8*a*cos(d*x + c)^3 + 4*(2*cos(d*x + c)^2 - cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c
) + a)/cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c) + a)/(cos(d*x + c) + 1))*sin(d*x + c) - 8*a^2*sqrt((a*cos
(d*x + c) + a)/cos(d*x + c))*cos(d*x + c))/(d*sin(d*x + c)), -(a^(5/2)*arctan(2*sqrt(a)*sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c)/(2*a*cos(d*x + c)^2 + a*cos(d*x + c) - a))*sin(d*x + c) + 4*a^2*sq
rt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c))/(d*sin(d*x + c))]

Sympy [F(-1)]

Timed out. \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)**2*(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(cot(d*x+c)^2*(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^(5/2)*cot(d*x + c)^2, x)

Giac [F]

\[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(cot(d*x+c)^2*(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

[In]

int(cot(c + d*x)^2*(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(cot(c + d*x)^2*(a + a/cos(c + d*x))^(5/2), x)