\(\int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\) [169]

Optimal result

Integrand size = 23, antiderivative size = 96 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^2 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 a \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2}}{3 d} \]

[Out]

2*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-2/3*a*cot(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)/d+2*a^
2*cot(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3972, 331, 209} \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^2 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{d}-\frac {2 a \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d} \]

[In]

Int[Cot[c + d*x]^4*(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 + (2*a^2*Cot[c + d*x]*Sqrt[a + a*Sec[c +
d*x]])/d - (2*a*Cot[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2))/(3*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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [C] (warning: unable to verify)

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

Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {2 \left (\frac {1}{1+\cos (c+d x)}\right )^{3/2} \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a (1+\sec (c+d x)))^{5/2}}{3 d \sqrt {\frac {1}{1+\sec (c+d x)}}} \]

[In]

Integrate[Cot[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(-2*((1 + Cos[c + d*x])^(-1))^(3/2)*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, -3/2, -1/2, 2*Sin[(c + d*x)/2]^2]*(
a*(1 + Sec[c + d*x]))^(5/2))/(3*d*Sqrt[(1 + Sec[c + d*x])^(-1)])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(179\) vs. \(2(84)=168\).

Time = 40.39 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.88

method	result	size
default	\(\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )-3 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+4 \cot \left (d x +c \right ) \cos \left (d x +c \right )-3 \cot \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )-1\right )}\)	\(180\)

[In]

int(cot(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/3/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)-1)*(3*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(co
s(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-3*(-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))+4*cot(d*x+c)*cos(d*x+c)-3*cot(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 355, normalized size of antiderivative = 3.70 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (d x + c\right )^{3} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) + 1}\right ) \sin \left (d x + c\right ) + 4 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{6 \, {\left (d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )}, \frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) - a}\right ) \sin \left (d x + c\right ) + 2 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{3 \, {\left (d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )}\right ] \]

[In]

integrate(cot(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^2*cos(d*x + c) - a^2)*sqrt(-a)*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) + 4*(4*a^2*cos(d*x + c)^2 - 3*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/((d*cos(d*x + c)
- d)*sin(d*x + c)), 1/3*(3*(a^2*cos(d*x + c) - a^2)*sqrt(a)*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) + 2*(4*a^2*cos(d*x +
c)^2 - 3*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/((d*cos(d*x + c) - d)*sin(d*x + c))]

Sympy [F(-1)]

Timed out. \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)**4*(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cot(d*x+c)^4*(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)^4, x)

Giac [F]

\[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cot(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

[In]

int(cot(c + d*x)^4*(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(cot(c + d*x)^4*(a + a/cos(c + d*x))^(5/2), x)