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

Optimal. Leaf size=66 \[ -\frac{4 a^2 \cot (c+d x) \sqrt{a \sec (c+d x)+a}}{d}-\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d} \]

[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

________________________________________________________________________________________

Rubi [A]  time = 0.0710611, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3887, 453, 203} \[ -\frac{4 a^2 \cot (c+d x) \sqrt{a \sec (c+d x)+a}}{d}-\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[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 3887

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]

Rule 453

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 203

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

Rubi steps

\begin{align*} \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx &=-\frac{\left (2 a^2\right ) \operatorname{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 ) \operatorname{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} \tan ^{-1}\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]  time = 0.714181, size = 124, normalized size = 1.88 \[ -\frac{\sqrt{2} \cot (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (\frac{1}{\sec (c+d x)+1}\right )^{3/2} (a (\sec (c+d x)+1))^{5/2} \left (2 \cos (c+d x)-\frac{(\cos (c+d x)-1) \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )}{\sqrt{1-\sec (c+d x)}}\right )}{d \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )}} \]

Antiderivative was successfully verified.

[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 [B]  time = 0.166, size = 192, normalized size = 2.9 \begin{align*}{\frac{{a}^{2}}{d \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}-1 \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( dx+c \right ) }{2\,\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) -\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( dx+c \right ) }{2\,\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) +4\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2*2^(1/2)*arctanh
(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))+4*cos(d*x+c)*sin(d*x+c
))/(cos(d*x+c)^2-1)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cot \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [B]  time = 1.99467, size = 693, normalized size = 10.5 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError