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

Optimal. Leaf size=109 \[ -\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d}+\frac{5 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} d}+\frac{a \sqrt{a \sec (c+d x)+a}}{2 d (1-\sec (c+d x))} \]

[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]))

________________________________________________________________________________________

Rubi [A]  time = 0.107463, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3880, 103, 156, 63, 207} \[ -\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d}+\frac{5 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} d}+\frac{a \sqrt{a \sec (c+d x)+a}}{2 d (1-\sec (c+d x))} \]

Antiderivative was successfully verified.

[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 3880

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]

Rule 103

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] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

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 63

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 207

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

Rubi steps

\begin{align*} \int \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2} \, dx &=\frac{a^4 \operatorname{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 \operatorname{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 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{d}-\frac{\left (5 a^3\right ) \operatorname{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) \operatorname{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 ) \operatorname{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} \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{5 a^{3/2} \tanh ^{-1}\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]  time = 0.339515, size = 99, normalized size = 0.91 \[ \frac{(a (\sec (c+d x)+1))^{3/2} \left (-\frac{2 \sqrt{\sec (c+d x)+1}}{\sec (c+d x)-1}-8 \tanh ^{-1}\left (\sqrt{\sec (c+d x)+1}\right )+5 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\sec (c+d x)+1}}{\sqrt{2}}\right )\right )}{4 d (\sec (c+d x)+1)^{3/2}} \]

Antiderivative was successfully verified.

[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)*(-8*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + 5*Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/S
qrt[2]] - (2*Sqrt[1 + Sec[c + d*x]])/(-1 + Sec[c + d*x])))/(4*d*(1 + Sec[c + d*x])^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.213, size = 258, normalized size = 2.4 \begin{align*} -{\frac{a}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) +5\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) -4\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) +2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-5\,\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) +2\,\cos \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d*a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(4*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arct
an(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+5*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arcta
n(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-4*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+2*cos(d*x+c)^2-5*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x
+c)/(cos(d*x+c)+1))^(1/2))+2*cos(d*x+c))/sin(d*x+c)^2

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Fricas [B]  time = 1.76525, size = 1003, normalized size = 9.2 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(-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)**3*(a+a*sec(d*x+c))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 5.00076, size = 167, normalized size = 1.53 \begin{align*} \frac{\sqrt{2} a^{2}{\left (\frac{4 \, \sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a}} - \frac{5 \, \arctan \left (\frac{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}\right )} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*a^2*(4*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) - 5*arcta
n(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/(a*tan(1/2*d*x
+ 1/2*c)^2))*sgn(cos(d*x + c))/d