3.138 \(\int \cot (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=73 \[ \frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d}-\frac{\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{d} \]

[Out]

(2*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d - (Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(S
qrt[2]*Sqrt[a])])/d

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Rubi [A]  time = 0.0713238, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3880, 86, 63, 207} \[ \frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d}-\frac{\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]*Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(2*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d - (Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(S
qrt[2]*Sqrt[a])])/d

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 86

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

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

Mathematica [A]  time = 0.0495493, size = 72, normalized size = 0.99 \[ \frac{\sqrt{a (\sec (c+d x)+1)} \left (2 \tanh ^{-1}\left (\sqrt{\sec (c+d x)+1}\right )-\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{\sec (c+d x)+1}}{\sqrt{2}}\right )\right )}{d \sqrt{\sec (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]*Sqrt[a + a*Sec[c + d*x]],x]

[Out]

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

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Maple [A]  time = 0.192, size = 98, normalized size = 1.3 \begin{align*} -{\frac{1}{d}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \sqrt{2}\arctan \left ({\frac{\sqrt{2}}{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) +\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sec(d*x + c) + a)*cot(d*x + c), x)

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Fricas [B]  time = 1.70235, size = 647, normalized size = 8.86 \begin{align*} \left [\frac{\sqrt{2} \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 ) + 2 \, \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 )}{2 \, d}, \frac{\sqrt{2} \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 ) - 2 \, \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 )}{d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/2*(sqrt(2)*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)) + 2*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))/d, (sqrt(2)*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)) - 2*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)))/d]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )} \cot{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(sec(c + d*x) + 1))*cot(c + d*x), x)

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Giac [A]  time = 5.02681, size = 117, normalized size = 1.6 \begin{align*} -\frac{\sqrt{2} a{\left (\frac{\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{\arctan \left (\frac{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}}\right )} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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