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

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Rubi [A]  time = 0.07, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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 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 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])

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]

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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.83, size = 242, normalized size = 3.32 \[ \left [\frac {\sqrt {2} \sqrt {a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) - 1}\right ) + 2 \, \sqrt {a} \log \left (-2 \, a \cos \left (d x + c\right ) - 2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - a\right )}{2 \, d}, \frac {\sqrt {2} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - 2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right )}{d}\right ] \]

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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giac [A]  time = 2.11, size = 88, normalized size = 1.21 \[ -\frac {\sqrt {2} {\left (\frac {\sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} - \frac {a \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{d} \]

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

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maple [A]  time = 1.14, size = 98, normalized size = 1.34 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+\sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )\right )}{d} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sec \left (d x + c\right ) + a} \cot \left (d x + c\right )\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {cot}\left (c+d\,x\right )\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \cot {\left (c + d x \right )}\, dx \]

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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