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

Optimal. Leaf size=92 \[ -\frac {1}{d \sqrt {a \sec (c+d x)+a}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3880, 85, 156, 63, 207} \[ -\frac {1}{d \sqrt {a \sec (c+d x)+a}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) - ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqrt[a])
]/(Sqrt[2]*Sqrt[a]*d) - 1/(d*Sqrt[a + a*Sec[c + d*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 85

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[(f*(e + f*x)^(p +
1))/((p + 1)*(b*e - a*f)*(d*e - c*f)), x] + Dist[1/((b*e - a*f)*(d*e - c*f)), Int[((b*d*e - b*c*f - a*d*f - b
*d*f*x)*(e + f*x)^(p + 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[p, -1]

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

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Mathematica [C]  time = 0.06, size = 57, normalized size = 0.62 \[ \frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {1}{2} (\sec (c+d x)+1)\right )-2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\sec (c+d x)+1\right )}{d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[-1/2, 1, 1/2, (1 + Sec[c + d*x])/2] - 2*Hypergeometric2F1[-1/2, 1, 1/2, 1 + Sec[c + d*x]])/
(d*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [B]  time = 0.73, size = 384, normalized size = 4.17 \[ \left [\frac {2 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (-8 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + \frac {\sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac {\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a}} - 3 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) - 1}\right )}{\sqrt {a}} - 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac {\sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, \sqrt {-a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) - 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \]

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]

[1/4*(2*sqrt(a)*(cos(d*x + c) + 1)*log(-8*a*cos(d*x + c)^2 - 4*(2*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*sqrt(
(a*cos(d*x + c) + a)/cos(d*x + c)) - 8*a*cos(d*x + c) - a) + sqrt(2)*(a*cos(d*x + c) + a)*log(-(2*sqrt(2)*sqrt
((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/sqrt(a) - 3*cos(d*x + c) - 1)/(cos(d*x + c) - 1))/sqrt(a) - 4
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c))/(a*d*cos(d*x + c) + a*d), 1/2*(sqrt(2)*(a*cos(d*x + c)
+ a)*sqrt(-1/a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)/(cos(d*x + c) +
1)) - 2*sqrt(-a)*(cos(d*x + c) + 1)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(2
*a*cos(d*x + c) + a)) - 2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c))/(a*d*cos(d*x + c) + a*d)]

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giac [A]  time = 1.23, size = 150, normalized size = 1.63 \[ \frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {\arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + \frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )}}{2 \, d} \]

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]

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

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maple [B]  time = 1.32, size = 259, normalized size = 2.82 \[ \frac {\left (2 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )+\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-2 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+2 \left (\cos ^{2}\left (d x +c \right )\right )-2 \cos \left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{2 d \sin \left (d x +c \right )^{2} a} \]

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]

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

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

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

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

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 \[ \int \frac {\cot {\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

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

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