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

Optimal. Leaf size=106 \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a-b}}\right )}{d \sqrt{a-b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )}{d \sqrt{a+b}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.134064, antiderivative size = 106, normalized size of antiderivative = 1., 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 = {3885, 898, 1170, 206, 207} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a-b}}\right )}{d \sqrt{a-b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )}{d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 898

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 + a*e^2)/e^2 - (2*c
*d*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1170

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(d + e*x
^2)^q/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && IntegerQ[q]

Rule 206

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

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 \frac{\cot (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx &=-\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+x} \left (b^2-x^2\right )} \, dx,x,b \sec (c+d x)\right )}{d}\\ &=-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-a+x^2\right ) \left (-a^2+b^2+2 a x^2-x^4\right )} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{d}\\ &=-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{b^2 \left (a-x^2\right )}+\frac{1}{2 b^2 \left (a+b-x^2\right )}-\frac{1}{2 b^2 \left (-a+b+x^2\right )}\right ) \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{d}+\frac{\operatorname{Subst}\left (\int \frac{1}{-a+b+x^2} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{d}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{d}\\ &=\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a-b}}\right )}{\sqrt{a-b} d}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )}{\sqrt{a+b} d}\\ \end{align*}

Mathematica [B]  time = 6.28259, size = 218, normalized size = 2.06 \[ \frac{\sqrt{a \cos (c+d x)+b} \left (\sqrt{a} \left (\sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{a \cos (c+d x)+b}}{\sqrt{a-b} \sqrt{-a \cos (c+d x)}}\right )+\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{a \cos (c+d x)+b}}{\sqrt{a+b} \sqrt{-a \cos (c+d x)}}\right )\right )-2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a \cos (c+d x)+b}}{\sqrt{-a \cos (c+d x)}}\right )\right )}{d \sqrt{a-b} \sqrt{a+b} \sqrt{-a \cos (c+d x)} \sqrt{a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[a - b]*Sqrt[a + b]*ArcTan[Sqrt[b + a*Cos[c + d*x]]/Sqrt[-(a*Cos[c + d*x])]] + Sqrt[a]*(Sqrt[a + b]*A
rcTan[(Sqrt[a]*Sqrt[b + a*Cos[c + d*x]])/(Sqrt[a - b]*Sqrt[-(a*Cos[c + d*x])])] + Sqrt[a - b]*ArcTan[(Sqrt[a]*
Sqrt[b + a*Cos[c + d*x]])/(Sqrt[a + b]*Sqrt[-(a*Cos[c + d*x])])]))*Sqrt[b + a*Cos[c + d*x]])/(Sqrt[a - b]*Sqrt
[a + b]*d*Sqrt[-(a*Cos[c + d*x])]*Sqrt[a + b*Sec[c + d*x]])

________________________________________________________________________________________

Maple [B]  time = 0.31, size = 691, normalized size = 6.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d/(a-b)^(3/2)/(a+b)/a*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*4^(1/2)*cos(d*x+c)*(2*(a-b)^(3/2)*a^(3/2)*ln(4*
cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x
+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)+2*(a-b)^(3/2)*a^(1/2)*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c
)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2
)+2*b)*b-(a-b)^(3/2)*(a+b)^(1/2)*ln(-2*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a+b)^(1/2)*cos
(d*x+c)+2*(a+b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*a*cos(d*x+c)+b*cos(d*x+c)+b)/(-1+
cos(d*x+c)))*a+ln(-1/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^
(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/
2)-b)/sin(d*x+c)^2)*a^3-ln(-1/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+
c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*
(a-b)^(1/2)-b)/sin(d*x+c)^2)*a*b^2)*(-1+cos(d*x+c))/sin(d*x+c)^2/((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2
)^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 11.2856, size = 5951, normalized size = 56.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(a^2 - b^2)*sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) - b^2 - 4*(2*a*cos(d*x + c)^2 + b*c
os(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))) + (a^2 + a*b)*sqrt(a - b)*log(-((8*a^2 - 8*a*b +
b^2)*cos(d*x + c)^2 + b^2 - 4*((2*a - b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a - b)*sqrt((a*cos(d*x + c) +
b)/cos(d*x + c)) + 2*(4*a*b - 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + (a^2 - a*b)*sqrt(a
+ b)*log(-((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^2 + b^2 - 4*((2*a + b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a
+ b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)) + 2*(4*a*b + 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 - 2*cos(d*x + c
) + 1)))/((a^3 - a*b^2)*d), -1/4*(4*(a^2 - b^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(d*x +
c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) - (a^2 + a*b)*sqrt(a - b)*log(-((8*a^2 - 8*a*b + b^2)*cos(d*x + c)^2
+ b^2 - 4*((2*a - b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a - b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)) + 2
*(4*a*b - 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - (a^2 - a*b)*sqrt(a + b)*log(-((8*a^2 +
8*a*b + b^2)*cos(d*x + c)^2 + b^2 - 4*((2*a + b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a + b)*sqrt((a*cos(d*x
+ c) + b)/cos(d*x + c)) + 2*(4*a*b + 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 - 2*cos(d*x + c) + 1)))/((a^3 - a*b
^2)*d), -1/4*(2*(a^2 + a*b)*sqrt(-a + b)*arctan(-2*sqrt(-a + b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*
x + c)/((2*a - b)*cos(d*x + c) + b)) - 2*(a^2 - b^2)*sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) -
b^2 - 4*(2*a*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))) - (a^2 - a*b)*s
qrt(a + b)*log(-((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^2 + b^2 - 4*((2*a + b)*cos(d*x + c)^2 + b*cos(d*x + c))*sq
rt(a + b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)) + 2*(4*a*b + 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 - 2*cos(d*
x + c) + 1)))/((a^3 - a*b^2)*d), -1/4*(4*(a^2 - b^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(
d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) + 2*(a^2 + a*b)*sqrt(-a + b)*arctan(-2*sqrt(-a + b)*sqrt((a*cos
(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/((2*a - b)*cos(d*x + c) + b)) - (a^2 - a*b)*sqrt(a + b)*log(-((8*a^2
+ 8*a*b + b^2)*cos(d*x + c)^2 + b^2 - 4*((2*a + b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a + b)*sqrt((a*cos(d
*x + c) + b)/cos(d*x + c)) + 2*(4*a*b + 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 - 2*cos(d*x + c) + 1)))/((a^3 - a
*b^2)*d), 1/4*(2*(a^2 - a*b)*sqrt(-a - b)*arctan(2*sqrt(-a - b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*
x + c)/((2*a + b)*cos(d*x + c) + b)) + 2*(a^2 - b^2)*sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) -
b^2 - 4*(2*a*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))) + (a^2 + a*b)*s
qrt(a - b)*log(-((8*a^2 - 8*a*b + b^2)*cos(d*x + c)^2 + b^2 - 4*((2*a - b)*cos(d*x + c)^2 + b*cos(d*x + c))*sq
rt(a - b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)) + 2*(4*a*b - 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 + 2*cos(d*
x + c) + 1)))/((a^3 - a*b^2)*d), -1/4*(4*(a^2 - b^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(
d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) - 2*(a^2 - a*b)*sqrt(-a - b)*arctan(2*sqrt(-a - b)*sqrt((a*cos(
d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/((2*a + b)*cos(d*x + c) + b)) - (a^2 + a*b)*sqrt(a - b)*log(-((8*a^2
- 8*a*b + b^2)*cos(d*x + c)^2 + b^2 - 4*((2*a - b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a - b)*sqrt((a*cos(d*
x + c) + b)/cos(d*x + c)) + 2*(4*a*b - 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)))/((a^3 - a*
b^2)*d), -1/2*((a^2 + a*b)*sqrt(-a + b)*arctan(-2*sqrt(-a + b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x
+ c)/((2*a - b)*cos(d*x + c) + b)) - (a^2 - a*b)*sqrt(-a - b)*arctan(2*sqrt(-a - b)*sqrt((a*cos(d*x + c) + b)
/cos(d*x + c))*cos(d*x + c)/((2*a + b)*cos(d*x + c) + b)) - (a^2 - b^2)*sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*
a*b*cos(d*x + c) - b^2 - 4*(2*a*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c
))))/((a^3 - a*b^2)*d), -1/2*(2*(a^2 - b^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))
*cos(d*x + c)/(2*a*cos(d*x + c) + b)) + (a^2 + a*b)*sqrt(-a + b)*arctan(-2*sqrt(-a + b)*sqrt((a*cos(d*x + c) +
b)/cos(d*x + c))*cos(d*x + c)/((2*a - b)*cos(d*x + c) + b)) - (a^2 - a*b)*sqrt(-a - b)*arctan(2*sqrt(-a - b)*
sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/((2*a + b)*cos(d*x + c) + b)))/((a^3 - a*b^2)*d)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot{\left (c + d x \right )}}{\sqrt{a + b \sec{\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+b*sec(d*x+c))**(1/2),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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