3.321 \(\int \cot (c+d x) \sqrt{a+b \sec (c+d x)} \, dx\)
Optimal. Leaf size=106 \[ \frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{d}-\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a-b}}\right )}{d}-\frac{\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )}{d} \]
[Out]
(2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/d - (Sqrt[a - b]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a
- b]])/d - (Sqrt[a + b]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]])/d
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Rubi [A] time = 0.158867, 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, 1287, 206, 207} \[ \frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{d}-\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a-b}}\right )}{d}-\frac{\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]*Sqrt[a + b*Sec[c + d*x]],x]
[Out]
(2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/d - (Sqrt[a - b]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a
- b]])/d - (Sqrt[a + b]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/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 1287
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[Ex
pandIntegrand[((f*x)^m*(d + e*x^2)^q)/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^
2 - 4*a*c, 0] && IntegerQ[q] && IntegerQ[m]
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 \cot (c+d x) \sqrt{a+b \sec (c+d x)} \, dx &=-\frac{b^2 \operatorname{Subst}\left (\int \frac{\sqrt{a+x}}{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{x^2}{\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{a}{b^2 \left (a-x^2\right )}+\frac{a+b}{2 b^2 \left (a+b-x^2\right )}+\frac{-a+b}{2 b^2 \left (-a+b+x^2\right )}\right ) \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{d}\\ &=\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{d}+\frac{(a-b) \operatorname{Subst}\left (\int \frac{1}{-a+b+x^2} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{d}-\frac{(a+b) \operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{d}\\ &=\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{d}-\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a-b}}\right )}{d}-\frac{\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )}{d}\\ \end{align*}
Mathematica [B] time = 5.66549, size = 343, normalized size = 3.24 \[ -\frac{2 \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{a+b \sec (c+d x)} \left (-\sqrt{b} \sqrt{b-a} \sqrt{a+b} \sqrt{\frac{a \cos (c+d x)+b}{b \cos (c+d x)+b}} \sin ^{-1}\left (\frac{\sqrt{b-a} \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}}}{\sqrt{b}}\right )-2 \sqrt{a} \sqrt{a+b} \sqrt{\frac{a \cos (c+d x)+b}{\cos (c+d x)+1}} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}}}{\sqrt{\frac{a \cos (c+d x)+b}{\cos (c+d x)+1}}}\right )+(a+b) \sqrt{\frac{-a \cos (c+d x)-b}{\cos (c+d x)+1}} \tan ^{-1}\left (\frac{\sqrt{a+b} \sqrt{\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )}}{\sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (-(a \cos (c+d x)+b))}}\right )\right )}{d \sqrt{a+b} (a \cos (c+d x)+b)} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]*Sqrt[a + b*Sec[c + d*x]],x]
[Out]
(-2*Cos[(c + d*x)/2]^2*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*((a + b)*ArcTan[(Sqrt[a + b]*Sqrt[Cos[c + d*x]*Se
c[(c + d*x)/2]^2])/Sqrt[-((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)]]*Sqrt[(-b - a*Cos[c + d*x])/(1 + Cos[c + d
*x])] - 2*Sqrt[a]*Sqrt[a + b]*ArcTanh[(Sqrt[a]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])])/Sqrt[(b + a*Cos[c + d*x]
)/(1 + Cos[c + d*x])]]*Sqrt[(b + a*Cos[c + d*x])/(1 + Cos[c + d*x])] - Sqrt[b]*Sqrt[-a + b]*Sqrt[a + b]*ArcSin
[(Sqrt[-a + b]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])])/Sqrt[b]]*Sqrt[(b + a*Cos[c + d*x])/(b + b*Cos[c + d*x])]
)*Sqrt[a + b*Sec[c + d*x]])/(Sqrt[a + b]*d*(b + a*Cos[c + d*x]))
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Maple [B] time = 0.298, size = 576, normalized size = 5.4 \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)^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*4^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))*(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)*(a-b)^(1/2)-(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-b)^(1/2)+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))*co
s(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*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)*b)/sin(d*x+c)^2/((b+a*cos(d*x+c))*cos(d*x
+c)/(cos(d*x+c)+1)^2)^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \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+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate(sqrt(b*sec(d*x + c) + a)*cot(d*x + c), x)
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Fricas [B] time = 3.97924, size = 5389, normalized size = 50.84 \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*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))) + 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)) + sqrt(a + b)*log(-((8*a^2 + 8*a*b + b^2)*co
s(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)))/d, 1/4*(2*sqrt(-a - b)*arct
an(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*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))) + 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)))/d, -1/4*(2*sqrt(-a + b)*arctan(-2*sqrt(-a + b)*sqrt((a*c
os(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/((2*a - b)*cos(d*x + c) + 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))) - 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)))/d, -1/2*(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)) - sqrt(-a - b)*arctan(2*sqrt(-a - b)*sqrt((a*cos(d*x + c) + b)/co
s(d*x + c))*cos(d*x + c)/((2*a + b)*cos(d*x + c) + b)) - 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))))/d, -1/4*(4
*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)) - sqr
t(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)) - 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*co
s(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)))/d, -1/4*(4*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*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)) - 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)))/d, -1/4*(4*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*co
s(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) + 2*sqrt(-a + b)*arctan(-2*sqrt(-a + b)*sqr
t((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/((2*a - b)*cos(d*x + c) + 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)))/d, -1/2*(2*sq
rt(-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)) + 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
)) - 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)))/d]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sec{\left (c + d x \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+b*sec(d*x+c))**(1/2),x)
[Out]
Integral(sqrt(a + b*sec(c + d*x))*cot(c + d*x), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \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+b*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(sqrt(b*sec(d*x + c) + a)*cot(d*x + c), x)