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((a+b*sec(d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)-arctanh((a+b*sec(d*x+c))^(1/2)/(a-b)^(1/2))/d/(a-b)^(1/2)-
arctanh((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2))/d/(a+b)^(1/2)
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Rubi [A] time = 0.13, antiderivative size = 106, 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 =
{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 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])
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 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]
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*}
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Mathematica [B] time = 5.98, 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]])
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fricas [B] time = 4.41, size = 2420, normalized size = 22.83 \[ \text {result too large to display} \]
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)]
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
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]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, integration of abs or sign assumes constant sign by intervals
(correct if the argument is real):Check [abs(cos(d*t_nostep+c))]Unable to check sign: (2*pi/t_nostep/2)>(-2*p
i/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)
>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nos
tep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi
/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Discontinuities at ze
roes of cos(d*t_nostep+c) were not checkedWarning, integration of abs or sign assumes constant sign by interva
ls (correct if the argument is real):Check [abs(t_nostep^2-1)]Evaluation time: 0.58Error: Bad Argument Type
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maple [B] time = 1.54, size = 690, normalized size = 6.51 \[ \frac {\sqrt {\frac {b +a \cos \left (d x +c \right )}{\cos \left (d x +c \right )}}\, \sqrt {4}\, \cos \left (d x +c \right ) \left (-2 \left (a -b \right )^{\frac {3}{2}} a^{\frac {3}{2}} \ln \left (4 \sqrt {a}\, \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+4 a \cos \left (d x +c \right )+2 b \right )+\left (a -b \right )^{\frac {3}{2}} \sqrt {a +b}\, \ln \left (-\frac {2 \left (2 \cos \left (d x +c \right ) \sqrt {a +b}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+2 a \cos \left (d x +c \right )+b \cos \left (d x +c \right )+2 \sqrt {a +b}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+b \right )}{-1+\cos \left (d x +c \right )}\right ) a -2 \left (a -b \right )^{\frac {3}{2}} \sqrt {a}\, \ln \left (4 \sqrt {a}\, \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}+4 a \cos \left (d x +c \right )+2 b \right ) b -a^{3} \ln \left (-\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (2 \cos \left (d x +c \right ) \sqrt {a -b}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}-2 a \cos \left (d x +c \right )+b \cos \left (d x +c \right )+2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {a -b}-b \right )}{\sin \left (d x +c \right )^{2} \sqrt {a -b}}\right )+\ln \left (-\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (2 \cos \left (d x +c \right ) \sqrt {a -b}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}-2 a \cos \left (d x +c \right )+b \cos \left (d x +c \right )+2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \sqrt {a -b}-b \right )}{\sin \left (d x +c \right )^{2} \sqrt {a -b}}\right ) a \,b^{2}\right ) \left (-1+\cos \left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{2} \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\, \left (a -b \right )^{\frac {3}{2}} a \left (a +b \right )} \]
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*((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*a^(1/2)*cos(d*x+c)*(
(b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(
1/2)+4*a*cos(d*x+c)+2*b)+(a-b)^(3/2)*(a+b)^(1/2)*ln(-2*(2*cos(d*x+c)*(a+b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/
(1+cos(d*x+c))^2)^(1/2)+2*a*cos(d*x+c)+b*cos(d*x+c)+2*(a+b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^
2)^(1/2)+b)/(-1+cos(d*x+c)))*a-2*(a-b)^(3/2)*a^(1/2)*ln(4*a^(1/2)*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+c
os(d*x+c))^2)^(1/2)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+4*a*cos(d*x+c)+2*b)*b-a^3*l
n(-(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+
c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2)
)+ln(-(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d
*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1
/2))*a*b^2)*(-1+cos(d*x+c))/sin(d*x+c)^2/((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)/(a-b)^(3/2)/a/(a
+b)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (d x + c\right )}{\sqrt {b \sec \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate(cot(d*x + c)/sqrt(b*sec(d*x + c) + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {cot}\left (c+d\,x\right )}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(c + d*x)/(a + b/cos(c + d*x))^(1/2),x)
[Out]
int(cot(c + d*x)/(a + b/cos(c + d*x))^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot {\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \]
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)
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