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*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d-arctanh((a+b*sec(d*x+c))^(1/2)/(a-b)^(1/2))*(a-b)^(1/2)/d-
arctanh((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2))*(a+b)^(1/2)/d
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Rubi [A] time = 0.16, 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, 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 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 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 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 \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*}
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Mathematica [B] time = 4.38, size = 333, normalized size = 3.14 \[ \frac {2 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {a+b \sec (c+d x)} \left (\sqrt {b} \sqrt {b-a} \sqrt {\frac {a \cos (c+d x)+b}{b \cos (c+d x)+b}} \sin ^{-1}\left (\frac {\sqrt {b-a} \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}}}{\sqrt {b}}\right )+2 \sqrt {a} \sqrt {\frac {a \cos (c+d x)+b}{\cos (c+d x)+1}} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}}}{\sqrt {\frac {a \cos (c+d x)+b}{\cos (c+d x)+1}}}\right )+\sqrt {-a-b} \sqrt {\frac {-a \cos (c+d x)-b}{\cos (c+d x)+1}} \tanh ^{-1}\left (\frac {\sqrt {-a-b} \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )}}{\sqrt {-\left (\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)\right )}}\right )\right )}{d (a \cos (c+d x)+b)} \]
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])]*(Sqrt[-a - b]*ArcTanh[(Sqrt[-a - b]*Sqrt[Cos[c + d
*x]*Sec[(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]*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]*ArcSin[(Sqrt[-a + b]*Sqr
t[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]])/(d*(b + a*Cos[c + d*x]))
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fricas [B] time = 1.08, size = 2132, normalized size = 20.11 \[ \text {result too large to display} \]
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*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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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]
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="giac")
[Out]
sage0*x
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maple [B] time = 1.52, size = 575, normalized size = 5.42 \[ \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 (-1+\cos \left (d x +c \right )\right ) \left (\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 ) \sqrt {a -b}-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 ) \sqrt {a -b}-a \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 ) b \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}}}\, \sqrt {a -b}} \]
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)*(-1+cos(d*x+c))*((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-b)^(1/2)-2*a^(1/2)*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)^(1/2)-a*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))+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))*b)/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)^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sec \left (d x + c\right ) + a} \cot \left (d x + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)*(a+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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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {cot}\left (c+d\,x\right )\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]
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 \[ \int \sqrt {a + b \sec {\left (c + d x \right )}} \cot {\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)*(a+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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