3.180 \(\int \frac {\cot ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=165 \[ -\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4 a d}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {7 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{4 \sqrt {2} \sqrt {a} d}-\frac {\cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a \sec (c+d x)+a}}{4 a d} \]

[Out]

-2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d/a^(1/2)+7/8*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*
sec(d*x+c))^(1/2))/d*2^(1/2)/a^(1/2)-1/4*cot(d*x+c)*(a+a*sec(d*x+c))^(1/2)/a/d-1/4*cos(d*x+c)*cot(d*x+c)*sec(1
/2*d*x+1/2*c)^2*(a+a*sec(d*x+c))^(1/2)/a/d

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Rubi [A]  time = 0.14, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3887, 472, 583, 522, 203} \[ -\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4 a d}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {7 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{4 \sqrt {2} \sqrt {a} d}-\frac {\cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a \sec (c+d x)+a}}{4 a d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(-2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d) + (7*ArcTan[(Sqrt[a]*Tan[c + d*x])/(S
qrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(4*Sqrt[2]*Sqrt[a]*d) - (Cot[c + d*x]*Sqrt[a + a*Sec[c + d*x]])/(4*a*d) - (
Cos[c + d*x]*Cot[c + d*x]*Sec[(c + d*x)/2]^2*Sqrt[a + a*Sec[c + d*x]])/(4*a*d)

Rule 203

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

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

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

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 3887

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

Rubi steps

\begin {align*} \int \frac {\cot ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx &=-\frac {2 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+a x^2\right ) \left (2+a x^2\right )^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a d}\\ &=-\frac {\cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{4 a d}-\frac {\operatorname {Subst}\left (\int \frac {a-3 a^2 x^2}{x^2 \left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{2 a^2 d}\\ &=-\frac {\cot (c+d x) \sqrt {a+a \sec (c+d x)}}{4 a d}-\frac {\cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{4 a d}+\frac {\operatorname {Subst}\left (\int \frac {9 a^2+a^3 x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^2 d}\\ &=-\frac {\cot (c+d x) \sqrt {a+a \sec (c+d x)}}{4 a d}-\frac {\cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{4 a d}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{2+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {7 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{4 \sqrt {2} \sqrt {a} d}-\frac {\cot (c+d x) \sqrt {a+a \sec (c+d x)}}{4 a d}-\frac {\cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{4 a d}\\ \end {align*}

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Mathematica [C]  time = 24.30, size = 5534, normalized size = 33.54 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[c + d*x]^2/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

Result too large to show

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fricas [A]  time = 0.58, size = 503, normalized size = 3.05 \[ \left [-\frac {7 \, \sqrt {2} \sqrt {-a} {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\frac {2 \, \sqrt {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 ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) \sin \left (d x + c\right ) + 8 \, \sqrt {-a} {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {8 \, a \cos \left (d x + c\right )^{3} - 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 )}} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) + 1}\right ) \sin \left (d x + c\right ) + 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{16 \, {\left (a d \cos \left (d x + c\right ) + a d\right )} \sin \left (d x + c\right )}, -\frac {7 \, \sqrt {2} \sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\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} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 8 \, \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 ) \sin \left (d x + c\right )}{2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) - a}\right ) \sin \left (d x + c\right ) + 2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{8 \, {\left (a d \cos \left (d x + c\right ) + a d\right )} \sin \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(7*sqrt(2)*sqrt(-a)*(cos(d*x + c) + 1)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*
cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))*
sin(d*x + c) + 8*sqrt(-a)*(cos(d*x + c) + 1)*log(-(8*a*cos(d*x + c)^3 - 4*(2*cos(d*x + c)^2 - cos(d*x + c))*sq
rt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c) + a)/(cos(d*x + c) + 1))*sin(d*
x + c) + 4*(3*cos(d*x + c)^2 + cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/((a*d*cos(d*x + c) + a*d
)*sin(d*x + c)), -1/8*(7*sqrt(2)*sqrt(a)*(cos(d*x + c) + 1)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin(d*x + c) + 8*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)*sin(d*x + c)/(2*a*cos(d*x + c)^2 + a*cos(d*x + c) - a))*sin(d*x
+ c) + 2*(3*cos(d*x + c)^2 + cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/((a*d*cos(d*x + c) + a*d)*
sin(d*x + c))]

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giac [A]  time = 1.42, size = 123, normalized size = 0.75 \[ -\frac {\sqrt {2} {\left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + \frac {4 \, \sqrt {-a}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*tan(1/2*d*x + 1/2*c)/(a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 4
*sqrt(-a)/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)*sgn(tan(1/2*d*x + 1/2
*c)^2 - 1)))/d

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maple [B]  time = 1.30, size = 374, normalized size = 2.27 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (8 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+7 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-8 \sqrt {2}\, \sin \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-6 \left (\cos ^{3}\left (d x +c \right )\right )-7 \sin \left (d x +c \right ) \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+4 \left (\cos ^{2}\left (d x +c \right )\right )+2 \cos \left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{3} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(8*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)^2*sin(d*x+c)+7*(-2*cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^2*sin
(d*x+c)-8*2^(1/2)*sin(d*x+c)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(
-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-6*cos(d*x+c)^3-7*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin
(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+4*cos(d*x+c)^2+2*cos(d*x+c))/sin(d*x+c)
^3/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (d x + c\right )^{2}}{\sqrt {a \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cot(d*x + c)^2/sqrt(a*sec(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cot}\left (c+d\,x\right )}^2}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(cot(c + d*x)**2/sqrt(a*(sec(c + d*x) + 1)), x)

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