3.175 \(\int \frac {\cot ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=152 \[ -\frac {a}{12 d (a \sec (c+d x)+a)^{3/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a \sec (c+d x)+a)^{3/2}}+\frac {7}{8 d \sqrt {a \sec (c+d x)+a}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {9 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d} \]

[Out]

-1/12*a/d/(a+a*sec(d*x+c))^(3/2)+1/2*a/d/(1-sec(d*x+c))/(a+a*sec(d*x+c))^(3/2)-2*arctanh((a+a*sec(d*x+c))^(1/2
)/a^(1/2))/d/a^(1/2)+9/16*arctanh(1/2*(a+a*sec(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)/a^(1/2)+7/8/d/(a+a*sec
(d*x+c))^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3880, 103, 152, 156, 63, 207} \[ -\frac {a}{12 d (a \sec (c+d x)+a)^{3/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a \sec (c+d x)+a)^{3/2}}+\frac {7}{8 d \sqrt {a \sec (c+d x)+a}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {9 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(-2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) + (9*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqrt
[a])])/(8*Sqrt[2]*Sqrt[a]*d) - a/(12*d*(a + a*Sec[c + d*x])^(3/2)) + a/(2*d*(1 - Sec[c + d*x])*(a + a*Sec[c +
d*x])^(3/2)) + 7/(8*d*Sqrt[a + a*Sec[c + d*x]])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

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 3880

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

Rubi steps

\begin {align*} \int \frac {\cot ^3(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx &=\frac {a^4 \operatorname {Subst}\left (\int \frac {1}{x (-a+a x)^2 (a+a x)^{5/2}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{3/2}}-\frac {a \operatorname {Subst}\left (\int \frac {2 a^2+\frac {5 a^2 x}{2}}{x (-a+a x) (a+a x)^{5/2}} \, dx,x,\sec (c+d x)\right )}{2 d}\\ &=-\frac {a}{12 d (a+a \sec (c+d x))^{3/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {-6 a^4-\frac {3 a^4 x}{4}}{x (-a+a x) (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{6 a^2 d}\\ &=-\frac {a}{12 d (a+a \sec (c+d x))^{3/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{3/2}}+\frac {7}{8 d \sqrt {a+a \sec (c+d x)}}-\frac {\operatorname {Subst}\left (\int \frac {6 a^6-\frac {21 a^6 x}{8}}{x (-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{6 a^5 d}\\ &=-\frac {a}{12 d (a+a \sec (c+d x))^{3/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{3/2}}+\frac {7}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d}-\frac {(9 a) \operatorname {Subst}\left (\int \frac {1}{(-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{16 d}\\ &=-\frac {a}{12 d (a+a \sec (c+d x))^{3/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{3/2}}+\frac {7}{8 d \sqrt {a+a \sec (c+d x)}}-\frac {9 \operatorname {Subst}\left (\int \frac {1}{-2 a+x^2} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{8 d}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{a d}\\ &=-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {9 \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d}-\frac {a}{12 d (a+a \sec (c+d x))^{3/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{3/2}}+\frac {7}{8 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.22, size = 90, normalized size = 0.59 \[ \frac {a \left (-9 (\sec (c+d x)-1) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {1}{2} (\sec (c+d x)+1)\right )+8 (\sec (c+d x)-1) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\sec (c+d x)+1\right )-6\right )}{12 d (\sec (c+d x)-1) (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(a*(-6 - 9*Hypergeometric2F1[-3/2, 1, -1/2, (1 + Sec[c + d*x])/2]*(-1 + Sec[c + d*x]) + 8*Hypergeometric2F1[-3
/2, 1, -1/2, 1 + Sec[c + d*x]]*(-1 + Sec[c + d*x])))/(12*d*(-1 + Sec[c + d*x])*(a*(1 + Sec[c + d*x]))^(3/2))

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fricas [B]  time = 0.59, size = 546, normalized size = 3.59 \[ \left [\frac {27 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sqrt {a} \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 ) + 3 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) - 1}\right ) + 48 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 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 )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (31 \, \cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} - 21 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{96 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d\right )}}, -\frac {27 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sqrt {-a} \arctan \left (\frac {\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 )}{a \cos \left (d x + c\right ) + a}\right ) - 48 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sqrt {-a} \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 )}{2 \, a \cos \left (d x + c\right ) + a}\right ) - 2 \, {\left (31 \, \cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} - 21 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{48 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/96*(27*sqrt(2)*(cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*sqrt(a)*log((2*sqrt(2)*sqrt(a)*sqrt((a*
cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) + 3*a*cos(d*x + c) + a)/(cos(d*x + c) - 1)) + 48*(cos(d*x + c)^3
+ cos(d*x + c)^2 - cos(d*x + c) - 1)*sqrt(a)*log(-8*a*cos(d*x + c)^2 + 4*(2*cos(d*x + c)^2 + cos(d*x + c))*sqr
t(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) - 8*a*cos(d*x + c) - a) + 4*(31*cos(d*x + c)^3 + 2*cos(d*x + c)^2
- 21*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*co
s(d*x + c) - a*d), -1/48*(27*sqrt(2)*(cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*sqrt(-a)*arctan(sqrt
(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c) + a)) - 48*(cos(d*x + c)^3 +
cos(d*x + c)^2 - cos(d*x + c) - 1)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x
+ c)/(2*a*cos(d*x + c) + a)) - 2*(31*cos(d*x + c)^3 + 2*cos(d*x + c)^2 - 21*cos(d*x + c))*sqrt((a*cos(d*x + c
) + a)/cos(d*x + c)))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*d)]

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giac [A]  time = 7.09, size = 229, normalized size = 1.51 \[ -\frac {\sqrt {2} {\left (\frac {48 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {27 \, \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {3 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {2 \, {\left ({\left (-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}} a^{4} + 12 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{5}\right )}}{a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/48*sqrt(2)*(48*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*sgn(tan(1
/2*d*x + 1/2*c)^2 - 1)) - 27*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*d*x +
1/2*c)^2 - 1)) - 3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/(a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*tan(1/2*d*x + 1/2*c)
^2) + 2*((-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a^4 + 12*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^5)/(a^6*sgn(tan(
1/2*d*x + 1/2*c)^2 - 1)))/d

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maple [B]  time = 1.52, size = 504, normalized size = 3.32 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right ) \left (-1+\cos \left (d x +c \right )\right )^{2} \left (48 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )+48 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )+27 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-48 \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )+27 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-48 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}+62 \left (\cos ^{3}\left (d x +c \right )\right )-27 \cos \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+4 \left (\cos ^{2}\left (d x +c \right )\right )-27 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-42 \cos \left (d x +c \right )\right )}{48 d \sin \left (d x +c \right )^{6} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3/(a+a*sec(d*x+c))^(1/2),x)

[Out]

-1/48/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(1+cos(d*x+c))*(-1+cos(d*x+c))^2*(48*cos(d*x+c)^3*2^(1/2)*(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+48*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)*cos(d*x+c)^2+27*cos(d*x+c)^3*
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-48*cos(d*x+c)*2^(1/2)*(-2*
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+27*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2-48*(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+62*cos(d*x+c)^3-27*cos(d*x+c)*(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+4*cos(d*x+c)^2-27*(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-42*cos(d*x+c))/sin(d*x+c)^6/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (d x + c\right )^{3}}{\sqrt {a \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cot(d*x + c)^3/sqrt(a*sec(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cot}\left (c+d\,x\right )}^3}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^3/(a + a/cos(c + d*x))^(1/2),x)

[Out]

int(cot(c + d*x)^3/(a + a/cos(c + d*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{3}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(cot(c + d*x)**3/sqrt(a*(sec(c + d*x) + 1)), x)

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