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

Optimal. Leaf size=131 \[ \frac {a}{4 d \sqrt {a \sec (c+d x)+a}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a \sec (c+d x)+a}}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {7 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} d} \]

[Out]

-2*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d+7/8*arctanh(1/2*(a+a*sec(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2
^(1/2)*a^(1/2)/d+1/4*a/d/(a+a*sec(d*x+c))^(1/2)+1/2*a/d/(1-sec(d*x+c))/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {a}{4 d \sqrt {a \sec (c+d x)+a}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a \sec (c+d x)+a}}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {7 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d + (7*Sqrt[a]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2
]*Sqrt[a])])/(4*Sqrt[2]*d) + a/(4*d*Sqrt[a + a*Sec[c + d*x]]) + a/(2*d*(1 - Sec[c + d*x])*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 \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)^{3/2}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}-\frac {a \operatorname {Subst}\left (\int \frac {2 a^2+\frac {3 a^2 x}{2}}{x (-a+a x) (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{2 d}\\ &=\frac {a}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}+\frac {\operatorname {Subst}\left (\int \frac {-2 a^4+\frac {a^4 x}{4}}{x (-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{2 a^2 d}\\ &=\frac {a}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}+\frac {a \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d}-\frac {\left (7 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{(-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{8 d}\\ &=\frac {a}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{d}-\frac {(7 a) \operatorname {Subst}\left (\int \frac {1}{-2 a+x^2} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{4 d}\\ &=-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {7 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} d}+\frac {a}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a}{2 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.30, size = 87, normalized size = 0.66 \[ \frac {\cot ^2(c+d x) \sqrt {a (\sec (c+d x)+1)} \left (-7 (\sec (c+d x)-1) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {1}{2} (\sec (c+d x)+1)\right )+8 (\sec (c+d x)-1) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\sec (c+d x)+1\right )-2\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[c + d*x]^2*(-2 - 7*Hypergeometric2F1[-1/2, 1, 1/2, (1 + Sec[c + d*x])/2]*(-1 + Sec[c + d*x]) + 8*Hypergeo
metric2F1[-1/2, 1, 1/2, 1 + Sec[c + d*x]]*(-1 + Sec[c + d*x]))*Sqrt[a*(1 + Sec[c + d*x])])/(4*d)

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fricas [A]  time = 1.13, size = 426, normalized size = 3.25 \[ \left [\frac {8 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sqrt {a} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 7 \, {\left (\sqrt {2} \cos \left (d x + c\right )^{2} - \sqrt {2}\right )} \sqrt {a} \log \left (\frac {2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{16 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}}, -\frac {7 \, {\left (\sqrt {2} \cos \left (d x + c\right )^{2} - \sqrt {2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - 8 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) - 2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{8 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}}\right ] \]

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]

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

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giac [A]  time = 0.79, size = 140, normalized size = 1.07 \[ \frac {\sqrt {2} {\left (\frac {8 \, \sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} - \frac {7 \, a \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} - \frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{8 \, d} \]

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]

1/8*sqrt(2)*(8*sqrt(2)*a*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) - 7*a*arcta
n(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) + 2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a) - sqrt(-a*tan
(1/2*d*x + 1/2*c)^2 + a)/tan(1/2*d*x + 1/2*c)^2)*sgn(cos(d*x + c))/d

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maple [B]  time = 1.38, size = 267, normalized size = 2.04 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (8 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )+7 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-8 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}-7 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+6 \left (\cos ^{2}\left (d x +c \right )\right )-2 \cos \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )-1\right )}{8 d \sin \left (d x +c \right )^{4}} \]

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]

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

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

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

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

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 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \cot ^{3}{\left (c + d x \right )}\, dx \]

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

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