3.169 \(\int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=96 \[ \frac {2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^2 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{d}-\frac {2 a \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d} \]

[Out]

2*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-2/3*a*cot(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)/d+2*a^
2*cot(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.08, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3887, 325, 203} \[ \frac {2 a^2 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{d}+\frac {2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 a \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(2*a^(5/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*Cot[c + d*x]*Sqrt[a + a*Sec[c +
d*x]])/d - (2*a*Cot[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2))/(3*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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx &=-\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{x^4 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac {2 a \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {2 a^2 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 a \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2}}{3 d}-\frac {\left (2 a^3\right ) \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 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^2 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 a \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2}}{3 d}\\ \end {align*}

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Mathematica [C]  time = 0.29, size = 81, normalized size = 0.84 \[ -\frac {2 \left (\frac {1}{\cos (c+d x)+1}\right )^{3/2} \cot ^3(c+d x) (a (\sec (c+d x)+1))^{5/2} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )}{3 d \sqrt {\frac {1}{\sec (c+d x)+1}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[c + d*x]^4*(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(-2*((1 + Cos[c + d*x])^(-1))^(3/2)*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, -3/2, -1/2, 2*Sin[(c + d*x)/2]^2]*(
a*(1 + Sec[c + d*x]))^(5/2))/(3*d*Sqrt[(1 + Sec[c + d*x])^(-1)])

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fricas [A]  time = 0.64, size = 355, normalized size = 3.70 \[ \left [\frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (d x + c\right )^{3} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) + 1}\right ) \sin \left (d x + c\right ) + 4 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{6 \, {\left (d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )}, \frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) - a}\right ) \sin \left (d x + c\right ) + 2 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{3 \, {\left (d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^2*cos(d*x + c) - a^2)*sqrt(-a)*log(-(8*a*cos(d*x + c)^3 - 4*(2*cos(d*x + c)^2 - cos(d*x + c))*sqrt(
-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*(4*a^2*cos(d*x + c)^2 - 3*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/((d*cos(d*x + c)
- d)*sin(d*x + c)), 1/3*(3*(a^2*cos(d*x + c) - a^2)*sqrt(a)*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*(4*a^2*cos(d*x +
c)^2 - 3*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/((d*cos(d*x + c) - d)*sin(d*x + c))]

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giac [B]  time = 5.97, size = 311, normalized size = 3.24 \[ -\frac {\frac {3 \, \sqrt {-a} a^{3} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} + \frac {\sqrt {2} {\left (9 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} \sqrt {-a} a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 12 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} \sqrt {-a} a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 7 \, \sqrt {-a} a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/3*(3*sqrt(-a)*a^3*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqr
t(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(2)*a
bs(a) - 6*a))*sgn(cos(d*x + c))/abs(a) + sqrt(2)*(9*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2
*c)^2 + a))^4*sqrt(-a)*a^3*sgn(cos(d*x + c)) - 12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c
)^2 + a))^2*sqrt(-a)*a^4*sgn(cos(d*x + c)) + 7*sqrt(-a)*a^5*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c)
- sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)^3)/d

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maple [B]  time = 1.18, size = 214, normalized size = 2.23 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )-3 \sqrt {2}\, \sin \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-8 \left (\cos ^{2}\left (d x +c \right )\right )+6 \cos \left (d x +c \right )\right ) a^{2}}{3 d \sin \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x)

[Out]

-1/3/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(3*cos(d*x+c)*sin(d*x+c)*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))-3*2^(1/2)*sin(d*x+c)*arctan
h(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)
-8*cos(d*x+c)^2+6*cos(d*x+c))/sin(d*x+c)/(-1+cos(d*x+c))*a^2

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^4*(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(cot(c + d*x)^4*(a + a/cos(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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