3.1.36 \(\int \frac {1}{(a+b \cot ^2(c+d x))^{5/2}} \, dx\) [36]

Optimal. Leaf size=135 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{(a-b)^{5/2} d}+\frac {b \cot (c+d x)}{3 a (a-b) d \left (a+b \cot ^2(c+d x)\right )^{3/2}}+\frac {(5 a-2 b) b \cot (c+d x)}{3 a^2 (a-b)^2 d \sqrt {a+b \cot ^2(c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3742, 425, 541, 12, 385, 209} \begin {gather*} \frac {b (5 a-2 b) \cot (c+d x)}{3 a^2 d (a-b)^2 \sqrt {a+b \cot ^2(c+d x)}}-\frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{d (a-b)^{5/2}}+\frac {b \cot (c+d x)}{3 a d (a-b) \left (a+b \cot ^2(c+d x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cot[c + d*x]^2)^(-5/2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 425

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

Rule 541

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

Rule 3742

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[c*(ff/f), Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 8.00, size = 367, normalized size = 2.72 \begin {gather*} -\frac {\cot ^5(c+d x) \left (24 (a-b)^3 \cos ^2(c+d x) \, _3F_2\left (2,2,2;1,\frac {9}{2};\frac {(a-b) \cos ^2(c+d x)}{a}\right ) \left (b+a \tan ^2(c+d x)\right )^2+24 (a-b)^3 \cos ^2(c+d x) \, _2F_1\left (2,2;\frac {9}{2};\frac {(a-b) \cos ^2(c+d x)}{a}\right ) \left (3 b^2+7 a b \tan ^2(c+d x)+4 a^2 \tan ^4(c+d x)\right )-\frac {35 a \left (8 b^2+20 a b \tan ^2(c+d x)+15 a^2 \tan ^4(c+d x)\right ) \left (-3 \text {ArcSin}\left (\sqrt {\frac {(a-b) \cos ^2(c+d x)}{a}}\right ) \left (b+a \tan ^2(c+d x)\right )^2+a \sec ^2(c+d x) \sqrt {\frac {(a-b) \cos ^4(c+d x) \left (b+a \tan ^2(c+d x)\right )}{a^2}} \left (4 b+a \left (-1+3 \tan ^2(c+d x)\right )\right )\right )}{\sqrt {\frac {(a-b) \cos ^4(c+d x) \left (b+a \tan ^2(c+d x)\right )}{a^2}}}\right )}{315 a^5 (a-b)^2 d \left (1+\cot ^2(c+d x)\right ) \sqrt {a+b \cot ^2(c+d x)} \left (1+\frac {b \cot ^2(c+d x)}{a}\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/315*(Cot[c + d*x]^5*(24*(a - b)^3*Cos[c + d*x]^2*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Cos[c + d*
x]^2)/a]*(b + a*Tan[c + d*x]^2)^2 + 24*(a - b)^3*Cos[c + d*x]^2*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Cos[c +
d*x]^2)/a]*(3*b^2 + 7*a*b*Tan[c + d*x]^2 + 4*a^2*Tan[c + d*x]^4) - (35*a*(8*b^2 + 20*a*b*Tan[c + d*x]^2 + 15*a
^2*Tan[c + d*x]^4)*(-3*ArcSin[Sqrt[((a - b)*Cos[c + d*x]^2)/a]]*(b + a*Tan[c + d*x]^2)^2 + a*Sec[c + d*x]^2*Sq
rt[((a - b)*Cos[c + d*x]^4*(b + a*Tan[c + d*x]^2))/a^2]*(4*b + a*(-1 + 3*Tan[c + d*x]^2))))/Sqrt[((a - b)*Cos[
c + d*x]^4*(b + a*Tan[c + d*x]^2))/a^2]))/(a^5*(a - b)^2*d*(1 + Cot[c + d*x]^2)*Sqrt[a + b*Cot[c + d*x]^2]*(1
+ (b*Cot[c + d*x]^2)/a))

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Maple [A]
time = 0.22, size = 162, normalized size = 1.20

method	result	size
derivativedivides	\(\frac {-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \cot \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}\right )}{\left (a -b \right )^{3} b^{2}}+\frac {b \left (\frac {\cot \left (d x +c \right )}{3 a \left (a +b \left (\cot ^{2}\left (d x +c \right )\right )\right )^{\frac {3}{2}}}+\frac {2 \cot \left (d x +c \right )}{3 a^{2} \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}\right )}{a -b}+\frac {b \cot \left (d x +c \right )}{\left (a -b \right )^{2} a \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}}{d}\)	\(162\)
default	\(\frac {-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \cot \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}\right )}{\left (a -b \right )^{3} b^{2}}+\frac {b \left (\frac {\cot \left (d x +c \right )}{3 a \left (a +b \left (\cot ^{2}\left (d x +c \right )\right )\right )^{\frac {3}{2}}}+\frac {2 \cot \left (d x +c \right )}{3 a^{2} \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}\right )}{a -b}+\frac {b \cot \left (d x +c \right )}{\left (a -b \right )^{2} a \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}}{d}\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cot(d*x+c)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/(a-b)^3*(b^4*(a-b))^(1/2)/b^2*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*cot(d*x+c)^2)^(1/2)*cot(d*x+c))+
1/(a-b)*b*(1/3*cot(d*x+c)/a/(a+b*cot(d*x+c)^2)^(3/2)+2/3/a^2*cot(d*x+c)/(a+b*cot(d*x+c)^2)^(1/2))+1/(a-b)^2*b*
cot(d*x+c)/a/(a+b*cot(d*x+c)^2)^(1/2))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (121) = 242\).
time = 3.34, size = 898, normalized size = 6.65 \begin {gather*} \left [-\frac {3 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )^{2} - 2 \, {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )\right )} \sqrt {-a + b} \log \left (-2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )^{2} - 2 \, {\left ({\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - b\right )} \sqrt {-a + b} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right ) + a^{2} - 2 \, b^{2} + 4 \, {\left (a b - b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )\right ) - 8 \, {\left (3 \, a^{3} b - 2 \, a^{2} b^{2} - 2 \, a b^{3} + b^{4} - {\left (3 \, a^{3} b - 7 \, a^{2} b^{2} + 5 \, a b^{3} - b^{4}\right )} \cos \left (2 \, d x + 2 \, c\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right )}{12 \, {\left ({\left (a^{7} - 5 \, a^{6} b + 10 \, a^{5} b^{2} - 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (2 \, d x + 2 \, c\right )^{2} - 2 \, {\left (a^{7} - 3 \, a^{6} b + 2 \, a^{5} b^{2} + 2 \, a^{4} b^{3} - 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} d \cos \left (2 \, d x + 2 \, c\right ) + {\left (a^{7} - a^{6} b - 2 \, a^{5} b^{2} + 2 \, a^{4} b^{3} + a^{3} b^{4} - a^{2} b^{5}\right )} d\right )}}, -\frac {3 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )^{2} - 2 \, {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )\right )} \sqrt {a - b} \arctan \left (-\frac {\sqrt {a - b} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right )}{{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - b}\right ) - 4 \, {\left (3 \, a^{3} b - 2 \, a^{2} b^{2} - 2 \, a b^{3} + b^{4} - {\left (3 \, a^{3} b - 7 \, a^{2} b^{2} + 5 \, a b^{3} - b^{4}\right )} \cos \left (2 \, d x + 2 \, c\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right )}{6 \, {\left ({\left (a^{7} - 5 \, a^{6} b + 10 \, a^{5} b^{2} - 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} - a^{2} b^{5}\right )} d \cos \left (2 \, d x + 2 \, c\right )^{2} - 2 \, {\left (a^{7} - 3 \, a^{6} b + 2 \, a^{5} b^{2} + 2 \, a^{4} b^{3} - 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} d \cos \left (2 \, d x + 2 \, c\right ) + {\left (a^{7} - a^{6} b - 2 \, a^{5} b^{2} + 2 \, a^{4} b^{3} + a^{3} b^{4} - a^{2} b^{5}\right )} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(a^4 + 2*a^3*b + a^2*b^2 + (a^4 - 2*a^3*b + a^2*b^2)*cos(2*d*x + 2*c)^2 - 2*(a^4 - a^2*b^2)*cos(2*d*
x + 2*c))*sqrt(-a + b)*log(-2*(a^2 - 2*a*b + b^2)*cos(2*d*x + 2*c)^2 - 2*((a - b)*cos(2*d*x + 2*c) - b)*sqrt(-
a + b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c) + a^2 - 2*b^2 + 4*(a*b
- b^2)*cos(2*d*x + 2*c)) - 8*(3*a^3*b - 2*a^2*b^2 - 2*a*b^3 + b^4 - (3*a^3*b - 7*a^2*b^2 + 5*a*b^3 - b^4)*cos
(2*d*x + 2*c))*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c))/((a^7 - 5*a^6
*b + 10*a^5*b^2 - 10*a^4*b^3 + 5*a^3*b^4 - a^2*b^5)*d*cos(2*d*x + 2*c)^2 - 2*(a^7 - 3*a^6*b + 2*a^5*b^2 + 2*a^
4*b^3 - 3*a^3*b^4 + a^2*b^5)*d*cos(2*d*x + 2*c) + (a^7 - a^6*b - 2*a^5*b^2 + 2*a^4*b^3 + a^3*b^4 - a^2*b^5)*d)
, -1/6*(3*(a^4 + 2*a^3*b + a^2*b^2 + (a^4 - 2*a^3*b + a^2*b^2)*cos(2*d*x + 2*c)^2 - 2*(a^4 - a^2*b^2)*cos(2*d*
x + 2*c))*sqrt(a - b)*arctan(-sqrt(a - b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(
2*d*x + 2*c)/((a - b)*cos(2*d*x + 2*c) - b)) - 4*(3*a^3*b - 2*a^2*b^2 - 2*a*b^3 + b^4 - (3*a^3*b - 7*a^2*b^2 +
5*a*b^3 - b^4)*cos(2*d*x + 2*c))*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x +
2*c))/((a^7 - 5*a^6*b + 10*a^5*b^2 - 10*a^4*b^3 + 5*a^3*b^4 - a^2*b^5)*d*cos(2*d*x + 2*c)^2 - 2*(a^7 - 3*a^6*b
+ 2*a^5*b^2 + 2*a^4*b^3 - 3*a^3*b^4 + a^2*b^5)*d*cos(2*d*x + 2*c) + (a^7 - a^6*b - 2*a^5*b^2 + 2*a^4*b^3 + a^
3*b^4 - a^2*b^5)*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \cot ^{2}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)**2)**(5/2),x)

[Out]

Integral((a + b*cot(c + d*x)**2)**(-5/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1160 vs. \(2 (121) = 242\).
time = 1.49, size = 1160, normalized size = 8.59 \begin {gather*} -\frac {\frac {{\left ({\left (\frac {{\left (5 \, a^{9} b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 42 \, a^{8} b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 156 \, a^{7} b^{4} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 336 \, a^{6} b^{5} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 462 \, a^{5} b^{6} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 420 \, a^{4} b^{7} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 252 \, a^{3} b^{8} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 96 \, a^{2} b^{9} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 21 \, a b^{10} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 2 \, b^{11} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{12} - 10 \, a^{11} b + 45 \, a^{10} b^{2} - 120 \, a^{9} b^{3} + 210 \, a^{8} b^{4} - 252 \, a^{7} b^{5} + 210 \, a^{6} b^{6} - 120 \, a^{5} b^{7} + 45 \, a^{4} b^{8} - 10 \, a^{3} b^{9} + a^{2} b^{10}} + \frac {3 \, {\left (8 \, a^{10} b \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 73 \, a^{9} b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 298 \, a^{8} b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 716 \, a^{7} b^{4} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 1120 \, a^{6} b^{5} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 1190 \, a^{5} b^{6} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 868 \, a^{4} b^{7} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 428 \, a^{3} b^{8} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 136 \, a^{2} b^{9} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 25 \, a b^{10} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 2 \, b^{11} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )\right )}}{a^{12} - 10 \, a^{11} b + 45 \, a^{10} b^{2} - 120 \, a^{9} b^{3} + 210 \, a^{8} b^{4} - 252 \, a^{7} b^{5} + 210 \, a^{6} b^{6} - 120 \, a^{5} b^{7} + 45 \, a^{4} b^{8} - 10 \, a^{3} b^{9} + a^{2} b^{10}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3 \, {\left (8 \, a^{10} b \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 73 \, a^{9} b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 298 \, a^{8} b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 716 \, a^{7} b^{4} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 1120 \, a^{6} b^{5} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 1190 \, a^{5} b^{6} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 868 \, a^{4} b^{7} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 428 \, a^{3} b^{8} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 136 \, a^{2} b^{9} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 25 \, a b^{10} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 2 \, b^{11} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )\right )}}{a^{12} - 10 \, a^{11} b + 45 \, a^{10} b^{2} - 120 \, a^{9} b^{3} + 210 \, a^{8} b^{4} - 252 \, a^{7} b^{5} + 210 \, a^{6} b^{6} - 120 \, a^{5} b^{7} + 45 \, a^{4} b^{8} - 10 \, a^{3} b^{9} + a^{2} b^{10}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {5 \, a^{9} b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 42 \, a^{8} b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 156 \, a^{7} b^{4} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 336 \, a^{6} b^{5} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 462 \, a^{5} b^{6} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 420 \, a^{4} b^{7} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 252 \, a^{3} b^{8} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 96 \, a^{2} b^{9} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 21 \, a b^{10} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 2 \, b^{11} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )}{a^{12} - 10 \, a^{11} b + 45 \, a^{10} b^{2} - 120 \, a^{9} b^{3} + 210 \, a^{8} b^{4} - 252 \, a^{7} b^{5} + 210 \, a^{6} b^{6} - 120 \, a^{5} b^{7} + 45 \, a^{4} b^{8} - 10 \, a^{3} b^{9} + a^{2} b^{10}}}{{\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b\right )}^{\frac {3}{2}}} - \frac {6 \, \arctan \left (-\frac {\sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b} + \sqrt {b}}{2 \, \sqrt {a - b}}\right )}{{\left (a^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 2 \, a b \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )\right )} \sqrt {a - b}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)^2)^(5/2),x, algorithm="giac")

[Out]

-1/3*(((((5*a^9*b^2*sgn(sin(d*x + c)) - 42*a^8*b^3*sgn(sin(d*x + c)) + 156*a^7*b^4*sgn(sin(d*x + c)) - 336*a^6
*b^5*sgn(sin(d*x + c)) + 462*a^5*b^6*sgn(sin(d*x + c)) - 420*a^4*b^7*sgn(sin(d*x + c)) + 252*a^3*b^8*sgn(sin(d
*x + c)) - 96*a^2*b^9*sgn(sin(d*x + c)) + 21*a*b^10*sgn(sin(d*x + c)) - 2*b^11*sgn(sin(d*x + c)))*tan(1/2*d*x
+ 1/2*c)^2/(a^12 - 10*a^11*b + 45*a^10*b^2 - 120*a^9*b^3 + 210*a^8*b^4 - 252*a^7*b^5 + 210*a^6*b^6 - 120*a^5*b
^7 + 45*a^4*b^8 - 10*a^3*b^9 + a^2*b^10) + 3*(8*a^10*b*sgn(sin(d*x + c)) - 73*a^9*b^2*sgn(sin(d*x + c)) + 298*
a^8*b^3*sgn(sin(d*x + c)) - 716*a^7*b^4*sgn(sin(d*x + c)) + 1120*a^6*b^5*sgn(sin(d*x + c)) - 1190*a^5*b^6*sgn(
sin(d*x + c)) + 868*a^4*b^7*sgn(sin(d*x + c)) - 428*a^3*b^8*sgn(sin(d*x + c)) + 136*a^2*b^9*sgn(sin(d*x + c))
- 25*a*b^10*sgn(sin(d*x + c)) + 2*b^11*sgn(sin(d*x + c)))/(a^12 - 10*a^11*b + 45*a^10*b^2 - 120*a^9*b^3 + 210*
a^8*b^4 - 252*a^7*b^5 + 210*a^6*b^6 - 120*a^5*b^7 + 45*a^4*b^8 - 10*a^3*b^9 + a^2*b^10))*tan(1/2*d*x + 1/2*c)^
2 - 3*(8*a^10*b*sgn(sin(d*x + c)) - 73*a^9*b^2*sgn(sin(d*x + c)) + 298*a^8*b^3*sgn(sin(d*x + c)) - 716*a^7*b^4
*sgn(sin(d*x + c)) + 1120*a^6*b^5*sgn(sin(d*x + c)) - 1190*a^5*b^6*sgn(sin(d*x + c)) + 868*a^4*b^7*sgn(sin(d*x
+ c)) - 428*a^3*b^8*sgn(sin(d*x + c)) + 136*a^2*b^9*sgn(sin(d*x + c)) - 25*a*b^10*sgn(sin(d*x + c)) + 2*b^11*
sgn(sin(d*x + c)))/(a^12 - 10*a^11*b + 45*a^10*b^2 - 120*a^9*b^3 + 210*a^8*b^4 - 252*a^7*b^5 + 210*a^6*b^6 - 1
20*a^5*b^7 + 45*a^4*b^8 - 10*a^3*b^9 + a^2*b^10))*tan(1/2*d*x + 1/2*c)^2 - (5*a^9*b^2*sgn(sin(d*x + c)) - 42*a
^8*b^3*sgn(sin(d*x + c)) + 156*a^7*b^4*sgn(sin(d*x + c)) - 336*a^6*b^5*sgn(sin(d*x + c)) + 462*a^5*b^6*sgn(sin
(d*x + c)) - 420*a^4*b^7*sgn(sin(d*x + c)) + 252*a^3*b^8*sgn(sin(d*x + c)) - 96*a^2*b^9*sgn(sin(d*x + c)) + 21
*a*b^10*sgn(sin(d*x + c)) - 2*b^11*sgn(sin(d*x + c)))/(a^12 - 10*a^11*b + 45*a^10*b^2 - 120*a^9*b^3 + 210*a^8*
b^4 - 252*a^7*b^5 + 210*a^6*b^6 - 120*a^5*b^7 + 45*a^4*b^8 - 10*a^3*b^9 + a^2*b^10))/(b*tan(1/2*d*x + 1/2*c)^4
+ 4*a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c)^2 + b)^(3/2) - 6*arctan(-1/2*(sqrt(b)*tan(1/2*d*x + 1
/2*c)^2 - sqrt(b*tan(1/2*d*x + 1/2*c)^4 + 4*a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c)^2 + b) + sqrt(
b))/sqrt(a - b))/((a^2*sgn(sin(d*x + c)) - 2*a*b*sgn(sin(d*x + c)) + b^2*sgn(sin(d*x + c)))*sqrt(a - b)))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (b\,{\mathrm {cot}\left (c+d\,x\right )}^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cot(c + d*x)^2)^(5/2),x)

[Out]

int(1/(a + b*cot(c + d*x)^2)^(5/2), x)

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