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

Optimal. Leaf size=135 \[ \frac {b (5 a-2 b) \cot (c+d x)}{3 a^2 d (a-b)^2 \sqrt {a+b \cot ^2(c+d x)}}+\frac {b \cot (c+d x)}{3 a d (a-b) \left (a+b \cot ^2(c+d x)\right )^{3/2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{d (a-b)^{5/2}} \]

[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 = 0.11, 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 = {3661, 414, 527, 12, 377, 203} \[ \frac {b (5 a-2 b) \cot (c+d x)}{3 a^2 d (a-b)^2 \sqrt {a+b \cot ^2(c+d x)}}+\frac {b \cot (c+d x)}{3 a d (a-b) \left (a+b \cot ^2(c+d x)\right )^{3/2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{d (a-b)^{5/2}} \]

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 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 377

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 414

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]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

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 3661

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 {\operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {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]  time = 7.94, size = 367, normalized size = 2.72 \[ -\frac {\cot ^5(c+d x) \left (24 (a-b)^3 \cos ^2(c+d x) \left (a \tan ^2(c+d x)+b\right )^2 \, _3F_2\left (2,2,2;1,\frac {9}{2};\frac {(a-b) \cos ^2(c+d x)}{a}\right )+24 (a-b)^3 \cos ^2(c+d x) \left (4 a^2 \tan ^4(c+d x)+7 a b \tan ^2(c+d x)+3 b^2\right ) \, _2F_1\left (2,2;\frac {9}{2};\frac {(a-b) \cos ^2(c+d x)}{a}\right )-\frac {35 a \left (15 a^2 \tan ^4(c+d x)+20 a b \tan ^2(c+d x)+8 b^2\right ) \left (a \sec ^2(c+d x) \left (a \left (3 \tan ^2(c+d x)-1\right )+4 b\right ) \sqrt {\frac {(a-b) \cos ^4(c+d x) \left (a \tan ^2(c+d x)+b\right )}{a^2}}-3 \left (a \tan ^2(c+d x)+b\right )^2 \sin ^{-1}\left (\sqrt {\frac {(a-b) \cos ^2(c+d x)}{a}}\right )\right )}{\sqrt {\frac {(a-b) \cos ^4(c+d x) \left (a \tan ^2(c+d x)+b\right )}{a^2}}}\right )}{315 a^5 d (a-b)^2 \left (\cot ^2(c+d x)+1\right ) \sqrt {a+b \cot ^2(c+d x)} \left (\frac {b \cot ^2(c+d x)}{a}+1\right )} \]

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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fricas [B]  time = 0.68, size = 898, normalized size = 6.65 \[ \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 ] \]

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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giac [B]  time = 13.77, size = 1341, normalized size = 9.93 \[ \text {result too large to display} \]

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*b - 2*b^2)*sgn(tan(1/2*d*x + 1/2*c))/(a^4*sqrt(b) - 2*a^3*b^(3/2) + a^2*b^(5/2)) + ((((5*a^9*b^2*sg
n(tan(1/2*d*x + 1/2*c)) - 42*a^8*b^3*sgn(tan(1/2*d*x + 1/2*c)) + 156*a^7*b^4*sgn(tan(1/2*d*x + 1/2*c)) - 336*a
^6*b^5*sgn(tan(1/2*d*x + 1/2*c)) + 462*a^5*b^6*sgn(tan(1/2*d*x + 1/2*c)) - 420*a^4*b^7*sgn(tan(1/2*d*x + 1/2*c
)) + 252*a^3*b^8*sgn(tan(1/2*d*x + 1/2*c)) - 96*a^2*b^9*sgn(tan(1/2*d*x + 1/2*c)) + 21*a*b^10*sgn(tan(1/2*d*x
+ 1/2*c)) - 2*b^11*sgn(tan(1/2*d*x + 1/2*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(tan(1/2*d*x + 1/2*c)) - 73*a^9*b^2*sgn(tan(1/2*d*x + 1/2*c)) + 298*a^8*b^3*sgn(tan(1/2*d*x + 1/2*c)) -
716*a^7*b^4*sgn(tan(1/2*d*x + 1/2*c)) + 1120*a^6*b^5*sgn(tan(1/2*d*x + 1/2*c)) - 1190*a^5*b^6*sgn(tan(1/2*d*x
+ 1/2*c)) + 868*a^4*b^7*sgn(tan(1/2*d*x + 1/2*c)) - 428*a^3*b^8*sgn(tan(1/2*d*x + 1/2*c)) + 136*a^2*b^9*sgn(ta
n(1/2*d*x + 1/2*c)) - 25*a*b^10*sgn(tan(1/2*d*x + 1/2*c)) + 2*b^11*sgn(tan(1/2*d*x + 1/2*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(tan(1/2*d*x + 1/2*c)) - 73*a^9*b^2*sgn(tan(1/2*d*x + 1
/2*c)) + 298*a^8*b^3*sgn(tan(1/2*d*x + 1/2*c)) - 716*a^7*b^4*sgn(tan(1/2*d*x + 1/2*c)) + 1120*a^6*b^5*sgn(tan(
1/2*d*x + 1/2*c)) - 1190*a^5*b^6*sgn(tan(1/2*d*x + 1/2*c)) + 868*a^4*b^7*sgn(tan(1/2*d*x + 1/2*c)) - 428*a^3*b
^8*sgn(tan(1/2*d*x + 1/2*c)) + 136*a^2*b^9*sgn(tan(1/2*d*x + 1/2*c)) - 25*a*b^10*sgn(tan(1/2*d*x + 1/2*c)) + 2
*b^11*sgn(tan(1/2*d*x + 1/2*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 + 2
10*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 - (5*a^9*b^2*sgn(tan(1/
2*d*x + 1/2*c)) - 42*a^8*b^3*sgn(tan(1/2*d*x + 1/2*c)) + 156*a^7*b^4*sgn(tan(1/2*d*x + 1/2*c)) - 336*a^6*b^5*s
gn(tan(1/2*d*x + 1/2*c)) + 462*a^5*b^6*sgn(tan(1/2*d*x + 1/2*c)) - 420*a^4*b^7*sgn(tan(1/2*d*x + 1/2*c)) + 252
*a^3*b^8*sgn(tan(1/2*d*x + 1/2*c)) - 96*a^2*b^9*sgn(tan(1/2*d*x + 1/2*c)) + 21*a*b^10*sgn(tan(1/2*d*x + 1/2*c)
) - 2*b^11*sgn(tan(1/2*d*x + 1/2*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(tan(1/2*d*x + 1/2*c)) - 2*a*b*sgn(tan(1/2*d*x + 1/2*c)) + b^2*sgn(tan(1/2*d*x + 1/2*c)))*sqrt(a - b
)))/d

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maple [A]  time = 0.38, size = 176, normalized size = 1.30 \[ \frac {b \cot \left (d x +c \right )}{d \left (a -b \right )^{2} a \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}+\frac {b \cot \left (d x +c \right )}{3 a \left (a -b \right ) d \left (a +b \left (\cot ^{2}\left (d x +c \right )\right )\right )^{\frac {3}{2}}}+\frac {2 b \cot \left (d x +c \right )}{3 d \left (a -b \right ) a^{2} \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \cot \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}\right )}{d \left (a -b \right )^{3} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 b-a positive or negative?

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

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

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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