\(\int \frac {1}{(a+b \csc ^2(c+d x))^{5/2}} \, dx\) [14]

Optimal result

Integrand size = 16, antiderivative size = 126 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b+b \cot ^2(c+d x)}}\right )}{a^{5/2} d}+\frac {b \cot (c+d x)}{3 a (a+b) d \left (a+b+b \cot ^2(c+d x)\right )^{3/2}}+\frac {b (5 a+3 b) \cot (c+d x)}{3 a^2 (a+b)^2 d \sqrt {a+b+b \cot ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4213, 425, 541, 12, 385, 209} \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)+b}}\right )}{a^{5/2} d}+\frac {b (5 a+3 b) \cot (c+d x)}{3 a^2 d (a+b)^2 \sqrt {a+b \cot ^2(c+d x)+b}}+\frac {b \cot (c+d x)}{3 a d (a+b) \left (a+b \cot ^2(c+d x)+b\right )^{3/2}} \]

[In]

Int[(a + b*Csc[c + d*x]^2)^(-5/2),x]

[Out]

-(ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[a + b + b*Cot[c + d*x]^2]]/(a^(5/2)*d)) + (b*Cot[c + d*x])/(3*a*(a + b)*d
*(a + b + b*Cot[c + d*x]^2)^(3/2)) + (b*(5*a + 3*b)*Cot[c + d*x])/(3*a^2*(a + b)^2*d*Sqrt[a + b + 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 4213

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

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

Mathematica [A] (verified)

Time = 1.97 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.37 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{5/2}} \, dx=\frac {\csc ^5(c+d x) \left (\frac {4 b \cos (c+d x) (a+2 b-a \cos (2 (c+d x))) \left (3 a^2+7 a b+3 b^2-a (3 a+2 b) \cos (2 (c+d x))\right )}{3 a^2 (a+b)^2}-\frac {\sqrt {2} (-a-2 b+a \cos (2 (c+d x)))^{5/2} \log \left (\sqrt {2} \sqrt {a} \cos (c+d x)+\sqrt {-a-2 b+a \cos (2 (c+d x))}\right )}{a^{5/2}}\right )}{8 d \left (a+b \csc ^2(c+d x)\right )^{5/2}} \]

[In]

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

[Out]

(Csc[c + d*x]^5*((4*b*Cos[c + d*x]*(a + 2*b - a*Cos[2*(c + d*x)])*(3*a^2 + 7*a*b + 3*b^2 - a*(3*a + 2*b)*Cos[2
*(c + d*x)]))/(3*a^2*(a + b)^2) - (Sqrt[2]*(-a - 2*b + a*Cos[2*(c + d*x)])^(5/2)*Log[Sqrt[2]*Sqrt[a]*Cos[c + d
*x] + Sqrt[-a - 2*b + a*Cos[2*(c + d*x)]]])/a^(5/2)))/(8*d*(a + b*Csc[c + d*x]^2)^(5/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2709\) vs. \(2(112)=224\).

Time = 7.28 (sec) , antiderivative size = 2710, normalized size of antiderivative = 21.51

method	result	size
default	\(\text {Expression too large to display}\)	\(2710\)

[In]

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

[Out]

-1/6/d*csc(d*x+c)*(3*cos(d*x+c)*a^4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)^4*ln(4*(-(a*cos(
d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)
^2)^(1/2)-4*cos(d*x+c)*a)-6*a^3*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^3*sin(d*x+c)^2*ln(4*
(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(
d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*b+3*b^2*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^5*ln(4*(-
(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*
x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a^2-6*sin(d*x+c)^4*cos(d*x+c)*(-a)^(1/2)*a^3*b+3*a^4*(-(a*cos(d*x+c)^2-a-b)/(
cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)^4*ln(4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+
4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)+4*sin(d*x+c)^2*cos(d*x+c)^3*(-a)^(
1/2)*a^2*b^2-6*a^3*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2*sin(d*x+c)^2*ln(4*(-(a*cos(d*x+
c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^
(1/2)-4*cos(d*x+c)*a)*b+3*b^2*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^4*ln(4*(-(a*cos(d*x+c)
^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1
/2)-4*cos(d*x+c)*a)*a^2+12*cos(d*x+c)*a^3*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)^2*ln(4*(-(
a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x
+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*b-18*b^2*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^3*ln(4*(-(a
*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+
c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a^2-6*b^3*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^3*ln(4*(-(a
*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+
c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a-15*sin(d*x+c)^2*cos(d*x+c)*(-a)^(1/2)*a^2*b^2+12*a^3*(-(a*cos(d*x+c)^2-a-b)/(
cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)^2*ln(4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+
4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*b+7*cos(d*x+c)^3*(-a)^(1/2)*a*b^3-
18*b^2*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2*ln(4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^
2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a^2
-6*b^3*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2*ln(4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^
2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a+1
8*cos(d*x+c)*b^2*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*ln(4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^
(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a^2+12
*cos(d*x+c)*b^3*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*ln(4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(
1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a+3*cos
(d*x+c)*b^4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*ln(4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)
*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)-12*cos(d*x+
c)*(-a)^(1/2)*a*b^3-3*cos(d*x+c)*(-a)^(1/2)*b^4+18*b^2*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*ln(4*(-(
a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x
+c)+1)^2)^(1/2)-4*cos(d*x+c)*a)*a^2+12*b^3*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*ln(4*(-(a*cos(d*x+c)
^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1
/2)-4*cos(d*x+c)*a)*a+3*b^4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)*ln(4*(-(a*cos(d*x+c)^2-a-b)/(cos(d*
x+c)+1)^2)^(1/2)*cos(d*x+c)*(-a)^(1/2)+4*(-a)^(1/2)*(-(a*cos(d*x+c)^2-a-b)/(cos(d*x+c)+1)^2)^(1/2)-4*cos(d*x+c
)*a))*b^2/(a+b*csc(d*x+c)^2)^(5/2)/(cos(d*x+c)-1)^2/(cos(d*x+c)+1)^2*4^(1/2)/((a*(a+b))^(1/2)-a)^2/((a*(a+b))^
(1/2)+a)^2/(-a)^(1/2)/(a+b)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 431 vs. \(2 (112) = 224\).

Time = 0.69 (sec) , antiderivative size = 973, normalized size of antiderivative = 7.72 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-1/24*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(d*x + c)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 2*(a^4 + 3*a
^3*b + 3*a^2*b^2 + a*b^3)*cos(d*x + c)^2)*sqrt(-a)*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 + a^3*b)*cos(d*x + c)
^6 + 160*(a^4 + 2*a^3*b + a^2*b^2)*cos(d*x + c)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^4 + 3*a^
3*b + 3*a^2*b^2 + a*b^3)*cos(d*x + c)^2 + 8*(16*a^3*cos(d*x + c)^7 - 24*(a^3 + a^2*b)*cos(d*x + c)^5 + 10*(a^3
+ 2*a^2*b + a*b^2)*cos(d*x + c)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x +
c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c)) + 8*(2*(3*a^3*b + 2*a^2*b^2)*cos(d*x + c)^3 - 3*(2*a^3*b + 3
*a^2*b^2 + a*b^3)*cos(d*x + c))*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c))/((a^7 + 2*
a^6*b + a^5*b^2)*d*cos(d*x + c)^4 - 2*(a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*d*cos(d*x + c)^2 + (a^7 + 4*a^6*b
+ 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d), 1/12*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(d*x + c)^4 + a^4 + 4*a^3*b + 6*a
^2*b^2 + 4*a*b^3 + b^4 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(d*x + c)^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(d
*x + c)^4 - 8*(a^2 + a*b)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)*sqrt(a)*sqrt((a*cos(d*x + c)^2 - a - b)/(cos(d*x
+ c)^2 - 1))*sin(d*x + c)/(2*a^3*cos(d*x + c)^5 - 3*(a^3 + a^2*b)*cos(d*x + c)^3 + (a^3 + 2*a^2*b + a*b^2)*co
s(d*x + c))) - 4*(2*(3*a^3*b + 2*a^2*b^2)*cos(d*x + c)^3 - 3*(2*a^3*b + 3*a^2*b^2 + a*b^3)*cos(d*x + c))*sqrt(
(a*cos(d*x + c)^2 - a - b)/(cos(d*x + c)^2 - 1))*sin(d*x + c))/((a^7 + 2*a^6*b + a^5*b^2)*d*cos(d*x + c)^4 - 2
*(a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*d*cos(d*x + c)^2 + (a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4)*d)
]

Sympy [F]

\[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 407 vs. \(2 (112) = 224\).

Time = 0.58 (sec) , antiderivative size = 407, normalized size of antiderivative = 3.23 \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\frac {{\left ({\left (\frac {{\left (5 \, a^{9} b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 3 \, a^{8} b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{12} + 2 \, a^{11} b + a^{10} b^{2}} + \frac {3 \, {\left (8 \, a^{10} b \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 7 \, a^{9} b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + a^{8} b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )\right )}}{a^{12} + 2 \, a^{11} b + a^{10} b^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3 \, {\left (8 \, a^{10} b \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 7 \, a^{9} b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + a^{8} b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )\right )}}{a^{12} + 2 \, a^{11} b + a^{10} b^{2}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {5 \, a^{9} b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + 3 \, a^{8} b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )}{a^{12} + 2 \, a^{11} b + a^{10} b^{2}}}{{\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b\right )}^{\frac {3}{2}}} - \frac {6 \, \arctan \left (-\frac {\sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b} + \sqrt {b}}{2 \, \sqrt {a}}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )}}{3 \, d} \]

[In]

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

[Out]

-1/3*(((((5*a^9*b^2*sgn(sin(d*x + c)) + 3*a^8*b^3*sgn(sin(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/(a^12 + 2*a^11*b +
a^10*b^2) + 3*(8*a^10*b*sgn(sin(d*x + c)) + 7*a^9*b^2*sgn(sin(d*x + c)) + a^8*b^3*sgn(sin(d*x + c)))/(a^12 +
2*a^11*b + a^10*b^2))*tan(1/2*d*x + 1/2*c)^2 - 3*(8*a^10*b*sgn(sin(d*x + c)) + 7*a^9*b^2*sgn(sin(d*x + c)) + a
^8*b^3*sgn(sin(d*x + c)))/(a^12 + 2*a^11*b + a^10*b^2))*tan(1/2*d*x + 1/2*c)^2 - (5*a^9*b^2*sgn(sin(d*x + c))
+ 3*a^8*b^3*sgn(sin(d*x + c)))/(a^12 + 2*a^11*b + a^10*b^2))/(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))/(a^(5/2)*sg
n(sin(d*x + c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \csc ^2(c+d x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\sin \left (c+d\,x\right )}^2}\right )}^{5/2}} \,d x \]

[In]

int(1/(a + b/sin(c + d*x)^2)^(5/2),x)

[Out]

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