Integrand size = 21, antiderivative size = 115 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {5 \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{3/2}}-\frac {5 \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{3/2}}+\frac {5}{2 b d \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d \sqrt {d \cos (a+b x)}} \]
5/4*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(3/2)-5/4*arctanh((d*cos(b*x+ a))^(1/2)/d^(1/2))/b/d^(3/2)+5/2/b/d/(d*cos(b*x+a))^(1/2)-1/2*csc(b*x+a)^2 /b/d/(d*cos(b*x+a))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.35 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.79 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {-\left (-\cot ^2(a+b x)\right )^{3/4} \left (-4+\cot ^2(a+b x)\right )+5 \cot ^2(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4},\frac {5}{4},\csc ^2(a+b x)\right )}{2 b d \sqrt {d \cos (a+b x)} \left (-\cot ^2(a+b x)\right )^{3/4}} \]
(-((-Cot[a + b*x]^2)^(3/4)*(-4 + Cot[a + b*x]^2)) + 5*Cot[a + b*x]^2*Hyper geometric2F1[1/4, 1/4, 5/4, Csc[a + b*x]^2])/(2*b*d*Sqrt[d*Cos[a + b*x]]*( -Cot[a + b*x]^2)^(3/4))
Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3045, 27, 253, 264, 266, 827, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (a+b x)^3 (d \cos (a+b x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3045 |
| \(\displaystyle -\frac {\int \frac {d^4}{(d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {d^3 \int \frac {1}{(d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle -\frac {d^3 \left (\frac {5 \int \frac {1}{(d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{4 d^2}+\frac {1}{2 d^2 \sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle -\frac {d^3 \left (\frac {5 \left (\frac {\int \frac {\sqrt {d \cos (a+b x)}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}\right )}{4 d^2}+\frac {1}{2 d^2 \sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {d^3 \left (\frac {5 \left (\frac {2 \int \frac {d^2 \cos ^2(a+b x)}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}\right )}{4 d^2}+\frac {1}{2 d^2 \sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle -\frac {d^3 \left (\frac {5 \left (\frac {2 \left (\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {1}{2} \int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}\right )}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}\right )}{4 d^2}+\frac {1}{2 d^2 \sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {d^3 \left (\frac {5 \left (\frac {2 \left (\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}\right )}{4 d^2}+\frac {1}{2 d^2 \sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {d^3 \left (\frac {5 \left (\frac {2 \left (\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}\right )}{4 d^2}+\frac {1}{2 d^2 \sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
-((d^3*(1/(2*d^2*Sqrt[d*Cos[a + b*x]]*(d^2 - d^2*Cos[a + b*x]^2)) + (5*((2 *(-1/2*ArcTan[Sqrt[d]*Cos[a + b*x]]/Sqrt[d] + ArcTanh[Sqrt[d]*Cos[a + b*x] ]/(2*Sqrt[d])))/d^2 - 2/(d^2*Sqrt[d*Cos[a + b*x]])))/(4*d^2)))/b)
3.3.49.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Leaf count of result is larger than twice the leaf count of optimal. \(688\) vs. \(2(91)=182\).
Time = 0.12 (sec) , antiderivative size = 689, normalized size of antiderivative = 5.99
| method | result | size |
| default | \(-\frac {\sqrt {-d}\, \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-\left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (-20 d^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )-10 \sqrt {-d}\, \ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) d -10 \sqrt {-d}\, \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) d \right )-5 \left (6 d^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )-4 \sqrt {-d}\, \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+3 \sqrt {-d}\, \ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) d +3 \sqrt {-d}\, \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) d \right ) \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+5 \left (2 d^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )-4 \sqrt {-d}\, \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+\sqrt {-d}\, \ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) d +\sqrt {-d}\, \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) d \right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 d^{\frac {5}{2}} \sqrt {-d}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-3 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1\right ) b}\) | \(689\) |
-1/8/d^(5/2)/(-d)^(1/2)/sin(1/2*b*x+1/2*a)^2/(2*sin(1/2*b*x+1/2*a)^4-3*sin (1/2*b*x+1/2*a)^2+1)*((-d)^(1/2)*d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/ 2)-sin(1/2*b*x+1/2*a)^6*(-20*d^(3/2)*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*( -2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-d))-10*(-d)^(1/2)*ln(-2/(cos(1/2*b*x+1/ 2*a)+1)*(2*d*cos(1/2*b*x+1/2*a)-d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2 )+d))*d-10*(-d)^(1/2)*ln(2/(cos(1/2*b*x+1/2*a)-1)*(2*d*cos(1/2*b*x+1/2*a)+ d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-d))*d)-5*(6*d^(3/2)*ln(2/cos(1 /2*b*x+1/2*a)*((-d)^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-d))-4*(-d)^( 1/2)*d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)+3*(-d)^(1/2)*ln(-2/(cos(1 /2*b*x+1/2*a)+1)*(2*d*cos(1/2*b*x+1/2*a)-d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^ 2+d)^(1/2)+d))*d+3*(-d)^(1/2)*ln(2/(cos(1/2*b*x+1/2*a)-1)*(2*d*cos(1/2*b*x +1/2*a)+d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-d))*d)*sin(1/2*b*x+1/2 *a)^4+5*(2*d^(3/2)*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*d*sin(1/2*b*x+1 /2*a)^2+d)^(1/2)-d))-4*(-d)^(1/2)*d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1 /2)+(-d)^(1/2)*ln(-2/(cos(1/2*b*x+1/2*a)+1)*(2*d*cos(1/2*b*x+1/2*a)-d^(1/2 )*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)+d))*d+(-d)^(1/2)*ln(2/(cos(1/2*b*x+1 /2*a)-1)*(2*d*cos(1/2*b*x+1/2*a)+d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/ 2)-d))*d)*sin(1/2*b*x+1/2*a)^2)/b
Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (91) = 182\).
Time = 0.34 (sec) , antiderivative size = 406, normalized size of antiderivative = 3.53 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\left [\frac {10 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) - 5 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {-d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} - 4\right )}}{16 \, {\left (b d^{2} \cos \left (b x + a\right )^{3} - b d^{2} \cos \left (b x + a\right )\right )}}, \frac {10 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) + 5 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} - 4\right )}}{16 \, {\left (b d^{2} \cos \left (b x + a\right )^{3} - b d^{2} \cos \left (b x + a\right )\right )}}\right ] \]
[1/16*(10*(cos(b*x + a)^3 - cos(b*x + a))*sqrt(-d)*arctan(1/2*sqrt(d*cos(b *x + a))*sqrt(-d)*(cos(b*x + a) + 1)/(d*cos(b*x + a))) - 5*(cos(b*x + a)^3 - cos(b*x + a))*sqrt(-d)*log((d*cos(b*x + a)^2 - 4*sqrt(d*cos(b*x + a))*s qrt(-d)*(cos(b*x + a) - 1) - 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 + 2*cos (b*x + a) + 1)) + 8*sqrt(d*cos(b*x + a))*(5*cos(b*x + a)^2 - 4))/(b*d^2*co s(b*x + a)^3 - b*d^2*cos(b*x + a)), 1/16*(10*(cos(b*x + a)^3 - cos(b*x + a ))*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*cos (b*x + a))) + 5*(cos(b*x + a)^3 - cos(b*x + a))*sqrt(d)*log((d*cos(b*x + a )^2 - 4*sqrt(d*cos(b*x + a))*sqrt(d)*(cos(b*x + a) + 1) + 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2 - 2*cos(b*x + a) + 1)) + 8*sqrt(d*cos(b*x + a))*(5*c os(b*x + a)^2 - 4))/(b*d^2*cos(b*x + a)^3 - b*d^2*cos(b*x + a))]
\[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int \frac {\csc ^{3}{\left (a + b x \right )}}{\left (d \cos {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {\frac {10 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{\sqrt {d}} + \frac {5 \, \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{\sqrt {d}} + \frac {4 \, {\left (5 \, d^{2} \cos \left (b x + a\right )^{2} - 4 \, d^{2}\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} - \sqrt {d \cos \left (b x + a\right )} d^{2}}}{8 \, b d} \]
1/8*(10*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/sqrt(d) + 5*log((sqrt(d*cos(b *x + a)) - sqrt(d))/(sqrt(d*cos(b*x + a)) + sqrt(d)))/sqrt(d) + 4*(5*d^2*c os(b*x + a)^2 - 4*d^2)/((d*cos(b*x + a))^(5/2) - sqrt(d*cos(b*x + a))*d^2) )/(b*d)
\[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int { \frac {\csc \left (b x + a\right )^{3}}{\left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int \frac {1}{{\sin \left (a+b\,x\right )}^3\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2}} \,d x \]