Integrand size = 21, antiderivative size = 115 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {5 \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{3/2}}-\frac {5 \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{3/2}}+\frac {5}{2 b d \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d \sqrt {d \cos (a+b x)}} \] Output:
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 = 0.54 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.79 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {-\left (-\cot ^2(a+b x)\right )^{3/4} \left (-4+\cot ^2(a+b x)\right )+5 \cot ^2(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4},\frac {5}{4},\csc ^2(a+b x)\right )}{2 b d \sqrt {d \cos (a+b x)} \left (-\cot ^2(a+b x)\right )^{3/4}} \] Input:
Integrate[Csc[a + b*x]^3/(d*Cos[a + b*x])^(3/2),x]
Output:
(-((-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 = 0.47 (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 |
\(\displaystyle -\frac {d^3 \left (\frac {5 \left (\frac {2 \left (\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}\right )}{4 d^2}+\frac {1}{2 d^2 \sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
Input:
Int[Csc[a + b*x]^3/(d*Cos[a + b*x])^(3/2),x]
Output:
-((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)
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 = 3.50 (sec) , antiderivative size = 689, normalized size of antiderivative = 5.99
method | result | size |
default | \(-\frac {\sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}\, \sqrt {-d}\, \sqrt {d}-\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{6} \left (-20 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) d^{\frac {3}{2}}-10 \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}\, d -10 \ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}\, d \right )-5 \left (6 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) d^{\frac {3}{2}}+3 \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}\, d -4 \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}\, \sqrt {-d}\, \sqrt {d}+3 \ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}\, d \right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+5 \left (2 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) d^{\frac {3}{2}}+\ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}\, d -4 \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}\, \sqrt {-d}\, \sqrt {d}+\ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} d +d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}\, d \right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{8 d^{\frac {5}{2}} \sqrt {-d}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-3 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1\right ) b}\) | \(689\) |
Input:
int(csc(b*x+a)^3/(d*cos(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
Output:
-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)*((-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*(-d)^(1/2)*d^(1/ 2)-sin(1/2*b*x+1/2*a)^6*(-20*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1 /2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(3/2)-10*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*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*(-d)^( 1/2)*d-10*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 *sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+d))*(-d)^(1/2)*d)-5*(6*ln(2/cos(1/2*b*x+1 /2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*d^(3/2)+3*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*sin(1/2*b*x+1/2* a)^2*d+d)^(1/2)-d))*(-d)^(1/2)*d-4*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)*(-d )^(1/2)*d^(1/2)+3*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*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+d))*(-d)^(1/2)*d)*sin(1/2*b*x+1/2 *a)^4+5*(2*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+ d)^(1/2)-d))*d^(3/2)+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*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))*(-d)^(1/2)*d-4*(-2*sin(1/2* b*x+1/2*a)^2*d+d)^(1/2)*(-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*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+d))*(-d) ^(1/2)*d)*sin(1/2*b*x+1/2*a)^2)/b
Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (91) = 182\).
Time = 0.14 (sec) , antiderivative size = 397, normalized size of antiderivative = 3.45 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\left [\frac {10 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d}}{d \cos \left (b x + a\right ) + d}\right ) - 5 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {-d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} - 4\right )}}{16 \, {\left (b d^{2} \cos \left (b x + a\right )^{3} - b d^{2} \cos \left (b x + a\right )\right )}}, -\frac {10 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d}}{d \cos \left (b x + a\right ) - d}\right ) - 5 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) - 8 \, \sqrt {d \cos \left (b x + a\right )} {\left (5 \, \cos \left (b x + a\right )^{2} - 4\right )}}{16 \, {\left (b d^{2} \cos \left (b x + a\right )^{3} - b d^{2} \cos \left (b x + a\right )\right )}}\right ] \] Input:
integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(3/2),x, algorithm="fricas")
Output:
[1/16*(10*(cos(b*x + a)^3 - cos(b*x + a))*sqrt(-d)*arctan(2*sqrt(d*cos(b*x + a))*sqrt(-d)/(d*cos(b*x + a) + d)) - 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*cos(b*x + a)^2 - 4))/(b*d^2*cos(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)*arcta n(2*sqrt(d*cos(b*x + a))*sqrt(d)/(d*cos(b*x + a) - d)) - 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))*sq rt(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*cos( b*x + a)^3 - b*d^2*cos(b*x + a))]
\[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int \frac {\csc ^{3}{\left (a + b x \right )}}{\left (d \cos {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(csc(b*x+a)**3/(d*cos(b*x+a))**(3/2),x)
Output:
Integral(csc(a + b*x)**3/(d*cos(a + b*x))**(3/2), x)
Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {\frac {10 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{\sqrt {d}} + \frac {5 \, \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{\sqrt {d}} + \frac {4 \, {\left (5 \, d^{2} \cos \left (b x + a\right )^{2} - 4 \, d^{2}\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} - \sqrt {d \cos \left (b x + a\right )} d^{2}}}{8 \, b d} \] Input:
integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(3/2),x, algorithm="maxima")
Output:
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)
\[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int { \frac {\csc \left (b x + a\right )^{3}}{\left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(csc(b*x+a)^3/(d*cos(b*x+a))^(3/2),x, algorithm="giac")
Output:
integrate(csc(b*x + a)^3/(d*cos(b*x + a))^(3/2), x)
Timed out. \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int \frac {1}{{\sin \left (a+b\,x\right )}^3\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2}} \,d x \] Input:
int(1/(sin(a + b*x)^3*(d*cos(a + b*x))^(3/2)),x)
Output:
int(1/(sin(a + b*x)^3*(d*cos(a + b*x))^(3/2)), x)
\[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\cos \left (b x +a \right )}\, \csc \left (b x +a \right )^{3}}{\cos \left (b x +a \right )^{2}}d x \right )}{d^{2}} \] Input:
int(csc(b*x+a)^3/(d*cos(b*x+a))^(3/2),x)
Output:
(sqrt(d)*int((sqrt(cos(a + b*x))*csc(a + b*x)**3)/cos(a + b*x)**2,x))/d**2