Integrand size = 21, antiderivative size = 137 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {9 \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}-\frac {9 \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b d^{7/2}}+\frac {9}{10 b d (d \cos (a+b x))^{5/2}}+\frac {9}{2 b d^3 \sqrt {d \cos (a+b x)}}-\frac {\csc ^2(a+b x)}{2 b d (d \cos (a+b x))^{5/2}} \]
9/4*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(7/2)-9/4*arctanh((d*cos(b*x+ a))^(1/2)/d^(1/2))/b/d^(7/2)+9/10/b/d/(d*cos(b*x+a))^(5/2)-1/2*csc(b*x+a)^ 2/b/d/(d*cos(b*x+a))^(5/2)+9/2/b/d^3/(d*cos(b*x+a))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.53 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.74 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {45 \cot ^2(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4},\frac {5}{4},\csc ^2(a+b x)\right )+\left (-\cot ^2(a+b x)\right )^{3/4} \left (40-5 \cot ^2(a+b x)+4 \sec ^2(a+b x)\right )}{10 b d^3 \sqrt {d \cos (a+b x)} \left (-\cot ^2(a+b x)\right )^{3/4}} \]
(45*Cot[a + b*x]^2*Hypergeometric2F1[1/4, 1/4, 5/4, Csc[a + b*x]^2] + (-Co t[a + b*x]^2)^(3/4)*(40 - 5*Cot[a + b*x]^2 + 4*Sec[a + b*x]^2))/(10*b*d^3* Sqrt[d*Cos[a + b*x]]*(-Cot[a + b*x]^2)^(3/4))
Time = 0.30 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3045, 27, 253, 264, 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.
| \(\displaystyle \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (a+b x)^3 (d \cos (a+b x))^{7/2}}dx\) |
| \(\Big \downarrow \) 3045 |
| \(\displaystyle -\frac {\int \frac {d^4}{(d \cos (a+b x))^{7/2} \left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {d^3 \int \frac {1}{(d \cos (a+b x))^{7/2} \left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle -\frac {d^3 \left (\frac {9 \int \frac {1}{(d \cos (a+b x))^{7/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{5/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle -\frac {d^3 \left (\frac {9 \left (\frac {\int \frac {1}{(d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{d^2}-\frac {2}{5 d^2 (d \cos (a+b x))^{5/2}}\right )}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{5/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle -\frac {d^3 \left (\frac {9 \left (\frac {\frac {\int \frac {\sqrt {d \cos (a+b x)}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}}{d^2}-\frac {2}{5 d^2 (d \cos (a+b x))^{5/2}}\right )}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{5/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {d^3 \left (\frac {9 \left (\frac {\frac {2 \int \frac {d^2 \cos ^2(a+b x)}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}}{d^2}-\frac {2}{5 d^2 (d \cos (a+b x))^{5/2}}\right )}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{5/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle -\frac {d^3 \left (\frac {9 \left (\frac {\frac {2 \left (\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {1}{2} \int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}\right )}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}}{d^2}-\frac {2}{5 d^2 (d \cos (a+b x))^{5/2}}\right )}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{5/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {d^3 \left (\frac {9 \left (\frac {\frac {2 \left (\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}}{d^2}-\frac {2}{5 d^2 (d \cos (a+b x))^{5/2}}\right )}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{5/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {d^3 \left (\frac {9 \left (\frac {\frac {2 \left (\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}}{d^2}-\frac {2}{5 d^2 (d \cos (a+b x))^{5/2}}\right )}{4 d^2}+\frac {1}{2 d^2 (d \cos (a+b x))^{5/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
-((d^3*(1/(2*d^2*(d*Cos[a + b*x])^(5/2)*(d^2 - d^2*Cos[a + b*x]^2)) + (9*( -2/(5*d^2*(d*Cos[a + b*x])^(5/2)) + ((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]]))/d^2))/(4*d^2)))/b)
3.3.51.3.1 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. \(1139\) vs. \(2(109)=218\).
Time = 0.18 (sec) , antiderivative size = 1140, normalized size of antiderivative = 8.32
-1/40/d^(9/2)/(-d)^(1/2)/sin(1/2*b*x+1/2*a)^2/(8*sin(1/2*b*x+1/2*a)^8-20*s in(1/2*b*x+1/2*a)^6+18*sin(1/2*b*x+1/2*a)^4-7*sin(1/2*b*x+1/2*a)^2+1)*(5*( -d)^(1/2)*d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)+360*(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))+(-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)^10+180*(-10*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)-5*(-d)^(1/2)*ln(2/(cos(1/2*b*x+1/2*a)-1)*(2*d*c os(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*(-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)^8-90*(-18*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) )+16*(-d)^(1/2)*d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-9*(-d)^(1/2)*l n(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-9*(-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)^6+9*(-70*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))+104*(-d)^(1/2)*d^(1/2)*(-2*d*sin(1/2*...
Time = 0.37 (sec) , antiderivative size = 438, normalized size of antiderivative = 3.20 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\left [\frac {90 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) - 45 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \sqrt {-d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, {\left (45 \, \cos \left (b x + a\right )^{4} - 36 \, \cos \left (b x + a\right )^{2} - 4\right )} \sqrt {d \cos \left (b x + a\right )}}{80 \, {\left (b d^{4} \cos \left (b x + a\right )^{5} - b d^{4} \cos \left (b x + a\right )^{3}\right )}}, \frac {90 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) + 45 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \sqrt {d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, {\left (45 \, \cos \left (b x + a\right )^{4} - 36 \, \cos \left (b x + a\right )^{2} - 4\right )} \sqrt {d \cos \left (b x + a\right )}}{80 \, {\left (b d^{4} \cos \left (b x + a\right )^{5} - b d^{4} \cos \left (b x + a\right )^{3}\right )}}\right ] \]
[1/80*(90*(cos(b*x + a)^5 - cos(b*x + a)^3)*sqrt(-d)*arctan(1/2*sqrt(d*cos (b*x + a))*sqrt(-d)*(cos(b*x + a) + 1)/(d*cos(b*x + a))) - 45*(cos(b*x + a )^5 - cos(b*x + a)^3)*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*(45*cos(b*x + a)^4 - 36*cos(b*x + a)^2 - 4)*sqrt( d*cos(b*x + a)))/(b*d^4*cos(b*x + a)^5 - b*d^4*cos(b*x + a)^3), 1/80*(90*( cos(b*x + a)^5 - cos(b*x + a)^3)*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*( cos(b*x + a) - 1)/(sqrt(d)*cos(b*x + a))) + 45*(cos(b*x + a)^5 - cos(b*x + a)^3)*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*(45*cos(b*x + a)^4 - 36*cos(b*x + a)^2 - 4)*sqrt(d*cos(b*x + a)))/ (b*d^4*cos(b*x + a)^5 - b*d^4*cos(b*x + a)^3)]
Timed out. \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.98 \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\frac {\frac {4 \, {\left (45 \, d^{4} \cos \left (b x + a\right )^{4} - 36 \, d^{4} \cos \left (b x + a\right )^{2} - 4 \, d^{4}\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} d^{2} - \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} d^{4}} + \frac {90 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {45 \, \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{d^{\frac {5}{2}}}}{40 \, b d} \]
1/40*(4*(45*d^4*cos(b*x + a)^4 - 36*d^4*cos(b*x + a)^2 - 4*d^4)/((d*cos(b* x + a))^(9/2)*d^2 - (d*cos(b*x + a))^(5/2)*d^4) + 90*arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(5/2) + 45*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d* cos(b*x + a)) + sqrt(d)))/d^(5/2))/(b*d)
\[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\int { \frac {\csc \left (b x + a\right )^{3}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc ^3(a+b x)}{(d \cos (a+b x))^{7/2}} \, dx=\int \frac {1}{{\sin \left (a+b\,x\right )}^3\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2}} \,d x \]