Integrand size = 21, antiderivative size = 135 \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\frac {9 d^{11/2} \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}+\frac {9 d^{11/2} \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}-\frac {9 d^5 \sqrt {d \cos (a+b x)}}{2 b}-\frac {9 d^3 (d \cos (a+b x))^{5/2}}{10 b}-\frac {d (d \cos (a+b x))^{9/2} \csc ^2(a+b x)}{2 b} \]
9/4*d^(11/2)*arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b+9/4*d^(11/2)*arctanh(( d*cos(b*x+a))^(1/2)/d^(1/2))/b-9/10*d^3*(d*cos(b*x+a))^(5/2)/b-1/2*d*(d*co s(b*x+a))^(9/2)*csc(b*x+a)^2/b-9/2*d^5*(d*cos(b*x+a))^(1/2)/b
Time = 1.87 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01 \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\frac {d (d \cos (a+b x))^{9/2} \left (45 \arctan \left (\sqrt {\cos (a+b x)}\right )+24 \text {arctanh}\left (\sqrt {\cos (a+b x)}\right )-2 \sqrt {\cos (a+b x)} \left (2 \cos (2 (a+b x))+5 \csc ^2(a+b x)\right )-\frac {21}{2} \left (8 \sqrt {\cos (a+b x)}+\log \left (1-\sqrt {\cos (a+b x)}\right )-\log \left (1+\sqrt {\cos (a+b x)}\right )\right )\right )}{20 b \cos ^{\frac {9}{2}}(a+b x)} \]
(d*(d*Cos[a + b*x])^(9/2)*(45*ArcTan[Sqrt[Cos[a + b*x]]] + 24*ArcTanh[Sqrt [Cos[a + b*x]]] - 2*Sqrt[Cos[a + b*x]]*(2*Cos[2*(a + b*x)] + 5*Csc[a + b*x ]^2) - (21*(8*Sqrt[Cos[a + b*x]] + Log[1 - Sqrt[Cos[a + b*x]]] - Log[1 + S qrt[Cos[a + b*x]]]))/2))/(20*b*Cos[a + b*x]^(9/2))
Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.99, 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, 252, 262, 262, 266, 756, 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.
| \(\displaystyle \int \csc ^3(a+b x) (d \cos (a+b x))^{11/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(d \cos (a+b x))^{11/2}}{\sin (a+b x)^3}dx\) |
| \(\Big \downarrow \) 3045 |
| \(\displaystyle -\frac {\int \frac {d^4 (d \cos (a+b x))^{11/2}}{\left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {d^3 \int \frac {(d \cos (a+b x))^{11/2}}{\left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \int \frac {(d \cos (a+b x))^{7/2}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))\right )}{b}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \int \frac {(d \cos (a+b x))^{3/2}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \left (d^2 \int \frac {1}{\sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))-2 \sqrt {d \cos (a+b x)}\right )-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \left (2 d^2 \int \frac {1}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}-2 \sqrt {d \cos (a+b x)}\right )-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \left (2 d^2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}}{2 d}\right )-2 \sqrt {d \cos (a+b x)}\right )-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \left (2 d^2 \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )-2 \sqrt {d \cos (a+b x)}\right )-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {d^3 \left (\frac {(d \cos (a+b x))^{9/2}}{2 \left (d^2-d^2 \cos ^2(a+b x)\right )}-\frac {9}{4} \left (d^2 \left (2 d^2 \left (\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}+\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )-2 \sqrt {d \cos (a+b x)}\right )-\frac {2}{5} (d \cos (a+b x))^{5/2}\right )\right )}{b}\) |
-((d^3*((d*Cos[a + b*x])^(9/2)/(2*(d^2 - d^2*Cos[a + b*x]^2)) - (9*((-2*(d *Cos[a + b*x])^(5/2))/5 + d^2*(2*d^2*(ArcTan[Sqrt[d]*Cos[a + b*x]]/(2*d^(3 /2)) + ArcTanh[Sqrt[d]*Cos[a + b*x]]/(2*d^(3/2))) - 2*Sqrt[d*Cos[a + b*x]] )))/4))/b)
3.3.42.3.1 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*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) 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. \(406\) vs. \(2(107)=214\).
Time = 6.10 (sec) , antiderivative size = 407, normalized size of antiderivative = 3.01
| method | result | size |
| default | \(\frac {-\frac {d^{5} \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{8 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}-\frac {9 d^{6} \ln \left (\frac {-2 d +2 \sqrt {-d}\, \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{4 \sqrt {-d}}-6 d^{5} \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}-\frac {8 d^{5} \left (\cos ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{5}+\frac {8 d^{5} \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{5}+\frac {8 d^{5} \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{5}+\frac {9 d^{\frac {11}{2}} \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right )}{8}+\frac {9 d^{\frac {11}{2}} \ln \left (\frac {-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right )}{8}+\frac {d^{5} \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}}{16 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-16}-\frac {d^{5} \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}}{16 \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}}{b}\) | \(407\) |
(-1/8*d^5/cos(1/2*b*x+1/2*a)^2*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2)-9/4*d^6/ (-d)^(1/2)*ln((-2*d+2*(-d)^(1/2)*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2))/cos(1 /2*b*x+1/2*a))-6*d^5*(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)-8/5*d^5*cos(1/2* b*x+1/2*a)^4*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2)+8/5*d^5*cos(1/2*b*x+1/2*a) ^2*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2)+8/5*d^5*(2*cos(1/2*b*x+1/2*a)^2*d-d) ^(1/2)+9/8*d^(11/2)*ln((4*d*cos(1/2*b*x+1/2*a)+2*d^(1/2)*(-2*d*sin(1/2*b*x +1/2*a)^2+d)^(1/2)-2*d)/(cos(1/2*b*x+1/2*a)-1))+9/8*d^(11/2)*ln((-4*d*cos( 1/2*b*x+1/2*a)+2*d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-2*d)/(cos(1/2 *b*x+1/2*a)+1))+1/16*d^5/(cos(1/2*b*x+1/2*a)-1)*(-2*d*sin(1/2*b*x+1/2*a)^2 +d)^(1/2)-1/16*d^5/(cos(1/2*b*x+1/2*a)+1)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1 /2))/b
Time = 0.47 (sec) , antiderivative size = 419, normalized size of antiderivative = 3.10 \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\left [-\frac {90 \, {\left (d^{5} \cos \left (b x + a\right )^{2} - d^{5}\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d}}{d \cos \left (b x + a\right ) + d}\right ) - 45 \, {\left (d^{5} \cos \left (b x + a\right )^{2} - d^{5}\right )} \sqrt {-d} \log \left (-\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, {\left (4 \, d^{5} \cos \left (b x + a\right )^{4} + 36 \, d^{5} \cos \left (b x + a\right )^{2} - 45 \, d^{5}\right )} \sqrt {d \cos \left (b x + a\right )}}{80 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}, -\frac {90 \, {\left (d^{5} \cos \left (b x + a\right )^{2} - d^{5}\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d}}{d \cos \left (b x + a\right ) - d}\right ) - 45 \, {\left (d^{5} \cos \left (b x + a\right )^{2} - d^{5}\right )} \sqrt {d} \log \left (-\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, {\left (4 \, d^{5} \cos \left (b x + a\right )^{4} + 36 \, d^{5} \cos \left (b x + a\right )^{2} - 45 \, d^{5}\right )} \sqrt {d \cos \left (b x + a\right )}}{80 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}\right ] \]
[-1/80*(90*(d^5*cos(b*x + a)^2 - d^5)*sqrt(-d)*arctan(2*sqrt(d*cos(b*x + a ))*sqrt(-d)/(d*cos(b*x + a) + d)) - 45*(d^5*cos(b*x + a)^2 - d^5)*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*(4*d ^5*cos(b*x + a)^4 + 36*d^5*cos(b*x + a)^2 - 45*d^5)*sqrt(d*cos(b*x + a)))/ (b*cos(b*x + a)^2 - b), -1/80*(90*(d^5*cos(b*x + a)^2 - d^5)*sqrt(d)*arcta n(2*sqrt(d*cos(b*x + a))*sqrt(d)/(d*cos(b*x + a) - d)) - 45*(d^5*cos(b*x + a)^2 - d^5)*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*(4*d^5*cos(b*x + a)^4 + 36*d^5*cos(b*x + a)^2 - 45*d^5)*sqr t(d*cos(b*x + a)))/(b*cos(b*x + a)^2 - b)]
Timed out. \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.99 \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\frac {\frac {20 \, \sqrt {d \cos \left (b x + a\right )} d^{8}}{d^{2} \cos \left (b x + a\right )^{2} - d^{2}} + 90 \, d^{\frac {13}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) - 45 \, d^{\frac {13}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right ) - 16 \, \left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}} d^{4} - 160 \, \sqrt {d \cos \left (b x + a\right )} d^{6}}{40 \, b d} \]
1/40*(20*sqrt(d*cos(b*x + a))*d^8/(d^2*cos(b*x + a)^2 - d^2) + 90*d^(13/2) *arctan(sqrt(d*cos(b*x + a))/sqrt(d)) - 45*d^(13/2)*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x + a)) + sqrt(d))) - 16*(d*cos(b*x + a))^(5/ 2)*d^4 - 160*sqrt(d*cos(b*x + a))*d^6)/(b*d)
\[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {11}{2}} \csc \left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int (d \cos (a+b x))^{11/2} \csc ^3(a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{11/2}}{{\sin \left (a+b\,x\right )}^3} \,d x \]