Integrand size = 21, antiderivative size = 93 \[ \int \sqrt {d \cos (a+b x)} \csc ^3(a+b x) \, dx=\frac {\sqrt {d} \arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}-\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{4 b}-\frac {(d \cos (a+b x))^{3/2} \csc ^2(a+b x)}{2 b d} \]
-1/2*(d*cos(b*x+a))^(3/2)*csc(b*x+a)^2/b/d+1/4*arctan((d*cos(b*x+a))^(1/2) /d^(1/2))*d^(1/2)/b-1/4*arctanh((d*cos(b*x+a))^(1/2)/d^(1/2))*d^(1/2)/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.34 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.67 \[ \int \sqrt {d \cos (a+b x)} \csc ^3(a+b x) \, dx=-\frac {d \left (\cot ^2(a+b x)+\sqrt [4]{-\cot ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4},\frac {5}{4},\csc ^2(a+b x)\right )\right )}{2 b \sqrt {d \cos (a+b x)}} \]
-1/2*(d*(Cot[a + b*x]^2 + (-Cot[a + b*x]^2)^(1/4)*Hypergeometric2F1[1/4, 1 /4, 5/4, Csc[a + b*x]^2]))/(b*Sqrt[d*Cos[a + b*x]])
Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3045, 27, 253, 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.
\(\displaystyle \int \csc ^3(a+b x) \sqrt {d \cos (a+b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {d \cos (a+b x)}}{\sin (a+b x)^3}dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \frac {d^4 \sqrt {d \cos (a+b x)}}{\left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {d^3 \int \frac {\sqrt {d \cos (a+b x)}}{\left (d^2-d^2 \cos ^2(a+b x)\right )^2}d(d \cos (a+b x))}{b}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle -\frac {d^3 \left (\frac {\int \frac {\sqrt {d \cos (a+b x)}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))}{4 d^2}+\frac {(d \cos (a+b x))^{3/2}}{2 d^2 \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {d^3 \left (\frac {\int \frac {d^2 \cos ^2(a+b x)}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d^2}+\frac {(d \cos (a+b x))^{3/2}}{2 d^2 \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {d^3 \left (\frac {\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {1}{2} \int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}}{2 d^2}+\frac {(d \cos (a+b x))^{3/2}}{2 d^2 \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {d^3 \left (\frac {\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}}{2 d^2}+\frac {(d \cos (a+b x))^{3/2}}{2 d^2 \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {d^3 \left (\frac {\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}}{2 d^2}+\frac {(d \cos (a+b x))^{3/2}}{2 d^2 \left (d^2-d^2 \cos ^2(a+b x)\right )}\right )}{b}\) |
-((d^3*((-1/2*ArcTan[Sqrt[d]*Cos[a + b*x]]/Sqrt[d] + ArcTanh[Sqrt[d]*Cos[a + b*x]]/(2*Sqrt[d]))/(2*d^2) + (d*Cos[a + b*x])^(3/2)/(2*d^2*(d^2 - d^2*C os[a + b*x]^2))))/b)
3.3.47.3.1 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] :> 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. \(274\) vs. \(2(73)=146\).
Time = 0.08 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.96
method | result | size |
default | \(\frac {\frac {\sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{8 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}-\frac {d \ln \left (\frac {-2 d +2 \sqrt {-d}\, \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d -d}}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{4 \sqrt {-d}}+\frac {\sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}}{16 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-16}-\frac {\sqrt {d}\, \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right )}{8}-\frac {\sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}}{16 \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}-\frac {\sqrt {d}\, \ln \left (\frac {-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right )}{8}}{b}\) | \(275\) |
(1/8/cos(1/2*b*x+1/2*a)^2*(2*cos(1/2*b*x+1/2*a)^2*d-d)^(1/2)-1/4*d/(-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))+1/16/(cos(1/2*b*x+1/2*a)-1)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-1/ 8*d^(1/2)*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/(cos(1/2*b*x+1/2*a)+1)*(-2*d*si n(1/2*b*x+1/2*a)^2+d)^(1/2)-1/8*d^(1/2)*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)))/b
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (73) = 146\).
Time = 0.36 (sec) , antiderivative size = 340, normalized size of antiderivative = 3.66 \[ \int \sqrt {d \cos (a+b x)} \csc ^3(a+b x) \, dx=\left [\frac {2 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) + {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {-d} \log \left (\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} \cos \left (b x + a\right )}{16 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}, \frac {2 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) + {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )} \cos \left (b x + a\right )}{16 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}}\right ] \]
[1/16*(2*(cos(b*x + a)^2 - 1)*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqr t(-d)*(cos(b*x + a) + 1)/(d*cos(b*x + a))) + (cos(b*x + a)^2 - 1)*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))*cos(b*x + a))/(b*cos(b*x + a)^2 - b), 1/16*(2*(cos(b*x + a )^2 - 1)*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt( d)*cos(b*x + a))) + (cos(b*x + a)^2 - 1)*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))*cos(b*x + a ))/(b*cos(b*x + a)^2 - b)]
\[ \int \sqrt {d \cos (a+b x)} \csc ^3(a+b x) \, dx=\int \sqrt {d \cos {\left (a + b x \right )}} \csc ^{3}{\left (a + b x \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.10 \[ \int \sqrt {d \cos (a+b x)} \csc ^3(a+b x) \, dx=\frac {\frac {4 \, \left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}} d^{2}}{d^{2} \cos \left (b x + a\right )^{2} - d^{2}} + 2 \, d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) + d^{\frac {3}{2}} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{8 \, b d} \]
1/8*(4*(d*cos(b*x + a))^(3/2)*d^2/(d^2*cos(b*x + a)^2 - d^2) + 2*d^(3/2)*a rctan(sqrt(d*cos(b*x + a))/sqrt(d)) + d^(3/2)*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*cos(b*x + a)) + sqrt(d))))/(b*d)
Time = 0.33 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02 \[ \int \sqrt {d \cos (a+b x)} \csc ^3(a+b x) \, dx=\frac {d^{3} {\left (\frac {2 \, \sqrt {d \cos \left (b x + a\right )} \cos \left (b x + a\right )}{{\left (d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} d} + \frac {\arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {-d}}\right )}{\sqrt {-d} d^{2}} + \frac {\arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {5}{2}}}\right )}}{4 \, b} \]
1/4*d^3*(2*sqrt(d*cos(b*x + a))*cos(b*x + a)/((d^2*cos(b*x + a)^2 - d^2)*d ) + arctan(sqrt(d*cos(b*x + a))/sqrt(-d))/(sqrt(-d)*d^2) + arctan(sqrt(d*c os(b*x + a))/sqrt(d))/d^(5/2))/b
Timed out. \[ \int \sqrt {d \cos (a+b x)} \csc ^3(a+b x) \, dx=\int \frac {\sqrt {d\,\cos \left (a+b\,x\right )}}{{\sin \left (a+b\,x\right )}^3} \,d x \]