Integrand size = 21, antiderivative size = 98 \[ \int (a+a \cos (c+d x))^{5/2} \sec (c+d x) \, dx=\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {14 a^3 \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d} \]
2*a^(5/2)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+14/3*a^3*si n(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/3*a^2*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2 )/d
Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.91 \[ \int (a+a \cos (c+d x))^{5/2} \sec (c+d x) \, dx=\frac {2 a^2 \sqrt {a (1+\cos (c+d x))} \left (3 \text {arctanh}\left (\sqrt {1-\cos (c+d x)}\right )+\sqrt {1-\cos (c+d x)} (8+\cos (c+d x))\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{3 d \sqrt {1-\cos (c+d x)}} \]
(2*a^2*Sqrt[a*(1 + Cos[c + d*x])]*(3*ArcTanh[Sqrt[1 - Cos[c + d*x]]] + Sqr t[1 - Cos[c + d*x]]*(8 + Cos[c + d*x]))*Tan[(c + d*x)/2])/(3*d*Sqrt[1 - Co s[c + d*x]])
Time = 0.53 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3242, 27, 3042, 3460, 3042, 3252, 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.
| \(\displaystyle \int \sec (c+d x) (a \cos (c+d x)+a)^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3242 |
| \(\displaystyle \frac {2}{3} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (7 \cos (c+d x) a^2+3 a^2\right ) \sec (c+d x)dx+\frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \sqrt {\cos (c+d x) a+a} \left (7 \cos (c+d x) a^2+3 a^2\right ) \sec (c+d x)dx+\frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (7 \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+3 a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {1}{3} \left (3 a^2 \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {14 a^3 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (3 a^2 \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {14 a^3 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {1}{3} \left (\frac {14 a^3 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {6 a^3 \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2 a^2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}+\frac {1}{3} \left (\frac {6 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {14 a^3 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )\) |
(2*a^2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ((6*a^(5/2)*ArcTanh[ (Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (14*a^3*Sin[c + d*x] )/(d*Sqrt[a + a*Cos[c + d*x]]))/3
3.2.16.3.1 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]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* (m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !LtQ[ n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ c, 0]))
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(245\) vs. \(2(84)=168\).
Time = 1.97 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.51
| method | result | size |
| default | \(\frac {a^{\frac {3}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+18 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+3 \ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +3 \ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a \right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(246\) |
1/3*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*2^(1/2)* (a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^2+18*2^(1/2)*(a* sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+3*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))* (2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/ 2)+2*a))*a+3*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1 /2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)/sin(1/2*d*x+ 1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.50 \[ \int (a+a \cos (c+d x))^{5/2} \sec (c+d x) \, dx=\frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (a^{2} \cos \left (d x + c\right ) + 8 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
1/6*(3*(a^2*cos(d*x + c) + a^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d* x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(a^2*cos(d*x + c) + 8*a^ 2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c) + d)
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \sec (c+d x) \, dx=\text {Timed out} \]
\[ \int (a+a \cos (c+d x))^{5/2} \sec (c+d x) \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right ) \,d x } \]
Time = 0.39 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.22 \[ \int (a+a \cos (c+d x))^{5/2} \sec (c+d x) \, dx=-\frac {\sqrt {2} {\left (8 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \sqrt {2} a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 36 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{6 \, d} \]
-1/6*sqrt(2)*(8*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 3*s qrt(2)*a^2*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4* sin(1/2*d*x + 1/2*c)))*sgn(cos(1/2*d*x + 1/2*c)) - 36*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \sec (c+d x) \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{\cos \left (c+d\,x\right )} \,d x \]