Integrand size = 25, antiderivative size = 151 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}-\frac {2 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}} \]
2/3*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+arctan(1/2*sin(d* x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/2)*cos( d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d/a^(1/2)-2/3*sin(d*x+c)*sec(d*x+c)^(1/2)/d/ (a+a*cos(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 6.46 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.15 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {2 \cot \left (\frac {c}{2}+\frac {d x}{2}\right ) \csc ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {1}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right )^{7/2} \left (12 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac {7}{2};1,\frac {9}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )+12 \operatorname {Hypergeometric2F1}\left (2,\frac {7}{2},\frac {9}{2},\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (4-7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+3 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )\right )+7 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3 \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \left (15-20 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+8 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (\text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \left (3-6 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )+\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \left (-3+7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )\right )}{63 d \sqrt {a (1+\cos (c+d x))}} \]
(-2*Cot[c/2 + (d*x)/2]*Csc[c/2 + (d*x)/2]^4*((1 - 2*Sin[c/2 + (d*x)/2]^2)^ (-1))^(7/2)*(12*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 7/2}, {1, 9/2} , Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^8 + 12*Hypergeometric2F1[2, 7/2, 9/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^8*(4 - 7*Sin[c/2 + (d*x)/2]^2 + 3*Sin[c/ 2 + (d*x)/2]^4) + 7*(1 - 2*Sin[c/2 + (d*x)/2]^2)^3*Sqrt[Sin[c/2 + (d*x)/2] ^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(15 - 20*Sin[c/2 + (d*x)/2]^2 + 8*Sin[c/ 2 + (d*x)/2]^4)*(ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x) /2]^2)]]*(3 - 6*Sin[c/2 + (d*x)/2]^2) + Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2* Sin[c/2 + (d*x)/2]^2)]*(-3 + 7*Sin[c/2 + (d*x)/2]^2))))/(63*d*Sqrt[a*(1 + Cos[c + d*x])])
Time = 0.72 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4710, 3042, 3258, 3042, 3463, 27, 3042, 3261, 218}
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.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a \cos (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 4710 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\) |
| \(\Big \downarrow \) 3258 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a-2 a \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a-2 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}\right )\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {3 a^2}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{3 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}\right )\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {6 a^2 \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {3 \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{3 a}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*Sin[c + d*x])/(3*d*Cos[c + d*x]^ (3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((-3*Sqrt[2]*Sqrt[a]*ArcTan[(Sqrt[a]*Sin [c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/d + (2* a*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/(3*a))
3.4.68.3.1 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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. ) + (f_.)*(x_)]], x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(2* b*(n + 1)*(c^2 - d^2)) Int[(c + d*Sin[e + f*x])^(n + 1)*(Simp[a*d - 2*b*c *(n + 1) + b*d*(2*n + 3)*Sin[e + f*x], x]/Sqrt[a + b*Sin[e + f*x]]), 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] && LtQ[n, -1] && IntegerQ[2*n]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Time = 6.80 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.11
| method | result | size |
| default | \(-\frac {\left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (3 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+\sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-\sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {2}}{3 d \left (1+\cos \left (d x +c \right )\right ) a}\) | \(168\) |
-1/3/d*sec(d*x+c)^(5/2)*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))*(3*cos(d*x +c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+2^(1 /2)*cos(d*x+c)^2*sin(d*x+c)+3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c) ^2*arcsin(cot(d*x+c)-csc(d*x+c))-sin(d*x+c)*cos(d*x+c)*2^(1/2))*2^(1/2)/a
Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.83 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\frac {3 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}} + \frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \]
-1/3*(3*sqrt(2)*(a*cos(d*x + c)^2 + a*cos(d*x + c))*arctan(sqrt(2)*sqrt(a* cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))/sqrt(a) + 2*s qrt(a*cos(d*x + c) + a)*(cos(d*x + c) - 1)*sin(d*x + c)/sqrt(cos(d*x + c)) )/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c))
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 0.60 (sec) , antiderivative size = 818, normalized size of antiderivative = 5.42 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \]
1/3*(3*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2 )*cos(2*d*x + 2*c) + sqrt(2))*arctan2(((abs(e^(I*d*x + I*c) + 1)^4 + cos(d *x + c)^4 + sin(d*x + c)^4 + 2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d* x + c) + 1)*abs(e^(I*d*x + I*c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c )^2 - 2*cos(d*x + c) + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1)^(1/4)*sin(1/2*arctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d *x + I*c) + 1)^2, (abs(e^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + sin(d*x + c))/a bs(e^(I*d*x + I*c) + 1), ((abs(e^(I*d*x + I*c) + 1)^4 + cos(d*x + c)^4 + s in(d*x + c)^4 + 2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)*a bs(e^(I*d*x + I*c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c)^2 - 2*cos(d *x + c) + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1)^(1/4) *sqrt(a)*cos(1/2*arctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d*x + I*c) + 1)^2, (abs(e^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + sqrt(a)*cos(d*x + c) - sqrt(a))/(sqrt(a)*abs(e^(I*d*x + I*c) + 1))) - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*(cos(1/2*arctan2(sin(2* d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - (cos(d*x + c) - 3)*sin(1 /2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 4*(cos(2*d*x + 2*c) ^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(3/2*arctan...
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {a \cos \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]