Integrand size = 23, antiderivative size = 134 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=-\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{1+\cos (c+d x)}\right )}{d}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {1+\cos (c+d x)}}-\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {1+\cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {1+\cos (c+d x)}} \]
-arcsin(sin(d*x+c)/(1+cos(d*x+c)))*2^(1/2)/d+2/5*sin(d*x+c)/d/cos(d*x+c)^( 5/2)/(1+cos(d*x+c))^(1/2)-2/15*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(1+cos(d*x+c) )^(1/2)+26/15*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(1+cos(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 7.17 (sec) , antiderivative size = 1538, normalized size of antiderivative = 11.48 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx =\text {Too large to display} \]
(-2*Cot[c/2 + (d*x)/2]*Csc[c/2 + (d*x)/2]^6*(4725*Sin[c/2 + (d*x)/2]^2 - 4 8825*Sin[c/2 + (d*x)/2]^4 + 210105*Sin[c/2 + (d*x)/2]^6 - 486630*Sin[c/2 + (d*x)/2]^8 + 655812*Sin[c/2 + (d*x)/2]^10 - 710*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x) /2]^10 - 40*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 9/2}, {1, 1, 11 /2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2 ]^10 - 518760*Sin[c/2 + (d*x)/2]^12 + 1770*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 226656*Sin[c/2 + (d*x)/2]^14 - 1500*Hypergeometric2F1[2, 9/2, 11/2, Sin [c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^14 - 4 2048*Sin[c/2 + (d*x)/2]^16 + 440*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 + 4725*Ar cTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sqrt[Sin[c /2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 56700*ArcTanh[Sqrt[Sin[c/ 2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^2*Sqrt[S in[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 291060*ArcTanh[Sqrt[S in[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^4*S qrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 833760*ArcTanh[S qrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2 ]^6*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 1458000*...
Time = 0.70 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3258, 3042, 3463, 27, 3042, 3463, 27, 3042, 3260, 223}
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.
| \(\displaystyle \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {\cos (c+d x)+1}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\) |
| \(\Big \downarrow \) 3258 |
| \(\displaystyle \frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}-\frac {1}{5} \int \frac {1-4 \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}-\frac {1}{5} \int \frac {1-4 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {1}{5} \left (-\frac {2}{3} \int -\frac {13-2 \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}dx-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {13-2 \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}dx-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {13-2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (2 \int -\frac {15}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx+\frac {26 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}\right )-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {26 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-15 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}dx\right )-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {26 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-15 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx\right )-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\) |
| \(\Big \downarrow \) 3260 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {15 \sqrt {2} \int \frac {1}{\sqrt {1-\frac {\sin ^2(c+d x)}{(\cos (c+d x)+1)^2}}}d\left (-\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}+\frac {26 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}\right )-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {26 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-\frac {15 \sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}\right )-\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}\) |
(2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[1 + Cos[c + d*x]]) + ((-2*Si n[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[1 + Cos[c + d*x]]) + ((-15*Sqrt[2 ]*ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])])/d + (26*Sin[c + d*x])/(d*Sqrt[C os[c + d*x]]*Sqrt[1 + Cos[c + d*x]]))/3)/5
3.3.37.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[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[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-Sqrt[2]/(Sqrt[a]*f) Subst[Int[1/Sqrt[1 - x^2], x], x, b*(Cos[e + f*x]/(a + b*Sin[e + f*x]))], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && EqQ[d, a/b] && GtQ[a, 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])
Time = 5.38 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.30
| method | result | size |
| default | \(\frac {\left (15 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+15 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {2}+26 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )+6 \sin \left (d x +c \right )\right ) \sqrt {2+2 \cos \left (d x +c \right )}\, \sqrt {2}}{30 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(174\) |
1/30/d*(15*cos(d*x+c)^3*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*(cos(d*x+c)/ (1+cos(d*x+c)))^(1/2)+15*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ar csin(cot(d*x+c)-csc(d*x+c))*2^(1/2)+26*cos(d*x+c)^2*sin(d*x+c)-2*cos(d*x+c )*sin(d*x+c)+6*sin(d*x+c))*(2+2*cos(d*x+c))^(1/2)/(1+cos(d*x+c))/cos(d*x+c )^(5/2)*2^(1/2)
Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=\frac {2 \, {\left (13 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 3\right )} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 15 \, {\left (\sqrt {2} \cos \left (d x + c\right )^{4} + \sqrt {2} \cos \left (d x + c\right )^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )}}\right )}{15 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]
1/15*(2*(13*cos(d*x + c)^2 - cos(d*x + c) + 3)*sqrt(cos(d*x + c) + 1)*sqrt (cos(d*x + c))*sin(d*x + c) - 15*(sqrt(2)*cos(d*x + c)^4 + sqrt(2)*cos(d*x + c)^3)*arctan(1/2*sqrt(2)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))*sin( d*x + c)/(cos(d*x + c)^2 + cos(d*x + c))))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)
Timed out. \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 0.59 (sec) , antiderivative size = 989, normalized size of antiderivative = 7.38 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=\text {Too large to display} \]
-1/15*(15*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqr t(2)*cos(2*d*x + 2*c) + sqrt(2))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*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)) /abs(e^(I*d*x + I*c) + 1), ((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)*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*c os(d*x + c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + cos(d*x + c) - 1)/abs(e^(I *d*x + I*c) + 1)) - 26*(cos(2*d*x + 2*c)^2*sin(d*x + c) + sin(2*d*x + 2*c) ^2*sin(d*x + c) + 2*cos(2*d*x + 2*c)*sin(d*x + c) + sin(d*x + c))*cos(1/2* arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 24*cos(5/2*arctan2(sin( 2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - 24*(cos(d*x + c) - ...
\[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {\cos \left (d x + c\right ) + 1} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {1+\cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {\cos \left (c+d\,x\right )+1}} \,d x \]