Integrand size = 23, antiderivative size = 107 \[ \int \frac {\sqrt {b \cos (c+d x)}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {3 \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{8 d \sqrt {\cos (c+d x)}}+\frac {\sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {3 \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d \cos ^{\frac {5}{2}}(c+d x)} \]
1/4*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(9/2)+3/8*sin(d*x+c)*(b*c os(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)+3/8*arctanh(sin(d*x+c))*(b*cos(d*x+c)) ^(1/2)/d/cos(d*x+c)^(1/2)
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {b \cos (c+d x)}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {\sqrt {b \cos (c+d x)} \left (3 \text {arctanh}(\sin (c+d x)) \cos ^4(c+d x)+\left (2+3 \cos ^2(c+d x)\right ) \sin (c+d x)\right )}{8 d \cos ^{\frac {9}{2}}(c+d x)} \]
(Sqrt[b*Cos[c + d*x]]*(3*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^4 + (2 + 3*Cos [c + d*x]^2)*Sin[c + d*x]))/(8*d*Cos[c + d*x]^(9/2))
Time = 0.38 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2031, 3042, 4255, 3042, 4255, 3042, 4257}
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.
\(\displaystyle \int \frac {\sqrt {b \cos (c+d x)}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \sec ^5(c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\sqrt {b \cos (c+d x)} \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
(Sqrt[b*Cos[c + d*x]]*((Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*(ArcTanh[S in[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4))/Sqrt[Cos[c + d*x]]
3.2.49.3.1 Defintions of rubi rules used
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.98 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.96
method | result | size |
default | \(\frac {\left (-3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+2 \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{8 d \cos \left (d x +c \right )^{\frac {9}{2}}}\) | \(103\) |
risch | \(-\frac {i \sqrt {\cos \left (d x +c \right ) b}\, \left (3 \,{\mathrm e}^{7 i \left (d x +c \right )}+11 \,{\mathrm e}^{5 i \left (d x +c \right )}-11 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 \sqrt {\cos \left (d x +c \right ) b}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 \sqrt {\cos \left (d x +c \right )}\, d}-\frac {3 \sqrt {\cos \left (d x +c \right ) b}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(156\) |
1/8/d*(-3*cos(d*x+c)^4*ln(-cot(d*x+c)+csc(d*x+c)-1)+3*cos(d*x+c)^4*ln(-cot (d*x+c)+csc(d*x+c)+1)+3*cos(d*x+c)^2*sin(d*x+c)+2*sin(d*x+c))*(cos(d*x+c)* b)^(1/2)/cos(d*x+c)^(9/2)
Time = 0.32 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.12 \[ \int \frac {\sqrt {b \cos (c+d x)}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\left [\frac {3 \, \sqrt {b} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} {\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{5}}, -\frac {3 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - \sqrt {b \cos \left (d x + c\right )} {\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{5}}\right ] \]
[1/16*(3*sqrt(b)*cos(d*x + c)^5*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3) + 2*sqrt(b*cos(d*x + c))*(3*cos(d*x + c)^2 + 2)*sqrt(cos(d*x + c)) *sin(d*x + c))/(d*cos(d*x + c)^5), -1/8*(3*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*cos(d*x + c)^5 - sqrt( b*cos(d*x + c))*(3*cos(d*x + c)^2 + 2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d *cos(d*x + c)^5)]
Timed out. \[ \int \frac {\sqrt {b \cos (c+d x)}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1656 vs. \(2 (89) = 178\).
Time = 0.48 (sec) , antiderivative size = 1656, normalized size of antiderivative = 15.48 \[ \int \frac {\sqrt {b \cos (c+d x)}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \]
-1/16*(12*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4* sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4 4*(sin(8*d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d* x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 44*(sin(8 *d*x + 8*c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c) )*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 12*(sin(8*d*x + 8 *c) + 4*sin(6*d*x + 6*c) + 6*sin(4*d*x + 4*c) + 4*sin(2*d*x + 2*c))*cos(1/ 2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d* x + 8*c)^2 + 8*(6*cos(4*d*x + 4*c) + 4*cos(2*d*x + 2*c) + 1)*cos(6*d*x + 6 *c) + 16*cos(6*d*x + 6*c)^2 + 12*(4*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 36*cos(4*d*x + 4*c)^2 + 16*cos(2*d*x + 2*c)^2 + 4*(2*sin(6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + sin(8*d*x + 8 *c)^2 + 16*(3*sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 16 *sin(6*d*x + 6*c)^2 + 36*sin(4*d*x + 4*c)^2 + 48*sin(4*d*x + 4*c)*sin(2*d* x + 2*c) + 16*sin(2*d*x + 2*c)^2 + 8*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arc tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 3*(2*(4*cos(6*d*x + 6*c) + 6*cos(4*d*x + 4*c) + 4*cos(2*d *x + 2*c) + 1)*cos(8*d*x + 8*c) + cos(8*d*x + 8*c)^2 + 8*(6*cos(4*d*x +...
\[ \int \frac {\sqrt {b \cos (c+d x)}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\int { \frac {\sqrt {b \cos \left (d x + c\right )}}{\cos \left (d x + c\right )^{\frac {11}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {b \cos (c+d x)}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\int \frac {\sqrt {b\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{11/2}} \,d x \]