Integrand size = 23, antiderivative size = 116 \[ \int \frac {(b \cos (c+d x))^{5/2}}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\frac {3 b^2 \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{8 d \sqrt {\cos (c+d x)}}+\frac {b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {3 b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d \cos ^{\frac {5}{2}}(c+d x)} \]
1/4*b^2*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(9/2)+3/8*b^2*sin(d*x +c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)+3/8*b^2*arctanh(sin(d*x+c))*(b *cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.57 \[ \int \frac {(b \cos (c+d x))^{5/2}}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{5/2} \left (3 \text {arctanh}(\sin (c+d x)) \cos ^4(c+d x)+\left (2+3 \cos ^2(c+d x)\right ) \sin (c+d x)\right )}{8 d \cos ^{\frac {13}{2}}(c+d x)} \]
((b*Cos[c + d*x])^(5/2)*(3*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^4 + (2 + 3*C os[c + d*x]^2)*Sin[c + d*x]))/(8*d*Cos[c + d*x]^(13/2))
Time = 0.39 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74, 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.
\(\displaystyle \int \frac {(b \cos (c+d x))^{5/2}}{\cos ^{\frac {15}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \sec ^5(c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
(b^2*Sqrt[b*Cos[c + d*x]]*((Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*(ArcTa nh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4))/Sqrt[Cos[ c + d*x]]
3.2.70.3.1 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 = 3.24 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91
method | result | size |
default | \(-\frac {\left (3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )-3 \left (\cos ^{4}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )-3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-2 \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}\, b^{2}}{8 d \cos \left (d x +c \right )^{\frac {9}{2}}}\) | \(106\) |
risch | \(-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (3 \,{\mathrm e}^{7 i \left (d x +c \right )}+11 \,{\mathrm e}^{5 i \left (d x +c \right )}-11 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 \sqrt {\cos \left (d x +c \right )}\, d}-\frac {3 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(165\) |
-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)*b^2/cos(d*x+c)^(9/2)
Time = 0.32 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.10 \[ \int \frac {(b \cos (c+d x))^{5/2}}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\left [\frac {3 \, b^{\frac {5}{2}} \cos \left (d x + c\right )^{5} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, {\left (3 \, b^{2} \cos \left (d x + c\right )^{2} + 2 \, b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{5}}, -\frac {3 \, \sqrt {-b} b^{2} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{5} - {\left (3 \, b^{2} \cos \left (d x + c\right )^{2} + 2 \, b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{5}}\right ] \]
[1/16*(3*b^(5/2)*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*(3*b^2*cos(d*x + c)^2 + 2*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d *x + c))*sin(d*x + c))/(d*cos(d*x + c)^5), -1/8*(3*sqrt(-b)*b^2*arctan(sqr t(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*sqrt(cos(d*x + c))))*cos(d*x + c)^5 - (3*b^2*cos(d*x + c)^2 + 2*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)]
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2}}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1914 vs. \(2 (98) = 196\).
Time = 0.51 (sec) , antiderivative size = 1914, normalized size of antiderivative = 16.50 \[ \int \frac {(b \cos (c+d x))^{5/2}}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\text {Too large to display} \]
-1/16*(12*(b^2*sin(8*d*x + 8*c) + 4*b^2*sin(6*d*x + 6*c) + 6*b^2*sin(4*d*x + 4*c) + 4*b^2*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2* d*x + 2*c))) + 44*(b^2*sin(8*d*x + 8*c) + 4*b^2*sin(6*d*x + 6*c) + 6*b^2*s in(4*d*x + 4*c) + 4*b^2*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(2*d*x + 2*c) , cos(2*d*x + 2*c))) - 44*(b^2*sin(8*d*x + 8*c) + 4*b^2*sin(6*d*x + 6*c) + 6*b^2*sin(4*d*x + 4*c) + 4*b^2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d* x + 2*c), cos(2*d*x + 2*c))) - 12*(b^2*sin(8*d*x + 8*c) + 4*b^2*sin(6*d*x + 6*c) + 6*b^2*sin(4*d*x + 4*c) + 4*b^2*sin(2*d*x + 2*c))*cos(1/2*arctan2( sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(b^2*cos(8*d*x + 8*c)^2 + 16*b^2* cos(6*d*x + 6*c)^2 + 36*b^2*cos(4*d*x + 4*c)^2 + 16*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(8*d*x + 8*c)^2 + 16*b^2*sin(6*d*x + 6*c)^2 + 36*b^2*sin(4*d*x + 4*c)^2 + 48*b^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 16*b^2*sin(2*d*x + 2* c)^2 + 8*b^2*cos(2*d*x + 2*c) + b^2 + 2*(4*b^2*cos(6*d*x + 6*c) + 6*b^2*co s(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c) + b^2)*cos(8*d*x + 8*c) + 8*(6*b^2 *cos(4*d*x + 4*c) + 4*b^2*cos(2*d*x + 2*c) + b^2)*cos(6*d*x + 6*c) + 12*(4 *b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c) + 4*(2*b^2*sin(6*d*x + 6*c) + 3*b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 16*( 3*b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*log(cos (1/2*arctan2(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)...
\[ \int \frac {(b \cos (c+d x))^{5/2}}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\int { \frac {\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {15}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2}}{\cos ^{\frac {15}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{15/2}} \,d x \]