Integrand size = 23, antiderivative size = 78 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx=\frac {\text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{2 b^2 d \sqrt {b \cos (c+d x)}}+\frac {\sin (c+d x)}{2 b^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \]
1/2*sin(d*x+c)/b^2/d/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2)+1/2*arctanh(sin (d*x+c))*cos(d*x+c)^(1/2)/b^2/d/(b*cos(d*x+c))^(1/2)
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {b \cos (c+d x)} \left (\text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+\sin (c+d x)\right )}{2 b^3 d \cos ^{\frac {5}{2}}(c+d x)} \]
(Sqrt[b*Cos[c + d*x]]*(ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + Sin[c + d*x] ))/(2*b^3*d*Cos[c + d*x]^(5/2))
Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.77, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2032, 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.
\(\displaystyle \int \frac {1}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2032 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \sec ^3(c+d x)dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
(Sqrt[Cos[c + d*x]]*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d *x])/(2*d)))/(b^2*Sqrt[b*Cos[c + d*x]])
3.2.99.3.1 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[a*v]/Sqrt[b*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && ILtQ[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 = 3.19 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.12
method | result | size |
default | \(\frac {-\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )+\left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right )+\sin \left (d x +c \right )}{2 d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \cos \left (d x +c \right )^{\frac {3}{2}}}\) | \(87\) |
risch | \(-\frac {i \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{2 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}\) | \(131\) |
1/2/d*(-cos(d*x+c)^2*ln(-cot(d*x+c)+csc(d*x+c)-1)+cos(d*x+c)^2*ln(-cot(d*x +c)+csc(d*x+c)+1)+sin(d*x+c))/b^2/(cos(d*x+c)*b)^(1/2)/cos(d*x+c)^(3/2)
Time = 0.30 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.65 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx=\left [\frac {\sqrt {b} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b^{3} d \cos \left (d x + c\right )^{3}}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{3} d \cos \left (d x + c\right )^{3}}\right ] \]
[1/4*(sqrt(b)*cos(d*x + c)^3*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))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*d*cos( d*x + c)^3), -1/2*(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)^3 - sqrt(b*cos(d*x + c))*sqrt(cos (d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^3)]
Timed out. \[ \int \frac {1}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 688 vs. \(2 (66) = 132\).
Time = 0.45 (sec) , antiderivative size = 688, normalized size of antiderivative = 8.82 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
-1/4*(4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos (1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4 *d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*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*s in(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), co s(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) - 4*(cos(4 *d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(3/2*arctan2(sin(2*d*x + 2*c), co s(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*a rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))/((b^2*cos(4*d*x + 4*c)^2 + 4* b^2*cos(2*d*x + 2*c)^2 + b^2*sin(4*d*x + 4*c)^2 + 4*b^2*sin(4*d*x + 4*c)*s in(2*d*x + 2*c) + 4*b^2*sin(2*d*x + 2*c)^2 + 4*b^2*cos(2*d*x + 2*c) + b^2 + 2*(2*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c))*sqrt(b)*d)
\[ \int \frac {1}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx=\int { \frac {1}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sqrt {\cos \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx=\int \frac {1}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]