Integrand size = 33, antiderivative size = 66 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\frac {B x \sqrt {\cos (c+d x)}}{b^2 \sqrt {b \cos (c+d x)}}+\frac {A \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b^2 d \sqrt {b \cos (c+d x)}} \]
B*x*cos(d*x+c)^(1/2)/b^2/(b*cos(d*x+c))^(1/2)+A*arctanh(sin(d*x+c))*cos(d* x+c)^(1/2)/b^2/d/(b*cos(d*x+c))^(1/2)
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.65 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\frac {(B d x+A \text {arctanh}(\sin (c+d x))) \sqrt {\cos (c+d x)}}{b^2 d \sqrt {b \cos (c+d x)}} \]
Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.64, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2031, 3042, 3214, 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.
\(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int (A+B \cos (c+d x)) \sec (c+d x)dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} (A \int \sec (c+d x)dx+B x)}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+B x\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {A \text {arctanh}(\sin (c+d x))}{d}+B x\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
3.9.83.3.1 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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 4.96 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.83
method | result | size |
default | \(-\frac {\left (2 A \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-B \left (d x +c \right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{b^{2} d \sqrt {\cos \left (d x +c \right ) b}}\) | \(55\) |
parts | \(-\frac {2 A \left (\sqrt {\cos }\left (d x +c \right )\right ) \operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right ) b}\, b^{2}}+\frac {B \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (d x +c \right )}{d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}\) | \(76\) |
risch | \(\frac {B x \left (\sqrt {\cos }\left (d x +c \right )\right )}{b^{2} \sqrt {\cos \left (d x +c \right ) b}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}-\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}\) | \(105\) |
-1/b^2/d*(2*A*arctanh(cot(d*x+c)-csc(d*x+c))-B*(d*x+c))*cos(d*x+c)^(1/2)/( cos(d*x+c)*b)^(1/2)
Time = 0.40 (sec) , antiderivative size = 215, normalized size of antiderivative = 3.26 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\left [-\frac {2 \, A \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) + B \sqrt {-b} \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{2 \, b^{3} d}, \frac {2 \, B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) + A \sqrt {b} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right )}{2 \, b^{3} d}\right ] \]
[-1/2*(2*A*sqrt(-b)*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/(b*s qrt(cos(d*x + c)))) + B*sqrt(-b)*log(2*b*cos(d*x + c)^2 + 2*sqrt(b*cos(d*x + c))*sqrt(-b)*sqrt(cos(d*x + c))*sin(d*x + c) - b))/(b^3*d), 1/2*(2*B*sq rt(b)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2) )) + A*sqrt(b)*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*sqr t(cos(d*x + c))*sin(d*x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3))/(b^3*d)]
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Time = 0.39 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.39 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\frac {A {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )}}{b^{\frac {5}{2}}} + \frac {4 \, B \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac {5}{2}}}}{2 \, d} \]
1/2*(A*(log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - log(co s(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/b^(5/2) + 4*B*arctan( sin(d*x + c)/(cos(d*x + c) + 1))/b^(5/2))/d
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]