Integrand size = 33, antiderivative size = 66 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {b^2 B x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {A b^2 \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \]
b^2*B*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+A*b^2*arctanh(sin(d*x+c))*(b *cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)
Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.61 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {(B d x+A \text {arctanh}(\sin (c+d x))) (b \cos (c+d x))^{5/2}}{d \cos ^{\frac {5}{2}}(c+d x)} \]
Time = 0.33 (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.
\(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int (A+B \cos (c+d x)) \sec (c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} (A \int \sec (c+d x)dx+B x)}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+B x\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {A \text {arctanh}(\sin (c+d x))}{d}+B x\right )}{\sqrt {\cos (c+d x)}}\) |
3.9.62.3.1 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 = 4.92 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.83
method | result | size |
default | \(-\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (2 A \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-B \left (d x +c \right )\right )}{d \sqrt {\cos \left (d x +c \right )}}\) | \(55\) |
parts | \(-\frac {2 A \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) b^{2} \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {B \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}\) | \(76\) |
risch | \(\frac {b^{2} B x \sqrt {\cos \left (d x +c \right ) b}}{\sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}-\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}\) | \(105\) |
Time = 0.36 (sec) , antiderivative size = 216, normalized size of antiderivative = 3.27 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\left [-\frac {2 \, A \sqrt {-b} b^{2} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) - B \sqrt {-b} b^{2} \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right )}{2 \, d}, \frac {2 \, B b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) + A b^{\frac {5}{2}} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right )}{2 \, d}\right ] \]
[-1/2*(2*A*sqrt(-b)*b^2*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin(d*x + c)/ (b*sqrt(cos(d*x + c)))) - B*sqrt(-b)*b^2*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))/d, 1/2*(2*B* b^(5/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/ 2))) + A*b^(5/2)*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*cos(d*x + c))*sqrt(b)*s qrt(cos(d*x + c))*sin(d*x + c) - 2*b*cos(d*x + c))/cos(d*x + c)^3))/d]
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
Time = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.50 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4 \, B b^{\frac {5}{2}} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + {\left (b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} A \sqrt {b}}{2 \, d} \]
1/2*(4*B*b^(5/2)*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) + (b^2*log(cos(d* x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - b^2*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1))*A*sqrt(b))/d
\[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^{7/2}} \,d x \]