Integrand size = 33, antiderivative size = 74 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {b^2 B \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)} \] Output:
b^2*B*arctanh(sin(d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)+A*b^2*(b *cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)
Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{5/2} \left (B \coth ^{-1}(\sin (c+d x)) \cos (c+d x)+A \sin (c+d x)\right )}{d \cos ^{\frac {7}{2}}(c+d x)} \] Input:
Integrate[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(9/2) ,x]
Output:
((b*Cos[c + d*x])^(5/2)*(B*ArcCoth[Sin[c + d*x]]*Cos[c + d*x] + A*Sin[c + d*x]))/(d*Cos[c + d*x]^(7/2))
Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {2031, 3042, 3227, 3042, 4254, 24, 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.
\(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int (A+B \cos (c+d x)) \sec ^2(c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (A \int \sec ^2(c+d x)dx+B \int \sec (c+d x)dx\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {A \int 1d(-\tan (c+d x))}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {A \tan (c+d x)}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {A \tan (c+d x)}{d}+\frac {B \text {arctanh}(\sin (c+d x))}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
Input:
Int[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(9/2),x]
Output:
(b^2*Sqrt[b*Cos[c + d*x]]*((B*ArcTanh[Sin[c + d*x]])/d + (A*Tan[c + d*x])/ d))/Sqrt[Cos[c + d*x]]
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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 10.16 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81
method | result | size |
default | \(\frac {b^{2} \left (-2 B \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+A \sin \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{d \cos \left (d x +c \right )^{\frac {3}{2}}}\) | \(60\) |
parts | \(\frac {A \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \sin \left (d x +c \right )}{d \cos \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 B \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}\, b^{2}}{d \sqrt {\cos \left (d x +c \right )}}\) | \(77\) |
risch | \(\frac {2 i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, A}{\sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}-\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}\) | \(122\) |
Input:
int((cos(d*x+c)*b)^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x,method=_RETUR NVERBOSE)
Output:
b^2/d*(-2*B*arctanh(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)+A*sin(d*x+c))*(cos(d *x+c)*b)^(1/2)/cos(d*x+c)^(3/2)
Time = 0.12 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.89 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\left [\frac {B b^{\frac {5}{2}} \cos \left (d x + c\right )^{2} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} A b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{2}}, -\frac {B \sqrt {-b} b^{2} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{2} - \sqrt {b \cos \left (d x + c\right )} A b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )^{2}}\right ] \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x, algori thm="fricas")
Output:
[1/2*(B*b^(5/2)*cos(d*x + c)^2*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))*A*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d* cos(d*x + c)^2), -(B*sqrt(-b)*b^2*arctan(sqrt(b*cos(d*x + c))*sqrt(-b)*sin (d*x + c)/(b*sqrt(cos(d*x + c))))*cos(d*x + c)^2 - sqrt(b*cos(d*x + c))*A* b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)]
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(9/2),x)
Output:
Timed out
Time = 0.31 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.72 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\frac {4 \, A b^{\frac {5}{2}} \sin \left (2 \, d x + 2 \, c\right )}{\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1} + {\left (b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} B \sqrt {b}}{2 \, d} \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x, algori thm="maxima")
Output:
1/2*(4*A*b^(5/2)*sin(2*d*x + 2*c)/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*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))*B*sqrt(b))/d
Time = 0.38 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {{\left (B b^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - B b^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {2 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )} \sqrt {b}}{d} \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x, algori thm="giac")
Output:
(B*b^2*log(tan(1/2*d*x + 1/2*c) + 1) - B*b^2*log(tan(1/2*d*x + 1/2*c) - 1) - 2*A*b^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1))*sqrt(b)/d
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^{9/2}} \,d x \] Input:
int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x)))/cos(c + d*x)^(9/2),x)
Output:
int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x)))/cos(c + d*x)^(9/2), x)
Time = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.91 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\sqrt {b}\, b^{2} \left (-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +\sin \left (d x +c \right ) a \right )}{\cos \left (d x +c \right ) d} \] Input:
int((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x)
Output:
(sqrt(b)*b**2*( - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + cos(c + d*x)* log(tan((c + d*x)/2) + 1)*b + sin(c + d*x)*a))/(cos(c + d*x)*d)