Integrand size = 33, antiderivative size = 74 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\frac {B \text {arctanh}(\sin (c+d x)) \sqrt {\cos (c+d x)}}{b^2 d \sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{b^2 d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \] Output:
B*arctanh(sin(d*x+c))*cos(d*x+c)^(1/2)/b^2/d/(b*cos(d*x+c))^(1/2)+A*sin(d* x+c)/b^2/d/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)
Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) \left (B \coth ^{-1}(\sin (c+d x)) \cos (c+d x)+A \sin (c+d x)\right )}{d (b \cos (c+d x))^{5/2}} \] Input:
Integrate[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x]))/(b*Cos[c + d*x])^(5/2) ,x]
Output:
(Cos[c + d*x]^(3/2)*(B*ArcCoth[Sin[c + d*x]]*Cos[c + d*x] + A*Sin[c + d*x] ))/(d*(b*Cos[c + d*x])^(5/2))
Time = 0.35 (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.
\(\displaystyle \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int (A+B \cos (c+d x)) \sec ^2(c+d x)dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (A \int \sec ^2(c+d x)dx+B \int \sec (c+d x)dx\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {A \int 1d(-\tan (c+d x))}{d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (B \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {A \tan (c+d x)}{d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {A \tan (c+d x)}{d}+\frac {B \text {arctanh}(\sin (c+d x))}{d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
Input:
Int[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x]))/(b*Cos[c + d*x])^(5/2),x]
Output:
(Sqrt[Cos[c + d*x]]*((B*ArcTanh[Sin[c + d*x]])/d + (A*Tan[c + d*x])/d))/(b ^2*Sqrt[b*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 = 9.80 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81
method | result | size |
default | \(\frac {-2 B \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+A \sin \left (d x +c \right )}{b^{2} d \sqrt {\cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b}}\) | \(60\) |
parts | \(\frac {A \sin \left (d x +c \right )}{b^{2} d \sqrt {\cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b}}-\frac {2 B \,\operatorname {arctanh}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}}{d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}\) | \(77\) |
risch | \(\frac {2 i \sqrt {\cos \left (d x +c \right )}\, A}{b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\sqrt {\cos \left (d x +c \right )}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}-\frac {\sqrt {\cos \left (d x +c \right )}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}\) | \(122\) |
Input:
int(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(cos(d*x+c)*b)^(5/2),x,method=_RETUR NVERBOSE)
Output:
1/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)^(1/2)/(cos(d*x+c)*b)^(1/2)
Time = 0.11 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.85 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\left [\frac {B \sqrt {b} \cos \left (d x + c\right )^{2} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{3} d \cos \left (d x + c\right )^{2}}, -\frac {B \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{2} - \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b^{3} d \cos \left (d x + c\right )^{2}}\right ] \] Input:
integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(5/2),x, algori thm="fricas")
Output:
[1/2*(B*sqrt(b)*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*sqrt(cos(d*x + c))*sin(d*x + c))/(b^3*d* cos(d*x + c)^2), -(B*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)^2 - sqrt(b*cos(d*x + c))*A*sqrt (cos(d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^2)]
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(1/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))**(5/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (66) = 132\).
Time = 0.33 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.80 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\frac {4 \, A \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{b^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{3} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{3} \cos \left (2 \, d x + 2 \, c\right ) + b^{3}} + \frac {B {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )}}{b^{\frac {5}{2}}}}{2 \, d} \] Input:
integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(5/2),x, algori thm="maxima")
Output:
1/2*(4*A*sqrt(b)*sin(2*d*x + 2*c)/(b^3*cos(2*d*x + 2*c)^2 + b^3*sin(2*d*x + 2*c)^2 + 2*b^3*cos(2*d*x + 2*c) + b^3) + B*(log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*si n(d*x + c) + 1))/b^(5/2))/d
Exception generated. \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(5/2),x, algori thm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int((cos(c + d*x)^(1/2)*(A + B*cos(c + d*x)))/(b*cos(c + d*x))^(5/2),x)
Output:
int((cos(c + d*x)^(1/2)*(A + B*cos(c + d*x)))/(b*cos(c + d*x))^(5/2), x)
Time = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {b}\, \left (-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +\sin \left (d x +c \right ) a \right )}{\cos \left (d x +c \right ) b^{3} d} \] Input:
int(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(5/2),x)
Output:
(sqrt(b)*( - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + cos(c + d*x)*log(t an((c + d*x)/2) + 1)*b + sin(c + d*x)*a))/(cos(c + d*x)*b**3*d)