\(\int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [310]

Optimal result

Integrand size = 43, antiderivative size = 102 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {b^2 B x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {A b^2 \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {b^2 C \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:

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)+b^2*C*(b*cos(d*x+c))^(1/2)*sin(d*x+c 
)/d/cos(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{5/2} \left (B c+B d x-A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+C \sin (c+d x)\right )}{d \cos ^{\frac {5}{2}}(c+d x)} \] Input:

Integrate[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) 
/Cos[c + d*x]^(7/2),x]
 

Output:

((b*Cos[c + d*x])^(5/2)*(B*c + B*d*x - A*Log[Cos[(c + d*x)/2] - Sin[(c + d 
*x)/2]] + A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + C*Sin[c + d*x]))/(d 
*Cos[c + d*x]^(5/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.52, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {2031, 3042, 3502, 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.

\(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 2031

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right ) \sec (c+d x)dx}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\int (A+B \cos (c+d x)) \sec (c+d x)dx+\frac {C \sin (c+d x)}{d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {C \sin (c+d x)}{d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (A \int \sec (c+d x)dx+B x+\frac {C \sin (c+d x)}{d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+B x+\frac {C \sin (c+d x)}{d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {A \text {arctanh}(\sin (c+d x))}{d}+B x+\frac {C \sin (c+d x)}{d}\right )}{\sqrt {\cos (c+d x)}}\)

Input:

Int[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c 
 + d*x]^(7/2),x]
 

Output:

(b^2*Sqrt[b*Cos[c + d*x]]*(B*x + (A*ArcTanh[Sin[c + d*x]])/d + (C*Sin[c + 
d*x])/d))/Sqrt[Cos[c + d*x]]
 

Defintions of rubi rules used

rule 2031

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3214

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]
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63

method	result	size
default	\(-\frac {b^{2} \left (2 A \,\operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )-B \left (d x +c \right )-C \sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}}{d \sqrt {\cos \left (d x +c \right )}}\)	\(64\)
parts	\(-\frac {2 A \sqrt {b \cos \left (d x +c \right )}\, \operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) b^{2}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {B \left (d x +c \right ) b^{2} \sqrt {b \cos \left (d x +c \right )}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} C \sqrt {b \cos \left (d x +c \right )}\, \sin \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}\)	\(108\)
risch	\(\frac {b^{2} B x \sqrt {b \cos \left (d x +c \right )}}{\sqrt {\cos \left (d x +c \right )}}-\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, C \,{\mathrm e}^{i \left (d x +c \right )}}{2 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, C \,{\mathrm e}^{-i \left (d x +c \right )}}{2 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}-\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}\)	\(179\)

Input:

int((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2), 
x,method=_RETURNVERBOSE)
 

Output:

-b^2/d*(2*A*arctanh(-csc(d*x+c)+cot(d*x+c))-B*(d*x+c)-C*sin(d*x+c))*(b*cos 
(d*x+c))^(1/2)/cos(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 316, normalized size of antiderivative = 3.10 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\left [-\frac {2 \, A \sqrt {-b} b^{2} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right ) - B \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, \sqrt {b \cos \left (d x + c\right )} C b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}, \frac {2 \, B b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + A b^{\frac {5}{2}} \cos \left (d x + c\right ) \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} C b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}\right ] \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ 
(7/2),x, algorithm="fricas")
 

Output:

[-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))))*cos(d*x + c) - B*sqrt(-b)*b^2*cos(d*x + c)*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) - 2*sqrt(b*cos(d*x + c))*C*b^2*sqrt(cos(d*x + c))*sin(d*x + c 
))/(d*cos(d*x + c)), 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)))*cos(d*x + c) + A*b^(5/2)*cos(d*x + c)*l 
og(-(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))* 
C*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c 
)**(7/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4 \, B b^{\frac {5}{2}} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, C b^{\frac {5}{2}} \sin \left (d x + c\right ) + {\left (b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b^{2} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} A \sqrt {b}}{2 \, d} \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ 
(7/2),x, algorithm="maxima")
 

Output:

1/2*(4*B*b^(5/2)*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) + 2*C*b^(5/2)*sin 
(d*x + c) + (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
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.11 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {{\left (A b^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - A b^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + i \, B b^{2} \log \left (i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - i \, B b^{2} \log \left (-i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + \frac {2 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}\right )} \sqrt {b}}{d} \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ 
(7/2),x, algorithm="giac")
 

Output:

(A*b^2*log(tan(1/2*d*x + 1/2*c) + 1) - A*b^2*log(tan(1/2*d*x + 1/2*c) - 1) 
 + I*B*b^2*log(I*tan(1/2*d*x + 1/2*c) - 1) - I*B*b^2*log(-I*tan(1/2*d*x + 
1/2*c) - 1) + 2*C*b^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1))*s 
qrt(b)/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^{7/2}} \,d x \] Input:

int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^(7/2),x)
 

Output:

int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x)^(7/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.50 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\sqrt {b}\, b^{2} \left (-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +\sin \left (d x +c \right ) c +b d x \right )}{d} \] Input:

int((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2), 
x)
 

Output:

(sqrt(b)*b**2*( - log(tan((c + d*x)/2) - 1)*a + log(tan((c + d*x)/2) + 1)* 
a + sin(c + d*x)*c + b*d*x))/d