\(\int \frac {(b \cos (c+d x))^{5/2} (A+C \cos ^2(c+d x))}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\) [113]

Optimal result

Integrand size = 35, antiderivative size = 84 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {b^2 (A+2 C) \text {arctanh}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{2 d \sqrt {\cos (c+d x)}}+\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x)} \] Output:

1/2*b^2*(A+2*C)*arctanh(sin(d*x+c))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2 
)+1/2*A*b^2*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.70 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{5/2} \left ((A+2 C) \text {arctanh}(\sin (c+d x)) \cos ^2(c+d x)+A \sin (c+d x)\right )}{2 d \cos ^{\frac {9}{2}}(c+d x)} \] Input:

Integrate[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11 
/2),x]
 

Output:

((b*Cos[c + d*x])^(5/2)*((A + 2*C)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + 
A*Sin[c + d*x]))/(2*d*Cos[c + d*x]^(9/2))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.79, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2031, 3042, 3491, 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.

\(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 2031

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \left (C \cos ^2(c+d x)+A\right ) \sec ^3(c+d x)dx}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3491

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{2} (A+2 C) \int \sec (c+d x)dx+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{2} (A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}\right )}{\sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 4257

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

Input:

Int[((b*Cos[c + d*x])^(5/2)*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(11/2),x]
 

Output:

(b^2*Sqrt[b*Cos[c + d*x]]*(((A + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (A*Se 
c[c + d*x]*Tan[c + d*x])/(2*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 3491

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x 
_)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m 
+ 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1))   Int[(b*Sin[e + f* 
x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, 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 = 0.36 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.42

method	result	size
default	\(-\frac {b^{2} \left (A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{2}-A \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}+4 C \,\operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}-A \sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}\)	\(119\)
parts	\(\frac {A \left (\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1\right ) \cos \left (d x +c \right )^{2}-\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}\, b^{2}}{2 d \cos \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 C \sqrt {b \cos \left (d x +c \right )}\, \operatorname {arctanh}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) b^{2}}{d \sqrt {\cos \left (d x +c \right )}}\)	\(132\)
risch	\(-\frac {i b^{2} \sqrt {b \cos \left (d x +c \right )}\, A \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{\sqrt {\cos \left (d x +c \right )}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {b \cos \left (d x +c \right )}\, \left (A +2 C \right ) \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 \sqrt {\cos \left (d x +c \right )}\, d}\)	\(152\)

Input:

int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x,method=_RE 
TURNVERBOSE)
 

Output:

-1/2*b^2/d*(A*ln(-cot(d*x+c)+csc(d*x+c)-1)*cos(d*x+c)^2-A*ln(-cot(d*x+c)+c 
sc(d*x+c)+1)*cos(d*x+c)^2+4*C*arctanh(-csc(d*x+c)+cot(d*x+c))*cos(d*x+c)^2 
-A*sin(d*x+c))*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.64 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\left [\frac {{\left (A + 2 \, C\right )} b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} A b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{3}}, -\frac {{\left (A + 2 \, C\right )} \sqrt {-b} b^{2} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right )^{3} - \sqrt {b \cos \left (d x + c\right )} A b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )^{3}}\right ] \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, alg 
orithm="fricas")
 

Output:

[1/4*((A + 2*C)*b^(5/2)*cos(d*x + c)^3*log(-(b*cos(d*x + c)^3 - 2*sqrt(b*c 
os(d*x + c))*sqrt(b)*sqrt(cos(d*x + c))*sin(d*x + c) - 2*b*cos(d*x + c))/c 
os(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)^3), -1/2*((A + 2*C)*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)^3 - sqr 
t(b*cos(d*x + c))*A*b^2*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3 
)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate((b*cos(d*x+c))**(5/2)*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(11/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 821 vs. \(2 (72) = 144\).

Time = 0.37 (sec) , antiderivative size = 821, normalized size of antiderivative = 9.77 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, alg 
orithm="maxima")
 

Output:

1/4*(2*(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))*C*sqrt(b) - ( 
4*(b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d* 
x + 2*c), cos(2*d*x + 2*c))) - 4*(b^2*sin(4*d*x + 4*c) + 2*b^2*sin(2*d*x + 
 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (b^2*cos(4*d 
*x + 4*c)^2 + 4*b^2*cos(2*d*x + 2*c)^2 + b^2*sin(4*d*x + 4*c)^2 + 4*b^2*si 
n(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b^2*sin(2*d*x + 2*c)^2 + 4*b^2*cos(2*d 
*x + 2*c) + b^2 + 2*(2*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c))*log(c 
os(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(si 
n(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), 
 cos(2*d*x + 2*c))) + 1) + (b^2*cos(4*d*x + 4*c)^2 + 4*b^2*cos(2*d*x + 2*c 
)^2 + b^2*sin(4*d*x + 4*c)^2 + 4*b^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4 
*b^2*sin(2*d*x + 2*c)^2 + 4*b^2*cos(2*d*x + 2*c) + b^2 + 2*(2*b^2*cos(2*d* 
x + 2*c) + b^2)*cos(4*d*x + 4*c))*log(cos(1/2*arctan2(sin(2*d*x + 2*c), co 
s(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^ 
2 - 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 4*(b^2*c 
os(4*d*x + 4*c) + 2*b^2*cos(2*d*x + 2*c) + b^2)*sin(3/2*arctan2(sin(2*d*x 
+ 2*c), cos(2*d*x + 2*c))) + 4*(b^2*cos(4*d*x + 4*c) + 2*b^2*cos(2*d*x + 2 
*c) + b^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*A*sqrt(b) 
/(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*...
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.44 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {{\left ({\left (A b^{2} + 2 \, C b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - {\left (A b^{2} + 2 \, C b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + \frac {2 \, {\left (A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}\right )} \sqrt {b}}{2 \, d} \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x, alg 
orithm="giac")
 

Output:

1/2*((A*b^2 + 2*C*b^2)*log(tan(1/2*d*x + 1/2*c) + 1) - (A*b^2 + 2*C*b^2)*l 
og(tan(1/2*d*x + 1/2*c) - 1) + 2*(A*b^2*tan(1/2*d*x + 1/2*c)^3 + A*b^2*tan 
(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 2*tan(1/2*d*x + 1/2*c)^2 + 1) 
)*sqrt(b)/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{11/2}} \,d x \] Input:

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^(11/2),x)
 

Output:

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(5/2))/cos(c + d*x)^(11/2), x 
)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.17 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {\sqrt {b}\, b^{2} \left (-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -\sin \left (d x +c \right ) a \right )}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int((b*cos(d*x+c))^(5/2)*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2),x)
 

Output:

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