3.9.16 \(\int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\) [816]

3.9.16.1 Optimal result
3.9.16.2 Mathematica [A] (verified)
3.9.16.3 Rubi [A] (verified)
3.9.16.4 Maple [A] (verified)
3.9.16.5 Fricas [C] (verification not implemented)
3.9.16.6 Sympy [F(-1)]
3.9.16.7 Maxima [F]
3.9.16.8 Giac [F]
3.9.16.9 Mupad [F(-1)]

3.9.16.1 Optimal result

Integrand size = 31, antiderivative size = 85 \[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {2 b^2 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 A b^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {b \cos (c+d x)}} \]

output
2*A*b^3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2* 
d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)+2*b^2*B*(cos(1 
/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^( 
1/2))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)
 
3.9.16.2 Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.64 \[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {2 (b \cos (c+d x))^{5/2} \left (B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+A \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )\right )}{d \cos ^{\frac {5}{2}}(c+d x)} \]

input
Integrate[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x]
 
output
(2*(b*Cos[c + d*x])^(5/2)*(B*EllipticE[(c + d*x)/2, 2] + A*EllipticF[(c + 
d*x)/2, 2]))/(d*Cos[c + d*x]^(5/2))
 
3.9.16.3 Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 2030, 3227, 3042, 3121, 3042, 3119, 3120}

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.

\(\displaystyle \int \sec ^3(c+d x) (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\)

\(\Big \downarrow \) 2030

\(\displaystyle b^3 \int \frac {A+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}}dx\)

\(\Big \downarrow \) 3227

\(\displaystyle b^3 \left (A \int \frac {1}{\sqrt {b \cos (c+d x)}}dx+\frac {B \int \sqrt {b \cos (c+d x)}dx}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^3 \left (A \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {B \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )\)

\(\Big \downarrow \) 3121

\(\displaystyle b^3 \left (\frac {A \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{\sqrt {b \cos (c+d x)}}+\frac {B \sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^3 \left (\frac {A \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {b \cos (c+d x)}}+\frac {B \sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle b^3 \left (\frac {A \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {b \cos (c+d x)}}+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle b^3 \left (\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {b \cos (c+d x)}}+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}\right )\)

input
Int[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^3,x]
 
output
b^3*((2*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(b*d*Sqrt[Cos[c 
+ d*x]]) + (2*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(d*Sqrt[b*Co 
s[c + d*x]]))
 

3.9.16.3.1 Defintions of rubi rules used

rule 2030
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3227
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]
 
3.9.16.4 Maple [A] (verified)

Time = 53.23 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.92

method result size
default \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-B E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(163\)
parts \(-\frac {2 A \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}+\frac {2 B \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(290\)
risch \(-\frac {i B \,b^{2} \sqrt {2}\, \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b \,{\mathrm e}^{-i \left (d x +c \right )}}}{d}-\frac {i \left (\frac {i A \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )}{\sqrt {b \,{\mathrm e}^{3 i \left (d x +c \right )}+b \,{\mathrm e}^{i \left (d x +c \right )}}}+B \left (-\frac {2 \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}{b \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i E\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {b \,{\mathrm e}^{3 i \left (d x +c \right )}+b \,{\mathrm e}^{i \left (d x +c \right )}}}\right )\right ) b^{2} \sqrt {2}\, \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b \,{\mathrm e}^{-i \left (d x +c \right )}}\, \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b \,{\mathrm e}^{i \left (d x +c \right )}}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\) \(414\)

input
int((cos(d*x+c)*b)^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^3,x,method=_RETURNVER 
BOSE)
 
output
-2*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*(sin(1/2* 
d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*(A*EllipticF(cos(1/2 
*d*x+1/2*c),2^(1/2))-B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-b*(2*sin(1 
/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/sin(1/2*d*x+1/2*c)/((2*cos(1/ 
2*d*x+1/2*c)^2-1)*b)^(1/2)/d
 
3.9.16.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.40 \[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {-i \, \sqrt {2} A b^{\frac {5}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} A b^{\frac {5}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} B b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - i \, \sqrt {2} B b^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{d} \]

input
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm= 
"fricas")
 
output
(-I*sqrt(2)*A*b^(5/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x 
+ c)) + I*sqrt(2)*A*b^(5/2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*si 
n(d*x + c)) + I*sqrt(2)*B*b^(5/2)*weierstrassZeta(-4, 0, weierstrassPInver 
se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - I*sqrt(2)*B*b^(5/2)*weierstras 
sZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
 
3.9.16.6 Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\text {Timed out} \]

input
integrate((b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**3,x)
 
output
Timed out
 
3.9.16.7 Maxima [F]

\[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]

input
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm= 
"maxima")
 
output
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^3, x)
 
3.9.16.8 Giac [F]

\[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]

input
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm= 
"giac")
 
output
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^3, x)
 
3.9.16.9 Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^3} \,d x \]

input
int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x)))/cos(c + d*x)^3,x)
 
output
int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x)))/cos(c + d*x)^3, x)