3.7.100 \(\int (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \, dx\) [700]

3.7.100.1 Optimal result
3.7.100.2 Mathematica [A] (verified)
3.7.100.3 Rubi [A] (verified)
3.7.100.4 Maple [A] (verified)
3.7.100.5 Fricas [C] (verification not implemented)
3.7.100.6 Sympy [F(-1)]
3.7.100.7 Maxima [F]
3.7.100.8 Giac [F]
3.7.100.9 Mupad [F(-1)]

3.7.100.1 Optimal result

Integrand size = 23, antiderivative size = 108 \[ \int (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {2 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {4 a b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]

output
2*a^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2*(a^2-b^2)*(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))*cos(d*x+c)^(1/ 
2)*sec(d*x+c)^(1/2)/d+4*a*b*(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)*sec(d*x+c)^(1/2)/ 
d
 
3.7.100.2 Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.77 \[ \int (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {2 \sqrt {\sec (c+d x)} \left (-\left (\left (a^2-b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )+a \left (2 b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+a \sin (c+d x)\right )\right )}{d} \]

input
Integrate[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^(3/2),x]
 
output
(2*Sqrt[Sec[c + d*x]]*(-((a^2 - b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x 
)/2, 2]) + a*(2*b*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + a*Sin[c + 
 d*x])))/d
 
3.7.100.3 Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3717, 3042, 4275, 3042, 4258, 3042, 3120, 4534, 3042, 4258, 3042, 3119}

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.

\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3717

\(\displaystyle \int \frac {(a \sec (c+d x)+b)^2}{\sqrt {\sec (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+b\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4275

\(\displaystyle \int \frac {b^2+a^2 \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}}dx+2 a b \int \sqrt {\sec (c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {b^2+a^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+2 a b \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle \int \frac {b^2+a^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {4 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\)

\(\Big \downarrow \) 4534

\(\displaystyle -\left (a^2-b^2\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {4 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\left (a^2-b^2\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {4 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\)

\(\Big \downarrow \) 4258

\(\displaystyle -\left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {4 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {4 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\)

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {2 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {4 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\)

input
Int[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^(3/2),x]
 
output
(-2*(a^2 - b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + 
d*x]])/d + (4*a*b*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c 
+ d*x]])/d + (2*a^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d
 

3.7.100.3.1 Defintions of rubi rules used

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 3717
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p)   Int[(d*Csc[e + f*x])^(m - n*p 
)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && 
  !IntegerQ[m] && IntegersQ[n, p]
 

rule 4258
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4275
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^2, x_Symbol] :> Simp[2*a*(b/d)   Int[(d*Csc[e + f*x])^(n + 1), x], x] 
 + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, 
 e, f, n}, x]
 

rule 4534
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) 
)), x] + Simp[(C*m + A*(m + 1))/(m + 1)   Int[(b*Csc[e + f*x])^m, x], x] /; 
 FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]
 
3.7.100.4 Maple [A] (verified)

Time = 8.49 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.87

method result size
default \(\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}-4 a b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(202\)
parts \(-\frac {2 a^{2} \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}+\frac {2 b^{2} \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \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 {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}-\frac {4 a b \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \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 {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(455\)

input
int((a+cos(d*x+c)*b)^2*sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 
output
2*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^2-2*a*b*(sin(1/2*d*x+1/2*c) 
^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^ 
(1/2))-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellip 
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1 
/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2)/sin(1/ 
2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 
3.7.100.5 Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.35 \[ \int (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {-2 i \, \sqrt {2} a b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 i \, \sqrt {2} a b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + \frac {2 \, a^{2} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + \sqrt {2} {\left (-i \, a^{2} + i \, b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (i \, a^{2} - i \, b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{d} \]

input
integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^(3/2),x, algorithm="fricas")
 
output
(-2*I*sqrt(2)*a*b*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c) 
) + 2*I*sqrt(2)*a*b*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + 
c)) + 2*a^2*sin(d*x + c)/sqrt(cos(d*x + c)) + sqrt(2)*(-I*a^2 + I*b^2)*wei 
erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + 
c))) + sqrt(2)*(I*a^2 - I*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse( 
-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
 
3.7.100.6 Sympy [F(-1)]

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

input
integrate((a+b*cos(d*x+c))**2*sec(d*x+c)**(3/2),x)
 
output
Timed out
 
3.7.100.7 Maxima [F]

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

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

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

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

Timed out. \[ \int (a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,d x \]

input
int((1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^2,x)
 
output
int((1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^2, x)