3.6.72 \(\int \frac {(a+b \cos (c+d x))^2}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [572]

3.6.72.1 Optimal result
3.6.72.2 Mathematica [A] (verified)
3.6.72.3 Rubi [A] (verified)
3.6.72.4 Maple [A] (verified)
3.6.72.5 Fricas [C] (verification not implemented)
3.6.72.6 Sympy [F(-1)]
3.6.72.7 Maxima [F]
3.6.72.8 Giac [F]
3.6.72.9 Mupad [B] (verification not implemented)

3.6.72.1 Optimal result

Integrand size = 23, antiderivative size = 68 \[ \int \frac {(a+b \cos (c+d x))^2}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \left (a^2-b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {4 a b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 a^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

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

Time = 0.44 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b \cos (c+d x))^2}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \left (\left (-a^2+b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+a \left (2 b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )\right )}{d} \]

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

Time = 0.41 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3268, 3042, 3120, 3491, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3268

\(\displaystyle \int \frac {a^2+b^2 \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx+2 a b \int \frac {1}{\sqrt {\cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle \int \frac {a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {4 a b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\)

\(\Big \downarrow \) 3491

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

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

3.6.72.3.1 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 3268
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)])^2, x_Symbol] :> Simp[2*c*(d/b)   Int[(b*Sin[e + f*x])^(m + 1), x], x] 
+ Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c, d, 
e, f, m}, 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]
 
3.6.72.4 Maple [A] (verified)

Time = 6.91 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.97

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 \(-\frac {2 a^{2} \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}+\frac {2 b^{2} \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}+\frac {4 a b \,\operatorname {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d}\) \(340\)

input
int((a+cos(d*x+c)*b)^2/cos(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
 
3.6.72.5 Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.62 \[ \int \frac {(a+b \cos (c+d x))^2}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-2 i \, \sqrt {2} a b \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 i \, \sqrt {2} a b \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, a^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, a^{2} + i \, b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (i \, a^{2} - i \, b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{d \cos \left (d x + c\right )} \]

input
integrate((a+b*cos(d*x+c))^2/cos(d*x+c)^(3/2),x, algorithm="fricas")
 
output
(-2*I*sqrt(2)*a*b*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I 
*sin(d*x + c)) + 2*I*sqrt(2)*a*b*cos(d*x + c)*weierstrassPInverse(-4, 0, c 
os(d*x + c) - I*sin(d*x + c)) + 2*a^2*sqrt(cos(d*x + c))*sin(d*x + c) + sq 
rt(2)*(-I*a^2 + I*b^2)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInv 
erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + sqrt(2)*(I*a^2 - I*b^2)*cos( 
d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - 
I*sin(d*x + c))))/(d*cos(d*x + c))
 
3.6.72.6 Sympy [F(-1)]

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

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

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

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

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

input
integrate((a+b*cos(d*x+c))^2/cos(d*x+c)^(3/2),x, algorithm="giac")
 
output
integrate((b*cos(d*x + c) + a)^2/cos(d*x + c)^(3/2), x)
 
3.6.72.9 Mupad [B] (verification not implemented)

Time = 15.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b \cos (c+d x))^2}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

input
int((a + b*cos(c + d*x))^2/cos(c + d*x)^(3/2),x)
 
output
(2*b^2*ellipticE(c/2 + (d*x)/2, 2))/d + (4*a*b*ellipticF(c/2 + (d*x)/2, 2) 
)/d + (2*a^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d* 
cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))