3.6.71 \(\int \frac {(a+b \cos (c+d x))^2}{\sqrt {\cos (c+d x)}} \, dx\) [571]

3.6.71.1 Optimal result

Integrand size = 23, antiderivative size = 72 \[ \int \frac {(a+b \cos (c+d x))^2}{\sqrt {\cos (c+d x)}} \, dx=\frac {4 a b E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (3 a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 b^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \]

output

4*a*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))/d+2/3*(3*a^2+b^2)*(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+2/3*b^2*sin(d*x+c)*cos(d 
*x+c)^(1/2)/d
 

3.6.71.2 Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b \cos (c+d x))^2}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \left (6 a b E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\left (3 a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+b^2 \sqrt {\cos (c+d x)} \sin (c+d x)\right )}{3 d} \]

input

Integrate[(a + b*Cos[c + d*x])^2/Sqrt[Cos[c + d*x]],x]
 

output

(2*(6*a*b*EllipticE[(c + d*x)/2, 2] + (3*a^2 + b^2)*EllipticF[(c + d*x)/2, 
 2] + b^2*Sqrt[Cos[c + d*x]]*Sin[c + d*x]))/(3*d)
 

3.6.71.3 Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 72, 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, 3119, 3493, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3268

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \int \frac {a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {4 a b E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\)

\(\Big \downarrow \) 3493

\(\displaystyle \frac {1}{3} \left (3 a^2+b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {4 a b E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (3 a^2+b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {4 a b E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 \left (3 a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a b E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

input

Int[(a + b*Cos[c + d*x])^2/Sqrt[Cos[c + d*x]],x]
 

output

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

3.6.71.3.1 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 3493

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

3.6.71.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(282\) vs. \(2(118)=236\).

Time = 7.04 (sec) , antiderivative size = 283, normalized size of antiderivative = 3.93

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}+3 a^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b \right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(283\)
parts	\(\frac {2 a^{2} \operatorname {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d}-\frac {2 b^{2} \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}+\frac {4 a b \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(337\)

input

int((a+cos(d*x+c)*b)^2/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

output

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2*d* 
x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^2-2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^ 
2*b^2+3*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)* 
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-6*(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))*a*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)^(1/2)/d
 

3.6.71.5 Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.04 \[ \int \frac {(a+b \cos (c+d x))^2}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \, b^{2} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 6 i \, \sqrt {2} a b {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 6 i \, \sqrt {2} a b {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (-3 i \, a^{2} - i \, b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (3 i \, a^{2} + i \, b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{3 \, d} \]

input

integrate((a+b*cos(d*x+c))^2/cos(d*x+c)^(1/2),x, algorithm="fricas")
 

output

1/3*(2*b^2*sqrt(cos(d*x + c))*sin(d*x + c) + 6*I*sqrt(2)*a*b*weierstrassZe 
ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 6*I 
*sqrt(2)*a*b*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c 
) - I*sin(d*x + c))) + sqrt(2)*(-3*I*a^2 - I*b^2)*weierstrassPInverse(-4, 
0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(3*I*a^2 + I*b^2)*weierstrassP 
Inverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)))/d
 

3.6.71.6 Sympy [F]

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

input

integrate((a+b*cos(d*x+c))**2/cos(d*x+c)**(1/2),x)
 

output

Integral((a + b*cos(c + d*x))**2/sqrt(cos(c + d*x)), x)
 

3.6.71.7 Maxima [F]

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

input

integrate((a+b*cos(d*x+c))^2/cos(d*x+c)^(1/2),x, algorithm="maxima")
 

output

integrate((b*cos(d*x + c) + a)^2/sqrt(cos(d*x + c)), x)
 

3.6.71.8 Giac [F]

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

input

integrate((a+b*cos(d*x+c))^2/cos(d*x+c)^(1/2),x, algorithm="giac")
 

output

integrate((b*cos(d*x + c) + a)^2/sqrt(cos(d*x + c)), x)
 

3.6.71.9 Mupad [B] (verification not implemented)

Time = 14.72 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b \cos (c+d x))^2}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,b^2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d}+\frac {4\,a\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d} \]

input

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

output

(2*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*b^2*ellipticF(c/2 + (d*x)/2, 2) 
)/(3*d) + (2*b^2*cos(c + d*x)^(1/2)*sin(c + d*x))/(3*d) + (4*a*b*ellipticE 
(c/2 + (d*x)/2, 2))/d