3.4.90 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [390]

3.4.90.1 Optimal result

Integrand size = 36, antiderivative size = 78 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}-\frac {2 a B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d}+\frac {2 a^2 B \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^2 (a+b) d} \]

output

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/d-2*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))/b^2/d+2*a^2*B*(cos(1/2*d*x+1/2*c)^2) 
^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2)) 
/b^2/(a+b)/d
 

3.4.90.2 Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 B \left (b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-(a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+a \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{b^2 d \sqrt {\sin ^2(c+d x)}} \]

input

Integrate[(Cos[c + d*x]^(3/2)*(a*B + b*B*Cos[c + d*x]))/(a + b*Cos[c + d*x 
])^2,x]
 

output

(-2*B*(b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] - (a + b)*EllipticF[Arc 
Sin[Sqrt[Cos[c + d*x]]], -1] + a*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d* 
x]]], -1])*Sin[c + d*x])/(b^2*d*Sqrt[Sin[c + d*x]^2])
 

3.4.90.3 Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2011, 3042, 3283, 3042, 3119, 3282, 3042, 3120, 3284}

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.

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3283

\(\displaystyle B \left (\frac {\int \sqrt {\cos (c+d x)}dx}{b}-\frac {a \int \frac {\sqrt {\cos (c+d x)}}{a+b \cos (c+d x)}dx}{b}\right )\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3282

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle B \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}-\frac {a \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}\right )\)

\(\Big \downarrow \) 3284

\(\displaystyle B \left (\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}-\frac {a \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}\right )}{b}\right )\)

input

Int[(Cos[c + d*x]^(3/2)*(a*B + b*B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^2,x 
]
 

output

B*((2*EllipticE[(c + d*x)/2, 2])/(b*d) - (a*((2*EllipticF[(c + d*x)/2, 2]) 
/(b*d) - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d)))/b 
)
 

3.4.90.3.1 Defintions of rubi rules used

rule 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

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 3282

Int[Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]/((a_.) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[d/b   Int[1/Sqrt[c + d*Sin[e + f*x]], x], x 
] + Simp[(b*c - a*d)/b   Int[1/((a + b*Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x 
]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 
 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3283

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[b/d   Int[Sqrt[a + b*Sin[e + f*x]], x], x 
] - Simp[(b*c - a*d)/d   Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), 
 x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b 
^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

3.4.90.4 Maple [A] (verified)

Time = 3.87 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.92

method	result	size
default	\(\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b +E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}-a^{2} \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right )\right )}{b^{2} \left (a -b \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(228\)

input

int(cos(d*x+c)^(3/2)*(B*a+b*B*cos(d*x+c))/(a+cos(d*x+c)*b)^2,x,method=_RET 
URNVERBOSE)
 

output

2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*B*(sin(1/2*d*x+1 
/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*(EllipticF(cos(1/2*d*x+1/ 
2*c),2^(1/2))*a^2-EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b+EllipticE(cos( 
1/2*d*x+1/2*c),2^(1/2))*a*b-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-a^2* 
EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))/b^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
 

3.4.90.5 Fricas [F(-1)]

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

input

integrate(cos(d*x+c)^(3/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algo 
rithm="fricas")
 

output

Timed out
 

3.4.90.6 Sympy [F(-1)]

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

input

integrate(cos(d*x+c)**(3/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))**2,x)
 

output

Timed out
 

3.4.90.7 Maxima [F]

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

input

integrate(cos(d*x+c)^(3/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algo 
rithm="maxima")
 

output

integrate((B*b*cos(d*x + c) + B*a)*cos(d*x + c)^(3/2)/(b*cos(d*x + c) + a) 
^2, x)
 

3.4.90.8 Giac [F]

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

input

integrate(cos(d*x+c)^(3/2)*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algo 
rithm="giac")
 

output

integrate((B*b*cos(d*x + c) + B*a)*cos(d*x + c)^(3/2)/(b*cos(d*x + c) + a) 
^2, x)
 

3.4.90.9 Mupad [F(-1)]

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

input

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

output

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