\(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [590]

Optimal result

Integrand size = 23, antiderivative size = 163 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b \left (a^2-b^2\right ) d}+\frac {\left (a^2-2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 \left (a^2-b^2\right ) d}-\frac {a \left (a^2-3 b^2\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^2 (a+b)^2 d}+\frac {a \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

-a*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b/(a^2-b^2)/d+(a^2-2*b^2)*Inverse 
JacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^2/(a^2-b^2)/d-a*(a^2-3*b^2)*EllipticPi(s 
in(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/(a-b)/b^2/(a+b)^2/d+a*cos(d*x+c)^(1/2 
)*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 3.66 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 a \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {8 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {10 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {2 \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{4 d} \] Input:

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

Output:

((4*a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(a + b*Cos[c + d*x])) 
- (8*EllipticF[(c + d*x)/2, 2] - (10*a*EllipticPi[(2*b)/(a + b), (c + d*x) 
/2, 2])/(a + b) + (2*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2 
*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*Elli 
pticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(b^2*Sqrt[Si 
n[c + d*x]^2]))/((a - b)*(a + b)))/(4*d)
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.96, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3278, 27, 3042, 3538, 25, 3042, 3119, 3481, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3278

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[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 3278

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*(a + b*Si 
n[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n - 1)/(f*(m + 1)*(a^2 - b^2))), 
 x] + Simp[1/((m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + 
d*Sin[e + f*x])^(n - 2)*Simp[c*(a*c - b*d)*(m + 1) + d*(b*c - a*d)*(n - 1) 
+ (d*(a*c - b*d)*(m + 1) - c*(b*c - a*d)*(m + 2))*Sin[e + f*x] - d*(b*c - a 
*d)*(m + n + 1)*Sin[e + f*x]^2, 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] && LtQ[m, -1 
] && LtQ[1, n, 2] && IntegersQ[2*m, 2*n]
 

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]
 

rule 3481

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

rule 3538

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(793\) vs. \(2(166)=332\).

Time = 5.81 (sec) , antiderivative size = 794, normalized size of antiderivative = 4.87

method	result	size
default	\(\text {Expression too large to display}\)	\(794\)

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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