\(\int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx\) [933]

Optimal result

Integrand size = 35, antiderivative size = 123 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx=-\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{a^{3/2} (c-d) f}+\frac {2 \sqrt {c+d} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{a^{3/2} (c-d) \sqrt {d} f} \] Output:

-2*2^(1/2)*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/ 
a^(3/2)/(c-d)/f+2*(c+d)^(1/2)*arctanh(a^(1/2)*d^(1/2)*cos(f*x+e)/(c+d)^(1/ 
2)/(a+a*sin(f*x+e))^(1/2))/a^(3/2)/(c-d)/d^(1/2)/f
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 4.79 (sec) , antiderivative size = 750, normalized size of antiderivative = 6.10 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*((-1)^(3/4)*((8 + 8*I)*Sqrt[d]*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + T 
an[(e + f*x)/4])] + (-1)^(1/4)*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + 
c*#1^4 & , (-(c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^(3/2)*Log[-#1 + T 
an[(e + f*x)/4]] - d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - 2*c*Sqrt[d] 
*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1 
 - c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d]*Log[-#1 + Tan[ 
(e + f*x)/4]]*#1^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + 3*d*Sqrt[c 
 + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Sqrt[c + d]*Log[-#1 + Tan[(e + 
f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ] + (-1)^(1/4)*RootSum[c 
+ 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & , (-(c*Sqrt[d]*Log[-#1 + Tan[(e 
+ f*x)/4]]) - d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]] + d*Sqrt[c + d]*Log[-#1 
+ Tan[(e + f*x)/4]] - 2*c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*d^(3/ 
2)*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/ 
4]]*#1 + c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + d^(3/2)*Log[-#1 + Ta 
n[(e + f*x)/4]]*#1^2 - 3*d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + 
c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c* 
#1^3) & ])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/(Sqrt[d]*(-c + d)*f*(a 
*(1 + Sin[e + f*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3395, 3042, 3464, 3042, 3128, 219, 3252, 221}

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.

\(\displaystyle \int \frac {\cos ^2(e+f x)}{(a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (e+f x)^2}{(a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}dx\)

\(\Big \downarrow \) 3395

\(\displaystyle \frac {\int \frac {a-a \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a-a \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{a^2}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {\frac {2 a \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{c-d}-\frac {(c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{c-d}-\frac {(c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{a^2}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {-\frac {4 a \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c-d)}-\frac {(c+d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{a^2}\)

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {2 a (c+d) \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c-d)}-\frac {2 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{a^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {2 \sqrt {a} \sqrt {c+d} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {d} f (c-d)}-\frac {2 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{a^2}\)

Input:

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

Output:

((-2*Sqrt[2]*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Si 
n[e + f*x]])])/((c - d)*f) + (2*Sqrt[a]*Sqrt[c + d]*ArcTanh[(Sqrt[a]*Sqrt[ 
d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/((c - d)*Sqrt[d] 
*f))/a^2
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128

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

rule 3252

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

rule 3395

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

rule 3464

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

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.30

method	result	size
default	\(-\frac {2 \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {a \left (c +d \right ) d}-\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sqrt {a}\, c -\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right ) \sqrt {a}\, d \right )}{a^{\frac {3}{2}} \left (c -d \right ) \sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(160\)

Input:

int(cos(f*x+e)^2/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x,method=_RETURNV 
ERBOSE)
 

Output:

-2/a^(3/2)*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(2^(1/2)*arctanh(1/2*( 
-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*(a*(c+d)*d)^(1/2)-arctanh((-a*(s 
in(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(1/2)*c-arctanh((-a*(sin(f*x+e) 
-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(1/2)*d)/(c-d)/(a*(c+d)*d)^(1/2)/cos(f*x 
+e)/(a+a*sin(f*x+e))^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (100) = 200\).

Time = 0.17 (sec) , antiderivative size = 667, normalized size of antiderivative = 5.42 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x, algorith 
m="fricas")
 

Output:

[-1/2*(sqrt((c + d)/(a*d))*log((d^2*cos(f*x + e)^3 - (6*c*d + 7*d^2)*cos(f 
*x + e)^2 - c^2 - 2*c*d - d^2 - 4*(d^2*cos(f*x + e)^2 - c*d - 3*d^2 - (c*d 
 + 2*d^2)*cos(f*x + e) + (d^2*cos(f*x + e) + c*d + 3*d^2)*sin(f*x + e))*sq 
rt(a*sin(f*x + e) + a)*sqrt((c + d)/(a*d)) - (c^2 + 8*c*d + 9*d^2)*cos(f*x 
 + e) + (d^2*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 + 2*(3*c*d + 4*d^2)*cos(f* 
x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - 
 c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c* 
d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 2*sqrt(2)*log(-(cos(f 
*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(a*sin(f*x + e 
) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(co 
s(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(a 
))/((a*c - a*d)*f), (sqrt(-(c + d)/(a*d))*arctan(1/2*sqrt(a*sin(f*x + e) + 
 a)*(d*sin(f*x + e) - c - 2*d)*sqrt(-(c + d)/(a*d))/((c + d)*cos(f*x + e)) 
) - sqrt(2)*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqr 
t(2)*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 
3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - co 
s(f*x + e) - 2))/sqrt(a))/((a*c - a*d)*f)]
 

Sympy [F(-1)]

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

integrate(cos(f*x+e)**2/(a+a*sin(f*x+e))**(3/2)/(c+d*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x, algorith 
m="maxima")
 

Output:

integrate(cos(f*x + e)^2/((a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c)) 
, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (100) = 200\).

Time = 0.16 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.75 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx=\frac {\sqrt {2} \sqrt {a} {\left (\frac {\sqrt {2} {\left (c + d\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{{\left (a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {-c d - d^{2}}} + \frac {\log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{f} \] Input:

integrate(cos(f*x+e)^2/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x, algorith 
m="giac")
 

Output:

sqrt(2)*sqrt(a)*(sqrt(2)*(c + d)*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 
1/2*e)/sqrt(-c*d - d^2))/((a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - a^2 
*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(-c*d - d^2)) + log(sin(-1/4*p 
i + 1/2*f*x + 1/2*e) + 1)/(a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - a^2 
*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - log(-sin(-1/4*pi + 1/2*f*x + 1/2 
*e) + 1)/(a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - a^2*d*sgn(cos(-1/4*p 
i + 1/2*f*x + 1/2*e))))/f
 

Mupad [F(-1)]

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

int(cos(e + f*x)^2/((a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x))),x)
                                                                                    
                                                                                    
 

Output:

int(cos(e + f*x)^2/((a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x))), x)
 

Reduce [F]

\[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3} d +\sin \left (f x +e \right )^{2} c +2 \sin \left (f x +e \right )^{2} d +2 \sin \left (f x +e \right ) c +\sin \left (f x +e \right ) d +c}d x \right )}{a^{2}} \] Input:

int(cos(f*x+e)^2/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x)
 

Output:

(sqrt(a)*int((sqrt(sin(e + f*x) + 1)*cos(e + f*x)**2)/(sin(e + f*x)**3*d + 
 sin(e + f*x)**2*c + 2*sin(e + f*x)**2*d + 2*sin(e + f*x)*c + sin(e + f*x) 
*d + c),x))/a**2