\(\int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^2} \, dx\) [249]

Optimal result

Integrand size = 25, antiderivative size = 116 \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^2} \, dx=\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 a^2 d \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a+a \sin (c+d x))^2}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e \left (a^2+a^2 \sin (c+d x)\right )} \] Output:

2/7*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^2/d/(e*cos(d 
*x+c))^(1/2)-2/7*(e*cos(d*x+c))^(1/2)/d/e/(a+a*sin(d*x+c))^2-2/7*(e*cos(d* 
x+c))^(1/2)/d/e/(a^2+a^2*sin(d*x+c))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^2} \, dx=-\frac {\sqrt {e \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {11}{4},\frac {5}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{2^{3/4} a^2 d e \sqrt [4]{1+\sin (c+d x)}} \] Input:

Integrate[1/(Sqrt[e*Cos[c + d*x]]*(a + a*Sin[c + d*x])^2),x]
 

Output:

-((Sqrt[e*Cos[c + d*x]]*Hypergeometric2F1[1/4, 11/4, 5/4, (1 - Sin[c + d*x 
])/2])/(2^(3/4)*a^2*d*e*(1 + Sin[c + d*x])^(1/4)))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3160, 3042, 3162, 3042, 3121, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3160

\(\displaystyle \frac {3 \int \frac {1}{\sqrt {e \cos (c+d x)} (\sin (c+d x) a+a)}dx}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int \frac {1}{\sqrt {e \cos (c+d x)} (\sin (c+d x) a+a)}dx}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3162

\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {e \cos (c+d x)}}dx}{3 a}-\frac {2 \sqrt {e \cos (c+d x)}}{3 d e (a \sin (c+d x)+a)}\right )}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}-\frac {2 \sqrt {e \cos (c+d x)}}{3 d e (a \sin (c+d x)+a)}\right )}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {3 \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 a \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{3 d e (a \sin (c+d x)+a)}\right )}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{3 d e (a \sin (c+d x)+a)}\right )}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {3 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{3 d e (a \sin (c+d x)+a)}\right )}{7 a}-\frac {2 \sqrt {e \cos (c+d x)}}{7 d e (a \sin (c+d x)+a)^2}\)

Input:

Int[1/(Sqrt[e*Cos[c + d*x]]*(a + a*Sin[c + d*x])^2),x]
 

Output:

(-2*Sqrt[e*Cos[c + d*x]])/(7*d*e*(a + a*Sin[c + d*x])^2) + (3*((2*Sqrt[Cos 
[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*a*d*Sqrt[e*Cos[c + d*x]]) - (2*Sq 
rt[e*Cos[c + d*x]])/(3*d*e*(a + a*Sin[c + d*x]))))/(7*a)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, 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 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3160

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x 
])^m/(a*f*g*(2*m + p + 1))), x] + Simp[(m + p + 1)/(a*(2*m + p + 1))   Int[ 
(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, 
f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] & 
& IntegersQ[2*m, 2*p]
 

rule 3162

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[b*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p - 1)*(a + b*S 
in[e + f*x]))), x] + Simp[p/(a*(p - 1))   Int[(g*Cos[e + f*x])^p, x], x] /; 
 FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] &&  !GeQ[p, 1] && Intege 
rQ[2*p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(371\) vs. \(2(103)=206\).

Time = 1.42 (sec) , antiderivative size = 372, normalized size of antiderivative = 3.21

method	result	size
default	\(-\frac {2 \left (8 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+6 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7 \left (8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, d}\)	\(372\)

Input:

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

Output:

-2/7/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^ 
2-1)/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(8*(2*sin( 
1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d 
*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^6+8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x 
+1/2*c)-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*E 
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-8*sin(1/2*d*x+1/ 
2*c)^4*cos(1/2*d*x+1/2*c)+6*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2 
+6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(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))-2*sin 
(1/2*d*x+1/2*c))/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\sqrt {\frac {1}{2}} {\left (-i \, \cos \left (d x + c\right )^{2} + 2 i \, \sin \left (d x + c\right ) + 2 i\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {\frac {1}{2}} {\left (i \, \cos \left (d x + c\right )^{2} - 2 i \, \sin \left (d x + c\right ) - 2 i\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + \sqrt {e \cos \left (d x + c\right )} {\left (\sin \left (d x + c\right ) + 2\right )}\right )}}{7 \, {\left (a^{2} d e \cos \left (d x + c\right )^{2} - 2 \, a^{2} d e \sin \left (d x + c\right ) - 2 \, a^{2} d e\right )}} \] Input:

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

Output:

2/7*(sqrt(1/2)*(-I*cos(d*x + c)^2 + 2*I*sin(d*x + c) + 2*I)*sqrt(e)*weiers 
trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(1/2)*(I*cos(d*x 
 + c)^2 - 2*I*sin(d*x + c) - 2*I)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d 
*x + c) - I*sin(d*x + c)) + sqrt(e*cos(d*x + c))*(sin(d*x + c) + 2))/(a^2* 
d*e*cos(d*x + c)^2 - 2*a^2*d*e*sin(d*x + c) - 2*a^2*d*e)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate(1/(sqrt(e*cos(d*x + c))*(a*sin(d*x + c) + a)^2), x)
 

Giac [F]

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

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

Output:

integrate(1/(sqrt(e*cos(d*x + c))*(a*sin(d*x + c) + a)^2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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