3.3.60 \(\int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^3} \, dx\) [260]

3.3.60.1 Optimal result
3.3.60.2 Mathematica [C] (verified)
3.3.60.3 Rubi [A] (verified)
3.3.60.4 Maple [B] (verified)
3.3.60.5 Fricas [C] (verification not implemented)
3.3.60.6 Sympy [F(-1)]
3.3.60.7 Maxima [F]
3.3.60.8 Giac [F]
3.3.60.9 Mupad [F(-1)]

3.3.60.1 Optimal result

Integrand size = 25, antiderivative size = 153 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^3 d \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{9 d e (a+a \sin (c+d x))^3}-\frac {2 (e \cos (c+d x))^{3/2}}{15 a d e (a+a \sin (c+d x))^2}-\frac {2 (e \cos (c+d x))^{3/2}}{15 d e \left (a^3+a^3 \sin (c+d x)\right )} \]

output
-2/9*(e*cos(d*x+c))^(3/2)/d/e/(a+a*sin(d*x+c))^3-2/15*(e*cos(d*x+c))^(3/2) 
/a/d/e/(a+a*sin(d*x+c))^2-2/15*(e*cos(d*x+c))^(3/2)/d/e/(a^3+a^3*sin(d*x+c 
))-2/15*(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))*(e*cos(d*x+c))^(1/2)/a^3/d/cos(d*x+c)^(1/2)
 
3.3.60.2 Mathematica [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.43 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^3} \, dx=-\frac {(e \cos (c+d x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {13}{4},\frac {7}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{6 \sqrt [4]{2} a^3 d e (1+\sin (c+d x))^{3/4}} \]

input
Integrate[Sqrt[e*Cos[c + d*x]]/(a + a*Sin[c + d*x])^3,x]
 
output
-1/6*((e*Cos[c + d*x])^(3/2)*Hypergeometric2F1[3/4, 13/4, 7/4, (1 - Sin[c 
+ d*x])/2])/(2^(1/4)*a^3*d*e*(1 + Sin[c + d*x])^(3/4))
 
3.3.60.3 Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3160, 3042, 3160, 3042, 3162, 3042, 3121, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3160

\(\displaystyle \frac {\int \frac {\sqrt {e \cos (c+d x)}}{(\sin (c+d x) a+a)^2}dx}{3 a}-\frac {2 (e \cos (c+d x))^{3/2}}{9 d e (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {e \cos (c+d x)}}{(\sin (c+d x) a+a)^2}dx}{3 a}-\frac {2 (e \cos (c+d x))^{3/2}}{9 d e (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3160

\(\displaystyle \frac {\frac {\int \frac {\sqrt {e \cos (c+d x)}}{\sin (c+d x) a+a}dx}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}}{3 a}-\frac {2 (e \cos (c+d x))^{3/2}}{9 d e (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\sqrt {e \cos (c+d x)}}{\sin (c+d x) a+a}dx}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}}{3 a}-\frac {2 (e \cos (c+d x))^{3/2}}{9 d e (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3162

\(\displaystyle \frac {\frac {-\frac {\int \sqrt {e \cos (c+d x)}dx}{a}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}}{3 a}-\frac {2 (e \cos (c+d x))^{3/2}}{9 d e (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {\int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}}{3 a}-\frac {2 (e \cos (c+d x))^{3/2}}{9 d e (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {\frac {-\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{a \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}}{3 a}-\frac {2 (e \cos (c+d x))^{3/2}}{9 d e (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {\sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \sqrt {\cos (c+d x)}}-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}}{3 a}-\frac {2 (e \cos (c+d x))^{3/2}}{9 d e (a \sin (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {-\frac {2 (e \cos (c+d x))^{3/2}}{d e (a \sin (c+d x)+a)}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{a d \sqrt {\cos (c+d x)}}}{5 a}-\frac {2 (e \cos (c+d x))^{3/2}}{5 d e (a \sin (c+d x)+a)^2}}{3 a}-\frac {2 (e \cos (c+d x))^{3/2}}{9 d e (a \sin (c+d x)+a)^3}\)

input
Int[Sqrt[e*Cos[c + d*x]]/(a + a*Sin[c + d*x])^3,x]
 
output
(-2*(e*Cos[c + d*x])^(3/2))/(9*d*e*(a + a*Sin[c + d*x])^3) + ((-2*(e*Cos[c 
 + d*x])^(3/2))/(5*d*e*(a + a*Sin[c + d*x])^2) + ((-2*Sqrt[e*Cos[c + d*x]] 
*EllipticE[(c + d*x)/2, 2])/(a*d*Sqrt[Cos[c + d*x]]) - (2*(e*Cos[c + d*x]) 
^(3/2))/(d*e*(a + a*Sin[c + d*x])))/(5*a))/(3*a)
 

3.3.60.3.1 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 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]
 
3.3.60.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(511\) vs. \(2(161)=322\).

Time = 7.74 (sec) , antiderivative size = 512, normalized size of antiderivative = 3.35

method result size
default \(\frac {2 \left (96 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-48 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-192 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+96 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+152 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-72 \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 ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-56 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+24 \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 ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+36 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+48 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \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 )-36 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-11 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}{45 \left (16 \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-32 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) \(512\)

input
int((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 
output
2/45/(16*sin(1/2*d*x+1/2*c)^8-32*sin(1/2*d*x+1/2*c)^6+24*sin(1/2*d*x+1/2*c 
)^4-8*sin(1/2*d*x+1/2*c)^2+1)/a^3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c 
)^2*e+e)^(1/2)*(96*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-48*EllipticE(c 
os(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1 
/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^8-192*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2 
*c)^8+96*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^ 
(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^6+152*sin(1/2*d*x+1/ 
2*c)^6*cos(1/2*d*x+1/2*c)-72*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(co 
s(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^ 
4-56*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+24*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)* 
sin(1/2*d*x+1/2*c)^2+36*sin(1/2*d*x+1/2*c)^5+48*sin(1/2*d*x+1/2*c)^2*cos(1 
/2*d*x+1/2*c)-3*(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))-36*sin(1/2*d*x+1/2*c)^3-11*sin(1 
/2*d*x+1/2*c))*e/d
 
3.3.60.5 Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.43 \[ \int \frac {\sqrt {e \cos (c+d x)}}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - 4 i \, \sqrt {2}\right )} \sin \left (d x + c\right ) - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - 4 i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + 4 i \, \sqrt {2}\right )} \sin \left (d x + c\right ) + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + 4 i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, \cos \left (d x + c\right )^{3} - 6 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} + 9 \, \cos \left (d x + c\right ) - 5\right )} \sin \left (d x + c\right ) - 14 \, \cos \left (d x + c\right ) - 5\right )} \sqrt {e \cos \left (d x + c\right )}}{45 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

input
integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")
 
output
-1/45*(3*(I*sqrt(2)*cos(d*x + c)^3 + 3*I*sqrt(2)*cos(d*x + c)^2 + (I*sqrt( 
2)*cos(d*x + c)^2 - 2*I*sqrt(2)*cos(d*x + c) - 4*I*sqrt(2))*sin(d*x + c) - 
 2*I*sqrt(2)*cos(d*x + c) - 4*I*sqrt(2))*sqrt(e)*weierstrassZeta(-4, 0, we 
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(-I*sqrt(2)*c 
os(d*x + c)^3 - 3*I*sqrt(2)*cos(d*x + c)^2 + (-I*sqrt(2)*cos(d*x + c)^2 + 
2*I*sqrt(2)*cos(d*x + c) + 4*I*sqrt(2))*sin(d*x + c) + 2*I*sqrt(2)*cos(d*x 
 + c) + 4*I*sqrt(2))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4 
, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*cos(d*x + c)^3 - 6*cos(d*x + c 
)^2 - (3*cos(d*x + c)^2 + 9*cos(d*x + c) - 5)*sin(d*x + c) - 14*cos(d*x + 
c) - 5)*sqrt(e*cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c) 
^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos( 
d*x + c) - 4*a^3*d)*sin(d*x + c))
 
3.3.60.6 Sympy [F(-1)]

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

input
integrate((e*cos(d*x+c))**(1/2)/(a+a*sin(d*x+c))**3,x)
 
output
Timed out
 
3.3.60.7 Maxima [F]

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

input
integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")
 
output
integrate(sqrt(e*cos(d*x + c))/(a*sin(d*x + c) + a)^3, x)
 
3.3.60.8 Giac [F]

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

input
integrate((e*cos(d*x+c))^(1/2)/(a+a*sin(d*x+c))^3,x, algorithm="giac")
 
output
integrate(sqrt(e*cos(d*x + c))/(a*sin(d*x + c) + a)^3, x)
 
3.3.60.9 Mupad [F(-1)]

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

input
int((e*cos(c + d*x))^(1/2)/(a + a*sin(c + d*x))^3,x)
 
output
int((e*cos(c + d*x))^(1/2)/(a + a*sin(c + d*x))^3, x)