3.3.1 \(\int \frac {a+a \sin (c+d x)}{(e \cos (c+d x))^{3/2}} \, dx\) [201]

3.3.1.1 Optimal result
3.3.1.2 Mathematica [C] (verified)
3.3.1.3 Rubi [A] (verified)
3.3.1.4 Maple [A] (verified)
3.3.1.5 Fricas [C] (verification not implemented)
3.3.1.6 Sympy [F]
3.3.1.7 Maxima [F]
3.3.1.8 Giac [F]
3.3.1.9 Mupad [F(-1)]

3.3.1.1 Optimal result

Integrand size = 23, antiderivative size = 91 \[ \int \frac {a+a \sin (c+d x)}{(e \cos (c+d x))^{3/2}} \, dx=\frac {2 a}{d e \sqrt {e \cos (c+d x)}}-\frac {2 a \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt {\cos (c+d x)}}+\frac {2 a \sin (c+d x)}{d e \sqrt {e \cos (c+d x)}} \]

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

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

Time = 0.31 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.68 \[ \int \frac {a+a \sin (c+d x)}{(e \cos (c+d x))^{3/2}} \, dx=\frac {2\ 2^{3/4} a \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {3}{4},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [4]{1+\sin (c+d x)}}{d e \sqrt {e \cos (c+d x)}} \]

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

Time = 0.42 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3148, 3042, 3116, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

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

3.3.1.3.1 Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3116
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

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 3148
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + 
 Simp[a   Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && 
 (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
 
3.3.1.4 Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.27

method result size
default \(\frac {2 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{e \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}\) \(116\)
parts \(-\frac {2 a \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{e \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}+\frac {2 a}{d e \sqrt {e \cos \left (d x +c \right )}}\) \(219\)

input
int((a+a*sin(d*x+c))/(e*cos(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 
output
2/e/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/sin(1/2*d*x+1/2*c)*(2*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+sin(1/2*d*x+1/2*c))* 
a/d
 
3.3.1.5 Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.93 \[ \int \frac {a+a \sin (c+d x)}{(e \cos (c+d x))^{3/2}} \, dx=-\frac {{\left (i \, \sqrt {2} a \cos \left (d x + c\right ) - i \, \sqrt {2} a \sin \left (d x + c\right ) + i \, \sqrt {2} a\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (-i \, \sqrt {2} a \cos \left (d x + c\right ) + i \, \sqrt {2} a \sin \left (d x + c\right ) - i \, \sqrt {2} a\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \sqrt {e \cos \left (d x + c\right )}}{d e^{2} \cos \left (d x + c\right ) - d e^{2} \sin \left (d x + c\right ) + d e^{2}} \]

input
integrate((a+a*sin(d*x+c))/(e*cos(d*x+c))^(3/2),x, algorithm="fricas")
 
output
-((I*sqrt(2)*a*cos(d*x + c) - I*sqrt(2)*a*sin(d*x + c) + I*sqrt(2)*a)*sqrt 
(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin 
(d*x + c))) + (-I*sqrt(2)*a*cos(d*x + c) + I*sqrt(2)*a*sin(d*x + c) - I*sq 
rt(2)*a)*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
 + c) - I*sin(d*x + c))) - 2*(a*cos(d*x + c) + a*sin(d*x + c) + a)*sqrt(e* 
cos(d*x + c)))/(d*e^2*cos(d*x + c) - d*e^2*sin(d*x + c) + d*e^2)
 
3.3.1.6 Sympy [F]

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

input
integrate((a+a*sin(d*x+c))/(e*cos(d*x+c))**(3/2),x)
 
output
a*(Integral((e*cos(c + d*x))**(-3/2), x) + Integral(sin(c + d*x)/(e*cos(c 
+ d*x))**(3/2), x))
 
3.3.1.7 Maxima [F]

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

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

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

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

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

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