3.2.22 \(\int \frac {(e \sin (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx\) [122]

3.2.22.1 Optimal result

Integrand size = 25, antiderivative size = 102 \[ \int \frac {(e \sin (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=-\frac {4 e^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 a d \sqrt {e \sin (c+d x)}}+\frac {2 e \sqrt {e \sin (c+d x)}}{a d}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 a d} \]

output

4/3*e^2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*Elli 
pticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/a/d/(e*sin(d*x+c 
))^(1/2)+2*e*(e*sin(d*x+c))^(1/2)/a/d-2/3*e*cos(d*x+c)*(e*sin(d*x+c))^(1/2 
)/a/d
 

3.2.22.2 Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.68 \[ \int \frac {(e \sin (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=-\frac {2 \left (-2 \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )+(-3+\cos (c+d x)) \sqrt {\sin (c+d x)}\right ) (e \sin (c+d x))^{3/2}}{3 a d \sin ^{\frac {3}{2}}(c+d x)} \]

input

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

output

(-2*(-2*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + (-3 + Cos[c + d*x])*Sqrt[Sin 
[c + d*x]])*(e*Sin[c + d*x])^(3/2))/(3*a*d*Sin[c + d*x]^(3/2))
 

3.2.22.3 Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4360, 25, 25, 3042, 3318, 3042, 3044, 15, 3049, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3044

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 3049

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 e \sqrt {e \sin (c+d x)}}{a d}-\frac {e^2 \left (\frac {2 \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d e}+\frac {4 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d \sqrt {e \sin (c+d x)}}\right )}{a}\)

input

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

output

(2*e*Sqrt[e*Sin[c + d*x]])/(a*d) - (e^2*((4*EllipticF[(c - Pi/2 + d*x)/2, 
2]*Sqrt[Sin[c + d*x]])/(3*d*Sqrt[e*Sin[c + d*x]]) + (2*Cos[c + d*x]*Sqrt[e 
*Sin[c + d*x]])/(3*d*e)))/a
 

3.2.22.3.1 Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 25

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

rule 3042

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

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3049

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

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 3318

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

rule 4360

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 

3.2.22.4 Maple [A] (verified)

Time = 3.80 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {2 e^{2} \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{3 a \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(112\)

input

int((e*sin(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
 

output

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

3.2.22.5 Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.86 \[ \int \frac {(e \sin (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx=-\frac {2 \, {\left (\sqrt {2} \sqrt {-i \, e} e {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} \sqrt {i \, e} e {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (e \cos \left (d x + c\right ) - 3 \, e\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{3 \, a d} \]

input

integrate((e*sin(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")
 

output

-2/3*(sqrt(2)*sqrt(-I*e)*e*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin( 
d*x + c)) + sqrt(2)*sqrt(I*e)*e*weierstrassPInverse(4, 0, cos(d*x + c) - I 
*sin(d*x + c)) + (e*cos(d*x + c) - 3*e)*sqrt(e*sin(d*x + c)))/(a*d)
 

3.2.22.6 Sympy [F]

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

input

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

output

Integral((e*sin(c + d*x))**(3/2)/(sec(c + d*x) + 1), x)/a
 

3.2.22.7 Maxima [F]

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

input

integrate((e*sin(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")
 

output

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

3.2.22.8 Giac [F]

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

input

integrate((e*sin(d*x+c))^(3/2)/(a+a*sec(d*x+c)),x, algorithm="giac")
 

output

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

3.2.22.9 Mupad [F(-1)]

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

input

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

output

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