\(\int \frac {(e \sin (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx\) [128]

Optimal result

Integrand size = 25, antiderivative size = 187 \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {4 e^3}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos (c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {2 e^3 \cos ^3(c+d x)}{a^2 d \sqrt {e \sin (c+d x)}}-\frac {44 e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 a^2 d \sqrt {\sin (c+d x)}}+\frac {4 e (e \sin (c+d x))^{3/2}}{3 a^2 d}-\frac {12 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 a^2 d} \] Output:

4*e^3/a^2/d/(e*sin(d*x+c))^(1/2)-2*e^3*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(1/ 
2)-2*e^3*cos(d*x+c)^3/a^2/d/(e*sin(d*x+c))^(1/2)+44/5*e^2*EllipticE(cos(1/ 
2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/a^2/d/sin(d*x+c)^(1/2)+4 
/3*e*(e*sin(d*x+c))^(3/2)/a^2/d-12/5*e*cos(d*x+c)*(e*sin(d*x+c))^(3/2)/a^2 
/d
 

Mathematica [C] (warning: unable to verify)

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

Time = 7.07 (sec) , antiderivative size = 446, normalized size of antiderivative = 2.39 \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc ^2(c+d x) \sec ^2(c+d x) \left (\frac {16 \cos (d x) \sin (c)}{3 d}-\frac {16 \sec \left (\frac {c}{2}\right ) \sec (c) \left (8 \sin \left (\frac {c}{2}\right )+3 \sin \left (\frac {3 c}{2}\right )\right )}{5 d}-\frac {4 \cos (2 d x) \sin (2 c)}{5 d}+\frac {16 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}+\frac {16 \cos (c) \sin (d x)}{3 d}-\frac {4 \cos (2 c) \sin (2 d x)}{5 d}\right ) (e \sin (c+d x))^{5/2}}{(a+a \sec (c+d x))^2}-\frac {88 \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (c) \sec ^2(c+d x) (e \sin (c+d x))^{5/2} \left (-\frac {\cot (c) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x-\arctan (\cot (c)))\right ) \sin (d x-\arctan (\cot (c)))}{\sqrt {1-\cos (d x-\arctan (\cot (c)))} \sqrt {1+\cos (d x-\arctan (\cot (c)))} \sqrt {1+\cot ^2(c)} \sqrt {\cos (d x-\arctan (\cot (c))) \sqrt {1+\cot ^2(c)} \sin (c)}}-\frac {\frac {2 \cos (d x-\arctan (\cot (c))) \sqrt {1+\cot ^2(c)} \sin ^2(c)}{\cos ^2(c)+\sin ^2(c)}-\frac {\cot (c) \sin (d x-\arctan (\cot (c)))}{\sqrt {1+\cot ^2(c)}}}{\sqrt {\cos (d x-\arctan (\cot (c))) \sqrt {1+\cot ^2(c)} \sin (c)}}\right )}{5 d (a+a \sec (c+d x))^2 \sin ^{\frac {5}{2}}(c+d x)} \] Input:

Integrate[(e*Sin[c + d*x])^(5/2)/(a + a*Sec[c + d*x])^2,x]
 

Output:

(Cos[c/2 + (d*x)/2]^4*Csc[c + d*x]^2*Sec[c + d*x]^2*((16*Cos[d*x]*Sin[c])/ 
(3*d) - (16*Sec[c/2]*Sec[c]*(8*Sin[c/2] + 3*Sin[(3*c)/2]))/(5*d) - (4*Cos[ 
2*d*x]*Sin[2*c])/(5*d) + (16*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2])/d + 
 (16*Cos[c]*Sin[d*x])/(3*d) - (4*Cos[2*c]*Sin[2*d*x])/(5*d))*(e*Sin[c + d* 
x])^(5/2))/(a + a*Sec[c + d*x])^2 - (88*Cos[c/2 + (d*x)/2]^4*Sec[c]*Sec[c 
+ d*x]^2*(e*Sin[c + d*x])^(5/2)*(-((Cot[c]*HypergeometricPFQ[{-1/2, -1/4}, 
 {3/4}, Cos[d*x - ArcTan[Cot[c]]]^2]*Sin[d*x - ArcTan[Cot[c]]])/(Sqrt[1 - 
Cos[d*x - ArcTan[Cot[c]]]]*Sqrt[1 + Cos[d*x - ArcTan[Cot[c]]]]*Sqrt[1 + Co 
t[c]^2]*Sqrt[Cos[d*x - ArcTan[Cot[c]]]*Sqrt[1 + Cot[c]^2]*Sin[c]])) - ((2* 
Cos[d*x - ArcTan[Cot[c]]]*Sqrt[1 + Cot[c]^2]*Sin[c]^2)/(Cos[c]^2 + Sin[c]^ 
2) - (Cot[c]*Sin[d*x - ArcTan[Cot[c]]])/Sqrt[1 + Cot[c]^2])/Sqrt[Cos[d*x - 
 ArcTan[Cot[c]]]*Sqrt[1 + Cot[c]^2]*Sin[c]]))/(5*d*(a + a*Sec[c + d*x])^2* 
Sin[c + d*x]^(5/2))
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4360, 3042, 3354, 3042, 3352, 2009}

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.

\(\displaystyle \int \frac {(e \sin (c+d x))^{5/2}}{(a \sec (c+d x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3354

\(\displaystyle \frac {e^4 \int \frac {\cos ^2(c+d x) (a-a \cos (c+d x))^2}{(e \sin (c+d x))^{3/2}}dx}{a^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^4 \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx}{a^4}\)

\(\Big \downarrow \) 3352

\(\displaystyle \frac {e^4 \int \left (\frac {a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{3/2}}-\frac {2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{3/2}}+\frac {a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{3/2}}\right )dx}{a^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^4 \left (\frac {4 a^2 (e \sin (c+d x))^{3/2}}{3 d e^3}-\frac {12 a^2 \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d e^3}-\frac {44 a^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d e^2 \sqrt {\sin (c+d x)}}+\frac {4 a^2}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos ^3(c+d x)}{d e \sqrt {e \sin (c+d x)}}-\frac {2 a^2 \cos (c+d x)}{d e \sqrt {e \sin (c+d x)}}\right )}{a^4}\)

Input:

Int[(e*Sin[c + d*x])^(5/2)/(a + a*Sec[c + d*x])^2,x]
 

Output:

(e^4*((4*a^2)/(d*e*Sqrt[e*Sin[c + d*x]]) - (2*a^2*Cos[c + d*x])/(d*e*Sqrt[ 
e*Sin[c + d*x]]) - (2*a^2*Cos[c + d*x]^3)/(d*e*Sqrt[e*Sin[c + d*x]]) - (44 
*a^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*d*e^2*Sqrt[ 
Sin[c + d*x]]) + (4*a^2*(e*Sin[c + d*x])^(3/2))/(3*d*e^3) - (12*a^2*Cos[c 
+ d*x]*(e*Sin[c + d*x])^(3/2))/(5*d*e^3)))/a^4
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3352

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig 
[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F 
reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 3354

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

Maple [A] (verified)

Time = 1.86 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {2 e^{3} \left (66 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-33 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+3 \cos \left (d x +c \right )^{4}-10 \cos \left (d x +c \right )^{3}-33 \cos \left (d x +c \right )^{2}+40 \cos \left (d x +c \right )\right )}{15 a^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(173\)

Input:

int((e*sin(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

2/15/a^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*e^3*(66*(-sin(d*x+c)+1)^(1/2)*(2* 
sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2 
^(1/2))-33*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*E 
llipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+3*cos(d*x+c)^4-10*cos(d*x+c)^3 
-33*cos(d*x+c)^2+40*cos(d*x+c))/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.85 \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left ({\left (3 \, e^{2} \cos \left (d x + c\right )^{2} - 7 \, e^{2} \cos \left (d x + c\right ) - 40 \, e^{2}\right )} \sqrt {e \sin \left (d x + c\right )} \sin \left (d x + c\right ) + 66 \, {\left (i \, e^{2} \cos \left (d x + c\right ) + i \, e^{2}\right )} \sqrt {-\frac {1}{2} i \, e} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 66 \, {\left (-i \, e^{2} \cos \left (d x + c\right ) - i \, e^{2}\right )} \sqrt {\frac {1}{2} i \, e} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )\right )}}{15 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

integrate((e*sin(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
 

Output:

-2/15*((3*e^2*cos(d*x + c)^2 - 7*e^2*cos(d*x + c) - 40*e^2)*sqrt(e*sin(d*x 
 + c))*sin(d*x + c) + 66*(I*e^2*cos(d*x + c) + I*e^2)*sqrt(-1/2*I*e)*weier 
strassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) 
 + 66*(-I*e^2*cos(d*x + c) - I*e^2)*sqrt(1/2*I*e)*weierstrassZeta(4, 0, we 
ierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + 
c) + a^2*d)
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate((e*sin(d*x+c))**(5/2)/(a+a*sec(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*sin(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
 

Output:

integrate((e*sin(d*x + c))^(5/2)/(a*sec(d*x + c) + a)^2, x)
 

Giac [F]

\[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((e*sin(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
 

Output:

integrate((e*sin(d*x + c))^(5/2)/(a*sec(d*x + c) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \] Input:

int((e*sin(c + d*x))^(5/2)/(a + a/cos(c + d*x))^2,x)
 

Output:

int((cos(c + d*x)^2*(e*sin(c + d*x))^(5/2))/(a^2*(cos(c + d*x) + 1)^2), x)
 

Reduce [F]

\[ \int \frac {(e \sin (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) e^{2}}{a^{2}} \] Input:

int((e*sin(d*x+c))^(5/2)/(a+a*sec(d*x+c))^2,x)
 

Output:

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