\(\int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx\) [115]

Optimal result

Integrand size = 25, antiderivative size = 192 \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\frac {2 a^2 e^{3/2} \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 e^{3/2} \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}-\frac {a^2 e^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d \sqrt {e \sin (c+d x)}}-\frac {4 a^2 e \sqrt {e \sin (c+d x)}}{d}-\frac {2 a^2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}+\frac {a^2 e \sec (c+d x) \sqrt {e \sin (c+d x)}}{d} \] Output:

2*a^2*e^(3/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d+2*a^2*e^(3/2)*arctanh 
((e*sin(d*x+c))^(1/2)/e^(1/2))/d-1/3*a^2*e^2*InverseJacobiAM(1/2*c-1/4*Pi+ 
1/2*d*x,2^(1/2))*sin(d*x+c)^(1/2)/d/(e*sin(d*x+c))^(1/2)-4*a^2*e*(e*sin(d* 
x+c))^(1/2)/d-2/3*a^2*e*cos(d*x+c)*(e*sin(d*x+c))^(1/2)/d+a^2*e*sec(d*x+c) 
*(e*sin(d*x+c))^(1/2)/d
 

Mathematica [C] (warning: unable to verify)

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

Time = 26.37 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.06 \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\frac {16 a^2 e \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {e \sin (c+d x)} \left (6 \arctan \left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+6 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+\sqrt {\sin (c+d x)}-12 \sqrt {\cos ^2(c+d x)} \sqrt {\sin (c+d x)}-\sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\sin ^2(c+d x)\right ) \sqrt {\sin (c+d x)}+2 \sin ^{\frac {5}{2}}(c+d x)\right ) \sin ^4\left (\frac {1}{2} \arcsin (\sin (c+d x))\right )}{3 d \sin ^{\frac {9}{2}}(c+d x)} \] Input:

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

Output:

(16*a^2*e*Cos[(c + d*x)/2]^4*Sec[c + d*x]*Sqrt[e*Sin[c + d*x]]*(6*ArcTan[S 
qrt[Sin[c + d*x]]]*Sqrt[Cos[c + d*x]^2] + 6*ArcTanh[Sqrt[Sin[c + d*x]]]*Sq 
rt[Cos[c + d*x]^2] + Sqrt[Sin[c + d*x]] - 12*Sqrt[Cos[c + d*x]^2]*Sqrt[Sin 
[c + d*x]] - Sqrt[Cos[c + d*x]^2]*Hypergeometric2F1[1/4, 1/2, 5/4, Sin[c + 
 d*x]^2]*Sqrt[Sin[c + d*x]] + 2*Sin[c + d*x]^(5/2))*Sin[ArcSin[Sin[c + d*x 
]]/2]^4)/(3*d*Sin[c + d*x]^(9/2))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4360, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3352

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 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 = 0.33 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {a^{2} \left (12 \cos \left (d x +c \right ) e^{\frac {3}{2}} \sqrt {e \sin \left (d x +c \right )}\, \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+12 \cos \left (d x +c \right ) e^{\frac {3}{2}} \sqrt {e \sin \left (d x +c \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e^{2}-4 e^{2} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-24 e^{2} \sin \left (d x +c \right ) \cos \left (d x +c \right )+6 e^{2} \sin \left (d x +c \right )\right )}{6 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(201\)
parts	\(-\frac {a^{2} e^{2} \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {a^{2} e^{2} \sqrt {\cos \left (d x +c \right )^{2} e \sin \left (d x +c \right )}\, \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \sin \left (d x +c \right )\right )}{2 \sqrt {-e \sin \left (d x +c \right ) \left (\sin \left (d x +c \right )-1\right ) \left (\sin \left (d x +c \right )+1\right )}\, \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 a^{2} \left (e^{\frac {3}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+e^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )-2 e \sqrt {e \sin \left (d x +c \right )}\right )}{d}\)	\(295\)

Input:

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

Output:

1/6/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*a^2*(12*cos(d*x+c)*e^(3/2)*(e*sin(d*x+ 
c))^(1/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))+12*cos(d*x+c)*e^(3/2)*(e*si 
n(d*x+c))^(1/2)*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))+(-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))*e^2-4*e^2*cos(d*x+c)^2*sin(d*x+c)-24*e^2*sin(d*x+c)*cos(d*x+c 
)+6*e^2*sin(d*x+c))/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 709, normalized size of antiderivative = 3.69 \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/12*(6*a^2*sqrt(-e)*e*arctan(1/4*(cos(d*x + c)^2 - 6*sin(d*x + c) - 2)* 
sqrt(e*sin(d*x + c))*sqrt(-e)/(e*cos(d*x + c)^2 - e*sin(d*x + c) - e))*cos 
(d*x + c) - 3*a^2*sqrt(-e)*e*cos(d*x + c)*log((e*cos(d*x + c)^4 - 72*e*cos 
(d*x + c)^2 + 8*(7*cos(d*x + c)^2 - (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8) 
*sqrt(e*sin(d*x + c))*sqrt(-e) + 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) 
+ 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d* 
x + c) + 8)) + 4*a^2*sqrt(-1/2*I*e)*e*cos(d*x + c)*weierstrassPInverse(4, 
0, cos(d*x + c) + I*sin(d*x + c)) + 4*a^2*sqrt(1/2*I*e)*e*cos(d*x + c)*wei 
erstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 4*(2*a^2*e*cos(d*x 
 + c)^2 + 12*a^2*e*cos(d*x + c) - 3*a^2*e)*sqrt(e*sin(d*x + c)))/(d*cos(d* 
x + c)), 1/12*(6*a^2*e^(3/2)*arctan(1/4*(cos(d*x + c)^2 + 6*sin(d*x + c) - 
 2)*sqrt(e*sin(d*x + c))*sqrt(e)/(e*cos(d*x + c)^2 + e*sin(d*x + c) - e))* 
cos(d*x + c) + 3*a^2*e^(3/2)*cos(d*x + c)*log((e*cos(d*x + c)^4 - 72*e*cos 
(d*x + c)^2 - 8*(7*cos(d*x + c)^2 + (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8) 
*sqrt(e*sin(d*x + c))*sqrt(e) - 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 
 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2)*sin(d*x 
 + c) + 8)) - 4*a^2*sqrt(-1/2*I*e)*e*cos(d*x + c)*weierstrassPInverse(4, 0 
, cos(d*x + c) + I*sin(d*x + c)) - 4*a^2*sqrt(1/2*I*e)*e*cos(d*x + c)*weie 
rstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 4*(2*a^2*e*cos(d*x 
+ c)^2 + 12*a^2*e*cos(d*x + c) - 3*a^2*e)*sqrt(e*sin(d*x + c)))/(d*cos(...
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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