\(\int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{3/2}} \, dx\) [209]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 25, antiderivative size = 85 \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{3/2}} \, dx=-\frac {6 a^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt {\cos (c+d x)}}+\frac {4 a^4 (e \cos (c+d x))^{3/2}}{d e^3 \left (a^2-a^2 \sin (c+d x)\right )} \] Output:

-6*a^2*(e*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/e^2/co 
s(d*x+c)^(1/2)+4*a^4*(e*cos(d*x+c))^(3/2)/d/e^3/(a^2-a^2*sin(d*x+c))
 

Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75 \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{3/2}} \, dx=\frac {4\ 2^{3/4} a^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{4},\frac {3}{4},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [4]{1+\sin (c+d x)}}{d e \sqrt {e \cos (c+d x)}} \] Input:

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

Output:

(4*2^(3/4)*a^2*Hypergeometric2F1[-3/4, -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]])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3149, 3042, 3159, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3149

\(\displaystyle \frac {a^4 \int \frac {(e \cos (c+d x))^{5/2}}{(a-a \sin (c+d x))^2}dx}{e^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^4 \int \frac {(e \cos (c+d x))^{5/2}}{(a-a \sin (c+d x))^2}dx}{e^4}\)

\(\Big \downarrow \) 3159

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {a^4 \left (\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {6 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{a^2 d \sqrt {\cos (c+d x)}}\right )}{e^4}\)

Input:

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

Output:

(a^4*((-6*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(a^2*d*Sqrt[ 
Cos[c + d*x]]) + (4*e*(e*Cos[c + d*x])^(3/2))/(d*(a^2 - a^2*Sin[c + d*x])) 
))/e^4
 

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 3149
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(a/g)^(2*m)   Int[(g*Cos[e + f*x])^(2*m + p)/( 
a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2 
, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]
 

rule 3159
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f 
*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 
)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; 
FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & 
& NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
 
Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.41

method result size
default \(\frac {2 \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{e \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}\) \(120\)
parts \(-\frac {2 a^{2} \left (-2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} e +\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} e +\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{e \sqrt {-e \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\, d}-\frac {4 a^{2} \left (-\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} e +\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} e +\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{e \sqrt {-e \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\, d}+\frac {4 a^{2}}{d \sqrt {e \cos \left (d x +c \right )}\, e}\) \(423\)

Input:

int((a+a*sin(d*x+c))^2/(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)*(4*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))+2*sin(1/2*d*x+1/2* 
c))*a^2/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.16 \[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{3/2}} \, dx=\frac {2 \, {\left (3 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{2} \cos \left (d x + c\right ) + i \, a^{2} \sin \left (d x + c\right ) - i \, a^{2}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {\frac {1}{2}} {\left (i \, a^{2} \cos \left (d x + c\right ) - i \, a^{2} \sin \left (d x + c\right ) + i \, a^{2}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2} \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {e \cos \left (d x + c\right )}\right )}}{d e^{2} \cos \left (d x + c\right ) - d e^{2} \sin \left (d x + c\right ) + d e^{2}} \] Input:

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

Output:

2*(3*sqrt(1/2)*(-I*a^2*cos(d*x + c) + I*a^2*sin(d*x + c) - I*a^2)*sqrt(e)* 
weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x 
 + c))) + 3*sqrt(1/2)*(I*a^2*cos(d*x + c) - I*a^2*sin(d*x + c) + I*a^2)*sq 
rt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*s 
in(d*x + c))) + 2*(a^2*cos(d*x + c) + a^2*sin(d*x + c) + a^2)*sqrt(e*cos(d 
*x + c)))/(d*e^2*cos(d*x + c) - d*e^2*sin(d*x + c) + d*e^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {e}\, a^{2} \left (\cos \left (d x +c \right ) \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) d +\cos \left (d x +c \right ) \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2}}d x \right ) d +4 \sqrt {\cos \left (d x +c \right )}\right )}{\cos \left (d x +c \right ) d \,e^{2}} \] Input:

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

Output:

(sqrt(e)*a**2*(cos(c + d*x)*int(sqrt(cos(c + d*x))/cos(c + d*x)**2,x)*d + 
cos(c + d*x)*int((sqrt(cos(c + d*x))*sin(c + d*x)**2)/cos(c + d*x)**2,x)*d 
 + 4*sqrt(cos(c + d*x))))/(cos(c + d*x)*d*e**2)