\(\int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx\) [47]

Optimal result

Integrand size = 25, antiderivative size = 124 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 \left (a^2-2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 a b \sqrt {e \sin (c+d x)}}{3 d e^3} \] Output:

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

Mathematica [A] (verified)

Time = 1.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.61 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=-\frac {2 \left (2 a b+\left (a^2+b^2\right ) \cos (c+d x)+\left (a^2-2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{3 d e (e \sin (c+d x))^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3170, 27, 3042, 3148, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

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 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])
 

rule 3170

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

Maple [A] (verified)

Time = 3.22 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.63

method	result	size
default	\(\frac {-\frac {4 a b}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}-2 b^{2} \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+2 a^{2} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+2 b^{2} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(202\)
parts	\(-\frac {a^{2} \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 b^{2} \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {5}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+\sin \left (d x +c \right )^{3}-\sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {4 a b}{3 \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}} e d}\)	\(234\)

Input:

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

Output:

(-4/3*a*b/e/(e*sin(d*x+c))^(3/2)-1/3/e^2*((1-sin(d*x+c))^(1/2)*(2+2*sin(d* 
x+c))^(1/2)*sin(d*x+c)^(5/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a 
^2-2*b^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(5/2)*Elli 
pticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))+2*a^2*cos(d*x+c)^2*sin(d*x+c)+2*b^ 
2*cos(d*x+c)^2*sin(d*x+c))/sin(d*x+c)^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.30 \[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\frac {2 \, {\left ({\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \sqrt {-\frac {1}{2} i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \sqrt {\frac {1}{2} i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (2 \, a b + {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{3 \, {\left (d e^{3} \cos \left (d x + c\right )^{2} - d e^{3}\right )}} \] Input:

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

Output:

2/3*(((a^2 - 2*b^2)*cos(d*x + c)^2 - a^2 + 2*b^2)*sqrt(-1/2*I*e)*weierstra 
ssPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + ((a^2 - 2*b^2)*cos(d*x + 
 c)^2 - a^2 + 2*b^2)*sqrt(1/2*I*e)*weierstrassPInverse(4, 0, cos(d*x + c) 
- I*sin(d*x + c)) + (2*a*b + (a^2 + b^2)*cos(d*x + c))*sqrt(e*sin(d*x + c) 
))/(d*e^3*cos(d*x + c)^2 - d*e^3)
 

Sympy [F]

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

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

Output:

Integral((a + b*cos(c + d*x))**2/(e*sin(c + d*x))**(5/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {e}\, \left (-4 \sqrt {\sin \left (d x +c \right )}\, a b +3 \left (\int \frac {\sqrt {\sin \left (d x +c \right )}}{\sin \left (d x +c \right )^{3}}d x \right ) \sin \left (d x +c \right )^{2} a^{2} d +3 \left (\int \frac {\sqrt {\sin \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\sin \left (d x +c \right )^{3}}d x \right ) \sin \left (d x +c \right )^{2} b^{2} d \right )}{3 \sin \left (d x +c \right )^{2} d \,e^{3}} \] Input:

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

Output:

(sqrt(e)*( - 4*sqrt(sin(c + d*x))*a*b + 3*int(sqrt(sin(c + d*x))/sin(c + d 
*x)**3,x)*sin(c + d*x)**2*a**2*d + 3*int((sqrt(sin(c + d*x))*cos(c + d*x)* 
*2)/sin(c + d*x)**3,x)*sin(c + d*x)**2*b**2*d))/(3*sin(c + d*x)**2*d*e**3)