\(\int \frac {(a+b \cos (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx\) [45]

Optimal result

Integrand size = 25, antiderivative size = 114 \[ \int \frac {(a+b \cos (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\frac {2 \left (3 a^2+2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d \sqrt {e \sin (c+d x)}}+\frac {10 a b \sqrt {e \sin (c+d x)}}{3 d e}+\frac {2 b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 d e} \] Output:

2/3*(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*sin(d*x+c))^(1/2)+10/3*a*b*(e*sin(d*x+c))^(1/2)/d/e+2/3*b*(a+b 
*cos(d*x+c))*(e*sin(d*x+c))^(1/2)/d/e
 

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b \cos (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\frac {-2 \left (3 a^2+2 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sqrt {\sin (c+d x)}+2 b (6 a+b \cos (c+d x)) \sin (c+d x)}{3 d \sqrt {e \sin (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 115, 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, 3171, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3171

\(\displaystyle \frac {2}{3} \int \frac {3 a^2+5 b \cos (c+d x) a+2 b^2}{2 \sqrt {e \sin (c+d x)}}dx+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {3 a^2+5 b \cos (c+d x) a+2 b^2}{\sqrt {e \sin (c+d x)}}dx+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {3 a^2-5 b \sin \left (c+d x-\frac {\pi }{2}\right ) a+2 b^2}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}dx+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e}\)

\(\Big \downarrow \) 3148

\(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}}dx+\frac {10 a b \sqrt {e \sin (c+d x)}}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\left (3 a^2+2 b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}}dx+\frac {10 a b \sqrt {e \sin (c+d x)}}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {1}{3} \left (\frac {\left (3 a^2+2 b^2\right ) \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}+\frac {10 a b \sqrt {e \sin (c+d x)}}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {\left (3 a^2+2 b^2\right ) \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}+\frac {10 a b \sqrt {e \sin (c+d x)}}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{3} \left (\frac {2 \left (3 a^2+2 b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}+\frac {10 a b \sqrt {e \sin (c+d x)}}{d e}\right )+\frac {2 b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e}\)

Input:

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

Output:

(2*b*(a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(3*d*e) + ((2*(3*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]]) + (10*a*b*Sqrt[e*Sin[c + d*x]])/(d*e))/3
 

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 3171

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

Maple [A] (verified)

Time = 2.74 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.49

method	result	size
default	\(-\frac {3 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}+2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) b^{2}-2 b^{2} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-12 a b \cos \left (d x +c \right ) \sin \left (d x +c \right )}{3 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(170\)
parts	\(-\frac {a^{2} \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {b^{2} \left (-\frac {2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )}{3}+\frac {2 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )}{3}\right )}{\cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {4 a b \sqrt {e \sin \left (d x +c \right )}}{d e}\)	\(192\)

Input:

int((a+cos(d*x+c)*b)^2/(e*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+ 
c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2 
+2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF( 
(1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2-2*b^2*cos(d*x+c)^2*sin(d*x+c)-12*a*b 
*cos(d*x+c)*sin(d*x+c))/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b \cos (c+d x))^2}{\sqrt {e \sin (c+d x)}} \, dx=\frac {2 \, {\left ({\left (3 \, a^{2} + 2 \, b^{2}\right )} \sqrt {-\frac {1}{2} i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (3 \, a^{2} + 2 \, b^{2}\right )} \sqrt {\frac {1}{2} i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (b^{2} \cos \left (d x + c\right ) + 6 \, a b\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{3 \, d e} \] Input:

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

2/3*((3*a^2 + 2*b^2)*sqrt(-1/2*I*e)*weierstrassPInverse(4, 0, cos(d*x + c) 
 + I*sin(d*x + c)) + (3*a^2 + 2*b^2)*sqrt(1/2*I*e)*weierstrassPInverse(4, 
0, cos(d*x + c) - I*sin(d*x + c)) + (b^2*cos(d*x + c) + 6*a*b)*sqrt(e*sin( 
d*x + c)))/(d*e)
 

Sympy [F]

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

integrate((a+b*cos(d*x+c))**2/(e*sin(d*x+c))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((a+b*cos(d*x+c))^2/(e*sin(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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