\(\int \frac {\sqrt {d \sin (e+f x)}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\) [1434]

Optimal result

Integrand size = 37, antiderivative size = 374 \[ \int \frac {\sqrt {d \sin (e+f x)}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=-\frac {2 \sqrt {2} a b d \operatorname {EllipticPi}\left (-\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{(-a+b)^{3/2} (a+b)^{3/2} f g^{3/2} \sqrt {d \sin (e+f x)}}+\frac {2 \sqrt {2} a b d \operatorname {EllipticPi}\left (\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{(-a+b)^{3/2} (a+b)^{3/2} f g^{3/2} \sqrt {d \sin (e+f x)}}-\frac {2 b \sqrt {d \sin (e+f x)}}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}}+\frac {2 a (d \sin (e+f x))^{3/2}}{\left (a^2-b^2\right ) d f g \sqrt {g \cos (e+f x)}}-\frac {2 a \sqrt {g \cos (e+f x)} E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {d \sin (e+f x)}}{\left (a^2-b^2\right ) f g^2 \sqrt {\sin (2 e+2 f x)}} \] Output:

-2*2^(1/2)*a*b*d*EllipticPi((g*cos(f*x+e))^(1/2)/g^(1/2)/(1+sin(f*x+e))^(1 
/2),-(-a+b)^(1/2)/(a+b)^(1/2),I)*sin(f*x+e)^(1/2)/(-a+b)^(3/2)/(a+b)^(3/2) 
/f/g^(3/2)/(d*sin(f*x+e))^(1/2)+2*2^(1/2)*a*b*d*EllipticPi((g*cos(f*x+e))^ 
(1/2)/g^(1/2)/(1+sin(f*x+e))^(1/2),(-a+b)^(1/2)/(a+b)^(1/2),I)*sin(f*x+e)^ 
(1/2)/(-a+b)^(3/2)/(a+b)^(3/2)/f/g^(3/2)/(d*sin(f*x+e))^(1/2)-2*b*(d*sin(f 
*x+e))^(1/2)/(a^2-b^2)/f/g/(g*cos(f*x+e))^(1/2)+2*a*(d*sin(f*x+e))^(3/2)/( 
a^2-b^2)/d/f/g/(g*cos(f*x+e))^(1/2)+2*a*(g*cos(f*x+e))^(1/2)*EllipticE(cos 
(e+1/4*Pi+f*x),2^(1/2))*(d*sin(f*x+e))^(1/2)/(a^2-b^2)/f/g^2/sin(2*f*x+2*e 
)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 25.53 (sec) , antiderivative size = 1274, normalized size of antiderivative = 3.41 \[ \int \frac {\sqrt {d \sin (e+f x)}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

Integrate[Sqrt[d*Sin[e + f*x]]/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x] 
)),x]
 

Output:

(2*Cos[e + f*x]*Sqrt[d*Sin[e + f*x]]*(-b + a*Sin[e + f*x]))/((a^2 - b^2)*f 
*(g*Cos[e + f*x])^(3/2)) + (a*Cos[e + f*x]^(3/2)*Sqrt[d*Sin[e + f*x]]*((4* 
a*(-(b*AppellF1[3/4, -1/4, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(- 
a^2 + b^2)]) + a*AppellF1[3/4, 1/4, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f 
*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^(3/2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])* 
Sin[e + f*x]^(3/2))/(3*(a^2 - b^2)*(1 - Cos[e + f*x]^2)^(3/4)*(a + b*Sin[e 
 + f*x])) + (Cos[2*(e + f*x)]*Sqrt[Tan[e + f*x]]*(b*Tan[e + f*x] + a*Sqrt[ 
1 + Tan[e + f*x]^2])*(56*b*(-3*a^2 + b^2)*AppellF1[3/4, 1/2, 1, 7/4, -Tan[ 
e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x]^(3/2) + 24*b*(-a^2 
 + b^2)*AppellF1[7/4, 1/2, 1, 11/4, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e 
+ f*x]^2]*Tan[e + f*x]^(7/2) + 21*a^(3/2)*(4*Sqrt[2]*a^(3/2)*ArcTan[1 - Sq 
rt[2]*Sqrt[Tan[e + f*x]]] - 4*Sqrt[2]*a^(3/2)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[ 
e + f*x]]] - (4*Sqrt[2]*a^2*ArcTan[1 - (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan 
[e + f*x]])/Sqrt[a]])/(a^2 - b^2)^(1/4) + (2*Sqrt[2]*b^2*ArcTan[1 - (Sqrt[ 
2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]])/(a^2 - b^2)^(1/4) + (4* 
Sqrt[2]*a^2*ArcTan[1 + (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt 
[a]])/(a^2 - b^2)^(1/4) - (2*Sqrt[2]*b^2*ArcTan[1 + (Sqrt[2]*(a^2 - b^2)^( 
1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]])/(a^2 - b^2)^(1/4) + 2*Sqrt[2]*a^(3/2)*L 
og[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - 2*Sqrt[2]*a^(3/2)*Log[ 
1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - (2*Sqrt[2]*a^2*Log[-a ...
 

Rubi [A] (verified)

Time = 2.22 (sec) , antiderivative size = 352, normalized size of antiderivative = 0.94, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.432, Rules used = {3042, 3382, 3042, 3317, 3042, 3043, 3051, 3042, 3052, 3042, 3119, 3385, 3042, 3384, 993, 1542}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3382

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3317

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3043

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

\(\Big \downarrow \) 3051

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {a b d \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {d \left (\frac {2 b \sqrt {d \sin (e+f x)}}{d f g \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{d}\right )}{a^2-b^2}\)

\(\Big \downarrow \) 3385

\(\displaystyle \frac {a b d \sqrt {\sin (e+f x)} \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}-\frac {d \left (\frac {2 b \sqrt {d \sin (e+f x)}}{d f g \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{d}\right )}{a^2-b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a b d \sqrt {\sin (e+f x)} \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}-\frac {d \left (\frac {2 b \sqrt {d \sin (e+f x)}}{d f g \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{d}\right )}{a^2-b^2}\)

\(\Big \downarrow \) 3384

\(\displaystyle -\frac {4 \sqrt {2} a b d \sqrt {\sin (e+f x)} \int \frac {g \cos (e+f x)}{(\sin (e+f x)+1) \sqrt {1-\frac {\cos ^2(e+f x)}{(\sin (e+f x)+1)^2}} \left ((a+b) g^2+\frac {(a-b) \cos ^2(e+f x) g^2}{(\sin (e+f x)+1)^2}\right )}d\frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)+1}}}{f g \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}-\frac {d \left (\frac {2 b \sqrt {d \sin (e+f x)}}{d f g \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{d}\right )}{a^2-b^2}\)

\(\Big \downarrow \) 993

\(\displaystyle -\frac {4 \sqrt {2} a b d \sqrt {\sin (e+f x)} \left (\frac {\int \frac {1}{\sqrt {1-\frac {\cos ^2(e+f x)}{(\sin (e+f x)+1)^2}} \left (\sqrt {a+b} g-\frac {\sqrt {b-a} g \cos (e+f x)}{\sin (e+f x)+1}\right )}d\frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)+1}}}{2 \sqrt {b-a}}-\frac {\int \frac {1}{\sqrt {1-\frac {\cos ^2(e+f x)}{(\sin (e+f x)+1)^2}} \left (\sqrt {a+b} g+\frac {\sqrt {b-a} \cos (e+f x) g}{\sin (e+f x)+1}\right )}d\frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)+1}}}{2 \sqrt {b-a}}\right )}{f g \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}-\frac {d \left (\frac {2 b \sqrt {d \sin (e+f x)}}{d f g \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{d}\right )}{a^2-b^2}\)

\(\Big \downarrow \) 1542

\(\displaystyle -\frac {4 \sqrt {2} a b d \sqrt {\sin (e+f x)} \left (\frac {\operatorname {EllipticPi}\left (\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right ),-1\right )}{2 \sqrt {g} \sqrt {b-a} \sqrt {a+b}}-\frac {\operatorname {EllipticPi}\left (-\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right ),-1\right )}{2 \sqrt {g} \sqrt {b-a} \sqrt {a+b}}\right )}{f g \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}-\frac {d \left (\frac {2 b \sqrt {d \sin (e+f x)}}{d f g \sqrt {g \cos (e+f x)}}-\frac {a \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{d}\right )}{a^2-b^2}\)

Input:

Int[Sqrt[d*Sin[e + f*x]]/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]
 

Output:

(-4*Sqrt[2]*a*b*d*(-1/2*EllipticPi[-(Sqrt[-a + b]/Sqrt[a + b]), ArcSin[Sqr 
t[g*Cos[e + f*x]]/(Sqrt[g]*Sqrt[1 + Sin[e + f*x]])], -1]/(Sqrt[-a + b]*Sqr 
t[a + b]*Sqrt[g]) + EllipticPi[Sqrt[-a + b]/Sqrt[a + b], ArcSin[Sqrt[g*Cos 
[e + f*x]]/(Sqrt[g]*Sqrt[1 + Sin[e + f*x]])], -1]/(2*Sqrt[-a + b]*Sqrt[a + 
 b]*Sqrt[g]))*Sqrt[Sin[e + f*x]])/((a^2 - b^2)*f*g*Sqrt[d*Sin[e + f*x]]) - 
 (d*((2*b*Sqrt[d*Sin[e + f*x]])/(d*f*g*Sqrt[g*Cos[e + f*x]]) - (a*((2*(d*S 
in[e + f*x])^(3/2))/(d*f*g*Sqrt[g*Cos[e + f*x]]) - (2*Sqrt[g*Cos[e + f*x]] 
*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Sin[e + f*x]])/(f*g^2*Sqrt[Sin[2*e + 
2*f*x]])))/d))/(a^2 - b^2)
 

Defintions of rubi rules used

rule 993

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> 
With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2* 
b)   Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Simp[s/(2*b)   Int[1/((r 
 - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
a*d, 0]
 

rule 1542

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ 
{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x 
], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]
 

rule 3042

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

rule 3043

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^( 
m_.), x_Symbol] :> Simp[(a*Sin[e + f*x])^(m + 1)*((b*Cos[e + f*x])^(n + 1)/ 
(a*b*f*(m + 1))), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2, 0] & 
& NeQ[m, -1]
 

rule 3051

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

rule 3052

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] 
, x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e 
 + 2*f*x]])   Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, 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 3317

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a   Int[(g*Co 
s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d   Int[(g*Cos[e + f*x])^ 
p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
 

rule 3382

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

rule 3384

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[sin[(e_.) + (f_.)*(x_)]]*((a_ 
) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[-4*Sqrt[2]*(g/f)   S 
ubst[Int[x^2/(((a + b)*g^2 + (a - b)*x^4)*Sqrt[1 - x^4/g^2]), x], x, Sqrt[g 
*Cos[e + f*x]]/Sqrt[1 + Sin[e + f*x]]], x] /; FreeQ[{a, b, e, f, g}, x] && 
NeQ[a^2 - b^2, 0]
 

rule 3385

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1137\) vs. \(2(317)=634\).

Time = 4.08 (sec) , antiderivative size = 1138, normalized size of antiderivative = 3.04

method	result	size
default	\(\text {Expression too large to display}\)	\(1138\)

Input:

int((d*sin(f*x+e))^(1/2)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x,method=_R 
ETURNVERBOSE)
 

Output:

1/f*csc(f*x+e)*((-1-cos(f*x+e))*(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*(-2*csc(f* 
x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*EllipticPi((csc( 
f*x+e)-cot(f*x+e)+1)^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*a*b+(1+c 
os(f*x+e))*(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^ 
(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*EllipticPi((csc(f*x+e)-cot(f*x+e)+1)^ 
(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*a*b-4*(csc(f*x+e)-cot(f*x+e)+ 
1)^(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/ 
2)*(1+cos(f*x+e))*a*EllipticE((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^(1/2)) 
*(-a^2+b^2)^(1/2)+2*(-csc(f*x+e)+cot(f*x+e))^(1/2)*(csc(f*x+e)-cot(f*x+e)+ 
1)^(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(1+cos(f*x+e))*a*EllipticF(( 
csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)+4*(-a^2+b^2)^ 
(1/2)*a*cos(f*x+e)+(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*(-2*csc(f*x+e)+2*cot(f* 
x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*(-1-cos(f*x+e))*b*EllipticPi( 
(csc(f*x+e)-cot(f*x+e)+1)^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-a 
^2+b^2)^(1/2)+(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+ 
2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*(-1-cos(f*x+e))*b*EllipticPi((csc( 
f*x+e)-cot(f*x+e)+1)^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*(-a^2+b^ 
2)^(1/2)+2*(-csc(f*x+e)+cot(f*x+e))^(1/2)*(csc(f*x+e)-cot(f*x+e)+1)^(1/2)* 
(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(1+cos(f*x+e))*b*EllipticF((csc(f*x+e 
)-cot(f*x+e)+1)^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)+4*sin(f*x+e)*(-a^2+...
 

Fricas [F(-1)]

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

integrate((d*sin(f*x+e))^(1/2)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, al 
gorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate((d*sin(f*x+e))**(1/2)/(g*cos(f*x+e))**(3/2)/(a+b*sin(f*x+e)),x)
                                                                                    
                                                                                    
 

Output:

Integral(sqrt(d*sin(e + f*x))/((g*cos(e + f*x))**(3/2)*(a + b*sin(e + f*x) 
)), x)
 

Maxima [F]

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

integrate((d*sin(f*x+e))^(1/2)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, al 
gorithm="maxima")
 

Output:

integrate(sqrt(d*sin(f*x + e))/((g*cos(f*x + e))^(3/2)*(b*sin(f*x + e) + a 
)), x)
 

Giac [F]

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

integrate((d*sin(f*x+e))^(1/2)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, al 
gorithm="giac")
 

Output:

integrate(sqrt(d*sin(f*x + e))/((g*cos(f*x + e))^(3/2)*(b*sin(f*x + e) + a 
)), x)
 

Mupad [F(-1)]

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

int((d*sin(e + f*x))^(1/2)/((g*cos(e + f*x))^(3/2)*(a + b*sin(e + f*x))),x 
)
 

Output:

int((d*sin(e + f*x))^(1/2)/((g*cos(e + f*x))^(3/2)*(a + b*sin(e + f*x))), 
x)
 

Reduce [F]

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

int((d*sin(f*x+e))^(1/2)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x)
 

Output:

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