\(\int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx\) [1374]

Optimal result

Integrand size = 33, antiderivative size = 514 \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a^2 \sqrt [4]{-a^2+b^2} g^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}-\frac {a^2 \sqrt [4]{-a^2+b^2} g^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}+\frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g}+\frac {2 a^3 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^4 f \sqrt {g \cos (e+f x)}}-\frac {2 a g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}-\frac {a^3 \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^3 \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 a g \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f} \] Output:

-a^2*(-a^2+b^2)^(1/4)*g^(3/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^ 
2)^(1/4)/g^(1/2))/b^(7/2)/f-a^2*(-a^2+b^2)^(1/4)*g^(3/2)*arctanh(b^(1/2)*( 
g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(7/2)/f+2*a^2*g*(g*cos(f*x 
+e))^(1/2)/b^3/f-2/5*(g*cos(f*x+e))^(5/2)/b/f/g+2*a^3*g^2*cos(f*x+e)^(1/2) 
*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))/b^4/f/(g*cos(f*x+e))^(1/2)-2/3*a*g 
^2*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))/b^2/f/(g*cos(f* 
x+e))^(1/2)-a^3*(a^2-b^2)*g^2*cos(f*x+e)^(1/2)*EllipticPi(sin(1/2*f*x+1/2* 
e),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))/b^4/(a^2-b*(b-(-a^2+b^2)^(1/2)))/f/(g 
*cos(f*x+e))^(1/2)-a^3*(a^2-b^2)*g^2*cos(f*x+e)^(1/2)*EllipticPi(sin(1/2*f 
*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))/b^4/(a^2-b*(b+(-a^2+b^2)^(1/2) 
))/f/(g*cos(f*x+e))^(1/2)-2/3*a*g*(g*cos(f*x+e))^(1/2)*sin(f*x+e)/b^2/f
 

Mathematica [C] (warning: unable to verify)

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

Time = 27.23 (sec) , antiderivative size = 1953, normalized size of antiderivative = 3.80 \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

((g*Cos[e + f*x])^(3/2)*Sec[e + f*x]*(-1/5*Cos[2*(e + f*x)]/b - (2*a*Sin[e 
 + f*x])/(3*b^2)))/f + ((g*Cos[e + f*x])^(3/2)*((-2*(10*a^2 + 3*b^2)*(a + 
b*Sqrt[1 - Cos[e + f*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, C 
os[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/(Sqr 
t[1 - Cos[e + f*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f* 
x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 
9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*App 
ellF1[5/4, 3/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] 
)*Cos[e + f*x]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*Sqrt[b 
]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 
 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] + L 
og[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x] 
] + I*b*Cos[e + f*x]] - Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2 
)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]]))/(-a^2 + b^2)^(3/4))*Sin[e 
 + f*x])/(Sqrt[1 - Cos[e + f*x]^2]*(a + b*Sin[e + f*x])) + ((30*a^2 - 3*b^ 
2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e + f*x)]*(((1/2 - I/2)*(-2*a^2 
 + b^2)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4) 
])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 + ( 
(1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + 
b^2)^(3/4)) + (4*Sqrt[Cos[e + f*x]])/b - (4*a*AppellF1[5/4, 1/2, 1, 9/4...
 

Rubi [A] (verified)

Time = 1.46 (sec) , antiderivative size = 514, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3042, 3377, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3377

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 a^3 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^4 f \sqrt {g \cos (e+f x)}}-\frac {a^2 g^{3/2} \sqrt [4]{b^2-a^2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}-\frac {a^2 g^{3/2} \sqrt [4]{b^2-a^2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}+\frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {a^3 g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^4 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^4 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {2 a g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}-\frac {2 a g \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 b^2 f}-\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g}\)

Input:

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

Output:

-((a^2*(-a^2 + b^2)^(1/4)*g^(3/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/(( 
-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(7/2)*f)) - (a^2*(-a^2 + b^2)^(1/4)*g^(3/2 
)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b 
^(7/2)*f) + (2*a^2*g*Sqrt[g*Cos[e + f*x]])/(b^3*f) - (2*(g*Cos[e + f*x])^( 
5/2))/(5*b*f*g) + (2*a^3*g^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]) 
/(b^4*f*Sqrt[g*Cos[e + f*x]]) - (2*a*g^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + 
 f*x)/2, 2])/(3*b^2*f*Sqrt[g*Cos[e + f*x]]) - (a^3*(a^2 - b^2)*g^2*Sqrt[Co 
s[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^4 
*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) - (a^3*(a^2 - b^ 
2)*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f* 
x)/2, 2])/(b^4*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) - 
(2*a*g*Sqrt[g*Cos[e + f*x]]*Sin[e + f*x])/(3*b^2*f)
 

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 3377

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a 
_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(g*cos[e + 
 f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, 
 g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/ 
2, 0])
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(957\) vs. \(2(456)=912\).

Time = 5.67 (sec) , antiderivative size = 958, normalized size of antiderivative = 1.86

method	result	size
default	\(\text {Expression too large to display}\)	\(958\)

Input:

int((g*cos(f*x+e))^(3/2)*sin(f*x+e)^2/(a+b*sin(f*x+e)),x,method=_RETURNVER 
BOSE)
 

Output:

(16*g^2*b*(-1/40/b^4/g*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(4*sin(1/2*f*x+ 
1/2*e)^4*b^2-4*sin(1/2*f*x+1/2*e)^2*b^2-5*a^2+b^2)+1/4*a^2*(a^2-b^2)/b^4*( 
g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(-ln((-(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos 
(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)-2*g*cos(1/2*f*x+1/2*e)^2-(g^2*(a^2-b^2) 
/b^2)^(1/2)+g)/((g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/ 
2)*2^(1/2)-2*g*cos(1/2*f*x+1/2*e)^2-(g^2*(a^2-b^2)/b^2)^(1/2)+g))-2*arctan 
(2^(1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+1)+2 
*arctan(-2^(1/2)/(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1 
/2)+1))/(16*a^2-16*b^2)/g)-8*(g*(-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/ 
2*e)^2)^(1/2)*a*g^2*(-1/12/b^4/(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/ 
2*e)^2)^(1/2)*(4*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^4*b^2-2*cos(1/2*f*x 
+1/2*e)*sin(1/2*f*x+1/2*e)^2*b^2-3*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*( 
-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*a^2+Elliptic 
F(cos(1/2*f*x+1/2*e),2^(1/2))*(-1+2*sin(1/2*f*x+1/2*e)^2)^(1/2)*(sin(1/2*f 
*x+1/2*e)^2)^(1/2)*b^2)-1/64*a^2*(a^2-b^2)/b^6*sum(1/_alpha/(2*_alpha^2-1) 
*(2^(1/2)/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*arctanh(1/2*g*(4*_alpha 
^2-3)/(4*a^2-3*b^2)*(b^2*_alpha^2+4*a^2*cos(1/2*f*x+1/2*e)^2-3*b^2*cos(1/2 
*f*x+1/2*e)^2-3*a^2+2*b^2)*2^(1/2)/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2 
)/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2))+8*b^2/a^2*_alp 
ha*(_alpha^2-1)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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