\(\int \frac {\sin ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [188]

Optimal result

Integrand size = 25, antiderivative size = 423 \[ \int \frac {\sin ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{\sqrt {b} d \left (\sqrt {a+b}+\sqrt {b} \cos ^2(c+d x)\right )}-\frac {(a+b)^{3/4} \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{b^{3/4} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}-\frac {\sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{2 b^{3/4} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 24.88 (sec) , antiderivative size = 2663, normalized size of antiderivative = 6.30 \[ \int \frac {\sin ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\text {Result too large to show} \] Input:

Integrate[Sin[c + d*x]^3/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

(Sqrt[a]*((3*Sin[c + d*x])/(Sqrt[2]*Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] 
+ b*Cos[4*(c + d*x)]]) - Sin[3*(c + d*x)]/(Sqrt[2]*Sqrt[8*a + 3*b - 4*b*Co 
s[2*(c + d*x)] + b*Cos[4*(c + d*x)]]))*Sqrt[((a + b)*(a + 2*a*Tan[c + d*x] 
^2 + (a + b)*Tan[c + d*x]^4))/(a*b)]*(-((Sqrt[a] - I*Sqrt[b])*EllipticE[Ar 
cSin[Sqrt[1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sq 
rt[b])]/Sqrt[2]], (2*Sqrt[a])/(Sqrt[a] - I*Sqrt[b])]*Sqrt[(1 - (I*Sqrt[a]) 
/Sqrt[b])*Sec[c + d*x]^2]) - I*Sqrt[b]*EllipticF[ArcSin[Sqrt[1 + (I*Sqrt[a 
])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]/Sqrt[2]], (2*Sq 
rt[a])/(Sqrt[a] - I*Sqrt[b])]*Sqrt[(1 - (I*Sqrt[a])/Sqrt[b])*Sec[c + d*x]^ 
2] + Sqrt[a]*Sqrt[((a + b)*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^ 
4))/(a*b)]))/((a + b)*d*((1 + Cos[2*(c + d*x)])^(-1))^(3/2)*Sqrt[8*a + 3*b 
 - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]]*(((4*a*Sec[c + d*x]^2*Tan[c 
+ d*x] + 4*(a + b)*Sec[c + d*x]^2*Tan[c + d*x]^3)*(-((Sqrt[a] - I*Sqrt[b]) 
*EllipticE[ArcSin[Sqrt[1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2 
)/(Sqrt[a]*Sqrt[b])]/Sqrt[2]], (2*Sqrt[a])/(Sqrt[a] - I*Sqrt[b])]*Sqrt[(1 
- (I*Sqrt[a])/Sqrt[b])*Sec[c + d*x]^2]) - I*Sqrt[b]*EllipticF[ArcSin[Sqrt[ 
1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]/Sq 
rt[2]], (2*Sqrt[a])/(Sqrt[a] - I*Sqrt[b])]*Sqrt[(1 - (I*Sqrt[a])/Sqrt[b])* 
Sec[c + d*x]^2] + Sqrt[a]*Sqrt[((a + b)*(a + 2*a*Tan[c + d*x]^2 + (a + b)* 
Tan[c + d*x]^4))/(a*b)]))/(2*Sqrt[a]*b*((1 + Cos[2*(c + d*x)])^(-1))^(3...
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3694, 1511, 1416, 1509}

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.

\(\displaystyle \int \frac {\sin ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^3}{\sqrt {a+b \sin (c+d x)^4}}dx\)

\(\Big \downarrow \) 3694

\(\displaystyle -\frac {\int \frac {1-\cos ^2(c+d x)}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 1511

\(\displaystyle -\frac {\left (1-\frac {\sqrt {a+b}}{\sqrt {b}}\right ) \int \frac {1}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)+\frac {\sqrt {a+b} \int \frac {1-\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)}{\sqrt {b}}}{d}\)

\(\Big \downarrow \) 1416

\(\displaystyle -\frac {\frac {\sqrt {a+b} \int \frac {1-\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)}{\sqrt {b}}+\frac {\sqrt [4]{a+b} \left (1-\frac {\sqrt {a+b}}{\sqrt {b}}\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 \sqrt [4]{b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}}{d}\)

\(\Big \downarrow \) 1509

\(\displaystyle -\frac {\frac {\sqrt [4]{a+b} \left (1-\frac {\sqrt {a+b}}{\sqrt {b}}\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 \sqrt [4]{b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}+\frac {\sqrt {a+b} \left (\frac {\sqrt [4]{a+b} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{\sqrt [4]{b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac {\cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )}\right )}{\sqrt {b}}}{d}\)

Input:

Int[Sin[c + d*x]^3/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

-(((Sqrt[a + b]*(-((Cos[c + d*x]*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c 
 + d*x]^4])/((a + b)*(1 + (Sqrt[b]*Cos[c + d*x]^2)/Sqrt[a + b]))) + ((a + 
b)^(1/4)*(1 + (Sqrt[b]*Cos[c + d*x]^2)/Sqrt[a + b])*Sqrt[(a + b - 2*b*Cos[ 
c + d*x]^2 + b*Cos[c + d*x]^4)/((a + b)*(1 + (Sqrt[b]*Cos[c + d*x]^2)/Sqrt 
[a + b])^2)]*EllipticE[2*ArcTan[(b^(1/4)*Cos[c + d*x])/(a + b)^(1/4)], (1 
+ Sqrt[b]/Sqrt[a + b])/2])/(b^(1/4)*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Co 
s[c + d*x]^4])))/Sqrt[b] + ((a + b)^(1/4)*(1 - Sqrt[a + b]/Sqrt[b])*(1 + ( 
Sqrt[b]*Cos[c + d*x]^2)/Sqrt[a + b])*Sqrt[(a + b - 2*b*Cos[c + d*x]^2 + b* 
Cos[c + d*x]^4)/((a + b)*(1 + (Sqrt[b]*Cos[c + d*x]^2)/Sqrt[a + b])^2)]*El 
lipticF[2*ArcTan[(b^(1/4)*Cos[c + d*x])/(a + b)^(1/4)], (1 + Sqrt[b]/Sqrt[ 
a + b])/2])/(2*b^(1/4)*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4] 
))/d)
 

Defintions of rubi rules used

rule 1416

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c 
/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1509

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q 
^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2* 
x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 
/(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 
- 4*a*c, 0] && PosQ[c/a]
 

rule 1511

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + b*x^2 + c*x^ 
4], x], x] - Simp[e/q   Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] /; 
NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && Pos 
Q[c/a]
 

rule 3042

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

rule 3694

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4)^p, 
 x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 
 1)/2]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.64 (sec) , antiderivative size = 398, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {\sqrt {1-\frac {\left (i \sqrt {a}\, \sqrt {b}+b \right ) \cos \left (d x +c \right )^{2}}{a +b}}\, \sqrt {1+\frac {\left (i \sqrt {a}\, \sqrt {b}-b \right ) \cos \left (d x +c \right )^{2}}{a +b}}\, \operatorname {EllipticF}\left (\cos \left (d x +c \right ) \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}, \sqrt {-1-\frac {2 \left (i \sqrt {a}\, \sqrt {b}-b \right )}{a +b}}\right )}{d \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}\, \sqrt {a +b -2 b \cos \left (d x +c \right )^{2}+b \cos \left (d x +c \right )^{4}}}-\frac {2 \left (a +b \right ) \sqrt {1-\frac {\left (i \sqrt {a}\, \sqrt {b}+b \right ) \cos \left (d x +c \right )^{2}}{a +b}}\, \sqrt {1+\frac {\left (i \sqrt {a}\, \sqrt {b}-b \right ) \cos \left (d x +c \right )^{2}}{a +b}}\, \left (\operatorname {EllipticF}\left (\cos \left (d x +c \right ) \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}, \sqrt {-1-\frac {2 \left (i \sqrt {a}\, \sqrt {b}-b \right )}{a +b}}\right )-\operatorname {EllipticE}\left (\cos \left (d x +c \right ) \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}, \sqrt {-1-\frac {2 \left (i \sqrt {a}\, \sqrt {b}-b \right )}{a +b}}\right )\right )}{d \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}\, \sqrt {a +b -2 b \cos \left (d x +c \right )^{2}+b \cos \left (d x +c \right )^{4}}\, \left (-2 b +2 i \sqrt {a}\, \sqrt {b}\right )}\)	\(398\)

Input:

int(sin(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/d/((I*a^(1/2)*b^(1/2)+b)/(a+b))^(1/2)*(1-(I*a^(1/2)*b^(1/2)+b)/(a+b)*co 
s(d*x+c)^2)^(1/2)*(1+(I*a^(1/2)*b^(1/2)-b)/(a+b)*cos(d*x+c)^2)^(1/2)/(a+b- 
2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)*EllipticF(cos(d*x+c)*((I*a^(1/2)*b^ 
(1/2)+b)/(a+b))^(1/2),(-1-2*(I*a^(1/2)*b^(1/2)-b)/(a+b))^(1/2))-2/d*(a+b)/ 
((I*a^(1/2)*b^(1/2)+b)/(a+b))^(1/2)*(1-(I*a^(1/2)*b^(1/2)+b)/(a+b)*cos(d*x 
+c)^2)^(1/2)*(1+(I*a^(1/2)*b^(1/2)-b)/(a+b)*cos(d*x+c)^2)^(1/2)/(a+b-2*b*c 
os(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)/(-2*b+2*I*a^(1/2)*b^(1/2))*(EllipticF(co 
s(d*x+c)*((I*a^(1/2)*b^(1/2)+b)/(a+b))^(1/2),(-1-2*(I*a^(1/2)*b^(1/2)-b)/( 
a+b))^(1/2))-EllipticE(cos(d*x+c)*((I*a^(1/2)*b^(1/2)+b)/(a+b))^(1/2),(-1- 
2*(I*a^(1/2)*b^(1/2)-b)/(a+b))^(1/2)))
 

Fricas [F]

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

integrate(sin(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")
 

Output:

integral(-(cos(d*x + c)^2 - 1)*sin(d*x + c)/sqrt(b*cos(d*x + c)^4 - 2*b*co 
s(d*x + c)^2 + a + b), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(sin(d*x+c)**3/(a+b*sin(d*x+c)**4)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(sin(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sin(d*x + c)^3/sqrt(b*sin(d*x + c)^4 + a), x)
 

Giac [F]

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

integrate(sin(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^3}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \] Input:

int(sin(c + d*x)^3/(a + b*sin(c + d*x)^4)^(1/2),x)
 

Output:

int(sin(c + d*x)^3/(a + b*sin(c + d*x)^4)^(1/2), x)
 

Reduce [F]

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

int(sin(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x)
 

Output:

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