\(\int \frac {\sin ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [187]

Optimal result

Integrand size = 25, antiderivative size = 476 \[ \int \frac {\sin ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{3 b d}+\frac {2 \cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{3 \sqrt {b} d \left (\sqrt {a+b}+\sqrt {b} \cos ^2(c+d x)\right )}-\frac {2 (a+b)^{3/4} \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{3 b^{3/4} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}+\frac {\sqrt [4]{a+b} \left (a-2 b+2 \sqrt {b} \sqrt {a+b}\right ) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{6 b^{5/4} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \] Output:

-1/3*cos(d*x+c)*(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)/b/d+2/3*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)-2/3*(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)+1/6*(a+b)^(1/4)*(a-2*b+2*b^(1/2)*(a+b)^(1/2))*(1+b^(1/2)*co 
s(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^(5/4)/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 = 28.84 (sec) , antiderivative size = 2854, normalized size of antiderivative = 6.00 \[ \int \frac {\sin ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\text {Result too large to show} \] Input:

Integrate[Sin[c + d*x]^5/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

-1/6*(Cos[c + d*x]*Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d* 
x)]])/(Sqrt[2]*b*d) + (2*Sqrt[2]*Sqrt[a]*((Sqrt[2]*Sin[c + d*x])/Sqrt[8*a 
+ 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]] - (2*Sqrt[2]*a*Sin[c + 
d*x])/(3*b*Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]]) - 
(Sqrt[2]*Sin[3*(c + d*x)])/(3*Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Co 
s[4*(c + d*x)]]))*Sqrt[((a + b)*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + 
d*x]^4))/(a*b)]*(-2*(Sqrt[a] - I*Sqrt[b])*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])*Se 
c[c + d*x]^2] + I*(a - 2*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*Sqrt[a])/(Sqr 
t[a] - I*Sqrt[b])]*Sqrt[(1 - (I*Sqrt[a])/Sqrt[b])*Sec[c + d*x]^2] + 2*Sqrt 
[a]*Sqrt[b]*Sqrt[((a + b)*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4 
))/(a*b)]))/(3*Sqrt[b]*(a + b)*d*Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b 
*Cos[4*(c + d*x)]]*(Sec[c + d*x]^2)^(3/2)*((Sqrt[2]*(4*a*Sec[c + d*x]^2*Ta 
n[c + d*x] + 4*(a + b)*Sec[c + d*x]^2*Tan[c + d*x]^3)*(-2*(Sqrt[a] - I*Sqr 
t[b])*Sqrt[b]*EllipticE[ArcSin[Sqrt[1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*T 
an[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*(a - 2*b)*Elliptic 
F[ArcSin[Sqrt[1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqr...
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 488, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3694, 1518, 25, 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.

\(\displaystyle \int \frac {\sin ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^5}{\sqrt {a+b \sin (c+d x)^4}}dx\)

\(\Big \downarrow \) 3694

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

\(\Big \downarrow \) 1518

\(\displaystyle -\frac {\frac {\int -\frac {2 b \cos ^2(c+d x)+a-2 b}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)}{3 b}+\frac {\cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{3 b}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{3 b}-\frac {\int \frac {2 b \cos ^2(c+d x)+a-2 b}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)}{3 b}}{d}\)

\(\Big \downarrow \) 1511

\(\displaystyle -\frac {\frac {\cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{3 b}-\frac {\left (2 \sqrt {b} \sqrt {a+b}+a-2 b\right ) \int \frac {1}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)-2 \sqrt {b} \sqrt {a+b} \int \frac {1-\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)}{3 b}}{d}\)

\(\Big \downarrow \) 1416

\(\displaystyle -\frac {\frac {\cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{3 b}-\frac {\frac {\sqrt [4]{a+b} \left (2 \sqrt {b} \sqrt {a+b}+a-2 b\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 \sqrt [4]{b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-2 \sqrt {b} \sqrt {a+b} \int \frac {1-\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)}{3 b}}{d}\)

\(\Big \downarrow \) 1509

\(\displaystyle -\frac {\frac {\cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{3 b}-\frac {\frac {\sqrt [4]{a+b} \left (2 \sqrt {b} \sqrt {a+b}+a-2 b\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 \sqrt [4]{b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-2 \sqrt {b} \sqrt {a+b} \left (\frac {\sqrt [4]{a+b} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{\sqrt [4]{b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac {\cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )}\right )}{3 b}}{d}\)

Input:

Int[Sin[c + d*x]^5/Sqrt[a + b*Sin[c + d*x]^4],x]
 

Output:

-(((Cos[c + d*x]*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])/(3*b 
) - (-2*Sqrt[b]*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*Cos[c + d*x]^4])) + ((a + b)^(1/4)*(a - 2*b + 2*Sqrt[b]*Sqrt[a + 
 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)]*EllipticF[2*ArcTan[(b^(1/4)*Cos[c + d*x])/(a + b)^(1/4)], (1 + Sq 
rt[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]))/(3*b))/d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 1518

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x 
_Symbol] :> Simp[e^q*x^(2*q - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(4*p + 2*q 
 + 1))), x] + Simp[1/(c*(4*p + 2*q + 1))   Int[(a + b*x^2 + c*x^4)^p*Expand 
ToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2* 
p + 2*q - 1)*e^q*x^(2*q - 2) - c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; 
 FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + 
 a*e^2, 0] && IGtQ[q, 1]
 

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 = 4.26 (sec) , antiderivative size = 837, normalized size of antiderivative = 1.76

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 {4 \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 )}-\frac {4 \left (\frac {\cos \left (d x +c \right ) \sqrt {a +b -2 b \cos \left (d x +c \right )^{2}+b \cos \left (d x +c \right )^{4}}}{12 b}-\frac {\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}}\, \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 )}{12 b \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 )}{3 \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 )}\right )}{d}\)	\(837\)

Input:

int(sin(d*x+c)^5/(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))-4/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)))-4/d*(1/12/b*cos(d*x+c)*(a+b-2*b*cos 
(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)-1/12*(a+b)/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*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))-2/3*(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*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)/ 
(-2*b+2*I*a^(1/2)*b^(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))-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))^...
 

Fricas [F]

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

integrate(sin(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(sin(d*x+c)**5/(a+b*sin(d*x+c)**4)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(sin(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sin(d*x + c)^5/sqrt(b*sin(d*x + c)^4 + a), x)
 

Giac [F]

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

integrate(sin(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^5(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^5}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \] Input:

int(sin(c + d*x)^5/(a + b*sin(c + d*x)^4)^(1/2),x)
 

Output:

int(sin(c + d*x)^5/(a + b*sin(c + d*x)^4)^(1/2), x)
 

Reduce [F]

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

int(sin(d*x+c)^5/(a+b*sin(d*x+c)^4)^(1/2),x)
 

Output:

int((sqrt(sin(c + d*x)**4*b + a)*sin(c + d*x)**5)/(sin(c + d*x)**4*b + a), 
x)