\(\int \frac {\sin ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [242]

Optimal result

Integrand size = 25, antiderivative size = 431 \[ \int \frac {\sin ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{\sqrt {b} \sqrt {a+b} d \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )}-\frac {(a+b)^{3/4} \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{b^{3/4} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}-\frac {\sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{2 b^{3/4} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \]

[Out]

cos(d*x+c)*(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)/d/b^(1/2)/(1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))/(a+b)^(1
/2)-(a+b)^(3/4)*(cos(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^
(1/4)))*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))*(1+cos(d*
x+c)^2*b^(1/2)/(a+b)^(1/2))*((a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)/(a+b)/(1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))^
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)*(cos(2*arctan(b^(1/4)*cos(d*x+c
)/(a+b)^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^(1/4)))*EllipticF(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))*(1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))*(b^(1/2)-(a+b)^(1/2
))*((a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)/(a+b)/(1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))^2)^(1/2)/b^(3/4)/d/(a+b-2
*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3294, 1211, 1117, 1209} \[ \int \frac {\sin ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\sqrt [4]{a+b} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 b^{3/4} d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac {(a+b)^{3/4} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{b^{3/4} d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}+\frac {\cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{\sqrt {b} d \sqrt {a+b} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )} \]

[In]

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

[Out]

(Cos[c + d*x]*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])/(Sqrt[b]*Sqrt[a + b]*d*(1 + (Sqrt[b]*Cos[c
+ d*x]^2)/Sqrt[a + b])) - ((a + b)^(3/4)*(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^(3/4)*d*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Co
s[c + d*x]^4]) - ((a + b)^(1/4)*(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 + Sqrt[b]/Sqrt[a + b])/2])/(2*b^(3/4)*d*Sqrt[a + b - 2*b*Cos[
c + d*x]^2 + b*Cos[c + d*x]^4])

Rule 1117

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 1209

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> 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 1211

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[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] && PosQ[c/a]

Rule 3294

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, Dist[-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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1-x^2}{\sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {\sqrt {a+b} \text {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a+b}}}{\sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{\sqrt {b} d}-\frac {\left (1-\frac {\sqrt {a+b}}{\sqrt {b}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{d} \\ & = \frac {\cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{\sqrt {b} \sqrt {a+b} d \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\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)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 23.80 (sec) , antiderivative size = 2663, normalized size of antiderivative = 6.18 \[ \int \frac {\sin ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

(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*Cos[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[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])]/Sqrt[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)]))/((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])]/Sqrt[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))^(3/2)*Sqrt[8*a + 3*b - 4*b*Cos[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)]) - (3*Sqrt[a]*Sin[2*(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[ArcSin[Sq
rt[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])/Sq
rt[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] + Sqrt[a]*Sqrt[((a + b)*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))
/(a*b)]))/((a + b)*Sqrt[(1 + Cos[2*(c + d*x)])^(-1)]*Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)
]]) - (Sqrt[a]*(8*b*Sin[2*(c + d*x)] - 4*b*Sin[4*(c + d*x)])*Sqrt[((a + b)*(a + 2*a*Tan[c + d*x]^2 + (a + b)*T
an[c + d*x]^4))/(a*b)]*(-((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])*Se
c[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*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*(a + b)*((1 + Cos[2*(c + d*x)]
)^(-1))^(3/2)*(8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)])^(3/2)) + (Sqrt[a]*Sqrt[((a + b)*(a + 2*a
*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))/(a*b)]*(-(((1 - (I*Sqrt[a])/Sqrt[b])*(Sqrt[a] - I*Sqrt[b])*Elliptic
E[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])/(S
qrt[a] - I*Sqrt[b])]*Sec[c + d*x]^2*Tan[c + d*x])/Sqrt[(1 - (I*Sqrt[a])/Sqrt[b])*Sec[c + d*x]^2]) - (I*(1 - (I
*Sqrt[a])/Sqrt[b])*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*Sqrt[a])/(Sqrt[a] - I*Sqrt[b])]*Sec[c + d*x]^2*Tan[c + d*x])/Sqrt[(1 - (I*Sqrt[a])/Sqr
t[b])*Sec[c + d*x]^2] + ((a + b)*(4*a*Sec[c + d*x]^2*Tan[c + d*x] + 4*(a + b)*Sec[c + d*x]^2*Tan[c + d*x]^3))/
(2*Sqrt[a]*b*Sqrt[((a + b)*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))/(a*b)]) + ((a + b)*Sec[c + d*x]^
2*Sqrt[(1 - (I*Sqrt[a])/Sqrt[b])*Sec[c + d*x]^2]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a]*Sqrt[1 + (I*Sqrt[a])/Sqrt[b] +
(I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]*Sqrt[1 + (-1 - (I*Sqrt[a])/Sqrt[b] - (I*(a + b)*Tan[c + d*x]^2)
/(Sqrt[a]*Sqrt[b]))/2]*Sqrt[1 - (Sqrt[a]*(1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b
])))/(Sqrt[a] - I*Sqrt[b])]) - (I*(Sqrt[a] - I*Sqrt[b])*(a + b)*Sec[c + d*x]^2*Sqrt[(1 - (I*Sqrt[a])/Sqrt[b])*
Sec[c + d*x]^2]*Tan[c + d*x]*Sqrt[1 - (Sqrt[a]*(1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*
Sqrt[b])))/(Sqrt[a] - I*Sqrt[b])])/(Sqrt[2]*Sqrt[a]*Sqrt[b]*Sqrt[1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Tan[c +
d*x]^2)/(Sqrt[a]*Sqrt[b])]*Sqrt[1 + (-1 - (I*Sqrt[a])/Sqrt[b] - (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b]))/
2])))/((a + b)*((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)]]
)))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.85 (sec) , antiderivative size = 398, normalized size of antiderivative = 0.92

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

[In]

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

[Out]

-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)*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/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*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))^(1/2)))

Fricas [F]

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

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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