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

Optimal. Leaf size=431 \[ -\frac{\sqrt [4]{a+b} \left (\sqrt{b}-\sqrt{a+b}\right ) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right ) \sqrt{\frac{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac{1}{2} \left (\frac{\sqrt{b}}{\sqrt{a+b}}+1\right )\right )}{2 b^{3/4} d \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac{(a+b)^{3/4} \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right ) \sqrt{\frac{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac{1}{2} \left (\frac{\sqrt{b}}{\sqrt{a+b}}+1\right )\right )}{b^{3/4} d \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}+\frac{\cos (c+d x) \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{\sqrt{b} d \sqrt{a+b} \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )} \]

[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])

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Rubi [A]  time = 0.302493, antiderivative size = 431, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3215, 1197, 1103, 1195} \[ -\frac{\sqrt [4]{a+b} \left (\sqrt{b}-\sqrt{a+b}\right ) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right ) \sqrt{\frac{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac{1}{2} \left (\frac{\sqrt{b}}{\sqrt{a+b}}+1\right )\right )}{2 b^{3/4} d \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac{(a+b)^{3/4} \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right ) \sqrt{\frac{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac{1}{2} \left (\frac{\sqrt{b}}{\sqrt{a+b}}+1\right )\right )}{b^{3/4} d \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}+\frac{\cos (c+d x) \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{\sqrt{b} d \sqrt{a+b} \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )} \]

Antiderivative was successfully verified.

[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 3215

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]

Rule 1197

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 1103

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)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

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)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{\sin ^3(c+d x)}{\sqrt{a+b \sin ^4(c+d x)}} \, dx &=-\frac{\operatorname{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} \operatorname{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 ) \operatorname{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 \tan ^{-1}\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}} F\left (2 \tan ^{-1}\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]  time = 31.8742, size = 89374, normalized size = 207.36 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [C]  time = 0.525, size = 398, normalized size = 0.9 \begin{align*} -{\frac{1}{d}\sqrt{1-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a+b} \left ( i\sqrt{a}\sqrt{b}+b \right ) }}\sqrt{1+{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a+b} \left ( i\sqrt{a}\sqrt{b}-b \right ) }}{\it EllipticF} \left ( \cos \left ( dx+c \right ) \sqrt{{\frac{1}{a+b} \left ( i\sqrt{a}\sqrt{b}+b \right ) }},\sqrt{-1-2\,{\frac{i\sqrt{a}\sqrt{b}-b}{a+b}}} \right ){\frac{1}{\sqrt{{\frac{1}{a+b} \left ( i\sqrt{a}\sqrt{b}+b \right ) }}}}{\frac{1}{\sqrt{a+b-2\,b \left ( \cos \left ( dx+c \right ) \right ) ^{2}+b \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}}}-2\,{\frac{a+b}{d\sqrt{a+b-2\,b \left ( \cos \left ( dx+c \right ) \right ) ^{2}+b \left ( \cos \left ( dx+c \right ) \right ) ^{4}} \left ( -2\,b+2\,i\sqrt{a}\sqrt{b} \right ) }\sqrt{1-{\frac{ \left ( i\sqrt{a}\sqrt{b}+b \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a+b}}}\sqrt{1+{\frac{ \left ( i\sqrt{a}\sqrt{b}-b \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{a+b}}} \left ({\it EllipticF} \left ( \cos \left ( dx+c \right ) \sqrt{{\frac{i\sqrt{a}\sqrt{b}+b}{a+b}}},\sqrt{-1-2\,{\frac{i\sqrt{a}\sqrt{b}-b}{a+b}}} \right ) -{\it EllipticE} \left ( \cos \left ( dx+c \right ) \sqrt{{\frac{i\sqrt{a}\sqrt{b}+b}{a+b}}},\sqrt{-1-2\,{\frac{i\sqrt{a}\sqrt{b}-b}{a+b}}} \right ) \right ){\frac{1}{\sqrt{{\frac{i\sqrt{a}\sqrt{b}+b}{a+b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )^{3}}{\sqrt{b \sin \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right )}{\sqrt{b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out