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

Optimal. Leaf size=484 \[ \frac{\sqrt [4]{a+b} \left (2 \sqrt{b} \sqrt{a+b}+a-2 b\right ) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right ) \sqrt{\frac{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac{1}{2} \left (\frac{\sqrt{b}}{\sqrt{a+b}}+1\right )\right )}{6 b^{5/4} d \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac{2 (a+b)^{3/4} \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right ) \sqrt{\frac{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac{1}{2} \left (\frac{\sqrt{b}}{\sqrt{a+b}}+1\right )\right )}{3 b^{3/4} d \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac{\cos (c+d x) \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{3 b d}+\frac{2 \cos (c+d x) \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{3 \sqrt{b} d \sqrt{a+b} \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )} \]

[Out]

-(Cos[c + d*x]*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])/(3*b*d) + (2*Cos[c + d*x]*Sqrt[a + b - 2*b
*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])/(3*Sqrt[b]*Sqrt[a + b]*d*(1 + (Sqrt[b]*Cos[c + d*x]^2)/Sqrt[a + b])) - (2
*(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])/(3*b^(3/4)*d*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 + Sqrt[b]/Sqrt[a + b])/2])/(6*b^(5/4)*d*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b
*Cos[c + d*x]^4])

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Rubi [A]  time = 0.425803, antiderivative size = 484, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3215, 1206, 1197, 1103, 1195} \[ \frac{\sqrt [4]{a+b} \left (2 \sqrt{b} \sqrt{a+b}+a-2 b\right ) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right ) \sqrt{\frac{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac{1}{2} \left (\frac{\sqrt{b}}{\sqrt{a+b}}+1\right )\right )}{6 b^{5/4} d \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac{2 (a+b)^{3/4} \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right ) \sqrt{\frac{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac{1}{2} \left (\frac{\sqrt{b}}{\sqrt{a+b}}+1\right )\right )}{3 b^{3/4} d \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac{\cos (c+d x) \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{3 b d}+\frac{2 \cos (c+d x) \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{3 \sqrt{b} d \sqrt{a+b} \left (\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^5/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])/(3*b*d) + (2*Cos[c + d*x]*Sqrt[a + b - 2*b
*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])/(3*Sqrt[b]*Sqrt[a + b]*d*(1 + (Sqrt[b]*Cos[c + d*x]^2)/Sqrt[a + b])) - (2
*(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])/(3*b^(3/4)*d*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 + Sqrt[b]/Sqrt[a + b])/2])/(6*b^(5/4)*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 1206

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] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + b*x^2 + c*x^4)^p*Ex
pandToSum[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 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 ^5(c+d x)}{\sqrt{a+b \sin ^4(c+d x)}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{\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)}}{3 b d}-\frac{\operatorname{Subst}\left (\int \frac{-a+2 b-2 b x^2}{\sqrt{a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{3 b d}\\ &=-\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{\left (2 \sqrt{a+b}\right ) \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 )}{3 \sqrt{b} d}+\frac{\left (a-2 b+2 \sqrt{b} \sqrt{a+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 )}{3 b d}\\ &=-\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} \sqrt{a+b} d \left (1+\frac{\sqrt{b} \cos ^2(c+d x)}{\sqrt{a+b}}\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 \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 )}{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}} 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 )}{6 b^{5/4} d \sqrt{a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 31.7204, size = 47246, normalized size = 97.62 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [C]  time = 0.468, size = 837, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^5/(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))-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*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)))-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))^(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 )^{5}}{\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)^5/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

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)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )^{5}}{\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)^5/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

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