3.148 \(\int \frac {\sin ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\)

Optimal. Leaf size=206 \[ \frac {a (2 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 b^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 b^2 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {\sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 b f} \]

[Out]

-1/3*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/b/f-2/3*(a-b)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e
)*(cos(f*x+e)^2)^(1/2)*(a+b*sin(f*x+e)^2)^(1/2)/b^2/f/(1+b*sin(f*x+e)^2/a)^(1/2)+1/3*a*(2*a-b)*EllipticF(sin(f
*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(1+b*sin(f*x+e)^2/a)^(1/2)/b^2/f/(a+b*sin(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3188, 479, 524, 426, 424, 421, 419} \[ \frac {a (2 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 b^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 b^2 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {\sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 b f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

-(Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*b*f) - (2*(a - b)*Sqrt[Cos[e + f*x]^2]*EllipticE[Ar
cSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*b^2*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a])
+ (a*(2*a - b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[1 + (b*Sin[e + f
*x]^2)/a])/(3*b^2*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 421

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 426

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b*x^2)/a]
, Int[Sqrt[1 + (b*x^2)/a]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 479

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(2*n
- 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q) + 1)), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 3188

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[(ff^(m + 1)*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(x^m*(a + b*ff^2*
x^2)^p)/Sqrt[1 - ff^2*x^2], x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] &&  !In
tegerQ[p]

Rubi steps

\begin {align*} \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx &=\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 b f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {a-2 (a-b) x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 b f}\\ &=-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 b f}-\frac {\left (2 (a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 b^2 f}+\frac {\left (a (2 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 b^2 f}\\ &=-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 b f}-\frac {\left (2 (a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left (a (2 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{3 b^2 f \sqrt {a+b \sin ^2(e+f x)}}\\ &=-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 b f}-\frac {2 (a-b) \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {a (2 a-b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 b^2 f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 163, normalized size = 0.79 \[ \frac {b \sin (2 (e+f x)) (-2 a+b \cos (2 (e+f x))-b)+2 \sqrt {2} a (2 a-b) \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )-4 \sqrt {2} a (a-b) \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{6 \sqrt {2} b^2 f \sqrt {2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

(-4*Sqrt[2]*a*(a - b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + 2*Sqrt[2]*a*(2*a - b
)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + b*(-2*a - b + b*Cos[2*(e + f*x)])*Sin[2*
(e + f*x)])/(6*Sqrt[2]*b^2*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{b \cos \left (f x + e\right )^{2} - a - b}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

integral(-(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-b*cos(f*x + e)^2 + a + b)/(b*cos(f*x + e)^2 - a - b),
x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{4}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^4/sqrt(b*sin(f*x + e)^2 + a), x)

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maple [A]  time = 1.44, size = 268, normalized size = 1.30 \[ \frac {b^{2} \left (\sin ^{5}\left (f x +e \right )\right )+2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}-a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b +a b \left (\sin ^{3}\left (f x +e \right )\right )-b^{2} \left (\sin ^{3}\left (f x +e \right )\right )-a b \sin \left (f x +e \right )}{3 b^{2} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2),x)

[Out]

1/3*(b^2*sin(f*x+e)^5+2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))
*a^2-a*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b-2*(cos(f*x+e)^
2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2+2*(cos(f*x+e)^2)^(1/2)*((a+b*si
n(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b+a*b*sin(f*x+e)^3-b^2*sin(f*x+e)^3-a*b*sin(f*x+e)
)/b^2/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{4}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^4/sqrt(b*sin(f*x + e)^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (e+f\,x\right )}^4}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^4/(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(sin(e + f*x)^4/(a + b*sin(e + f*x)^2)^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**4/(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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