3.167 \(\int \frac{\sin ^4(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=269 \[ \frac{(2 a+3 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^2 f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (a+2 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^2 f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}-\frac{2 (a+2 b) \sin (e+f x) \cos (e+f x)}{3 b f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}+\frac{a \sin (e+f x) \cos (e+f x)}{3 b f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.275601, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3188, 470, 527, 524, 426, 424, 421, 419} \[ \frac{(2 a+3 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^2 f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (a+2 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^2 f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}-\frac{2 (a+2 b) \sin (e+f x) \cos (e+f x)}{3 b f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}+\frac{a \sin (e+f x) \cos (e+f x)}{3 b f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

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

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]

Rule 470

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

Rule 527

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

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 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 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 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 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)])

Rubi steps

\begin{align*} \int \frac{\sin ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, 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} \left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{a \cos (e+f x) \sin (e+f x)}{3 b (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{a+(-2 a-3 b) x^2}{\sqrt{1-x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 b (a+b) f}\\ &=\frac{a \cos (e+f x) \sin (e+f x)}{3 b (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{2 (a+2 b) \cos (e+f x) \sin (e+f x)}{3 b (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{a (a+3 b)-2 a (a+2 b) x^2}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 a b (a+b)^2 f}\\ &=\frac{a \cos (e+f x) \sin (e+f x)}{3 b (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{2 (a+2 b) \cos (e+f x) \sin (e+f x)}{3 b (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left (2 (a+2 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 (a+b)^2 f}+\frac{\left ((2 a+3 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 (a+b) f}\\ &=\frac{a \cos (e+f x) \sin (e+f x)}{3 b (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{2 (a+2 b) \cos (e+f x) \sin (e+f x)}{3 b (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left (2 (a+2 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 (a+b)^2 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{\left ((2 a+3 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 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}\\ &=\frac{a \cos (e+f x) \sin (e+f x)}{3 b (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{2 (a+2 b) \cos (e+f x) \sin (e+f x)}{3 b (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (a+2 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 (a+b)^2 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{(2 a+3 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 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.63713, size = 182, normalized size = 0.68 \[ -\frac{-\sqrt{2} b \sin (2 (e+f x)) \left (-a^2+b (a+2 b) \cos (2 (e+f x))-4 a b-2 b^2\right )-a \left (2 a^2+5 a b+3 b^2\right ) \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} F\left (e+f x\left |-\frac{b}{a}\right .\right )+2 a^2 (a+2 b) \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 b^2 f (a+b)^2 (2 a-b \cos (2 (e+f x))+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

-(2*a^2*(a + 2*b)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticE[e + f*x, -(b/a)] - a*(2*a^2 + 5*a*b + 3*b
^2)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticF[e + f*x, -(b/a)] - Sqrt[2]*b*(-a^2 - 4*a*b - 2*b^2 + b*
(a + 2*b)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)])/(3*b^2*(a + b)^2*f*(2*a + b - b*Cos[2*(e + f*x)])^(3/2))

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Maple [B]  time = 1.527, size = 623, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),x)

[Out]

1/3*((2*a*b^2+4*b^3)*sin(f*x+e)*cos(f*x+e)^4+(-a^2*b-5*a*b^2-4*b^3)*cos(f*x+e)^2*sin(f*x+e)-(cos(f*x+e)^2)^(1/
2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*b*(2*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2+5*EllipticF(sin(f*x+e),(-1/
a*b)^(1/2))*a*b+3*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b^2-2*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2-4*Ellipt
icE(sin(f*x+e),(-1/a*b)^(1/2))*a*b)*cos(f*x+e)^2+2*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*Elli
pticF(sin(f*x+e),(-1/a*b)^(1/2))*a^3+7*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*
x+e),(-1/a*b)^(1/2))*a^2*b+8*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a
*b)^(1/2))*a*b^2+3*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))
*b^3-2*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^3-6*(cos(
f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b-4*(cos(f*x+e)^2)^
(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2)/(a+b*sin(f*x+e)^2)^(3/2)/(
a+b)^2/b^2/cos(f*x+e)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^4/(b*sin(f*x + e)^2 + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{3} \cos \left (f x + e\right )^{6} - 3 \,{\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3} + 3 \,{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),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^3*cos(f*x + e)^6 - 3*(a*b
^2 + b^3)*cos(f*x + e)^4 - a^3 - 3*a^2*b - 3*a*b^2 - b^3 + 3*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^2), 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(f*x+e)**4/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^4/(b*sin(f*x + e)^2 + a)^(5/2), x)