3.166 \(\int \frac{\sin ^6(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=285 \[ \frac{\left (8 a^2+13 a b+3 b^2\right ) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^3 f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}+\frac{2 a (2 a+3 b) \sin (e+f x) \cos (e+f x)}{3 b^2 f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}-\frac{a (8 a+9 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^3 f (a+b) \sqrt{a+b \sin ^2(e+f x)}}+\frac{a \sin ^3(e+f x) \cos (e+f x)}{3 b f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

(a*Cos[e + f*x]*Sin[e + f*x]^3)/(3*b*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*a*(2*a + 3*b)*Cos[e + f*x]*S
in[e + f*x])/(3*b^2*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) + ((8*a^2 + 13*a*b + 3*b^2)*Sqrt[Cos[e + f*x]^2]*E
llipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*b^3*(a + b)^2*f*Sqrt[1 + (b
*Sin[e + f*x]^2)/a]) - (a*(8*a + 9*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^3*(a + b)*f*Sqrt[a + b*Sin[e + f*x]^2])

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Rubi [A]  time = 0.332362, antiderivative size = 285, 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, 578, 524, 426, 424, 421, 419} \[ \frac{\left (8 a^2+13 a b+3 b^2\right ) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^3 f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}+\frac{2 a (2 a+3 b) \sin (e+f x) \cos (e+f x)}{3 b^2 f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}-\frac{a (8 a+9 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 b^3 f (a+b) \sqrt{a+b \sin ^2(e+f x)}}+\frac{a \sin ^3(e+f x) \cos (e+f x)}{3 b f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^6/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

(a*Cos[e + f*x]*Sin[e + f*x]^3)/(3*b*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*a*(2*a + 3*b)*Cos[e + f*x]*S
in[e + f*x])/(3*b^2*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) + ((8*a^2 + 13*a*b + 3*b^2)*Sqrt[Cos[e + f*x]^2]*E
llipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*b^3*(a + b)^2*f*Sqrt[1 + (b
*Sin[e + f*x]^2)/a]) - (a*(8*a + 9*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^3*(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 578

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

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 ^6(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^6}{\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 ^3(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{x^2 \left (3 a+(-4 a-3 b) x^2\right )}{\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 ^3(e+f x)}{3 b (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 a (2 a+3 b) \cos (e+f x) \sin (e+f x)}{3 b^2 (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{2 a (2 a+3 b)+\left (-8 a^2-13 a b-3 b^2\right ) x^2}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 b^2 (a+b)^2 f}\\ &=\frac{a \cos (e+f x) \sin ^3(e+f x)}{3 b (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 a (2 a+3 b) \cos (e+f x) \sin (e+f x)}{3 b^2 (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left (a (8 a+9 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^3 (a+b) f}-\frac{\left (\left (-8 a^2-13 a b-3 b^2\right ) \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^3 (a+b)^2 f}\\ &=\frac{a \cos (e+f x) \sin ^3(e+f x)}{3 b (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 a (2 a+3 b) \cos (e+f x) \sin (e+f x)}{3 b^2 (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left (\left (-8 a^2-13 a b-3 b^2\right ) \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^3 (a+b)^2 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{\left (a (8 a+9 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^3 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}\\ &=\frac{a \cos (e+f x) \sin ^3(e+f x)}{3 b (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 a (2 a+3 b) \cos (e+f x) \sin (e+f x)}{3 b^2 (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2+13 a b+3 b^2\right ) \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^3 (a+b)^2 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{a (8 a+9 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^3 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.01392, size = 192, normalized size = 0.67 \[ -\frac{a \left (\sqrt{2} b \sin (2 (e+f x)) \left (-8 a^2+b (5 a+7 b) \cos (2 (e+f x))-17 a b-7 b^2\right )+2 a \left (8 a^2+17 a b+9 b^2\right ) \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} F\left (e+f x\left |-\frac{b}{a}\right .\right )-2 a \left (8 a^2+13 a b+3 b^2\right ) \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} E\left (e+f x\left |-\frac{b}{a}\right .\right )\right )}{6 b^3 f (a+b)^2 (2 a-b \cos (2 (e+f x))+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^6/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

-(a*(-2*a*(8*a^2 + 13*a*b + 3*b^2)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticE[e + f*x, -(b/a)] + 2*a*(
8*a^2 + 17*a*b + 9*b^2)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticF[e + f*x, -(b/a)] + Sqrt[2]*b*(-8*a^
2 - 17*a*b - 7*b^2 + b*(5*a + 7*b)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)]))/(6*b^3*(a + b)^2*f*(2*a + b - b*Cos[2*
(e + f*x)])^(3/2))

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Maple [B]  time = 1.5, size = 698, normalized size = 2.5 \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)^6/(a+b*sin(f*x+e)^2)^(5/2),x)

[Out]

-1/3*((5*a*b^2+7*b^3)*sin(f*x+e)*cos(f*x+e)^4+(-4*a^2*b-11*a*b^2-7*b^3)*cos(f*x+e)^2*sin(f*x+e)-(-b/a*cos(f*x+
e)^2+(a+b)/a)^(1/2)*(cos(f*x+e)^2)^(1/2)*b*(8*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2+17*EllipticF(sin(f*x+e)
,(-1/a*b)^(1/2))*a*b+9*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b^2-8*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2-13*
EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b-3*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*b^2)*cos(f*x+e)^2+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^3+25*(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+26*(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+9*(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^3-8*(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^3-21*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*Elliptic
E(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b-16*(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^2-3*(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))*b^3)*a/(a+b*sin(f*x+e)^2)^(3/2)/(a+b)^2/b^3/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 )^{6}}{{\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)^6/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^6/(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 )^{6} - 3 \, \cos \left (f x + e\right )^{4} + 3 \, \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)^6/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

integral((cos(f*x + e)^6 - 3*cos(f*x + e)^4 + 3*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)**6/(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 )^{6}}{{\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)^6/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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