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

Optimal. Leaf size=217 \[ \frac{(a+2 b) \sin (e+f x) \cos (e+f x)}{3 a^2 f (a+b) \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sin (e+f x) \cos (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 a b f \sqrt{a+b \sin ^2(e+f x)}}+\frac{(a+2 b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 a b f (a+b) \sqrt{a+b \sin ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.241758, antiderivative size = 257, normalized size of antiderivative = 1.18, 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 = {3192, 412, 527, 524, 426, 424, 421, 419} \[ \frac{(a+2 b) \sin (e+f x) \cos (e+f x)}{3 a^2 f (a+b) \sqrt{a+b \sin ^2(e+f x)}}+\frac{(a+2 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 a^2 b f (a+b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}+\frac{\sin (e+f x) \cos (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 a b f \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3192

Int[cos[(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*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(1 - ff^2*x^2)^((m - 1)/2
)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] &&  !IntegerQ
[p]

Rule 412

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] + Dist[1/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(n*(p
+ 1) + 1) + d*(n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p,
-1] && LtQ[0, q, 1] && IntBinomialQ[a, b, c, d, 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{\cos ^2(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{\sqrt{1-x^2}}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\cos (e+f x) \sin (e+f x)}{3 a 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{-2+x^2}{\sqrt{1-x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 a f}\\ &=\frac{\cos (e+f x) \sin (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{(a+2 b) \cos (e+f x) \sin (e+f x)}{3 a^2 (a+b) 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+2 b) x^2}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 a^2 (a+b) f}\\ &=\frac{\cos (e+f x) \sin (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{(a+2 b) \cos (e+f x) \sin (e+f x)}{3 a^2 (a+b) 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{1}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 a b f}+\frac{\left ((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 a^2 b (a+b) f}\\ &=\frac{\cos (e+f x) \sin (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{(a+2 b) \cos (e+f x) \sin (e+f x)}{3 a^2 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left ((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 a^2 b (a+b) f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{\left (\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 a b f \sqrt{a+b \sin ^2(e+f x)}}\\ &=\frac{\cos (e+f x) \sin (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{(a+2 b) \cos (e+f x) \sin (e+f x)}{3 a^2 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}+\frac{(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 a^2 b (a+b) f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{\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 a b f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.42125, size = 175, normalized size = 0.81 \[ \frac{-\sqrt{2} b \sin (2 (e+f x)) \left (-4 a^2+b (a+2 b) \cos (2 (e+f x))-7 a b-2 b^2\right )-2 a^2 (a+b) \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} F\left (e+f x\left |-\frac{b}{a}\right .\right )+2 a^2 (a+2 b) \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{6 a^2 b f (a+b) (2 a-b \cos (2 (e+f x))+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2/(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)] - 2*a^2*(a + b)*((2*a + b
- b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticF[e + f*x, -(b/a)] - Sqrt[2]*b*(-4*a^2 - 7*a*b - 2*b^2 + b*(a + 2*b)*C
os[2*(e + f*x)])*Sin[2*(e + f*x)])/(6*a^2*b*(a + b)*f*(2*a + b - b*Cos[2*(e + f*x)])^(3/2))

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Maple [B]  time = 1.753, size = 549, 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(cos(f*x+e)^2/(a+b*sin(f*x+e)^2)^(5/2),x)

[Out]

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

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

[Out]

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

[Out]

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

[Out]

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