3.362 \(\int \frac{\sec ^2(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=240 \[ \frac{\tan (e+f x)}{f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{b (a-b) \sin (e+f x) \cos (e+f x)}{a f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{(a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{a f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

[Out]

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

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Rubi [A]  time = 0.239238, antiderivative size = 240, 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 = {3192, 414, 527, 524, 426, 424, 421, 419} \[ \frac{\tan (e+f x)}{f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{b (a-b) \sin (e+f x) \cos (e+f x)}{a f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{(a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{a f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

-(((a - b)*b*Cos[e + f*x]*Sin[e + f*x])/(a*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2])) - ((a - 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])/(a*(a + b)^2*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])/((a + b)*f*Sqrt[a + b*Sin[e + f*x]^2]) + Tan[e + f*x]/((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 414

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

Mathematica [A]  time = 1.19473, size = 167, normalized size = 0.7 \[ \frac{\tan (e+f x) \left (2 a^2+b (b-a) \cos (2 (e+f x))+a b+b^2\right )+\sqrt{2} a (a+b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )-\sqrt{2} a (a-b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{\sqrt{2} a f (a+b)^2 \sqrt{2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(-(Sqrt[2]*a*(a - b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)]) + Sqrt[2]*a*(a + b)*Sq
rt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + (2*a^2 + a*b + b^2 + b*(-a + b)*Cos[2*(e + f
*x)])*Tan[e + f*x])/(Sqrt[2]*a*(a + b)^2*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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Maple [B]  time = 2.345, size = 468, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)^2/(b*sin(f*x + e)^2 + a)^(3/2), 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} \sec \left (f x + e\right )^{2}}{b^{2} \cos \left (f x + e\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**2/(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Integral(sec(e + f*x)**2/(a + b*sin(e + f*x)**2)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^2/(b*sin(f*x + e)^2 + a)^(3/2), x)