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

Optimal. Leaf size=229 \[ \frac{\sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \text{EllipticF}\left (\sin ^{-1}(\sin (e+f x)),\frac{a}{a+b}\right )}{a f \sqrt{\cos ^2(e+f x)} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}+\frac{\sin (e+f x)}{f (a+b) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}-\frac{\left (-a \sin ^2(e+f x)+a+b\right ) E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{a f (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}} \]

[Out]

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

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Rubi [A]  time = 0.447886, antiderivative size = 284, normalized size of antiderivative = 1.24, number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {4148, 6722, 1974, 412, 493, 426, 424, 421, 419} \[ \frac{\sin (e+f x) \sqrt{a \cos ^2(e+f x)+b}}{f (a+b) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}+\frac{\sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a \cos ^2(e+f x)+b} F\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{a f \sqrt{\cos ^2(e+f x)} \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}-\frac{\sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a \cos ^2(e+f x)+b} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{a f (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4148

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + b/(1 - ff^2*x^2)^(n/2))^p/(1 - ff^2*x^2)^((m + 1)/2), x
], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] &&  !IntegerQ
[p]

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rule 1974

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  !BinomialMatchQ[{u, v}, x]

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 493

Int[(x_)^(n_)/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[1/b, Int[Sqrt[a +
b*x^n]/Sqrt[c + d*x^n], x], x] - Dist[a/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b, c,
d}, x] && NeQ[b*c - a*d, 0] && (EqQ[n, 2] || EqQ[n, 4]) &&  !(EqQ[n, 2] && 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 (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \left (a+\frac{b}{1-x^2}\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{\sqrt{1-x^2}}{\left (b+a \left (1-x^2\right )\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{\sqrt{1-x^2}}{\left (a+b-a x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{(a+b) f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{(a+b) f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{a f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}-\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{\sqrt{a+b-a x^2}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{a (a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{(a+b) f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}-\frac{\left (\sqrt{b+a \cos ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{a x^2}{a+b}}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{a (a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}+\frac{\left (\sqrt{b+a \cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1-\frac{a x^2}{a+b}}} \, dx,x,\sin (e+f x)\right )}{a f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{(a+b) f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}-\frac{\sqrt{b+a \cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right ) \sqrt{a+b-a \sin ^2(e+f x)}}{a (a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}+\frac{\sqrt{b+a \cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right ) \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}{a f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 9.4885, size = 822, normalized size = 3.59 \[ \frac{(\cos (2 e+2 f x) a+a+2 b)^{3/2} \sec ^3(e+f x) \left (\frac{\sqrt{-\frac{1}{a+b}} \cos (2 (e+f x)) \left (\sqrt{-\frac{1}{a+b}} (\cos (2 e+2 f x) a+a) \left (-2 a^2+(\cos (2 e+2 f x) a+a-4 b) a+2 b (\cos (2 e+2 f x) a+a)\right )-i b (a+2 b) \sqrt{\frac{a-a \cos (2 e+2 f x)}{a+b}} \sqrt{\cos (2 e+2 f x) a+a+2 b} \sqrt{4-\frac{2 (\cos (2 e+2 f x) a+a+2 b)}{b}} E\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{1}{a+b}} \sqrt{\cos (2 e+2 f x) a+a+2 b}}{\sqrt{2}}\right )|\frac{a+b}{b}\right )-i a b \sqrt{\cos (2 e+2 f x) a+a+2 b} \sqrt{\frac{4 a+4 b-2 (\cos (2 e+2 f x) a+a+2 b)}{a+b}} \sqrt{2-\frac{\cos (2 e+2 f x) a+a+2 b}{b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{1}{a+b}} \sqrt{\cos (2 e+2 f x) a+a+2 b}}{\sqrt{2}}\right ),\frac{a+b}{b}\right )\right ) \sec \left (2 \left (e+\frac{1}{2} \left (\cos ^{-1}(\cos (2 e+2 f x))-2 e\right )\right )\right ) \sin (2 e+2 f x)}{4 a^2 b f \sqrt{\frac{(a-a \cos (2 e+2 f x)) (\cos (2 e+2 f x) a+a)}{a^2}} \sqrt{\cos (2 e+2 f x) a+a+2 b} \sqrt{1-\cos ^2(2 e+2 f x)}}-\frac{\left (\frac{2 (a-a \cos (2 e+2 f x)) (\cos (2 e+2 f x) a+a)}{b \sqrt{\cos (2 e+2 f x) a+a+2 b}}+\frac{2 i \sqrt{\frac{a-a \cos (2 e+2 f x)}{a+b}} \sqrt{4-\frac{2 (\cos (2 e+2 f x) a+a+2 b)}{b}} \left (E\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{1}{a+b}} \sqrt{\cos (2 e+2 f x) a+a+2 b}}{\sqrt{2}}\right )|\frac{a+b}{b}\right )-\text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{1}{a+b}} \sqrt{\cos (2 e+2 f x) a+a+2 b}}{\sqrt{2}}\right ),\frac{a+b}{b}\right )\right )}{\sqrt{-\frac{1}{a+b}}}\right ) \sin (2 e+2 f x)}{8 a (a+b) f \sqrt{\frac{(a-a \cos (2 e+2 f x)) (\cos (2 e+2 f x) a+a)}{a^2}} \sqrt{1-\cos ^2(2 e+2 f x)}}\right )}{2 \left (b \sec ^2(e+f x)+a\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e + f*x]^3*(-(((2*(a - a*Cos[2*e + 2*f*x])*(a + a*Cos[2*e + 2*f*x]))
/(b*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]) + ((2*I)*Sqrt[(a - a*Cos[2*e + 2*f*x])/(a + b)]*Sqrt[4 - (2*(a + 2*b +
a*Cos[2*e + 2*f*x]))/b]*(EllipticE[I*ArcSinh[(Sqrt[-(a + b)^(-1)]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]])/Sqrt[2]
], (a + b)/b] - EllipticF[I*ArcSinh[(Sqrt[-(a + b)^(-1)]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]])/Sqrt[2]], (a + b)
/b]))/Sqrt[-(a + b)^(-1)])*Sin[2*e + 2*f*x])/(8*a*(a + b)*f*Sqrt[((a - a*Cos[2*e + 2*f*x])*(a + a*Cos[2*e + 2*
f*x]))/a^2]*Sqrt[1 - Cos[2*e + 2*f*x]^2]) + (Sqrt[-(a + b)^(-1)]*Cos[2*(e + f*x)]*(Sqrt[-(a + b)^(-1)]*(a + a*
Cos[2*e + 2*f*x])*(-2*a^2 + 2*b*(a + a*Cos[2*e + 2*f*x]) + a*(a - 4*b + a*Cos[2*e + 2*f*x])) - I*b*(a + 2*b)*S
qrt[(a - a*Cos[2*e + 2*f*x])/(a + b)]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sqrt[4 - (2*(a + 2*b + a*Cos[2*e + 2*
f*x]))/b]*EllipticE[I*ArcSinh[(Sqrt[-(a + b)^(-1)]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]])/Sqrt[2]], (a + b)/b] -
I*a*b*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sqrt[(4*a + 4*b - 2*(a + 2*b + a*Cos[2*e + 2*f*x]))/(a + b)]*Sqrt[2 -
(a + 2*b + a*Cos[2*e + 2*f*x])/b]*EllipticF[I*ArcSinh[(Sqrt[-(a + b)^(-1)]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]
)/Sqrt[2]], (a + b)/b])*Sec[2*(e + (-2*e + ArcCos[Cos[2*e + 2*f*x]])/2)]*Sin[2*e + 2*f*x])/(4*a^2*b*f*Sqrt[((a
- a*Cos[2*e + 2*f*x])*(a + a*Cos[2*e + 2*f*x]))/a^2]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sqrt[1 - Cos[2*e + 2*
f*x]^2])))/(2*(a + b*Sec[e + f*x]^2)^(3/2))

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Maple [C]  time = 0.5, size = 6593, normalized size = 28.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral(sec(e + f*x)/(a + b*sec(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 )}{{\left (b \sec \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)/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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