3.213 \(\int \frac{\sec ^4(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=123 \[ \frac{(a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 b^{3/2} f (a+b)^{5/2}}+\frac{(a+4 b) \tan (e+f x)}{8 b f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{a \tan (e+f x)}{4 b f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[Out]

((a + 4*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*b^(3/2)*(a + b)^(5/2)*f) - (a*Tan[e + f*x])/(4*b*(a
+ b)*f*(a + b + b*Tan[e + f*x]^2)^2) + ((a + 4*b)*Tan[e + f*x])/(8*b*(a + b)^2*f*(a + b + b*Tan[e + f*x]^2))

________________________________________________________________________________________

Rubi [A]  time = 0.101683, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4146, 385, 199, 205} \[ \frac{(a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 b^{3/2} f (a+b)^{5/2}}+\frac{(a+4 b) \tan (e+f x)}{8 b f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac{a \tan (e+f x)}{4 b f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^4/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

((a + 4*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*b^(3/2)*(a + b)^(5/2)*f) - (a*Tan[e + f*x])/(4*b*(a
+ b)*f*(a + b + b*Tan[e + f*x]^2)^2) + ((a + 4*b)*Tan[e + f*x])/(8*b*(a + b)^2*f*(a + b + b*Tan[e + f*x]^2))

Rule 4146

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{\left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a \tan (e+f x)}{4 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{(a+4 b) \operatorname{Subst}\left (\int \frac{1}{\left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 b (a+b) f}\\ &=-\frac{a \tan (e+f x)}{4 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{(a+4 b) \tan (e+f x)}{8 b (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{(a+4 b) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 b (a+b)^2 f}\\ &=\frac{(a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 b^{3/2} (a+b)^{5/2} f}-\frac{a \tan (e+f x)}{4 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{(a+4 b) \tan (e+f x)}{8 b (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 3.8305, size = 283, normalized size = 2.3 \[ \frac{\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac{((a+4 b) \sin (2 e)-(a-2 b) \sin (2 f x)) (a \cos (2 (e+f x))+a+2 b)}{b (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}-\frac{4 (a+b) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{a (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}-\frac{(a+4 b) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{b \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{64 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^4/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*(-(((a + 4*b)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a +
2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x
)])^2*(Cos[2*e] - I*Sin[2*e]))/(b*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])) - (4*(a + b)*((a + 2*b)*Sin[2*e]
- a*Sin[2*f*x]))/(a*(Cos[e] - Sin[e])*(Cos[e] + Sin[e])) + ((a + 2*b + a*Cos[2*(e + f*x)])*((a + 4*b)*Sin[2*e
] - (a - 2*b)*Sin[2*f*x]))/(b*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(64*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^3)

________________________________________________________________________________________

Maple [B]  time = 0.071, size = 238, normalized size = 1.9 \begin{align*}{\frac{a \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}b}{2\,f \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-{\frac{a\tan \left ( fx+e \right ) }{8\, \left ( a+b \right ) bf \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{\tan \left ( fx+e \right ) }{ \left ( 2\,a+2\,b \right ) f \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{a}{8\,f \left ({a}^{2}+2\,ab+{b}^{2} \right ) b}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}+{\frac{1}{2\,f \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x)

[Out]

1/8/f/(a+b+b*tan(f*x+e)^2)^2*a/(a^2+2*a*b+b^2)*tan(f*x+e)^3+1/2/f/(a+b+b*tan(f*x+e)^2)^2/(a^2+2*a*b+b^2)*tan(f
*x+e)^3*b-1/8*a*tan(f*x+e)/b/(a+b)/f/(a+b+b*tan(f*x+e)^2)^2+1/2*tan(f*x+e)/(a+b)/f/(a+b+b*tan(f*x+e)^2)^2+1/8/
f/(a^2+2*a*b+b^2)/b/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))*a+1/2/f/(a^2+2*a*b+b^2)/((a+b)*b)^(1/
2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 0.667593, size = 1449, normalized size = 11.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[-1/32*(((a^3 + 4*a^2*b)*cos(f*x + e)^4 + a*b^2 + 4*b^3 + 2*(a^2*b + 4*a*b^2)*cos(f*x + e)^2)*sqrt(-a*b - b^2)
*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a + 2*b)*cos(f*x + e)^3 -
b*cos(f*x + e))*sqrt(-a*b - b^2)*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) + 4*((
a^3*b - a^2*b^2 - 2*a*b^3)*cos(f*x + e)^3 - (a^2*b^2 + 5*a*b^3 + 4*b^4)*cos(f*x + e))*sin(f*x + e))/((a^5*b^2
+ 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 + 2*(a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 + a*b^6)*f*cos(f*x +
e)^2 + (a^3*b^4 + 3*a^2*b^5 + 3*a*b^6 + b^7)*f), -1/16*(((a^3 + 4*a^2*b)*cos(f*x + e)^4 + a*b^2 + 4*b^3 + 2*(a
^2*b + 4*a*b^2)*cos(f*x + e)^2)*sqrt(a*b + b^2)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)/(sqrt(a*b + b^2)*cos
(f*x + e)*sin(f*x + e))) + 2*((a^3*b - a^2*b^2 - 2*a*b^3)*cos(f*x + e)^3 - (a^2*b^2 + 5*a*b^3 + 4*b^4)*cos(f*x
+ e))*sin(f*x + e))/((a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 + 2*(a^4*b^3 + 3*a^3*b^4 +
3*a^2*b^5 + a*b^6)*f*cos(f*x + e)^2 + (a^3*b^4 + 3*a^2*b^5 + 3*a*b^6 + b^7)*f)]

________________________________________________________________________________________

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(sec(f*x+e)**4/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.39014, size = 231, normalized size = 1.88 \begin{align*} \frac{\frac{{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )}{\left (a + 4 \, b\right )}}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sqrt{a b + b^{2}}} + \frac{a b \tan \left (f x + e\right )^{3} + 4 \, b^{2} \tan \left (f x + e\right )^{3} - a^{2} \tan \left (f x + e\right ) + 3 \, a b \tan \left (f x + e\right ) + 4 \, b^{2} \tan \left (f x + e\right )}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

1/8*((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*(a + 4*b)/((a^2*b + 2*a*b^
2 + b^3)*sqrt(a*b + b^2)) + (a*b*tan(f*x + e)^3 + 4*b^2*tan(f*x + e)^3 - a^2*tan(f*x + e) + 3*a*b*tan(f*x + e)
+ 4*b^2*tan(f*x + e))/((a^2*b + 2*a*b^2 + b^3)*(b*tan(f*x + e)^2 + a + b)^2))/f