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]

1/8*(a+4*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/b^(3/2)/(a+b)^(5/2)/f-1/4*a*tan(f*x+e)/b/(a+b)/f/(a+b+b*tan
(f*x+e)^2)^2+1/8*(a+4*b)*tan(f*x+e)/b/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.10, antiderivative size = 123, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [C]  time = 3.50, size = 283, normalized size = 2.30 \[ \frac {\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac {((a+4 b) \sin (2 e)-(a-2 b) \sin (2 f x)) (a \cos (2 (e+f x))+a+2 b)}{b (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}-\frac {4 (a+b) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{a (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}-\frac {(a+4 b) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{b \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{64 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^3} \]

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)

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fricas [B]  time = 0.88, size = 654, normalized size = 5.32 \[ \left [-\frac {{\left ({\left (a^{3} + 4 \, a^{2} b\right )} \cos \left (f x + e\right )^{4} + a b^{2} + 4 \, b^{3} + 2 \, {\left (a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 4 \, {\left ({\left (a^{3} b - a^{2} b^{2} - 2 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{2} b^{2} + 5 \, a b^{3} + 4 \, b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\left ({\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{3} + 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} + a b^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{4} + 3 \, a^{2} b^{5} + 3 \, a b^{6} + b^{7}\right )} f\right )}}, -\frac {{\left ({\left (a^{3} + 4 \, a^{2} b\right )} \cos \left (f x + e\right )^{4} + a b^{2} + 4 \, b^{3} + 2 \, {\left (a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 2 \, {\left ({\left (a^{3} b - a^{2} b^{2} - 2 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{2} b^{2} + 5 \, a b^{3} + 4 \, b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{3} + 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} + a b^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{4} + 3 \, a^{2} b^{5} + 3 \, a b^{6} + b^{7}\right )} f\right )}}\right ] \]

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)]

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giac [A]  time = 1.36, size = 171, normalized size = 1.39 \[ \frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (a + 4 \, b\right )}}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sqrt {a b + b^{2}}} + \frac {a b \tan \left (f x + e\right )^{3} + 4 \, b^{2} \tan \left (f x + e\right )^{3} - a^{2} \tan \left (f x + e\right ) + 3 \, a b \tan \left (f x + e\right ) + 4 \, b^{2} \tan \left (f x + e\right )}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}}}{8 \, f} \]

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

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maple [B]  time = 0.62, size = 238, normalized size = 1.93 \[ \frac {a \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (\tan ^{3}\left (f x +e \right )\right ) b}{2 f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} \left (a^{2}+2 a b +b^{2}\right )}-\frac {a \tan \left (f x +e \right )}{8 b \left (a +b \right ) f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {\tan \left (f x +e \right )}{2 \left (a +b \right ) f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {\arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) a}{8 f \left (a^{2}+2 a b +b^{2}\right ) b \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f \left (a^{2}+2 a b +b^{2}\right ) \sqrt {\left (a +b \right ) b}} \]

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))

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maxima [A]  time = 0.46, size = 187, normalized size = 1.52 \[ \frac {\frac {{\left (a + 4 \, b\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {{\left (a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{3} - {\left (a^{2} - 3 \, a b - 4 \, b^{2}\right )} \tan \left (f x + e\right )}{a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5} + {\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{2}}}{8 \, f} \]

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]

1/8*((a + 4*b)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^2*b + 2*a*b^2 + b^3)*sqrt((a + b)*b)) + ((a*b + 4*b^
2)*tan(f*x + e)^3 - (a^2 - 3*a*b - 4*b^2)*tan(f*x + e))/(a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5 + (a^2*
b^3 + 2*a*b^4 + b^5)*tan(f*x + e)^4 + 2*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*tan(f*x + e)^2))/f

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mupad [B]  time = 5.10, size = 125, normalized size = 1.02 \[ \frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (a+4\,b\right )}{8\,{\left (a+b\right )}^2}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a-4\,b\right )}{8\,b\,\left (a+b\right )}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (a+4\,b\right )}{8\,b^{3/2}\,f\,{\left (a+b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)^4*(a + b/cos(e + f*x)^2)^3),x)

[Out]

((tan(e + f*x)^3*(a + 4*b))/(8*(a + b)^2) - (tan(e + f*x)*(a - 4*b))/(8*b*(a + b)))/(f*(2*a*b + a^2 + b^2 + ta
n(e + f*x)^2*(2*a*b + 2*b^2) + b^2*tan(e + f*x)^4)) + (atan((b^(1/2)*tan(e + f*x))/(a + b)^(1/2))*(a + 4*b))/(
8*b^(3/2)*f*(a + b)^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \]

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]

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

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