3.212 \(\int \frac {\sec ^6(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=142 \[ \frac {\left (3 a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 b^{5/2} f (a+b)^{5/2}}-\frac {3 a (a+2 b) \tan (e+f x)}{8 b^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {a \tan (e+f x) \sec ^2(e+f x)}{4 b f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 142, 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, 413, 385, 205} \[ \frac {\left (3 a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 b^{5/2} f (a+b)^{5/2}}-\frac {3 a (a+2 b) \tan (e+f x)}{8 b^2 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {a \tan (e+f x) \sec ^2(e+f x)}{4 b f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

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

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 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

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 ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{\left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {a \sec ^2(e+f x) \tan (e+f x)}{4 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {a+4 b+(3 a+4 b) x^2}{\left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 b (a+b) f}\\ &=-\frac {a \sec ^2(e+f x) \tan (e+f x)}{4 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {3 a (a+2 b) \tan (e+f x)}{8 b^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\left (3 a^2+8 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 b^2 (a+b)^2 f}\\ &=\frac {\left (3 a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 b^{5/2} (a+b)^{5/2} f}-\frac {a \sec ^2(e+f x) \tan (e+f x)}{4 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {3 a (a+2 b) \tan (e+f x)}{8 b^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.90, size = 125, normalized size = 0.88 \[ \frac {\frac {\left (3 a^2+8 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}-\frac {a \sqrt {b} \sin (2 (e+f x)) \left (3 a^2+3 a (a+2 b) \cos (2 (e+f x))+16 a b+16 b^2\right )}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)^2}}{8 b^{5/2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

(((3*a^2 + 8*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(5/2) - (a*Sqrt[b]*(3*a^2 + 16*a
*b + 16*b^2 + 3*a*(a + 2*b)*Cos[2*(e + f*x)])*Sin[2*(e + f*x)])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2))/
(8*b^(5/2)*f)

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fricas [B]  time = 1.45, size = 722, normalized size = 5.08 \[ \left [-\frac {{\left ({\left (3 \, a^{4} + 8 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} + 8 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (3 \, a^{3} b + 8 \, a^{2} b^{2} + 8 \, a b^{3}\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 (3 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 2 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{3} b^{2} + 13 \, a^{2} b^{3} + 8 \, a b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\left ({\left (a^{5} b^{3} + 3 \, a^{4} b^{4} + 3 \, a^{3} b^{5} + a^{2} b^{6}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{4} + 3 \, a^{3} b^{5} + 3 \, a^{2} b^{6} + a b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{5} + 3 \, a^{2} b^{6} + 3 \, a b^{7} + b^{8}\right )} f\right )}}, -\frac {{\left ({\left (3 \, a^{4} + 8 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} + 8 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (3 \, a^{3} b + 8 \, a^{2} b^{2} + 8 \, a b^{3}\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 (3 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 2 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{3} b^{2} + 13 \, a^{2} b^{3} + 8 \, a b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{5} b^{3} + 3 \, a^{4} b^{4} + 3 \, a^{3} b^{5} + a^{2} b^{6}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{4} + 3 \, a^{3} b^{5} + 3 \, a^{2} b^{6} + a b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{5} + 3 \, a^{2} b^{6} + 3 \, a b^{7} + b^{8}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[-1/32*(((3*a^4 + 8*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 + 3*a^2*b^2 + 8*a*b^3 + 8*b^4 + 2*(3*a^3*b + 8*a^2*b^2 +
8*a*b^3)*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*(3*(a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*cos(f*x + e)^3 + (5*a^3*b^2 + 13*a^
2*b^3 + 8*a*b^4)*cos(f*x + e))*sin(f*x + e))/((a^5*b^3 + 3*a^4*b^4 + 3*a^3*b^5 + a^2*b^6)*f*cos(f*x + e)^4 + 2
*(a^4*b^4 + 3*a^3*b^5 + 3*a^2*b^6 + a*b^7)*f*cos(f*x + e)^2 + (a^3*b^5 + 3*a^2*b^6 + 3*a*b^7 + b^8)*f), -1/16*
(((3*a^4 + 8*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 + 3*a^2*b^2 + 8*a*b^3 + 8*b^4 + 2*(3*a^3*b + 8*a^2*b^2 + 8*a*b^
3)*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*(3*(a^4*b + 3*a^3*b^2 + 2*a^2*b^3)*cos(f*x + e)^3 + (5*a^3*b^2 + 13*a^2*b^3 + 8*a*b^4)*cos(f*x
+ e))*sin(f*x + e))/((a^5*b^3 + 3*a^4*b^4 + 3*a^3*b^5 + a^2*b^6)*f*cos(f*x + e)^4 + 2*(a^4*b^4 + 3*a^3*b^5 +
3*a^2*b^6 + a*b^7)*f*cos(f*x + e)^2 + (a^3*b^5 + 3*a^2*b^6 + 3*a*b^7 + b^8)*f)]

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giac [A]  time = 0.36, size = 193, normalized size = 1.36 \[ \frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (3 \, a^{2} + 8 \, a b + 8 \, b^{2}\right )}}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \sqrt {a b + b^{2}}} - \frac {5 \, a^{2} b \tan \left (f x + e\right )^{3} + 8 \, a b^{2} \tan \left (f x + e\right )^{3} + 3 \, a^{3} \tan \left (f x + e\right ) + 11 \, a^{2} b \tan \left (f x + e\right ) + 8 \, a b^{2} \tan \left (f x + e\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6/(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)))*(3*a^2 + 8*a*b + 8*b^2)/((
a^2*b^2 + 2*a*b^3 + b^4)*sqrt(a*b + b^2)) - (5*a^2*b*tan(f*x + e)^3 + 8*a*b^2*tan(f*x + e)^3 + 3*a^3*tan(f*x +
e) + 11*a^2*b*tan(f*x + e) + 8*a*b^2*tan(f*x + e))/((a^2*b^2 + 2*a*b^3 + b^4)*(b*tan(f*x + e)^2 + a + b)^2))/
f

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maple [B]  time = 0.64, size = 294, normalized size = 2.07 \[ -\frac {5 a^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} b \left (a^{2}+2 a b +b^{2}\right )}-\frac {a \left (\tan ^{3}\left (f x +e \right )\right )}{f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} \left (a^{2}+2 a b +b^{2}\right )}-\frac {3 a^{2} \tan \left (f x +e \right )}{8 f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2} b^{2} \left (a +b \right )}-\frac {a \tan \left (f x +e \right )}{b \left (a +b \right ) f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {3 \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) a^{2}}{8 f \left (a^{2}+2 a b +b^{2}\right ) b^{2} \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) a}{f \left (a^{2}+2 a b +b^{2}\right ) b \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \left (a^{2}+2 a b +b^{2}\right ) \sqrt {\left (a +b \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x)

[Out]

-5/8/f/(a+b+b*tan(f*x+e)^2)^2*a^2/b/(a^2+2*a*b+b^2)*tan(f*x+e)^3-1/f/(a+b+b*tan(f*x+e)^2)^2*a/(a^2+2*a*b+b^2)*
tan(f*x+e)^3-3/8/f/(a+b+b*tan(f*x+e)^2)^2*a^2/b^2/(a+b)*tan(f*x+e)-a*tan(f*x+e)/b/(a+b)/f/(a+b+b*tan(f*x+e)^2)
^2+3/8/f/(a^2+2*a*b+b^2)/b^2/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))*a^2+1/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/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 = 0.45, size = 210, normalized size = 1.48 \[ \frac {\frac {{\left (3 \, a^{2} + 8 \, a b + 8 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \sqrt {{\left (a + b\right )} b}} - \frac {{\left (5 \, a^{2} b + 8 \, a b^{2}\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a^{3} + 11 \, a^{2} b + 8 \, a b^{2}\right )} \tan \left (f x + e\right )}{a^{4} b^{2} + 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} + 4 \, a b^{5} + b^{6} + {\left (a^{2} b^{4} + 2 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{2}}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

1/8*((3*a^2 + 8*a*b + 8*b^2)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^2*b^2 + 2*a*b^3 + b^4)*sqrt((a + b)*b)
) - ((5*a^2*b + 8*a*b^2)*tan(f*x + e)^3 + (3*a^3 + 11*a^2*b + 8*a*b^2)*tan(f*x + e))/(a^4*b^2 + 4*a^3*b^3 + 6*
a^2*b^4 + 4*a*b^5 + b^6 + (a^2*b^4 + 2*a*b^5 + b^6)*tan(f*x + e)^4 + 2*(a^3*b^3 + 3*a^2*b^4 + 3*a*b^5 + b^6)*t
an(f*x + e)^2))/f

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mupad [B]  time = 5.21, size = 149, normalized size = 1.05 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{8\,b^{5/2}\,f\,{\left (a+b\right )}^{5/2}}-\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (5\,a^2+8\,b\,a\right )}{8\,b\,{\left (a+b\right )}^2}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (3\,a^2+8\,b\,a\right )}{8\,b^2\,\left (a+b\right )}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)^3),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**6/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Integral(sec(e + f*x)**6/(a + b*sec(e + f*x)**2)**3, x)

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