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]

((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))

________________________________________________________________________________________

Rubi [A]  time = 0.156647, antiderivative size = 142, 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, 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 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 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 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 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 ^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*}

Mathematica [A]  time = 0.99519, 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)

________________________________________________________________________________________

Maple [B]  time = 0.073, size = 294, normalized size = 2.1 \begin{align*} -{\frac{5\,{a}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}b \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-{\frac{a \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{f \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-{\frac{3\,{a}^{2}\tan \left ( fx+e \right ) }{8\,f \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}{b}^{2} \left ( a+b \right ) }}-{\frac{a\tan \left ( fx+e \right ) }{ \left ( a+b \right ) bf \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{3\,{a}^{2}}{8\,f \left ({a}^{2}+2\,ab+{b}^{2} \right ){b}^{2}}\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{a}{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}{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)^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))

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 0.697212, size = 1597, normalized size = 11.25 \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)^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)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.4028, size = 261, normalized size = 1.84 \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 (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} \end{align*}

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