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

Optimal. Leaf size=125 \[ \frac {(4 a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{3/2} f (a+b)^{5/2}}+\frac {(4 a+b) \sin (e+f x)}{8 a f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}-\frac {b \sin (e+f x)}{4 a f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2} \]

[Out]

1/8*(4*a+b)*arctanh(sin(f*x+e)*a^(1/2)/(a+b)^(1/2))/a^(3/2)/(a+b)^(5/2)/f-1/4*b*sin(f*x+e)/a/(a+b)/f/(a+b-a*si
n(f*x+e)^2)^2+1/8*(4*a+b)*sin(f*x+e)/a/(a+b)^2/f/(a+b-a*sin(f*x+e)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 125, 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 = {4147, 385, 199, 208} \[ \frac {(4 a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{3/2} f (a+b)^{5/2}}+\frac {(4 a+b) \sin (e+f x)}{8 a f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}-\frac {b \sin (e+f x)}{4 a f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a + b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(8*a^(3/2)*(a + b)^(5/2)*f) - (b*Sin[e + f*x])/(4*a*(a
+ b)*f*(a + b - a*Sin[e + f*x]^2)^2) + ((4*a + b)*Sin[e + f*x])/(8*a*(a + b)^2*f*(a + b - a*Sin[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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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 4147

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[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^
((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n
/2] && IntegerQ[p]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.61, size = 163, normalized size = 1.30 \[ -\frac {\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b)^3 \left (\frac {8 \sin (e+f x)}{\left (-a \sin ^2(e+f x)+a+b\right )^2}-(4 a+b) \left (\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {a} (a+b)^{5/2}}+\frac {4 \sin (e+f x) \left (5 (a+b)-3 a \sin ^2(e+f x)\right )}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)^2}\right )\right )}{192 a f \left (a+b \sec ^2(e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/192*((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*((8*Sin[e + f*x])/(a + b - a*Sin[e + f*x]^2)^2 - (4*a
+ b)*((3*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(Sqrt[a]*(a + b)^(5/2)) + (4*Sin[e + f*x]*(5*(a + b) - 3
*a*Sin[e + f*x]^2))/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2))))/(a*f*(a + b*Sec[e + f*x]^2)^3)

________________________________________________________________________________________

fricas [B]  time = 1.28, size = 544, normalized size = 4.35 \[ \left [\frac {{\left ({\left (4 \, a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{4} + 4 \, a b^{2} + b^{3} + 2 \, {\left (4 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a^{2} + a b} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, {\left (2 \, a^{3} b + a^{2} b^{2} - a b^{3} + {\left (4 \, a^{4} + 5 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}, -\frac {{\left ({\left (4 \, a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{4} + 4 \, a b^{2} + b^{3} + 2 \, {\left (4 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a^{2} - a b} \arctan \left (\frac {\sqrt {-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) - {\left (2 \, a^{3} b + a^{2} b^{2} - a b^{3} + {\left (4 \, a^{4} + 5 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.74, size = 162, normalized size = 1.30 \[ -\frac {\frac {{\left (4 \, a + b\right )} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {-a^{2} - a b}} + \frac {4 \, a^{2} \sin \left (f x + e\right )^{3} + a b \sin \left (f x + e\right )^{3} - 4 \, a^{2} \sin \left (f x + e\right ) - 3 \, a b \sin \left (f x + e\right ) + b^{2} \sin \left (f x + e\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} {\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2}}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((4*a + b)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a^3 + 2*a^2*b + a*b^2)*sqrt(-a^2 - a*b)) + (4*a^2*si
n(f*x + e)^3 + a*b*sin(f*x + e)^3 - 4*a^2*sin(f*x + e) - 3*a*b*sin(f*x + e) + b^2*sin(f*x + e))/((a^3 + 2*a^2*
b + a*b^2)*(a*sin(f*x + e)^2 - a - b)^2))/f

________________________________________________________________________________________

maple [A]  time = 0.93, size = 124, normalized size = 0.99 \[ \frac {\frac {-\frac {\left (4 a +b \right ) \left (\sin ^{3}\left (f x +e \right )\right )}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 a -b \right ) \sin \left (f x +e \right )}{8 \left (a +b \right ) a}}{\left (-a -b +a \left (\sin ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {\left (4 a +b \right ) \arctanh \left (\frac {a \sin \left (f x +e \right )}{\sqrt {\left (a +b \right ) a}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) a \sqrt {\left (a +b \right ) a}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*((-1/8*(4*a+b)/(a^2+2*a*b+b^2)*sin(f*x+e)^3+1/8*(4*a-b)/(a+b)/a*sin(f*x+e))/(-a-b+a*sin(f*x+e)^2)^2+1/8*(4
*a+b)/(a^2+2*a*b+b^2)/a/((a+b)*a)^(1/2)*arctanh(a*sin(f*x+e)/((a+b)*a)^(1/2)))

________________________________________________________________________________________

maxima [A]  time = 0.45, size = 212, normalized size = 1.70 \[ -\frac {\frac {{\left (4 \, a + b\right )} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {2 \, {\left ({\left (4 \, a^{2} + a b\right )} \sin \left (f x + e\right )^{3} - {\left (4 \, a^{2} + 3 \, a b - b^{2}\right )} \sin \left (f x + e\right )\right )}}{a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4} + {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sin \left (f x + e\right )^{4} - 2 \, {\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} \sin \left (f x + e\right )^{2}}}{16 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((4*a + b)*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) + sqrt((a + b)*a)))/((a^3 + 2*a^2*b +
a*b^2)*sqrt((a + b)*a)) + 2*((4*a^2 + a*b)*sin(f*x + e)^3 - (4*a^2 + 3*a*b - b^2)*sin(f*x + e))/(a^5 + 4*a^4*b
+ 6*a^3*b^2 + 4*a^2*b^3 + a*b^4 + (a^5 + 2*a^4*b + a^3*b^2)*sin(f*x + e)^4 - 2*(a^5 + 3*a^4*b + 3*a^3*b^2 + a
^2*b^3)*sin(f*x + e)^2))/f

________________________________________________________________________________________

mupad [B]  time = 4.68, size = 129, normalized size = 1.03 \[ \frac {\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (4\,a+b\right )}{8\,a^{3/2}\,f\,{\left (a+b\right )}^{5/2}}-\frac {\frac {{\sin \left (e+f\,x\right )}^3\,\left (4\,a+b\right )}{8\,{\left (a+b\right )}^2}-\frac {\sin \left (e+f\,x\right )\,\left (4\,a-b\right )}{8\,a\,\left (a+b\right )}}{f\,\left (2\,a\,b+a^2+b^2-{\sin \left (e+f\,x\right )}^2\,\left (2\,a^2+2\,b\,a\right )+a^2\,{\sin \left (e+f\,x\right )}^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________