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

Optimal. Leaf size=144 \[ -\frac {3 b (2 a+b) \sin (e+f x)}{8 a^2 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac {\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{5/2} f (a+b)^{5/2}}-\frac {b \sin (e+f x) \cos ^2(e+f x)}{4 a f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4147, 413, 385, 208} \[ \frac {\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{5/2} f (a+b)^{5/2}}-\frac {3 b (2 a+b) \sin (e+f x)}{8 a^2 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}-\frac {b \sin (e+f x) \cos ^2(e+f x)}{4 a f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [C]  time = 6.82, size = 927, normalized size = 6.44 \[ \frac {(\cos (2 (e+f x)) a+a+2 b) \sec ^5(e+f x) \left (32 \sqrt {a} (a+b)^{3/2} \sqrt {(\cos (e)-i \sin (e))^2} \tan (e+f x) b^2-8 \sqrt {a} \sqrt {a+b} (8 a+5 b) (\cos (2 (e+f x)) a+a+2 b) \sqrt {(\cos (e)-i \sin (e))^2} \tan (e+f x) b-2 i \left (8 a^2+8 b a+3 b^2\right ) \tan ^{-1}\left (\frac {(a+b) \sin (e)}{(a+b) \cos (e)-\sqrt {a} \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} (\cos (2 e)+i \sin (2 e)) \sin (e+f x)}\right ) (\cos (2 (e+f x)) a+a+2 b)^2 \sec (e+f x) (\cos (e)-i \sin (e))+\left (8 a^2+8 b a+3 b^2\right ) (\cos (2 (e+f x)) a+a+2 b)^2 \log \left (-\cos (2 (e+f x)) a-2 i \sin (2 e) a+a+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt {a}+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt {a}+2 (a+b) \cos (2 e)-2 i b \sin (2 e)\right ) \sec (e+f x) (\cos (e)-i \sin (e))-\left (8 a^2+8 b a+3 b^2\right ) (\cos (2 (e+f x)) a+a+2 b)^2 \log \left (\cos (2 (e+f x)) a+2 i \sin (2 e) a-a+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt {a}+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt {a}-2 (a+b) \cos (2 e)+2 i b \sin (2 e)\right ) \sec (e+f x) (\cos (e)-i \sin (e))+2 \left (8 a^2+8 b a+3 b^2\right ) \tan ^{-1}\left (\frac {2 \sin (e) \left (\sin (2 e) a+i a-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt {a}+\sqrt {a+b} \cos (f x) \sqrt {(\cos (e)-i \sin (e))^2} \sqrt {a}-\sqrt {a+b} \cos (2 e+f x) \sqrt {(\cos (e)-i \sin (e))^2} \sqrt {a}+i b+i (a+b) \cos (2 e)+b \sin (2 e)\right )}{i (a+3 b) \cos (e)+i (a+b) \cos (3 e)+i a \cos (e+2 f x)+i a \cos (3 e+2 f x)+3 a \sin (e)+b \sin (e)+a \sin (3 e)+b \sin (3 e)+a \sin (e+2 f x)-a \sin (3 e+2 f x)}\right ) (\cos (2 (e+f x)) a+a+2 b)^2 \sec (e+f x) (i \cos (e)+\sin (e))\right )}{256 a^{5/2} (a+b)^{5/2} f \left (b \sec ^2(e+f x)+a\right )^3 \sqrt {(\cos (e)-i \sin (e))^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^5*((-2*I)*(8*a^2 + 8*a*b + 3*b^2)*ArcTan[((a + b)*Sin[e])/((a + b
)*Cos[e] - Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*(Cos[2*e] + I*Sin[2*e])*Sin[e + f*x])]*(a + 2*b + a
*Cos[2*(e + f*x)])^2*Sec[e + f*x]*(Cos[e] - I*Sin[e]) + (8*a^2 + 8*a*b + 3*b^2)*(a + 2*b + a*Cos[2*(e + f*x)])
^2*Log[a + 2*(a + b)*Cos[2*e] - a*Cos[2*(e + f*x)] - (2*I)*a*Sin[2*e] - (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a +
b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]]*Se
c[e + f*x]*(Cos[e] - I*Sin[e]) - (8*a^2 + 8*a*b + 3*b^2)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Log[-a - 2*(a + b)*C
os[2*e] + a*Cos[2*(e + f*x)] + (2*I)*a*Sin[2*e] + (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Si
n[e])^2]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]]*Sec[e + f*x]*(Cos[e] - I
*Sin[e]) + 2*(8*a^2 + 8*a*b + 3*b^2)*ArcTan[(2*Sin[e]*(I*a + I*b + I*(a + b)*Cos[2*e] + Sqrt[a]*Sqrt[a + b]*Co
s[f*x]*Sqrt[(Cos[e] - I*Sin[e])^2] - Sqrt[a]*Sqrt[a + b]*Cos[2*e + f*x]*Sqrt[(Cos[e] - I*Sin[e])^2] + a*Sin[2*
e] + b*Sin[2*e] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[f*x] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos
[e] - I*Sin[e])^2]*Sin[2*e + f*x]))/(I*(a + 3*b)*Cos[e] + I*(a + b)*Cos[3*e] + I*a*Cos[e + 2*f*x] + I*a*Cos[3*
e + 2*f*x] + 3*a*Sin[e] + b*Sin[e] + a*Sin[3*e] + b*Sin[3*e] + a*Sin[e + 2*f*x] - a*Sin[3*e + 2*f*x])]*(a + 2*
b + a*Cos[2*(e + f*x)])^2*Sec[e + f*x]*(I*Cos[e] + Sin[e]) + 32*Sqrt[a]*b^2*(a + b)^(3/2)*Sqrt[(Cos[e] - I*Sin
[e])^2]*Tan[e + f*x] - 8*Sqrt[a]*b*Sqrt[a + b]*(8*a + 5*b)*(a + 2*b + a*Cos[2*(e + f*x)])*Sqrt[(Cos[e] - I*Sin
[e])^2]*Tan[e + f*x]))/(256*a^(5/2)*(a + b)^(5/2)*f*(a + b*Sec[e + f*x]^2)^3*Sqrt[(Cos[e] - I*Sin[e])^2])

________________________________________________________________________________________

fricas [B]  time = 1.20, size = 613, normalized size = 4.26 \[ \left [\frac {{\left ({\left (8 \, a^{4} + 8 \, a^{3} b + 3 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{2} b^{2} + 8 \, a b^{3} + 3 \, b^{4} + 2 \, {\left (8 \, a^{3} b + 8 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a^{2} + a b} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) - 2 \, {\left (6 \, a^{3} b^{2} + 9 \, a^{2} b^{3} + 3 \, a b^{4} + {\left (8 \, a^{4} b + 13 \, a^{3} b^{2} + 5 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{7} b + 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} + a^{4} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} f\right )}}, -\frac {{\left ({\left (8 \, a^{4} + 8 \, a^{3} b + 3 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{2} b^{2} + 8 \, a b^{3} + 3 \, b^{4} + 2 \, {\left (8 \, a^{3} b + 8 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a^{2} - a b} \arctan \left (\frac {\sqrt {-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) + {\left (6 \, a^{3} b^{2} + 9 \, a^{2} b^{3} + 3 \, a b^{4} + {\left (8 \, a^{4} b + 13 \, a^{3} b^{2} + 5 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{7} b + 3 \, a^{6} b^{2} + 3 \, a^{5} b^{3} + a^{4} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{6} b^{2} + 3 \, a^{5} b^{3} + 3 \, a^{4} b^{4} + a^{3} b^{5}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(((8*a^4 + 8*a^3*b + 3*a^2*b^2)*cos(f*x + e)^4 + 8*a^2*b^2 + 8*a*b^3 + 3*b^4 + 2*(8*a^3*b + 8*a^2*b^2 +
3*a*b^3)*cos(f*x + e)^2)*sqrt(a^2 + a*b)*log(-(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*(6*a^3*b^2 + 9*a^2*b^3 + 3*a*b^4 + (8*a^4*b + 13*a^3*b^2 + 5*a^2*b^3)*cos(f*x + e)^2
)*sin(f*x + e))/((a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a
^4*b^4)*f*cos(f*x + e)^2 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f), -1/8*(((8*a^4 + 8*a^3*b + 3*a^2*b^2
)*cos(f*x + e)^4 + 8*a^2*b^2 + 8*a*b^3 + 3*b^4 + 2*(8*a^3*b + 8*a^2*b^2 + 3*a*b^3)*cos(f*x + e)^2)*sqrt(-a^2 -
a*b)*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a + b)) + (6*a^3*b^2 + 9*a^2*b^3 + 3*a*b^4 + (8*a^4*b + 13*a^3*b^2
+ 5*a^2*b^3)*cos(f*x + e)^2)*sin(f*x + e))/((a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^7*b
+ 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*f*cos(f*x + e)^2 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f)]

________________________________________________________________________________________

giac [A]  time = 1.10, size = 185, normalized size = 1.28 \[ -\frac {\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {-a^{2} - a b}} - \frac {8 \, a^{2} b \sin \left (f x + e\right )^{3} + 5 \, a b^{2} \sin \left (f x + e\right )^{3} - 8 \, a^{2} b \sin \left (f x + e\right ) - 11 \, a b^{2} \sin \left (f x + e\right ) - 3 \, b^{3} \sin \left (f x + e\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2}}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.96, size = 142, normalized size = 0.99 \[ \frac {-\frac {-\frac {b \left (8 a +5 b \right ) \left (\sin ^{3}\left (f x +e \right )\right )}{8 a \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (8 a +3 b \right ) b \sin \left (f x +e \right )}{8 a^{2} \left (a +b \right )}}{\left (-a -b +a \left (\sin ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {\left (8 a^{2}+8 a b +3 b^{2}\right ) \arctanh \left (\frac {a \sin \left (f x +e \right )}{\sqrt {\left (a +b \right ) a}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) a^{2} \sqrt {\left (a +b \right ) a}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.45, size = 233, normalized size = 1.62 \[ -\frac {\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {2 \, {\left ({\left (8 \, a^{2} b + 5 \, a b^{2}\right )} \sin \left (f x + e\right )^{3} - {\left (8 \, a^{2} b + 11 \, a b^{2} + 3 \, b^{3}\right )} \sin \left (f x + e\right )\right )}}{a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4} + {\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} \sin \left (f x + e\right )^{4} - 2 \, {\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \sin \left (f x + e\right )^{2}}}{16 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((8*a^2 + 8*a*b + 3*b^2)*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) + sqrt((a + b)*a)))/((a^
4 + 2*a^3*b + a^2*b^2)*sqrt((a + b)*a)) - 2*((8*a^2*b + 5*a*b^2)*sin(f*x + e)^3 - (8*a^2*b + 11*a*b^2 + 3*b^3)
*sin(f*x + e))/(a^6 + 4*a^5*b + 6*a^4*b^2 + 4*a^3*b^3 + a^2*b^4 + (a^6 + 2*a^5*b + a^4*b^2)*sin(f*x + e)^4 - 2
*(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*sin(f*x + e)^2))/f

________________________________________________________________________________________

mupad [B]  time = 0.25, size = 149, normalized size = 1.03 \[ \frac {\frac {{\sin \left (e+f\,x\right )}^3\,\left (5\,b^2+8\,a\,b\right )}{8\,a\,{\left (a+b\right )}^2}-\frac {\sin \left (e+f\,x\right )\,\left (3\,b^2+8\,a\,b\right )}{8\,a^2\,\left (a+b\right )}}{f\,\left (2\,a\,b+a^2+b^2-{\sin \left (e+f\,x\right )}^2\,\left (2\,a^2+2\,b\,a\right )+a^2\,{\sin \left (e+f\,x\right )}^4\right )}+\frac {\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (8\,a^2+8\,a\,b+3\,b^2\right )}{8\,a^{5/2}\,f\,{\left (a+b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________