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

Optimal. Leaf size=156 \[ -\frac {3 b \left (4 (a+b)^2+(2 a+b)^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{7/2} f (a+b)^{5/2}}-\frac {b^3 \sin (e+f x)}{4 a^3 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}+\frac {3 b^2 (4 a+3 b) \sin (e+f x)}{8 a^3 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac {\sin (e+f x)}{a^3 f} \]

[Out]

-3/8*b*(4*(a+b)^2+(2*a+b)^2)*arctanh(sin(f*x+e)*a^(1/2)/(a+b)^(1/2))/a^(7/2)/(a+b)^(5/2)/f+sin(f*x+e)/a^3/f-1/
4*b^3*sin(f*x+e)/a^3/(a+b)/f/(a+b-a*sin(f*x+e)^2)^2+3/8*b^2*(4*a+3*b)*sin(f*x+e)/a^3/(a+b)^2/f/(a+b-a*sin(f*x+
e)^2)

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Rubi [A]  time = 0.19, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4147, 390, 1157, 385, 208} \[ -\frac {b^3 \sin (e+f x)}{4 a^3 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}+\frac {3 b^2 (4 a+3 b) \sin (e+f x)}{8 a^3 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}-\frac {3 b \left (4 (a+b)^2+(2 a+b)^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{7/2} f (a+b)^{5/2}}+\frac {\sin (e+f x)}{a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(4*(a + b)^2 + (2*a + b)^2)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(8*a^(7/2)*(a + b)^(5/2)*f) + S
in[e + f*x]/(a^3*f) - (b^3*Sin[e + f*x])/(4*a^3*(a + b)*f*(a + b - a*Sin[e + f*x]^2)^2) + (3*b^2*(4*a + 3*b)*S
in[e + f*x])/(8*a^3*(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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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 {\cos (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 )^3}{\left (a+b-a x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a^3}-\frac {b \left (3 a^2+3 a b+b^2\right )-3 a b (2 a+b) x^2+3 a^2 b x^4}{a^3 \left (a+b-a x^2\right )^3}\right ) \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\sin (e+f x)}{a^3 f}-\frac {\operatorname {Subst}\left (\int \frac {b \left (3 a^2+3 a b+b^2\right )-3 a b (2 a+b) x^2+3 a^2 b x^4}{\left (a+b-a x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{a^3 f}\\ &=\frac {\sin (e+f x)}{a^3 f}-\frac {b^3 \sin (e+f x)}{4 a^3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {-3 b (2 a+b)^2+12 a b (a+b) x^2}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 a^3 (a+b) f}\\ &=\frac {\sin (e+f x)}{a^3 f}-\frac {b^3 \sin (e+f x)}{4 a^3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {3 b^2 (4 a+3 b) \sin (e+f x)}{8 a^3 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )}-\frac {\left (3 b \left (4 (a+b)^2+(2 a+b)^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{8 a^3 (a+b)^2 f}\\ &=-\frac {3 b \left (4 (a+b)^2+(2 a+b)^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{7/2} (a+b)^{5/2} f}+\frac {\sin (e+f x)}{a^3 f}-\frac {b^3 \sin (e+f x)}{4 a^3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {3 b^2 (4 a+3 b) \sin (e+f x)}{8 a^3 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [C]  time = 7.51, size = 2382, normalized size = 15.27 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[f*x]*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*Sin[e])/(8*a^3*f*(a + b*Sec[e + f*x]^2)^3) + ((8*a^2
*b + 12*a*b^2 + 5*b^3)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((((-3*I)/128)*ArcTan[((-I)*a*Cos[e] -
I*b*Cos[e] + I*a*Cos[3*e] + I*b*Cos[3*e] + a*Sin[e] + b*Sin[e] - Sqrt[a]*Sqrt[a + b]*Cos[e - f*x]*Sqrt[Cos[2*e
] - I*Sin[2*e]] + Sqrt[a]*Sqrt[a + b]*Cos[3*e + f*x]*Sqrt[Cos[2*e] - I*Sin[2*e]] + a*Sin[3*e] + b*Sin[3*e] - I
*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[e - f*x] - (2*I)*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Si
n[2*e]]*Sin[e + f*x] + I*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[3*e + f*x])/(a*Cos[e] + 3*b*Cos[e
] + a*Cos[3*e] + b*Cos[3*e] + a*Cos[e + 2*f*x] + a*Cos[3*e + 2*f*x] - (3*I)*a*Sin[e] - I*b*Sin[e] - I*a*Sin[3*
e] - I*b*Sin[3*e] - I*a*Sin[e + 2*f*x] + I*a*Sin[3*e + 2*f*x])]*Cos[e])/(a^(7/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] -
I*Sin[2*e]]) - (3*ArcTan[((-I)*a*Cos[e] - I*b*Cos[e] + I*a*Cos[3*e] + I*b*Cos[3*e] + a*Sin[e] + b*Sin[e] - Sq
rt[a]*Sqrt[a + b]*Cos[e - f*x]*Sqrt[Cos[2*e] - I*Sin[2*e]] + Sqrt[a]*Sqrt[a + b]*Cos[3*e + f*x]*Sqrt[Cos[2*e]
- I*Sin[2*e]] + a*Sin[3*e] + b*Sin[3*e] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[e - f*x] - (2*
I)*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[e + f*x] + I*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[
2*e]]*Sin[3*e + f*x])/(a*Cos[e] + 3*b*Cos[e] + a*Cos[3*e] + b*Cos[3*e] + a*Cos[e + 2*f*x] + a*Cos[3*e + 2*f*x]
- (3*I)*a*Sin[e] - I*b*Sin[e] - I*a*Sin[3*e] - I*b*Sin[3*e] - I*a*Sin[e + 2*f*x] + I*a*Sin[3*e + 2*f*x])]*Sin
[e])/(128*a^(7/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]])))/((a + b)^2*(a + b*Sec[e + f*x]^2)^3) + ((8*a^2*
b + 12*a*b^2 + 5*b^3)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((3*ArcTanh[(2*(a + b)*Sin[e])/((-2*I)*a
*Cos[e] - (2*I)*b*Cos[e] - Sqrt[a]*Sqrt[a + b]*Cos[e - f*x]*Sqrt[Cos[2*e] - I*Sin[2*e]] + Sqrt[a]*Sqrt[a + b]*
Cos[3*e + f*x]*Sqrt[Cos[2*e] - I*Sin[2*e]] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[e - f*x] +
I*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[3*e + f*x])]*Cos[e])/(128*a^(7/2)*Sqrt[a + b]*f*Sqrt[Cos
[2*e] - I*Sin[2*e]]) - (((3*I)/128)*ArcTanh[(2*(a + b)*Sin[e])/((-2*I)*a*Cos[e] - (2*I)*b*Cos[e] - Sqrt[a]*Sqr
t[a + b]*Cos[e - f*x]*Sqrt[Cos[2*e] - I*Sin[2*e]] + Sqrt[a]*Sqrt[a + b]*Cos[3*e + f*x]*Sqrt[Cos[2*e] - I*Sin[2
*e]] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[e - f*x] + I*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] -
I*Sin[2*e]]*Sin[3*e + f*x])]*Sin[e])/(a^(7/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]])))/((a + b)^2*(a + b*S
ec[e + f*x]^2)^3) + ((8*a^2*b + 12*a*b^2 + 5*b^3)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((-3*Cos[e]*
Log[a + 2*a*Cos[2*e] + 2*b*Cos[2*e] - a*Cos[2*e + 2*f*x] - (2*I)*a*Sin[2*e] - (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqr
t[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[2*e + f*
x]])/(256*a^(7/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]]) + (((3*I)/256)*Log[a + 2*a*Cos[2*e] + 2*b*Cos[2*e
] - a*Cos[2*e + 2*f*x] - (2*I)*a*Sin[2*e] - (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e
]]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[2*e + f*x]]*Sin[e])/(a^(7/2)*Sqrt[a + b]*f
*Sqrt[Cos[2*e] - I*Sin[2*e]])))/((a + b)^2*(a + b*Sec[e + f*x]^2)^3) + ((8*a^2*b + 12*a*b^2 + 5*b^3)*(a + 2*b
+ a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((3*Cos[e]*Log[-a - 2*a*Cos[2*e] - 2*b*Cos[2*e] + a*Cos[2*e + 2*f*x] +
(2*I)*a*Sin[2*e] + (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[f*x] + 2*Sqrt[a]*S
qrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[2*e + f*x]])/(256*a^(7/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]]
) - (((3*I)/256)*Log[-a - 2*a*Cos[2*e] - 2*b*Cos[2*e] + a*Cos[2*e + 2*f*x] + (2*I)*a*Sin[2*e] + (2*I)*b*Sin[2*
e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[
2*e]]*Sin[2*e + f*x]]*Sin[e])/(a^(7/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]])))/((a + b)^2*(a + b*Sec[e +
f*x]^2)^3) + (Cos[e]*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*Sin[f*x])/(8*a^3*f*(a + b*Sec[e + f*x]^2)
^3) + (3*(a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^6*(4*a*b^2*Sin[e + f*x] + 3*b^3*Sin[e + f*x]))/(32*a^3*
(a + b)^2*f*(a + b*Sec[e + f*x]^2)^3) - (b^3*(a + 2*b + a*Cos[2*e + 2*f*x])*Sec[e + f*x]^5*Tan[e + f*x])/(8*a^
3*(a + b)*f*(a + b*Sec[e + f*x]^2)^3)

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fricas [B]  time = 1.64, size = 727, normalized size = 4.66 \[ \left [\frac {3 \, {\left (8 \, a^{2} b^{3} + 12 \, a b^{4} + 5 \, b^{5} + {\left (8 \, a^{4} b + 12 \, a^{3} b^{2} + 5 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (8 \, a^{3} b^{2} + 12 \, a^{2} b^{3} + 5 \, a b^{4}\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 (8 \, a^{4} b^{2} + 34 \, a^{3} b^{3} + 41 \, a^{2} b^{4} + 15 \, a b^{5} + 8 \, {\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \cos \left (f x + e\right )^{4} + {\left (16 \, a^{5} b + 60 \, a^{4} b^{2} + 69 \, a^{3} b^{3} + 25 \, a^{2} b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{9} + 3 \, a^{8} b + 3 \, a^{7} b^{2} + a^{6} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{8} b + 3 \, a^{7} b^{2} + 3 \, a^{6} b^{3} + a^{5} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{7} b^{2} + 3 \, a^{6} b^{3} + 3 \, a^{5} b^{4} + a^{4} b^{5}\right )} f\right )}}, \frac {3 \, {\left (8 \, a^{2} b^{3} + 12 \, a b^{4} + 5 \, b^{5} + {\left (8 \, a^{4} b + 12 \, a^{3} b^{2} + 5 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (8 \, a^{3} b^{2} + 12 \, a^{2} b^{3} + 5 \, a b^{4}\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 (8 \, a^{4} b^{2} + 34 \, a^{3} b^{3} + 41 \, a^{2} b^{4} + 15 \, a b^{5} + 8 \, {\left (a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3}\right )} \cos \left (f x + e\right )^{4} + {\left (16 \, a^{5} b + 60 \, a^{4} b^{2} + 69 \, a^{3} b^{3} + 25 \, a^{2} b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{9} + 3 \, a^{8} b + 3 \, a^{7} b^{2} + a^{6} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{8} b + 3 \, a^{7} b^{2} + 3 \, a^{6} b^{3} + a^{5} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{7} b^{2} + 3 \, a^{6} b^{3} + 3 \, a^{5} b^{4} + a^{4} b^{5}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(8*a^2*b^3 + 12*a*b^4 + 5*b^5 + (8*a^4*b + 12*a^3*b^2 + 5*a^2*b^3)*cos(f*x + e)^4 + 2*(8*a^3*b^2 + 12
*a^2*b^3 + 5*a*b^4)*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*(8*a^4*b^2 + 34*a^3*b^3 + 41*a^2*b^4 + 15*a*b^5 + 8*(a^6 + 3*a^5*b + 3*a^
4*b^2 + a^3*b^3)*cos(f*x + e)^4 + (16*a^5*b + 60*a^4*b^2 + 69*a^3*b^3 + 25*a^2*b^4)*cos(f*x + e)^2)*sin(f*x +
e))/((a^9 + 3*a^8*b + 3*a^7*b^2 + a^6*b^3)*f*cos(f*x + e)^4 + 2*(a^8*b + 3*a^7*b^2 + 3*a^6*b^3 + a^5*b^4)*f*co
s(f*x + e)^2 + (a^7*b^2 + 3*a^6*b^3 + 3*a^5*b^4 + a^4*b^5)*f), 1/8*(3*(8*a^2*b^3 + 12*a*b^4 + 5*b^5 + (8*a^4*b
+ 12*a^3*b^2 + 5*a^2*b^3)*cos(f*x + e)^4 + 2*(8*a^3*b^2 + 12*a^2*b^3 + 5*a*b^4)*cos(f*x + e)^2)*sqrt(-a^2 - a
*b)*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a + b)) + (8*a^4*b^2 + 34*a^3*b^3 + 41*a^2*b^4 + 15*a*b^5 + 8*(a^6 +
3*a^5*b + 3*a^4*b^2 + a^3*b^3)*cos(f*x + e)^4 + (16*a^5*b + 60*a^4*b^2 + 69*a^3*b^3 + 25*a^2*b^4)*cos(f*x + e
)^2)*sin(f*x + e))/((a^9 + 3*a^8*b + 3*a^7*b^2 + a^6*b^3)*f*cos(f*x + e)^4 + 2*(a^8*b + 3*a^7*b^2 + 3*a^6*b^3
+ a^5*b^4)*f*cos(f*x + e)^2 + (a^7*b^2 + 3*a^6*b^3 + 3*a^5*b^4 + a^4*b^5)*f)]

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giac [A]  time = 0.48, size = 205, normalized size = 1.31 \[ \frac {\frac {3 \, {\left (8 \, a^{2} b + 12 \, a b^{2} + 5 \, b^{3}\right )} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {-a^{2} - a b}} - \frac {12 \, a^{2} b^{2} \sin \left (f x + e\right )^{3} + 9 \, a b^{3} \sin \left (f x + e\right )^{3} - 12 \, a^{2} b^{2} \sin \left (f x + e\right ) - 19 \, a b^{3} \sin \left (f x + e\right ) - 7 \, b^{4} \sin \left (f x + e\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} {\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2}} + \frac {8 \, \sin \left (f x + e\right )}{a^{3}}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*(8*a^2*b + 12*a*b^2 + 5*b^3)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a^5 + 2*a^4*b + a^3*b^2)*sqrt(-a
^2 - a*b)) - (12*a^2*b^2*sin(f*x + e)^3 + 9*a*b^3*sin(f*x + e)^3 - 12*a^2*b^2*sin(f*x + e) - 19*a*b^3*sin(f*x
+ e) - 7*b^4*sin(f*x + e))/((a^5 + 2*a^4*b + a^3*b^2)*(a*sin(f*x + e)^2 - a - b)^2) + 8*sin(f*x + e)/a^3)/f

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maple [A]  time = 1.40, size = 149, normalized size = 0.96 \[ \frac {\frac {\sin \left (f x +e \right )}{a^{3}}+\frac {b \left (\frac {-\frac {3 a b \left (4 a +3 b \right ) \left (\sin ^{3}\left (f x +e \right )\right )}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (12 a +7 b \right ) b \sin \left (f x +e \right )}{8 a +8 b}}{\left (-a -b +a \left (\sin ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {3 \left (8 a^{2}+12 a b +5 b^{2}\right ) \arctanh \left (\frac {a \sin \left (f x +e \right )}{\sqrt {\left (a +b \right ) a}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {\left (a +b \right ) a}}\right )}{a^{3}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(1/a^3*sin(f*x+e)+1/a^3*b*((-3/8*a*b*(4*a+3*b)/(a^2+2*a*b+b^2)*sin(f*x+e)^3+1/8*(12*a+7*b)*b/(a+b)*sin(f*x
+e))/(-a-b+a*sin(f*x+e)^2)^2-3/8*(8*a^2+12*a*b+5*b^2)/(a^2+2*a*b+b^2)/((a+b)*a)^(1/2)*arctanh(a*sin(f*x+e)/((a
+b)*a)^(1/2))))

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maxima [A]  time = 0.44, size = 253, normalized size = 1.62 \[ \frac {\frac {3 \, {\left (8 \, a^{2} b + 12 \, a b^{2} + 5 \, b^{3}\right )} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {2 \, {\left (3 \, {\left (4 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \sin \left (f x + e\right )^{3} - {\left (12 \, a^{2} b^{2} + 19 \, a b^{3} + 7 \, b^{4}\right )} \sin \left (f x + e\right )\right )}}{a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4} + {\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} \sin \left (f x + e\right )^{4} - 2 \, {\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac {16 \, \sin \left (f x + e\right )}{a^{3}}}{16 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*(8*a^2*b + 12*a*b^2 + 5*b^3)*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) + sqrt((a + b)*a))
)/((a^5 + 2*a^4*b + a^3*b^2)*sqrt((a + b)*a)) - 2*(3*(4*a^2*b^2 + 3*a*b^3)*sin(f*x + e)^3 - (12*a^2*b^2 + 19*a
*b^3 + 7*b^4)*sin(f*x + e))/(a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4 + (a^7 + 2*a^6*b + a^5*b^2)*sin(f
*x + e)^4 - 2*(a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*sin(f*x + e)^2) + 16*sin(f*x + e)/a^3)/f

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mupad [B]  time = 4.74, size = 175, normalized size = 1.12 \[ \frac {\sin \left (e+f\,x\right )}{a^3\,f}+\frac {\frac {\sin \left (e+f\,x\right )\,\left (7\,b^3+12\,a\,b^2\right )}{8\,\left (a+b\right )}-\frac {3\,{\sin \left (e+f\,x\right )}^3\,\left (4\,a^2\,b^2+3\,a\,b^3\right )}{8\,{\left (a+b\right )}^2}}{f\,\left (2\,a^4\,b-{\sin \left (e+f\,x\right )}^2\,\left (2\,a^5+2\,b\,a^4\right )+a^5+a^3\,b^2+a^5\,{\sin \left (e+f\,x\right )}^4\right )}-\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (8\,a^2+12\,a\,b+5\,b^2\right )}{8\,a^{7/2}\,f\,{\left (a+b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(e + f*x)/(a^3*f) + ((sin(e + f*x)*(12*a*b^2 + 7*b^3))/(8*(a + b)) - (3*sin(e + f*x)^3*(3*a*b^3 + 4*a^2*b^2
))/(8*(a + b)^2))/(f*(2*a^4*b - sin(e + f*x)^2*(2*a^4*b + 2*a^5) + a^5 + a^3*b^2 + a^5*sin(e + f*x)^4)) - (3*b
*atanh((a^(1/2)*sin(e + f*x))/(a + b)^(1/2))*(12*a*b + 8*a^2 + 5*b^2))/(8*a^(7/2)*f*(a + b)^(5/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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