\(\int \frac {\cos ^5(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [211]

Optimal result

Integrand size = 23, antiderivative size = 214 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {b^3 \left (80 a^2+140 a b+63 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{11/2} (a+b)^{5/2} f}+\frac {\left (a^2-3 a b+6 b^2\right ) \sin (e+f x)}{a^5 f}-\frac {(2 a-3 b) \sin ^3(e+f x)}{3 a^4 f}+\frac {\sin ^5(e+f x)}{5 a^3 f}-\frac {b^5 \sin (e+f x)}{4 a^5 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {b^4 (20 a+17 b) \sin (e+f x)}{8 a^5 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )} \]

[Out]

-1/8*b^3*(80*a^2+140*a*b+63*b^2)*arctanh(sin(f*x+e)*a^(1/2)/(a+b)^(1/2))/a^(11/2)/(a+b)^(5/2)/f+(a^2-3*a*b+6*b
^2)*sin(f*x+e)/a^5/f-1/3*(2*a-3*b)*sin(f*x+e)^3/a^4/f+1/5*sin(f*x+e)^5/a^3/f-1/4*b^5*sin(f*x+e)/a^5/(a+b)/f/(a
+b-a*sin(f*x+e)^2)^2+1/8*b^4*(20*a+17*b)*sin(f*x+e)/a^5/(a+b)^2/f/(a+b-a*sin(f*x+e)^2)

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4232, 398, 1171, 393, 214} \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {b^5 \sin (e+f x)}{4 a^5 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}+\frac {b^4 (20 a+17 b) \sin (e+f x)}{8 a^5 f (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}-\frac {(2 a-3 b) \sin ^3(e+f x)}{3 a^4 f}+\frac {\sin ^5(e+f x)}{5 a^3 f}-\frac {b^3 \left (80 a^2+140 a b+63 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{11/2} f (a+b)^{5/2}}+\frac {\left (a^2-3 a b+6 b^2\right ) \sin (e+f x)}{a^5 f} \]

[In]

Int[Cos[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

-1/8*(b^3*(80*a^2 + 140*a*b + 63*b^2)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(a^(11/2)*(a + b)^(5/2)*f)
+ ((a^2 - 3*a*b + 6*b^2)*Sin[e + f*x])/(a^5*f) - ((2*a - 3*b)*Sin[e + f*x]^3)/(3*a^4*f) + Sin[e + f*x]^5/(5*a^
3*f) - (b^5*Sin[e + f*x])/(4*a^5*(a + b)*f*(a + b - a*Sin[e + f*x]^2)^2) + (b^4*(20*a + 17*b)*Sin[e + f*x])/(8
*a^5*(a + b)^2*f*(a + b - a*Sin[e + f*x]^2))

Rule 214

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

Rule 393

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 398

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 1171

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] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 4232

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^5}{\left (a+b-a x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a^2-3 a b+6 b^2}{a^5}-\frac {(2 a-3 b) x^2}{a^4}+\frac {x^4}{a^3}-\frac {b^3 \left (10 a^2+15 a b+6 b^2\right )-5 a b^3 (4 a+3 b) x^2+10 a^2 b^3 x^4}{a^5 \left (a+b-a x^2\right )^3}\right ) \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (a^2-3 a b+6 b^2\right ) \sin (e+f x)}{a^5 f}-\frac {(2 a-3 b) \sin ^3(e+f x)}{3 a^4 f}+\frac {\sin ^5(e+f x)}{5 a^3 f}-\frac {\text {Subst}\left (\int \frac {b^3 \left (10 a^2+15 a b+6 b^2\right )-5 a b^3 (4 a+3 b) x^2+10 a^2 b^3 x^4}{\left (a+b-a x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{a^5 f} \\ & = \frac {\left (a^2-3 a b+6 b^2\right ) \sin (e+f x)}{a^5 f}-\frac {(2 a-3 b) \sin ^3(e+f x)}{3 a^4 f}+\frac {\sin ^5(e+f x)}{5 a^3 f}-\frac {b^5 \sin (e+f x)}{4 a^5 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {\text {Subst}\left (\int \frac {-b^3 \left (40 a^2+60 a b+23 b^2\right )+40 a b^3 (a+b) x^2}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 a^5 (a+b) f} \\ & = \frac {\left (a^2-3 a b+6 b^2\right ) \sin (e+f x)}{a^5 f}-\frac {(2 a-3 b) \sin ^3(e+f x)}{3 a^4 f}+\frac {\sin ^5(e+f x)}{5 a^3 f}-\frac {b^5 \sin (e+f x)}{4 a^5 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {b^4 (20 a+17 b) \sin (e+f x)}{8 a^5 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )}-\frac {\left (b^3 \left (80 a^2+140 a b+63 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{8 a^5 (a+b)^2 f} \\ & = -\frac {b^3 \left (80 a^2+140 a b+63 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{11/2} (a+b)^{5/2} f}+\frac {\left (a^2-3 a b+6 b^2\right ) \sin (e+f x)}{a^5 f}-\frac {(2 a-3 b) \sin ^3(e+f x)}{3 a^4 f}+\frac {\sin ^5(e+f x)}{5 a^3 f}-\frac {b^5 \sin (e+f x)}{4 a^5 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {b^4 (20 a+17 b) \sin (e+f x)}{8 a^5 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.35 (sec) , antiderivative size = 2670, normalized size of antiderivative = 12.48 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Result too large to show} \]

[In]

Integrate[Cos[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

((5*a^2 - 18*a*b + 48*b^2)*Cos[f*x]*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*Sin[e])/(64*a^5*f*(a + b*S
ec[e + f*x]^2)^3) + ((-80*a^2*b^3 - 140*a*b^4 - 63*b^5)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*(((I/1
28)*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*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])]*Cos[e])/(a^(11/2)*
Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]]) + (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]*Co
s[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*S
in[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^(11/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]])))/((a + b)^2*(a + b*S
ec[e + f*x]^2)^3) + ((80*a^2*b^3 + 140*a*b^4 + 63*b^5)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((ArcTa
nh[(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*S
in[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^(11/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]]) - ((I/128)*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])]*Sin[e])/(a^(11/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin
[2*e]])))/((a + b)^2*(a + b*Sec[e + f*x]^2)^3) + ((-80*a^2*b^3 - 140*a*b^4 - 63*b^5)*(a + 2*b + a*Cos[2*e + 2*
f*x])^3*Sec[e + f*x]^6*((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]*Sqrt[a + b]*Sqrt[Cos
[2*e] - I*Sin[2*e]]*Sin[2*e + f*x]])/(256*a^(11/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]]) - ((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]]*S
in[e])/(a^(11/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]])))/((a + b)^2*(a + b*Sec[e + f*x]^2)^3) + ((80*a^2*
b^3 + 140*a*b^4 + 63*b^5)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((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]*Sqrt[a + b]*Sqrt[Cos[2*e] - I*Sin[2*e]]*Sin[2*e + f*x]])/(256*a^(11/2)*Sqrt[a
+ b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]]) - ((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^(11/2)*Sqrt[a + b]*f*Sqrt[Cos[2*e] - I*Sin[2*e]
])))/((a + b)^2*(a + b*Sec[e + f*x]^2)^3) + ((5*a - 12*b)*Cos[3*f*x]*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e +
f*x]^6*Sin[3*e])/(384*a^4*f*(a + b*Sec[e + f*x]^2)^3) + (Cos[5*f*x]*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f
*x]^6*Sin[5*e])/(640*a^3*f*(a + b*Sec[e + f*x]^2)^3) + ((5*a^2 - 18*a*b + 48*b^2)*Cos[e]*(a + 2*b + a*Cos[2*e
+ 2*f*x])^3*Sec[e + f*x]^6*Sin[f*x])/(64*a^5*f*(a + b*Sec[e + f*x]^2)^3) + ((5*a - 12*b)*Cos[3*e]*(a + 2*b + a
*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*Sin[3*f*x])/(384*a^4*f*(a + b*Sec[e + f*x]^2)^3) + (Cos[5*e]*(a + 2*b + a*
Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*Sin[5*f*x])/(640*a^3*f*(a + b*Sec[e + f*x]^2)^3) + ((a + 2*b + a*Cos[2*e +
2*f*x])^2*Sec[e + f*x]^6*(20*a*b^4*Sin[e + f*x] + 17*b^5*Sin[e + f*x]))/(32*a^5*(a + b)^2*f*(a + b*Sec[e + f*x
]^2)^3) - (b^5*(a + 2*b + a*Cos[2*e + 2*f*x])*Sec[e + f*x]^5*Tan[e + f*x])/(8*a^5*(a + b)*f*(a + b*Sec[e + f*x
]^2)^3)

Maple [A] (verified)

Time = 10.27 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {\frac {\frac {a^{2} \sin \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \sin \left (f x +e \right )^{3}}{3}+a \sin \left (f x +e \right )^{3} b +\sin \left (f x +e \right ) a^{2}-3 \sin \left (f x +e \right ) a b +6 \sin \left (f x +e \right ) b^{2}}{a^{5}}+\frac {b^{3} \left (\frac {-\frac {a b \left (20 a +17 b \right ) \sin \left (f x +e \right )^{3}}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {5 \left (4 a +3 b \right ) b \sin \left (f x +e \right )}{8 \left (a +b \right )}}{\left (a \sin \left (f x +e \right )^{2}-a -b \right )^{2}}-\frac {\left (80 a^{2}+140 a b +63 b^{2}\right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{a^{5}}}{f}\)	\(214\)
default	\(\frac {\frac {\frac {a^{2} \sin \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \sin \left (f x +e \right )^{3}}{3}+a \sin \left (f x +e \right )^{3} b +\sin \left (f x +e \right ) a^{2}-3 \sin \left (f x +e \right ) a b +6 \sin \left (f x +e \right ) b^{2}}{a^{5}}+\frac {b^{3} \left (\frac {-\frac {a b \left (20 a +17 b \right ) \sin \left (f x +e \right )^{3}}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {5 \left (4 a +3 b \right ) b \sin \left (f x +e \right )}{8 \left (a +b \right )}}{\left (a \sin \left (f x +e \right )^{2}-a -b \right )^{2}}-\frac {\left (80 a^{2}+140 a b +63 b^{2}\right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{a^{5}}}{f}\)	\(214\)
risch	\(-\frac {i {\mathrm e}^{-3 i \left (f x +e \right )} b}{8 a^{4} f}-\frac {5 i {\mathrm e}^{3 i \left (f x +e \right )}}{96 a^{3} f}+\frac {3 i {\mathrm e}^{-i \left (f x +e \right )} b^{2}}{f \,a^{5}}-\frac {i {\mathrm e}^{5 i \left (f x +e \right )}}{160 a^{3} f}-\frac {9 i {\mathrm e}^{-i \left (f x +e \right )} b}{8 a^{4} f}+\frac {i {\mathrm e}^{-5 i \left (f x +e \right )}}{160 a^{3} f}-\frac {i b^{4} \left (20 a^{2} {\mathrm e}^{7 i \left (f x +e \right )}+17 a b \,{\mathrm e}^{7 i \left (f x +e \right )}+20 a^{2} {\mathrm e}^{5 i \left (f x +e \right )}+89 a b \,{\mathrm e}^{5 i \left (f x +e \right )}+60 b^{2} {\mathrm e}^{5 i \left (f x +e \right )}-20 a^{2} {\mathrm e}^{3 i \left (f x +e \right )}-89 a b \,{\mathrm e}^{3 i \left (f x +e \right )}-60 b^{2} {\mathrm e}^{3 i \left (f x +e \right )}-20 a^{2} {\mathrm e}^{i \left (f x +e \right )}-17 a b \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 a^{5} \left (a +b \right )^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}-\frac {3 i {\mathrm e}^{i \left (f x +e \right )} b^{2}}{f \,a^{5}}+\frac {5 i {\mathrm e}^{-i \left (f x +e \right )}}{16 a^{3} f}+\frac {i {\mathrm e}^{3 i \left (f x +e \right )} b}{8 a^{4} f}-\frac {5 i {\mathrm e}^{i \left (f x +e \right )}}{16 a^{3} f}+\frac {9 i {\mathrm e}^{i \left (f x +e \right )} b}{8 a^{4} f}+\frac {5 i {\mathrm e}^{-3 i \left (f x +e \right )}}{96 a^{3} f}+\frac {5 b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{\sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{3}}+\frac {35 b^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{4 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{4}}+\frac {63 b^{5} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{5}}-\frac {5 b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{\sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{3}}-\frac {35 b^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{4 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{4}}-\frac {63 b^{5} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{5}}\)	\(787\)

[In]

int(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)

[Out]

1/f*(1/a^5*(1/5*a^2*sin(f*x+e)^5-2/3*a^2*sin(f*x+e)^3+a*sin(f*x+e)^3*b+sin(f*x+e)*a^2-3*sin(f*x+e)*a*b+6*sin(f
*x+e)*b^2)+b^3/a^5*((-1/8*a*b*(20*a+17*b)/(a^2+2*a*b+b^2)*sin(f*x+e)^3+5/8*(4*a+3*b)*b/(a+b)*sin(f*x+e))/(a*si
n(f*x+e)^2-a-b)^2-1/8*(80*a^2+140*a*b+63*b^2)/(a^2+2*a*b+b^2)/(a*(a+b))^(1/2)*arctanh(a*sin(f*x+e)/(a*(a+b))^(
1/2))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 485 vs. \(2 (202) = 404\).

Time = 0.35 (sec) , antiderivative size = 995, normalized size of antiderivative = 4.65 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [\frac {15 \, {\left (80 \, a^{2} b^{5} + 140 \, a b^{6} + 63 \, b^{7} + {\left (80 \, a^{4} b^{3} + 140 \, a^{3} b^{4} + 63 \, a^{2} b^{5}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (80 \, a^{3} b^{4} + 140 \, a^{2} b^{5} + 63 \, a b^{6}\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 (24 \, {\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3}\right )} \cos \left (f x + e\right )^{8} + 64 \, a^{6} b^{2} - 48 \, a^{5} b^{3} + 192 \, a^{4} b^{4} + 1774 \, a^{3} b^{5} + 2415 \, a^{2} b^{6} + 945 \, a b^{7} + 8 \, {\left (4 \, a^{8} + 3 \, a^{7} b - 15 \, a^{6} b^{2} - 23 \, a^{5} b^{3} - 9 \, a^{4} b^{4}\right )} \cos \left (f x + e\right )^{6} + 8 \, {\left (8 \, a^{8} + 2 \, a^{7} b + 21 \, a^{6} b^{2} + 131 \, a^{5} b^{3} + 167 \, a^{4} b^{4} + 63 \, a^{3} b^{5}\right )} \cos \left (f x + e\right )^{4} + {\left (128 \, a^{7} b - 64 \, a^{6} b^{2} + 360 \, a^{5} b^{3} + 3044 \, a^{4} b^{4} + 4067 \, a^{3} b^{5} + 1575 \, a^{2} b^{6}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{240 \, {\left ({\left (a^{11} + 3 \, a^{10} b + 3 \, a^{9} b^{2} + a^{8} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{10} b + 3 \, a^{9} b^{2} + 3 \, a^{8} b^{3} + a^{7} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{9} b^{2} + 3 \, a^{8} b^{3} + 3 \, a^{7} b^{4} + a^{6} b^{5}\right )} f\right )}}, \frac {15 \, {\left (80 \, a^{2} b^{5} + 140 \, a b^{6} + 63 \, b^{7} + {\left (80 \, a^{4} b^{3} + 140 \, a^{3} b^{4} + 63 \, a^{2} b^{5}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (80 \, a^{3} b^{4} + 140 \, a^{2} b^{5} + 63 \, a b^{6}\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 (24 \, {\left (a^{8} + 3 \, a^{7} b + 3 \, a^{6} b^{2} + a^{5} b^{3}\right )} \cos \left (f x + e\right )^{8} + 64 \, a^{6} b^{2} - 48 \, a^{5} b^{3} + 192 \, a^{4} b^{4} + 1774 \, a^{3} b^{5} + 2415 \, a^{2} b^{6} + 945 \, a b^{7} + 8 \, {\left (4 \, a^{8} + 3 \, a^{7} b - 15 \, a^{6} b^{2} - 23 \, a^{5} b^{3} - 9 \, a^{4} b^{4}\right )} \cos \left (f x + e\right )^{6} + 8 \, {\left (8 \, a^{8} + 2 \, a^{7} b + 21 \, a^{6} b^{2} + 131 \, a^{5} b^{3} + 167 \, a^{4} b^{4} + 63 \, a^{3} b^{5}\right )} \cos \left (f x + e\right )^{4} + {\left (128 \, a^{7} b - 64 \, a^{6} b^{2} + 360 \, a^{5} b^{3} + 3044 \, a^{4} b^{4} + 4067 \, a^{3} b^{5} + 1575 \, a^{2} b^{6}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{120 \, {\left ({\left (a^{11} + 3 \, a^{10} b + 3 \, a^{9} b^{2} + a^{8} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{10} b + 3 \, a^{9} b^{2} + 3 \, a^{8} b^{3} + a^{7} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{9} b^{2} + 3 \, a^{8} b^{3} + 3 \, a^{7} b^{4} + a^{6} b^{5}\right )} f\right )}}\right ] \]

[In]

integrate(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[1/240*(15*(80*a^2*b^5 + 140*a*b^6 + 63*b^7 + (80*a^4*b^3 + 140*a^3*b^4 + 63*a^2*b^5)*cos(f*x + e)^4 + 2*(80*a
^3*b^4 + 140*a^2*b^5 + 63*a*b^6)*cos(f*x + e)^2)*sqrt(a^2 + a*b)*log(-(a*cos(f*x + e)^2 + 2*sqrt(a^2 + a*b)*si
n(f*x + e) - 2*a - b)/(a*cos(f*x + e)^2 + b)) + 2*(24*(a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*cos(f*x + e)^8 + 6
4*a^6*b^2 - 48*a^5*b^3 + 192*a^4*b^4 + 1774*a^3*b^5 + 2415*a^2*b^6 + 945*a*b^7 + 8*(4*a^8 + 3*a^7*b - 15*a^6*b
^2 - 23*a^5*b^3 - 9*a^4*b^4)*cos(f*x + e)^6 + 8*(8*a^8 + 2*a^7*b + 21*a^6*b^2 + 131*a^5*b^3 + 167*a^4*b^4 + 63
*a^3*b^5)*cos(f*x + e)^4 + (128*a^7*b - 64*a^6*b^2 + 360*a^5*b^3 + 3044*a^4*b^4 + 4067*a^3*b^5 + 1575*a^2*b^6)
*cos(f*x + e)^2)*sin(f*x + e))/((a^11 + 3*a^10*b + 3*a^9*b^2 + a^8*b^3)*f*cos(f*x + e)^4 + 2*(a^10*b + 3*a^9*b
^2 + 3*a^8*b^3 + a^7*b^4)*f*cos(f*x + e)^2 + (a^9*b^2 + 3*a^8*b^3 + 3*a^7*b^4 + a^6*b^5)*f), 1/120*(15*(80*a^2
*b^5 + 140*a*b^6 + 63*b^7 + (80*a^4*b^3 + 140*a^3*b^4 + 63*a^2*b^5)*cos(f*x + e)^4 + 2*(80*a^3*b^4 + 140*a^2*b
^5 + 63*a*b^6)*cos(f*x + e)^2)*sqrt(-a^2 - a*b)*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a + b)) + (24*(a^8 + 3*a
^7*b + 3*a^6*b^2 + a^5*b^3)*cos(f*x + e)^8 + 64*a^6*b^2 - 48*a^5*b^3 + 192*a^4*b^4 + 1774*a^3*b^5 + 2415*a^2*b
^6 + 945*a*b^7 + 8*(4*a^8 + 3*a^7*b - 15*a^6*b^2 - 23*a^5*b^3 - 9*a^4*b^4)*cos(f*x + e)^6 + 8*(8*a^8 + 2*a^7*b
+ 21*a^6*b^2 + 131*a^5*b^3 + 167*a^4*b^4 + 63*a^3*b^5)*cos(f*x + e)^4 + (128*a^7*b - 64*a^6*b^2 + 360*a^5*b^3
+ 3044*a^4*b^4 + 4067*a^3*b^5 + 1575*a^2*b^6)*cos(f*x + e)^2)*sin(f*x + e))/((a^11 + 3*a^10*b + 3*a^9*b^2 + a
^8*b^3)*f*cos(f*x + e)^4 + 2*(a^10*b + 3*a^9*b^2 + 3*a^8*b^3 + a^7*b^4)*f*cos(f*x + e)^2 + (a^9*b^2 + 3*a^8*b^
3 + 3*a^7*b^4 + a^6*b^5)*f)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]

[In]

integrate(cos(f*x+e)**5/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.42 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {15 \, {\left (80 \, a^{2} b^{3} + 140 \, a b^{4} + 63 \, b^{5}\right )} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {30 \, {\left ({\left (20 \, a^{2} b^{4} + 17 \, a b^{5}\right )} \sin \left (f x + e\right )^{3} - 5 \, {\left (4 \, a^{2} b^{4} + 7 \, a b^{5} + 3 \, b^{6}\right )} \sin \left (f x + e\right )\right )}}{a^{9} + 4 \, a^{8} b + 6 \, a^{7} b^{2} + 4 \, a^{6} b^{3} + a^{5} b^{4} + {\left (a^{9} + 2 \, a^{8} b + a^{7} b^{2}\right )} \sin \left (f x + e\right )^{4} - 2 \, {\left (a^{9} + 3 \, a^{8} b + 3 \, a^{7} b^{2} + a^{6} b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac {16 \, {\left (3 \, a^{2} \sin \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} - 3 \, a b\right )} \sin \left (f x + e\right )^{3} + 15 \, {\left (a^{2} - 3 \, a b + 6 \, b^{2}\right )} \sin \left (f x + e\right )\right )}}{a^{5}}}{240 \, f} \]

[In]

integrate(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

1/240*(15*(80*a^2*b^3 + 140*a*b^4 + 63*b^5)*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) + sqrt((a +
b)*a)))/((a^7 + 2*a^6*b + a^5*b^2)*sqrt((a + b)*a)) - 30*((20*a^2*b^4 + 17*a*b^5)*sin(f*x + e)^3 - 5*(4*a^2*b
^4 + 7*a*b^5 + 3*b^6)*sin(f*x + e))/(a^9 + 4*a^8*b + 6*a^7*b^2 + 4*a^6*b^3 + a^5*b^4 + (a^9 + 2*a^8*b + a^7*b^
2)*sin(f*x + e)^4 - 2*(a^9 + 3*a^8*b + 3*a^7*b^2 + a^6*b^3)*sin(f*x + e)^2) + 16*(3*a^2*sin(f*x + e)^5 - 5*(2*
a^2 - 3*a*b)*sin(f*x + e)^3 + 15*(a^2 - 3*a*b + 6*b^2)*sin(f*x + e))/a^5)/f

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {15 \, {\left (80 \, a^{2} b^{3} + 140 \, a b^{4} + 63 \, b^{5}\right )} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} \sqrt {-a^{2} - a b}} - \frac {15 \, {\left (20 \, a^{2} b^{4} \sin \left (f x + e\right )^{3} + 17 \, a b^{5} \sin \left (f x + e\right )^{3} - 20 \, a^{2} b^{4} \sin \left (f x + e\right ) - 35 \, a b^{5} \sin \left (f x + e\right ) - 15 \, b^{6} \sin \left (f x + e\right )\right )}}{{\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} {\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2}} + \frac {8 \, {\left (3 \, a^{12} \sin \left (f x + e\right )^{5} - 10 \, a^{12} \sin \left (f x + e\right )^{3} + 15 \, a^{11} b \sin \left (f x + e\right )^{3} + 15 \, a^{12} \sin \left (f x + e\right ) - 45 \, a^{11} b \sin \left (f x + e\right ) + 90 \, a^{10} b^{2} \sin \left (f x + e\right )\right )}}{a^{15}}}{120 \, f} \]

[In]

integrate(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

1/120*(15*(80*a^2*b^3 + 140*a*b^4 + 63*b^5)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a^7 + 2*a^6*b + a^5*b^2)
*sqrt(-a^2 - a*b)) - 15*(20*a^2*b^4*sin(f*x + e)^3 + 17*a*b^5*sin(f*x + e)^3 - 20*a^2*b^4*sin(f*x + e) - 35*a*
b^5*sin(f*x + e) - 15*b^6*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2)*(a*sin(f*x + e)^2 - a - b)^2) + 8*(3*a^12*s
in(f*x + e)^5 - 10*a^12*sin(f*x + e)^3 + 15*a^11*b*sin(f*x + e)^3 + 15*a^12*sin(f*x + e) - 45*a^11*b*sin(f*x +
e) + 90*a^10*b^2*sin(f*x + e))/a^15)/f

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {5\,\sin \left (e+f\,x\right )\,\left (3\,b^5+4\,a\,b^4\right )}{8\,\left (a+b\right )}-\frac {{\sin \left (e+f\,x\right )}^3\,\left (20\,a^2\,b^4+17\,a\,b^5\right )}{8\,{\left (a+b\right )}^2}}{f\,\left (2\,a^6\,b-{\sin \left (e+f\,x\right )}^2\,\left (2\,a^7+2\,b\,a^6\right )+a^7+a^5\,b^2+a^7\,{\sin \left (e+f\,x\right )}^4\right )}+\frac {{\sin \left (e+f\,x\right )}^5}{5\,a^3\,f}+\frac {{\sin \left (e+f\,x\right )}^3\,\left (\frac {a+b}{a^4}-\frac {5}{3\,a^3}\right )}{f}+\frac {\sin \left (e+f\,x\right )\,\left (\frac {10}{a^3}-\frac {3\,{\left (a+b\right )}^2}{a^5}+\frac {3\,\left (a+b\right )\,\left (\frac {3\,\left (a+b\right )}{a^4}-\frac {5}{a^3}\right )}{a}\right )}{f}+\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\sqrt {a+b}}\right )\,\left (80\,a^2+140\,a\,b+63\,b^2\right )\,1{}\mathrm {i}}{8\,a^{11/2}\,f\,{\left (a+b\right )}^{5/2}} \]

[In]

int(cos(e + f*x)^5/(a + b/cos(e + f*x)^2)^3,x)

[Out]

((5*sin(e + f*x)*(4*a*b^4 + 3*b^5))/(8*(a + b)) - (sin(e + f*x)^3*(17*a*b^5 + 20*a^2*b^4))/(8*(a + b)^2))/(f*(
2*a^6*b - sin(e + f*x)^2*(2*a^6*b + 2*a^7) + a^7 + a^5*b^2 + a^7*sin(e + f*x)^4)) + sin(e + f*x)^5/(5*a^3*f) +
(sin(e + f*x)^3*((a + b)/a^4 - 5/(3*a^3)))/f + (sin(e + f*x)*(10/a^3 - (3*(a + b)^2)/a^5 + (3*(a + b)*((3*(a
+ b))/a^4 - 5/a^3))/a))/f + (b^3*atan((a^(1/2)*sin(e + f*x)*1i)/(a + b)^(1/2))*(140*a*b + 80*a^2 + 63*b^2)*1i)
/(8*a^(11/2)*f*(a + b)^(5/2))