\(\int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [191]

Optimal result

Integrand size = 23, antiderivative size = 117 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\left (3 a^2-4 a b+8 b^2\right ) x}{8 a^3}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b} f}+\frac {(3 a-4 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4231, 425, 541, 536, 209, 211} \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^3 f \sqrt {a+b}}+\frac {(3 a-4 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f}+\frac {x \left (3 a^2-4 a b+8 b^2\right )}{8 a^3}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 a f} \]

[In]

Int[Cos[e + f*x]^4/(a + b*Sec[e + f*x]^2),x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 211

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

Rule 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 4231

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f}-\frac {\text {Subst}\left (\int \frac {-3 a+b-3 b x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{4 a f} \\ & = \frac {(3 a-4 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f}+\frac {\text {Subst}\left (\int \frac {3 a^2-a b+4 b^2+(3 a-4 b) b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 f} \\ & = \frac {(3 a-4 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f}-\frac {b^3 \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{a^3 f}+\frac {\left (3 a^2-4 a b+8 b^2\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 f} \\ & = \frac {\left (3 a^2-4 a b+8 b^2\right ) x}{8 a^3}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b} f}+\frac {(3 a-4 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 a f} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {4 \left (3 a^2-4 a b+8 b^2\right ) (e+f x)-\frac {32 b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+8 a (a-b) \sin (2 (e+f x))+a^2 \sin (4 (e+f x))}{32 a^3 f} \]

[In]

Integrate[Cos[e + f*x]^4/(a + b*Sec[e + f*x]^2),x]

[Out]

(4*(3*a^2 - 4*a*b + 8*b^2)*(e + f*x) - (32*b^(5/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/Sqrt[a + b] + 8
*a*(a - b)*Sin[2*(e + f*x)] + a^2*Sin[4*(e + f*x)])/(32*a^3*f)

Maple [A] (verified)

Time = 1.37 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {-\frac {b^{3} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}+\frac {\frac {\left (\frac {3}{8} a^{2}-\frac {1}{2} a b \right ) \tan \left (f x +e \right )^{3}+\left (-\frac {1}{2} a b +\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{3}}}{f}\)	\(116\)
default	\(\frac {-\frac {b^{3} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}+\frac {\frac {\left (\frac {3}{8} a^{2}-\frac {1}{2} a b \right ) \tan \left (f x +e \right )^{3}+\left (-\frac {1}{2} a b +\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{a^{3}}}{f}\)	\(116\)
risch	\(\frac {3 x}{8 a}-\frac {x b}{2 a^{2}}+\frac {x \,b^{2}}{a^{3}}-\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{8 a^{2} f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{8 a^{2} f}+\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right ) f \,a^{3}}-\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right ) f \,a^{3}}+\frac {\sin \left (4 f x +4 e \right )}{32 a f}\)	\(227\)

[In]

int(cos(f*x+e)^4/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.93 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {2 \, b^{2} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} f x + {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, a^{3} f}, \frac {4 \, b^{2} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} f x + {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, a^{3} f}\right ] \]

[In]

integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/8*(2*b^2*sqrt(-b/(a + b))*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 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*co
s(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) + (3*a^2 - 4*a*b + 8*b^2)*f*x + (2*a^2*cos(f*x + e)^3 + (3*a^2 - 4
*a*b)*cos(f*x + e))*sin(f*x + e))/(a^3*f), 1/8*(4*b^2*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b
)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e))) + (3*a^2 - 4*a*b + 8*b^2)*f*x + (2*a^2*cos(f*x + e)^3 + (3*a^
2 - 4*a*b)*cos(f*x + e))*sin(f*x + e))/(a^3*f)]

Sympy [F]

\[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\cos ^{4}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

[In]

integrate(cos(f*x+e)**4/(a+b*sec(f*x+e)**2),x)

[Out]

Integral(cos(e + f*x)**4/(a + b*sec(e + f*x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {8 \, b^{3} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3}} - \frac {{\left (3 \, a - 4 \, b\right )} \tan \left (f x + e\right )^{3} + {\left (5 \, a - 4 \, b\right )} \tan \left (f x + e\right )}{a^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{2} \tan \left (f x + e\right )^{2} + a^{2}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{3}}}{8 \, f} \]

[In]

integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.21 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {8 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b^{3}}{\sqrt {a b + b^{2}} a^{3}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{3}} - \frac {3 \, a \tan \left (f x + e\right )^{3} - 4 \, b \tan \left (f x + e\right )^{3} + 5 \, a \tan \left (f x + e\right ) - 4 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{2}}}{8 \, f} \]

[In]

integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

-1/8*(8*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*b^3/(sqrt(a*b + b^2)*a^
3) - (3*a^2 - 4*a*b + 8*b^2)*(f*x + e)/a^3 - (3*a*tan(f*x + e)^3 - 4*b*tan(f*x + e)^3 + 5*a*tan(f*x + e) - 4*b
*tan(f*x + e))/((tan(f*x + e)^2 + 1)^2*a^2))/f

Mupad [B] (verification not implemented)

Time = 20.26 (sec) , antiderivative size = 1114, normalized size of antiderivative = 9.52 \[ \int \frac {\cos ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\text {Too large to display} \]

[In]

int(cos(e + f*x)^4/(a + b/cos(e + f*x)^2),x)

[Out]

((tan(e + f*x)*(5*a - 4*b))/(8*a^2) + (tan(e + f*x)^3*(3*a - 4*b))/(8*a^2))/(f*(2*tan(e + f*x)^2 + tan(e + f*x
)^4 + 1)) - (atan((((-b^5*(a + b))^(1/2)*((tan(e + f*x)*(128*b^7 - 64*a*b^6 + 64*a^2*b^5 - 24*a^3*b^4 + 9*a^4*
b^3))/(64*a^4) - ((-b^5*(a + b))^(1/2)*((2*a^6*b^4 - (a^7*b^3)/2 + (3*a^8*b^2)/2)/(2*a^6) - (tan(e + f*x)*(512
*a^6*b^3 + 256*a^7*b^2)*(-b^5*(a + b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4)))*1i)/(a^3*b + a^4) +
((-b^5*(a + b))^(1/2)*((tan(e + f*x)*(128*b^7 - 64*a*b^6 + 64*a^2*b^5 - 24*a^3*b^4 + 9*a^4*b^3))/(64*a^4) + (
(-b^5*(a + b))^(1/2)*((2*a^6*b^4 - (a^7*b^3)/2 + (3*a^8*b^2)/2)/(2*a^6) + (tan(e + f*x)*(512*a^6*b^3 + 256*a^7
*b^2)*(-b^5*(a + b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4)))*1i)/(a^3*b + a^4))/(((5*a*b^7)/4 - b^
8 - (3*a^2*b^6)/4 + (9*a^3*b^5)/32)/a^6 + ((-b^5*(a + b))^(1/2)*((tan(e + f*x)*(128*b^7 - 64*a*b^6 + 64*a^2*b^
5 - 24*a^3*b^4 + 9*a^4*b^3))/(64*a^4) - ((-b^5*(a + b))^(1/2)*((2*a^6*b^4 - (a^7*b^3)/2 + (3*a^8*b^2)/2)/(2*a^
6) - (tan(e + f*x)*(512*a^6*b^3 + 256*a^7*b^2)*(-b^5*(a + b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4
))))/(a^3*b + a^4) - ((-b^5*(a + b))^(1/2)*((tan(e + f*x)*(128*b^7 - 64*a*b^6 + 64*a^2*b^5 - 24*a^3*b^4 + 9*a^
4*b^3))/(64*a^4) + ((-b^5*(a + b))^(1/2)*((2*a^6*b^4 - (a^7*b^3)/2 + (3*a^8*b^2)/2)/(2*a^6) + (tan(e + f*x)*(5
12*a^6*b^3 + 256*a^7*b^2)*(-b^5*(a + b))^(1/2))/(128*a^4*(a^3*b + a^4))))/(2*(a^3*b + a^4))))/(a^3*b + a^4)))*
(-b^5*(a + b))^(1/2)*1i)/(f*(a^3*b + a^4)) - (atan((63*b^4*tan(e + f*x))/(64*((63*b^4)/64 - (81*a*b^3)/256 + (
27*a^2*b^2)/256 - (35*b^5)/(32*a) + (5*b^6)/(4*a^2))) - (81*b^3*tan(e + f*x))/(256*((27*a*b^2)/256 - (81*b^3)/
256 + (63*b^4)/(64*a) - (35*b^5)/(32*a^2) + (5*b^6)/(4*a^3))) - (35*b^5*tan(e + f*x))/(32*((63*a*b^4)/64 - (35
*b^5)/32 - (81*a^2*b^3)/256 + (27*a^3*b^2)/256 + (5*b^6)/(4*a))) + (5*b^6*tan(e + f*x))/(4*((5*b^6)/4 - (35*a*
b^5)/32 + (63*a^2*b^4)/64 - (81*a^3*b^3)/256 + (27*a^4*b^2)/256)) + (27*b^2*tan(e + f*x))/(256*((27*b^2)/256 -
(81*b^3)/(256*a) + (63*b^4)/(64*a^2) - (35*b^5)/(32*a^3) + (5*b^6)/(4*a^4))))*(a^2*3i - a*b*4i + b^2*8i)*1i)/
(8*a^3*f)