\(\int \sec (e+f x) (a+b \sec ^2(e+f x))^2 \, dx\) [167]

Optimal result

Integrand size = 21, antiderivative size = 91 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}(\sin (e+f x))}{8 f}+\frac {3 b (2 a+b) \sec (e+f x) \tan (e+f x)}{8 f}+\frac {b \sec ^3(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{4 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4232, 424, 393, 212} \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}(\sin (e+f x))}{8 f}+\frac {3 b (2 a+b) \tan (e+f x) \sec (e+f x)}{8 f}+\frac {b \tan (e+f x) \sec ^3(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{4 f} \]

[In]

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

[Out]

((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sin[e + f*x]])/(8*f) + (3*b*(2*a + b)*Sec[e + f*x]*Tan[e + f*x])/(8*f) + (b*S
ec[e + f*x]^3*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x])/(4*f)

Rule 212

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

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 424

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 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 (a+b-a x^2\right )^2}{\left (1-x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {b \sec ^3(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{4 f}-\frac {\text {Subst}\left (\int \frac {-((a+b) (4 a+3 b))+a (4 a+b) x^2}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 f} \\ & = \frac {3 b (2 a+b) \sec (e+f x) \tan (e+f x)}{8 f}+\frac {b \sec ^3(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{4 f}+\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (e+f x)\right )}{8 f} \\ & = \frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}(\sin (e+f x))}{8 f}+\frac {3 b (2 a+b) \sec (e+f x) \tan (e+f x)}{8 f}+\frac {b \sec ^3(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{4 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.20 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {a^2 \text {arctanh}(\sin (e+f x))}{f}+\frac {a b \text {arctanh}(\sin (e+f x))}{f}+\frac {3 b^2 \text {arctanh}(\sin (e+f x))}{8 f}+\frac {a b \sec (e+f x) \tan (e+f x)}{f}+\frac {3 b^2 \sec (e+f x) \tan (e+f x)}{8 f}+\frac {b^2 \sec ^3(e+f x) \tan (e+f x)}{4 f} \]

[In]

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

[Out]

(a^2*ArcTanh[Sin[e + f*x]])/f + (a*b*ArcTanh[Sin[e + f*x]])/f + (3*b^2*ArcTanh[Sin[e + f*x]])/(8*f) + (a*b*Sec
[e + f*x]*Tan[e + f*x])/f + (3*b^2*Sec[e + f*x]*Tan[e + f*x])/(8*f) + (b^2*Sec[e + f*x]^3*Tan[e + f*x])/(4*f)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.18

method	result	size
derivativedivides	\(\frac {a^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+2 a b \left (\frac {\tan \left (f x +e \right ) \sec \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )+b^{2} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )}{f}\)	\(107\)
default	\(\frac {a^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+2 a b \left (\frac {\tan \left (f x +e \right ) \sec \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )+b^{2} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )}{f}\)	\(107\)
parts	\(\frac {a^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {b^{2} \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )}{f}+\frac {a b \tan \left (f x +e \right ) \sec \left (f x +e \right )}{f}+\frac {a b \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\)	\(112\)
parallelrisch	\(\frac {-4 \left (a^{2}+a b +\frac {3}{8} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 f x +4 e \right )}{4}+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+4 \left (a^{2}+a b +\frac {3}{8} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 f x +4 e \right )}{4}+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+2 b \left (\left (a +\frac {3 b}{8}\right ) \sin \left (3 f x +3 e \right )+\sin \left (f x +e \right ) \left (a +\frac {11 b}{8}\right )\right )}{f \left (\cos \left (4 f x +4 e \right )+4 \cos \left (2 f x +2 e \right )+3\right )}\)	\(157\)
norman	\(\frac {-\frac {b \left (8 a -3 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{4 f}-\frac {b \left (8 a -3 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{4 f}+\frac {b \left (8 a +5 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}+\frac {b \left (8 a +5 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{4 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{4}}-\frac {\left (8 a^{2}+8 a b +3 b^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{8 f}+\frac {\left (8 a^{2}+8 a b +3 b^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{8 f}\)	\(177\)
risch	\(-\frac {i b \,{\mathrm e}^{i \left (f x +e \right )} \left (8 a \,{\mathrm e}^{6 i \left (f x +e \right )}+3 b \,{\mathrm e}^{6 i \left (f x +e \right )}+8 a \,{\mathrm e}^{4 i \left (f x +e \right )}+11 b \,{\mathrm e}^{4 i \left (f x +e \right )}-8 a \,{\mathrm e}^{2 i \left (f x +e \right )}-11 b \,{\mathrm e}^{2 i \left (f x +e \right )}-8 a -3 b \right )}{4 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) a^{2}}{f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) a b}{f}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) b^{2}}{8 f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a^{2}}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a b}{f}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) b^{2}}{8 f}\)	\(246\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.27 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left ({\left (8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, b^{2}\right )} \sin \left (f x + e\right )}{16 \, f \cos \left (f x + e\right )^{4}} \]

[In]

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

[Out]

1/16*((8*a^2 + 8*a*b + 3*b^2)*cos(f*x + e)^4*log(sin(f*x + e) + 1) - (8*a^2 + 8*a*b + 3*b^2)*cos(f*x + e)^4*lo
g(-sin(f*x + e) + 1) + 2*((8*a*b + 3*b^2)*cos(f*x + e)^2 + 2*b^2)*sin(f*x + e))/(f*cos(f*x + e)^4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.31 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) - 1\right ) - \frac {2 \, {\left ({\left (8 \, a b + 3 \, b^{2}\right )} \sin \left (f x + e\right )^{3} - {\left (8 \, a b + 5 \, b^{2}\right )} \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1}}{16 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.32 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (f x + e\right ) + 1 \right |}\right ) - {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (f x + e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (8 \, a b \sin \left (f x + e\right )^{3} + 3 \, b^{2} \sin \left (f x + e\right )^{3} - 8 \, a b \sin \left (f x + e\right ) - 5 \, b^{2} \sin \left (f x + e\right )\right )}}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{2}}}{16 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 18.69 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )\,\left (a^2+a\,b+\frac {3\,b^2}{8}\right )}{f}+\frac {\sin \left (e+f\,x\right )\,\left (\frac {5\,b^2}{8}+a\,b\right )-{\sin \left (e+f\,x\right )}^3\,\left (\frac {3\,b^2}{8}+a\,b\right )}{f\,\left ({\sin \left (e+f\,x\right )}^4-2\,{\sin \left (e+f\,x\right )}^2+1\right )} \]

[In]

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

[Out]

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