\(\int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x)) \, dx\) [247]

Optimal result

Integrand size = 27, antiderivative size = 61 \[ \int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x)) \, dx=\frac {(2 a c+b d) \text {arctanh}(\sin (e+f x))}{2 f}+\frac {(b c+a d) \tan (e+f x)}{f}+\frac {b d \sec (e+f x) \tan (e+f x)}{2 f} \]

[Out]

1/2*(2*a*c+b*d)*arctanh(sin(f*x+e))/f+(a*d+b*c)*tan(f*x+e)/f+1/2*b*d*sec(f*x+e)*tan(f*x+e)/f

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4082, 3872, 3855, 3852, 8} \[ \int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x)) \, dx=\frac {(2 a c+b d) \text {arctanh}(\sin (e+f x))}{2 f}+\frac {(a d+b c) \tan (e+f x)}{f}+\frac {b d \tan (e+f x) \sec (e+f x)}{2 f} \]

[In]

Int[Sec[e + f*x]*(a + b*Sec[e + f*x])*(c + d*Sec[e + f*x]),x]

[Out]

((2*a*c + b*d)*ArcTanh[Sin[e + f*x]])/(2*f) + ((b*c + a*d)*Tan[e + f*x])/f + (b*d*Sec[e + f*x]*Tan[e + f*x])/(
2*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3872

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

Rule 4082

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Dist[1/(n + 1), Int[(d
*Csc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e,
f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {b d \sec (e+f x) \tan (e+f x)}{2 f}+\frac {1}{2} \int \sec (e+f x) (2 a c+b d+2 (b c+a d) \sec (e+f x)) \, dx \\ & = \frac {b d \sec (e+f x) \tan (e+f x)}{2 f}+(b c+a d) \int \sec ^2(e+f x) \, dx+\frac {1}{2} (2 a c+b d) \int \sec (e+f x) \, dx \\ & = \frac {(2 a c+b d) \text {arctanh}(\sin (e+f x))}{2 f}+\frac {b d \sec (e+f x) \tan (e+f x)}{2 f}-\frac {(b c+a d) \text {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{f} \\ & = \frac {(2 a c+b d) \text {arctanh}(\sin (e+f x))}{2 f}+\frac {(b c+a d) \tan (e+f x)}{f}+\frac {b d \sec (e+f x) \tan (e+f x)}{2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x)) \, dx=\frac {a c \text {arctanh}(\sin (e+f x))}{f}+\frac {b d \text {arctanh}(\sin (e+f x))}{2 f}+\frac {b c \tan (e+f x)}{f}+\frac {a d \tan (e+f x)}{f}+\frac {b d \sec (e+f x) \tan (e+f x)}{2 f} \]

[In]

Integrate[Sec[e + f*x]*(a + b*Sec[e + f*x])*(c + d*Sec[e + f*x]),x]

[Out]

(a*c*ArcTanh[Sin[e + f*x]])/f + (b*d*ArcTanh[Sin[e + f*x]])/(2*f) + (b*c*Tan[e + f*x])/f + (a*d*Tan[e + f*x])/
f + (b*d*Sec[e + f*x]*Tan[e + f*x])/(2*f)

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {a c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+a d \tan \left (f x +e \right )+b c \tan \left (f x +e \right )+b d \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}\)	\(75\)
default	\(\frac {a c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+a d \tan \left (f x +e \right )+b c \tan \left (f x +e \right )+b d \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}\)	\(75\)
parts	\(\frac {\left (a d +b c \right ) \tan \left (f x +e \right )}{f}+\frac {a c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {b d \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}\)	\(76\)
parallelrisch	\(\frac {-\left (a c +\frac {b d}{2}\right ) \left (1+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+\left (a c +\frac {b d}{2}\right ) \left (1+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\left (a d +b c \right ) \sin \left (2 f x +2 e \right )+\sin \left (f x +e \right ) b d}{f \left (1+\cos \left (2 f x +2 e \right )\right )}\)	\(110\)
norman	\(\frac {\frac {\left (2 a d +2 b c +b d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {\left (2 a d +2 b c -b d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2}}-\frac {\left (2 a c +b d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 f}+\frac {\left (2 a c +b d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 f}\)	\(123\)
risch	\(-\frac {i \left (b d \,{\mathrm e}^{3 i \left (f x +e \right )}-2 a d \,{\mathrm e}^{2 i \left (f x +e \right )}-2 b c \,{\mathrm e}^{2 i \left (f x +e \right )}-b d \,{\mathrm e}^{i \left (f x +e \right )}-2 a d -2 b c \right )}{f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )^{2}}-\frac {a c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) b d}{2 f}+\frac {a c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) b d}{2 f}\)	\(160\)

[In]

int(sec(f*x+e)*(a+b*sec(f*x+e))*(c+d*sec(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/f*(a*c*ln(sec(f*x+e)+tan(f*x+e))+a*d*tan(f*x+e)+b*c*tan(f*x+e)+b*d*(1/2*sec(f*x+e)*tan(f*x+e)+1/2*ln(sec(f*x
+e)+tan(f*x+e))))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.57 \[ \int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x)) \, dx=\frac {{\left (2 \, a c + b d\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (2 \, a c + b d\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (b d + 2 \, {\left (b c + a d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, f \cos \left (f x + e\right )^{2}} \]

[In]

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

[Out]

1/4*((2*a*c + b*d)*cos(f*x + e)^2*log(sin(f*x + e) + 1) - (2*a*c + b*d)*cos(f*x + e)^2*log(-sin(f*x + e) + 1)
+ 2*(b*d + 2*(b*c + a*d)*cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x)) \, dx=-\frac {b d {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 4 \, a c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 4 \, b c \tan \left (f x + e\right ) - 4 \, a d \tan \left (f x + e\right )}{4 \, f} \]

[In]

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

[Out]

-1/4*(b*d*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) - 4*a*c*log(se
c(f*x + e) + tan(f*x + e)) - 4*b*c*tan(f*x + e) - 4*a*d*tan(f*x + e))/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (57) = 114\).

Time = 0.31 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.51 \[ \int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x)) \, dx=\frac {{\left (2 \, a c + b d\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - {\left (2 \, a c + b d\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, b c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - b d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, b c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2}}}{2 \, f} \]

[In]

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

[Out]

1/2*((2*a*c + b*d)*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - (2*a*c + b*d)*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(
2*b*c*tan(1/2*f*x + 1/2*e)^3 + 2*a*d*tan(1/2*f*x + 1/2*e)^3 - b*d*tan(1/2*f*x + 1/2*e)^3 - 2*b*c*tan(1/2*f*x +
1/2*e) - 2*a*d*tan(1/2*f*x + 1/2*e) - b*d*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^2)/f

Mupad [B] (verification not implemented)

Time = 14.56 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.70 \[ \int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x)) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (2\,a\,c+b\,d\right )}{f}+\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,a\,d+2\,b\,c+b\,d\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,a\,d+2\,b\,c-b\,d\right )}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \]

[In]

int(((a + b/cos(e + f*x))*(c + d/cos(e + f*x)))/cos(e + f*x),x)

[Out]

(atanh(tan(e/2 + (f*x)/2))*(2*a*c + b*d))/f + (tan(e/2 + (f*x)/2)*(2*a*d + 2*b*c + b*d) - tan(e/2 + (f*x)/2)^3
*(2*a*d + 2*b*c - b*d))/(f*(tan(e/2 + (f*x)/2)^4 - 2*tan(e/2 + (f*x)/2)^2 + 1))