3.247 \(\int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x)) \, dx\)

Optimal. Leaf size=61 \[ \frac{(a d+b c) \tan (e+f x)}{f}+\frac{(2 a c+b d) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{b d \tan (e+f x) \sec (e+f x)}{2 f} \]

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.0750499, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {3997, 3787, 3770, 3767, 8} \[ \frac{(a d+b c) \tan (e+f x)}{f}+\frac{(2 a c+b d) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{b d \tan (e+f x) \sec (e+f x)}{2 f} \]

Antiderivative was successfully verified.

[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 3997

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*C
sc[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]

Rule 3787

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 3770

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

Rule 3767

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 8

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

Rubi steps

\begin{align*} \int \sec (e+f x) (a+b \sec (e+f x)) (c+d \sec (e+f x)) \, dx &=\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) \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{b d \sec (e+f x) \tan (e+f x)}{2 f}-\frac{(b c+a d) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{f}\\ &=\frac{(2 a c+b d) \tanh ^{-1}(\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]  time = 0.0269827, size = 75, normalized size = 1.23 \[ \frac{a c \tanh ^{-1}(\sin (e+f x))}{f}+\frac{a d \tan (e+f x)}{f}+\frac{b c \tan (e+f x)}{f}+\frac{b d \tanh ^{-1}(\sin (e+f x))}{2 f}+\frac{b d \tan (e+f x) \sec (e+f x)}{2 f} \]

Antiderivative was successfully verified.

[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]  time = 0.026, size = 86, normalized size = 1.4 \begin{align*}{\frac{ac\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+{\frac{ad\tan \left ( fx+e \right ) }{f}}+{\frac{bc\tan \left ( fx+e \right ) }{f}}+{\frac{db\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}+{\frac{db\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.01459, size = 119, normalized size = 1.95 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]  time = 0.535161, size = 247, normalized size = 4.05 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [B]  time = 1.34489, size = 219, normalized size = 3.59 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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