\(\int \sec (e+f x) (a+b \sec ^2(e+f x)) \, dx\) [154]

Optimal result

Integrand size = 19, antiderivative size = 40 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {(2 a+b) \text {arctanh}(\sin (e+f x))}{2 f}+\frac {b \sec (e+f x) \tan (e+f x)}{2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {4131, 3855} \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {(2 a+b) \text {arctanh}(\sin (e+f x))}{2 f}+\frac {b \tan (e+f x) \sec (e+f x)}{2 f} \]

[In]

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

[Out]

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

Rule 3855

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

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {a \text {arctanh}(\sin (e+f x))}{f}+\frac {b \text {arctanh}(\sin (e+f x))}{2 f}+\frac {b \sec (e+f x) \tan (e+f x)}{2 f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(\frac {a \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+b \left (\frac {\tan \left (f x +e \right ) \sec \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}\)	\(55\)
default	\(\frac {a \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+b \left (\frac {\tan \left (f x +e \right ) \sec \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}\)	\(55\)
parts	\(\frac {b \left (\frac {\tan \left (f x +e \right ) \sec \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}+\frac {a \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\)	\(57\)
parallelrisch	\(\frac {-\left (a +\frac {b}{2}\right ) \left (1+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+\left (a +\frac {b}{2}\right ) \left (1+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\sin \left (f x +e \right ) b}{f \left (1+\cos \left (2 f x +2 e \right )\right )}\)	\(86\)
norman	\(\frac {\frac {b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2}}-\frac {\left (2 a +b \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 f}+\frac {\left (2 a +b \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 f}\)	\(93\)
risch	\(-\frac {i b \left ({\mathrm e}^{3 i \left (f x +e \right )}-{\mathrm e}^{i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) b}{2 f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) a}{f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) b}{2 f}\)	\(118\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.80 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {{\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, b \sin \left (f x + e\right )}{4 \, f \cos \left (f x + e\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.45 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {{\left (2 \, a + b\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (2 \, a + b\right )} \log \left (\sin \left (f x + e\right ) - 1\right ) - \frac {2 \, b \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1}}{4 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.50 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {{\left (2 \, a + b\right )} \log \left ({\left | \sin \left (f x + e\right ) + 1 \right |}\right ) - {\left (2 \, a + b\right )} \log \left ({\left | \sin \left (f x + e\right ) - 1 \right |}\right ) - \frac {2 \, b \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1}}{4 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 18.46 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.02 \[ \int \sec (e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )\,\left (a+\frac {b}{2}\right )}{f}-\frac {b\,\sin \left (e+f\,x\right )}{2\,f\,\left ({\sin \left (e+f\,x\right )}^2-1\right )} \]

[In]

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

[Out]

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