\(\int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx\) [221]

Optimal result

Integrand size = 29, antiderivative size = 65 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx=\frac {(c-d) \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {(c+2 d) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {4085, 3879} \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx=\frac {(c+2 d) \tan (e+f x)}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {(c-d) \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x]))/(a + a*Sec[e + f*x])^2,x]

[Out]

((c - d)*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) + ((c + 2*d)*Tan[e + f*x])/(3*f*(a^2 + a^2*Sec[e + f*x]))

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 4085

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

Rubi steps \begin{align*} \text {integral}& = \frac {(c-d) \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {(c+2 d) \int \frac {\sec (e+f x)}{a+a \sec (e+f x)} \, dx}{3 a} \\ & = \frac {(c-d) \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {(c+2 d) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.17 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx=\frac {\cos \left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {e}{2}\right ) \left (3 (c+d) \sin \left (\frac {f x}{2}\right )-3 c \sin \left (e+\frac {f x}{2}\right )+(2 c+d) \sin \left (e+\frac {3 f x}{2}\right )\right )}{3 a^2 f (1+\cos (e+f x))^2} \]

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x]))/(a + a*Sec[e + f*x])^2,x]

[Out]

(Cos[(e + f*x)/2]*Sec[e/2]*(3*(c + d)*Sin[(f*x)/2] - 3*c*Sin[e + (f*x)/2] + (2*c + d)*Sin[e + (3*f*x)/2]))/(3*
a^2*f*(1 + Cos[e + f*x])^2)

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(-\frac {\left (-3 c -3 d +\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{6 a^{2} f}\)	\(42\)
derivativedivides	\(\frac {-\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+\frac {d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 f \,a^{2}}\)	\(60\)
default	\(\frac {-\frac {c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+\frac {d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 f \,a^{2}}\)	\(60\)
risch	\(\frac {2 i \left (3 \,{\mathrm e}^{2 i \left (f x +e \right )} c +3 \,{\mathrm e}^{i \left (f x +e \right )} c +3 d \,{\mathrm e}^{i \left (f x +e \right )}+2 c +d \right )}{3 f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3}}\)	\(64\)
norman	\(\frac {-\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{6 a f}-\frac {\left (c +d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 a f}+\frac {\left (2 c +d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}}{a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )}\)	\(89\)

[In]

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

[Out]

-1/6*(-3*c-3*d+(c-d)*tan(1/2*f*x+1/2*e)^2)*tan(1/2*f*x+1/2*e)/a^2/f

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx=\frac {{\left ({\left (2 \, c + d\right )} \cos \left (f x + e\right ) + c + 2 \, d\right )} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}} \]

[In]

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

[Out]

1/3*((2*c + d)*cos(f*x + e) + c + 2*d)*sin(f*x + e)/(a^2*f*cos(f*x + e)^2 + 2*a^2*f*cos(f*x + e) + a^2*f)

Sympy [F]

\[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx=\frac {\int \frac {c \sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))/(a+a*sec(f*x+e))**2,x)

[Out]

(Integral(c*sec(e + f*x)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(d*sec(e + f*x)**2/(sec(e + f*x)
**2 + 2*sec(e + f*x) + 1), x))/a**2

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.43 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx=\frac {\frac {d {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} + \frac {c {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \]

[In]

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

[Out]

1/6*(d*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 + c*(3*sin(f*x + e)/(cos(
f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2)/f

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx=-\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{6 \, a^{2} f} \]

[In]

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

[Out]

-1/6*(c*tan(1/2*f*x + 1/2*e)^3 - d*tan(1/2*f*x + 1/2*e)^3 - 3*c*tan(1/2*f*x + 1/2*e) - 3*d*tan(1/2*f*x + 1/2*e
))/(a^2*f)

Mupad [B] (verification not implemented)

Time = 13.58 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (c+d\right )}{2\,a^2\,f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (c-d\right )}{6\,a^2\,f} \]

[In]

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

[Out]

(tan(e/2 + (f*x)/2)*(c + d))/(2*a^2*f) - (tan(e/2 + (f*x)/2)^3*(c - d))/(6*a^2*f)