\(\int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [55]

Optimal result

Integrand size = 21, antiderivative size = 55 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {2 \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3882, 3879} \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {2 \tan (c+d x)}{3 d \left (a^2 \sec (c+d x)+a^2\right )}-\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[In]

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

[Out]

-1/3*Tan[c + d*x]/(d*(a + a*Sec[c + d*x])^2) + (2*Tan[c + d*x])/(3*d*(a^2 + a^2*Sec[c + d*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 3882

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (3 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )}{12 a^2 d} \]

[In]

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

[Out]

(Sec[(c + d*x)/2]^3*(3*Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2]))/(12*a^2*d)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.56

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3\right )}{6 a^{2} d}\)	\(31\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}\)	\(32\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}\)	\(32\)
risch	\(\frac {2 i \left (1+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(36\)
norman	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 a d}}{a \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\)	\(76\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {{\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{6 \, a^{2} d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.56 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{6 \, a^{2} d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.48 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.55 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\right )}{6\,a^2\,d} \]

[In]

int(1/(cos(c + d*x)^2*(a + a/cos(c + d*x))^2),x)

[Out]

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