3.1.57 \(\int \frac {1}{(a+a \sec (c+d x))^2} \, dx\) [57]

Optimal. Leaf size=57 \[ \frac {x}{a^2}-\frac {4 \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

x/a^2-4/3*tan(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*tan(d*x+c)/d/(a+a*sec(d*x+c))^2

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Rubi [A]
time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3862, 4004, 3879} \begin {gather*} -\frac {4 \tan (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {x}{a^2}-\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(-2),x]

[Out]

x/a^2 - (4*Tan[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - Tan[c + d*x]/(3*d*(a + a*Sec[c + d*x])^2)

Rule 3862

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-Cot[c + d*x])*((a + b*Csc[c + d*x])^n/(d*
(2*n + 1))), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*
x]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

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 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a),
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sec (c+d x))^2} \, dx &=-\frac {\tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {-3 a+a \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=\frac {x}{a^2}-\frac {\tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {4 \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{3 a}\\ &=\frac {x}{a^2}-\frac {\tan (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {4 \tan (c+d x)}{3 d \left (a^2+a^2 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.30, size = 112, normalized size = 1.96 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (9 d x \cos \left (\frac {d x}{2}\right )+9 d x \cos \left (c+\frac {d x}{2}\right )+3 d x \cos \left (c+\frac {3 d x}{2}\right )+3 d x \cos \left (2 c+\frac {3 d x}{2}\right )-18 \sin \left (\frac {d x}{2}\right )+12 \sin \left (c+\frac {d x}{2}\right )-10 \sin \left (c+\frac {3 d x}{2}\right )\right )}{24 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(9*d*x*Cos[(d*x)/2] + 9*d*x*Cos[c + (d*x)/2] + 3*d*x*Cos[c + (3*d*x)/2] + 3*d*x*C
os[2*c + (3*d*x)/2] - 18*Sin[(d*x)/2] + 12*Sin[c + (d*x)/2] - 10*Sin[c + (3*d*x)/2]))/(24*a^2*d)

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Maple [A]
time = 0.05, size = 46, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(46\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(46\)
norman	\(\frac {\frac {x}{a}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a d}}{a}\)	\(47\)
risch	\(\frac {x}{a^{2}}-\frac {2 i \left (6 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/2/d/a^2*(1/3*tan(1/2*d*x+1/2*c)^3-3*tan(1/2*d*x+1/2*c)+4*arctan(tan(1/2*d*x+1/2*c)))

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Maxima [A]
time = 0.50, size = 72, normalized size = 1.26 \begin {gather*} -\frac {\frac {\frac {9 \, \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}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/6*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c)/(
cos(d*x + c) + 1))/a^2)/d

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Fricas [A]
time = 2.99, size = 80, normalized size = 1.40 \begin {gather*} \frac {3 \, d x \cos \left (d x + c\right )^{2} + 6 \, d x \cos \left (d x + c\right ) + 3 \, d x - {\left (5 \, \cos \left (d x + c\right ) + 4\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(3*d*x*cos(d*x + c)^2 + 6*d*x*cos(d*x + c) + 3*d*x - (5*cos(d*x + c) + 4)*sin(d*x + c))/(a^2*d*cos(d*x + c
)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.44, size = 50, normalized size = 0.88 \begin {gather*} \frac {\frac {6 \, {\left (d x + c\right )}}{a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)/a^2 + (a^4*tan(1/2*d*x + 1/2*c)^3 - 9*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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Mupad [B]
time = 0.63, size = 35, normalized size = 0.61 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,d\,x}{6\,a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a/cos(c + d*x))^2,x)

[Out]

(tan(c/2 + (d*x)/2)^3 - 9*tan(c/2 + (d*x)/2) + 6*d*x)/(6*a^2*d)

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