3.67 \(\int \frac{1}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=88 \[ -\frac{22 \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{x}{a^3}-\frac{7 \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

x/a^3 - Tan[c + d*x]/(5*d*(a + a*Sec[c + d*x])^3) - (7*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - (22*Tan
[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.112323, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3777, 3922, 3919, 3794} \[ -\frac{22 \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{x}{a^3}-\frac{7 \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(-3),x]

[Out]

x/a^3 - Tan[c + d*x]/(5*d*(a + a*Sec[c + d*x])^3) - (7*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - (22*Tan
[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

Rule 3777

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 3922

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

Rule 3919

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]

Rule 3794

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]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sec (c+d x))^3} \, dx &=-\frac{\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{\int \frac{-5 a+2 a \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{7 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{15 a^2-7 a^2 \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=\frac{x}{a^3}-\frac{\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{7 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{22 \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac{x}{a^3}-\frac{\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{7 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{22 \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.271052, size = 162, normalized size = 1.84 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (60 d x \cos ^5\left (\frac{1}{2} (c+d x)\right )+26 \tan \left (\frac{c}{2}\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right )-3 \tan \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right )-3 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )-128 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right )+26 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )\right )}{15 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^(-3),x]

[Out]

(2*Cos[(c + d*x)/2]*Sec[c + d*x]^3*(60*d*x*Cos[(c + d*x)/2]^5 - 3*Sec[c/2]*Sin[(d*x)/2] + 26*Cos[(c + d*x)/2]^
2*Sec[c/2]*Sin[(d*x)/2] - 128*Cos[(c + d*x)/2]^4*Sec[c/2]*Sin[(d*x)/2] - 3*Cos[(c + d*x)/2]*Tan[c/2] + 26*Cos[
(c + d*x)/2]^3*Tan[c/2]))/(15*a^3*d*(1 + Sec[c + d*x])^3)

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Maple [A]  time = 0.04, size = 75, normalized size = 0.9 \begin{align*} -{\frac{1}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{1}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(d*x+c))^3,x)

[Out]

-1/20/d/a^3*tan(1/2*d*x+1/2*c)^5+1/3/d/a^3*tan(1/2*d*x+1/2*c)^3-7/4/d/a^3*tan(1/2*d*x+1/2*c)+2/d/a^3*arctan(ta
n(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.71245, size = 124, normalized size = 1.41 \begin{align*} -\frac{\frac{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{120 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]  time = 1.66019, size = 300, normalized size = 3.41 \begin{align*} \frac{15 \, d x \cos \left (d x + c\right )^{3} + 45 \, d x \cos \left (d x + c\right )^{2} + 45 \, d x \cos \left (d x + c\right ) + 15 \, d x -{\left (32 \, \cos \left (d x + c\right )^{2} + 51 \, \cos \left (d x + c\right ) + 22\right )} \sin \left (d x + c\right )}{15 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(15*d*x*cos(d*x + c)^3 + 45*d*x*cos(d*x + c)^2 + 45*d*x*cos(d*x + c) + 15*d*x - (32*cos(d*x + c)^2 + 51*c
os(d*x + c) + 22)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(1/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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Giac [A]  time = 1.34064, size = 92, normalized size = 1.05 \begin{align*} \frac{\frac{60 \,{\left (d x + c\right )}}{a^{3}} - \frac{3 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 20 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*(d*x + c)/a^3 - (3*a^12*tan(1/2*d*x + 1/2*c)^5 - 20*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*a^12*tan(1/2*d*
x + 1/2*c))/a^15)/d