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

Optimal. Leaf size=83 \[ \frac{2 \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{2 \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}+\frac{\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.080896, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3796, 3794} \[ \frac{2 \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{2 \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}+\frac{\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/(a + a*Sec[c + d*x])^3,x]

[Out]

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

Rule 3796

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*Cot[e + f*x]*(a
+ b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(m + 1)/(a*(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)] && IntegerQ[2*m]

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

Mathematica [A]  time = 0.2189, size = 86, normalized size = 1.04 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-30 \sin \left (c+\frac{d x}{2}\right )+20 \sin \left (c+\frac{3 d x}{2}\right )-15 \sin \left (2 c+\frac{3 d x}{2}\right )+7 \sin \left (2 c+\frac{5 d x}{2}\right )+40 \sin \left (\frac{d x}{2}\right )\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right )}{240 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(40*Sin[(d*x)/2] - 30*Sin[c + (d*x)/2] + 20*Sin[c + (3*d*x)/2] - 15*Sin[2*c + (3*
d*x)/2] + 7*Sin[2*c + (5*d*x)/2]))/(240*a^3*d)

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Maple [A]  time = 0.039, size = 45, normalized size = 0.5 \begin{align*}{\frac{1}{4\,d{a}^{3}} \left ({\frac{1}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{2}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.11988, size = 90, normalized size = 1.08 \begin{align*} \frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{10 \, \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}}}{60 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 10*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*d)

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Fricas [A]  time = 1.60977, size = 186, normalized size = 2.24 \begin{align*} \frac{{\left (7 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 2\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(sec(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(7*cos(d*x + c)^2 + 6*cos(d*x + c) + 2)*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{\sec{\left (c + d x \right )}}{\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(sec(d*x+c)/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)/(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 = 1.34133, size = 62, normalized size = 0.75 \begin{align*} \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{60 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(3*tan(1/2*d*x + 1/2*c)^5 - 10*tan(1/2*d*x + 1/2*c)^3 + 15*tan(1/2*d*x + 1/2*c))/(a^3*d)