3.65 \(\int \frac{\sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0967388, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3797, 3796, 3794} \[ \frac{\tan (c+d x)}{5 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{\tan (c+d x)}{5 a d (a \sec (c+d x)+a)^2}-\frac{\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^3,x]

[Out]

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

Rule 3797

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)]

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

Mathematica [A]  time = 0.159927, size = 71, normalized size = 0.86 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-5 \sin \left (c+\frac{d x}{2}\right )+5 \sin \left (c+\frac{3 d x}{2}\right )+\sin \left (2 c+\frac{5 d x}{2}\right )+5 \sin \left (\frac{d x}{2}\right )\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right )}{80 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^3,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(5*Sin[(d*x)/2] - 5*Sin[c + (d*x)/2] + 5*Sin[c + (3*d*x)/2] + Sin[2*c + (5*d*x)/2
]))/(80*a^3*d)

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Maple [A]  time = 0.031, size = 32, normalized size = 0.4 \begin{align*}{\frac{1}{4\,d{a}^{3}} \left ( -{\frac{1}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+\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)^2/(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+tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.16613, size = 63, normalized size = 0.76 \begin{align*} \frac{\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{20 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.59893, size = 182, normalized size = 2.19 \begin{align*} \frac{{\left (\cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{5 \,{\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)^2/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/5*(cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*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 ^{2}{\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)**2/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**2/(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 = 1.39828, size = 42, normalized size = 0.51 \begin{align*} -\frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 5 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{20 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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