3.75 \(\int \frac{\sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=112 \[ \frac{8 \tan (c+d x)}{105 d \left (a^4 \sec (c+d x)+a^4\right )}+\frac{8 \tan (c+d x)}{105 d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac{4 \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac{\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

-Tan[c + d*x]/(7*d*(a + a*Sec[c + d*x])^4) + (4*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3) + (8*Tan[c + d*x
])/(105*d*(a^2 + a^2*Sec[c + d*x])^2) + (8*Tan[c + d*x])/(105*d*(a^4 + a^4*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.125312, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3797, 3796, 3794} \[ \frac{8 \tan (c+d x)}{105 d \left (a^4 \sec (c+d x)+a^4\right )}+\frac{8 \tan (c+d x)}{105 d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac{4 \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac{\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^4,x]

[Out]

-Tan[c + d*x]/(7*d*(a + a*Sec[c + d*x])^4) + (4*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3) + (8*Tan[c + d*x
])/(105*d*(a^2 + a^2*Sec[c + d*x])^2) + (8*Tan[c + d*x])/(105*d*(a^4 + a^4*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))^4} \, dx &=-\frac{\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{4 \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx}{7 a}\\ &=-\frac{\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{4 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{8 \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{35 a^2}\\ &=-\frac{\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{4 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{8 \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac{8 \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{105 a^3}\\ &=-\frac{\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{4 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{8 \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac{8 \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.258605, size = 99, normalized size = 0.88 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-175 \sin \left (c+\frac{d x}{2}\right )+168 \sin \left (c+\frac{3 d x}{2}\right )-105 \sin \left (2 c+\frac{3 d x}{2}\right )+91 \sin \left (2 c+\frac{5 d x}{2}\right )+13 \sin \left (3 c+\frac{7 d x}{2}\right )+280 \sin \left (\frac{d x}{2}\right )\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right )}{6720 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^4,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(280*Sin[(d*x)/2] - 175*Sin[c + (d*x)/2] + 168*Sin[c + (3*d*x)/2] - 105*Sin[2*c +
(3*d*x)/2] + 91*Sin[2*c + (5*d*x)/2] + 13*Sin[3*c + (7*d*x)/2]))/(6720*a^4*d)

________________________________________________________________________________________

Maple [A]  time = 0.035, size = 58, normalized size = 0.5 \begin{align*}{\frac{1}{8\,d{a}^{4}} \left ({\frac{1}{7} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{1}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{1}{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)^2/(a+a*sec(d*x+c))^4,x)

[Out]

1/8/d/a^4*(1/7*tan(1/2*d*x+1/2*c)^7-1/5*tan(1/2*d*x+1/2*c)^5-1/3*tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

Maxima [A]  time = 1.32243, size = 117, normalized size = 1.04 \begin{align*} \frac{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{840 \, a^{4} 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))^4,x, algorithm="maxima")

[Out]

1/840*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^4*d)

________________________________________________________________________________________

Fricas [A]  time = 1.5864, size = 251, normalized size = 2.24 \begin{align*} \frac{{\left (13 \, \cos \left (d x + c\right )^{3} + 52 \, \cos \left (d x + c\right )^{2} + 32 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right )}{105 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} 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))^4,x, algorithm="fricas")

[Out]

1/105*(13*cos(d*x + c)^3 + 52*cos(d*x + c)^2 + 32*cos(d*x + c) + 8)*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4
*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{2}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec{\left (c + d x \right )} + 1}\, dx}{a^{4}} \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))**4,x)

[Out]

Integral(sec(c + d*x)**2/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)/a*
*4

________________________________________________________________________________________

Giac [A]  time = 1.35183, size = 80, normalized size = 0.71 \begin{align*} \frac{15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 21 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 35 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{840 \, a^{4} 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))^4,x, algorithm="giac")

[Out]

1/840*(15*tan(1/2*d*x + 1/2*c)^7 - 21*tan(1/2*d*x + 1/2*c)^5 - 35*tan(1/2*d*x + 1/2*c)^3 + 105*tan(1/2*d*x + 1
/2*c))/(a^4*d)