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

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

[Out]

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

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Rubi [A]  time = 0.110848, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3796, 3794} \[ \frac{2 \tan (c+d x)}{35 d \left (a^4 \sec (c+d x)+a^4\right )}+\frac{2 \tan (c+d x)}{35 d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac{3 \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3}+\frac{\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Tan[c + d*x]/(7*d*(a + a*Sec[c + d*x])^4) + (3*Tan[c + d*x])/(35*a*d*(a + a*Sec[c + d*x])^3) + (2*Tan[c + d*x]
)/(35*d*(a^2 + a^2*Sec[c + d*x])^2) + (2*Tan[c + d*x])/(35*d*(a^4 + a^4*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))^4} \, dx &=\frac{\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{3 \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{3 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{6 \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{3 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{2 \tan (c+d x)}{35 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac{2 \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{35 a^3}\\ &=\frac{\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{3 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{2 \tan (c+d x)}{35 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac{2 \tan (c+d x)}{35 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.237308, size = 112, normalized size = 1. \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-210 \sin \left (c+\frac{d x}{2}\right )+147 \sin \left (c+\frac{3 d x}{2}\right )-105 \sin \left (2 c+\frac{3 d x}{2}\right )+49 \sin \left (2 c+\frac{5 d x}{2}\right )-35 \sin \left (3 c+\frac{5 d x}{2}\right )+12 \sin \left (3 c+\frac{7 d x}{2}\right )+210 \sin \left (\frac{d x}{2}\right )\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right )}{2240 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(210*Sin[(d*x)/2] - 210*Sin[c + (d*x)/2] + 147*Sin[c + (3*d*x)/2] - 105*Sin[2*c +
(3*d*x)/2] + 49*Sin[2*c + (5*d*x)/2] - 35*Sin[3*c + (5*d*x)/2] + 12*Sin[3*c + (7*d*x)/2]))/(2240*a^4*d)

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Maple [A]  time = 0.036, 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{3}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}- \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))^4,x)

[Out]

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

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Maxima [A]  time = 1.30959, size = 117, normalized size = 1.04 \begin{align*} \frac{\frac{35 \, \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{5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{280 \, a^{4} 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))^4,x, algorithm="maxima")

[Out]

1/280*(35*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 - 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a^4*d)

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Fricas [A]  time = 1.59648, size = 248, normalized size = 2.21 \begin{align*} \frac{{\left (12 \, \cos \left (d x + c\right )^{3} + 13 \, \cos \left (d x + c\right )^{2} + 8 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{35 \,{\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)/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/35*(12*cos(d*x + c)^3 + 13*cos(d*x + c)^2 + 8*cos(d*x + c) + 2)*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)

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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 ^{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)/(a+a*sec(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)/(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

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Giac [A]  time = 1.36765, size = 80, normalized size = 0.71 \begin{align*} -\frac{5 \, \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} - 35 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{280 \, a^{4} 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))^4,x, algorithm="giac")

[Out]

-1/280*(5*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 - 35*tan(1/2*d*x + 1/
2*c))/(a^4*d)