3.85 \(\int \frac{\sec ^2(c+d x)}{(a+a \sec (c+d x))^5} \, dx\)

Optimal. Leaf size=143 \[ \frac{2 \tan (c+d x)}{63 d \left (a^5 \sec (c+d x)+a^5\right )}+\frac{2 \tan (c+d x)}{63 a d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac{\tan (c+d x)}{21 a^2 d (a \sec (c+d x)+a)^3}+\frac{5 \tan (c+d x)}{63 a d (a \sec (c+d x)+a)^4}-\frac{\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5} \]

[Out]

-Tan[c + d*x]/(9*d*(a + a*Sec[c + d*x])^5) + (5*Tan[c + d*x])/(63*a*d*(a + a*Sec[c + d*x])^4) + Tan[c + d*x]/(
21*a^2*d*(a + a*Sec[c + d*x])^3) + (2*Tan[c + d*x])/(63*a*d*(a^2 + a^2*Sec[c + d*x])^2) + (2*Tan[c + d*x])/(63
*d*(a^5 + a^5*Sec[c + d*x]))

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Rubi [A]  time = 0.157324, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3797, 3796, 3794} \[ \frac{2 \tan (c+d x)}{63 d \left (a^5 \sec (c+d x)+a^5\right )}+\frac{2 \tan (c+d x)}{63 a d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac{\tan (c+d x)}{21 a^2 d (a \sec (c+d x)+a)^3}+\frac{5 \tan (c+d x)}{63 a d (a \sec (c+d x)+a)^4}-\frac{\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^5,x]

[Out]

-Tan[c + d*x]/(9*d*(a + a*Sec[c + d*x])^5) + (5*Tan[c + d*x])/(63*a*d*(a + a*Sec[c + d*x])^4) + Tan[c + d*x]/(
21*a^2*d*(a + a*Sec[c + d*x])^3) + (2*Tan[c + d*x])/(63*a*d*(a^2 + a^2*Sec[c + d*x])^2) + (2*Tan[c + d*x])/(63
*d*(a^5 + a^5*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))^5} \, dx &=-\frac{\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac{5 \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^4} \, dx}{9 a}\\ &=-\frac{\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac{5 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac{5 \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx}{21 a^2}\\ &=-\frac{\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac{5 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac{\tan (c+d x)}{21 a^2 d (a+a \sec (c+d x))^3}+\frac{2 \int \frac{\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{21 a^3}\\ &=-\frac{\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac{5 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac{\tan (c+d x)}{21 a^2 d (a+a \sec (c+d x))^3}+\frac{2 \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac{2 \int \frac{\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{63 a^4}\\ &=-\frac{\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac{5 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac{\tan (c+d x)}{21 a^2 d (a+a \sec (c+d x))^3}+\frac{2 \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac{2 \tan (c+d x)}{63 d \left (a^5+a^5 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.22972, size = 125, normalized size = 0.87 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-315 \sin \left (c+\frac{d x}{2}\right )+273 \sin \left (c+\frac{3 d x}{2}\right )-147 \sin \left (2 c+\frac{3 d x}{2}\right )+117 \sin \left (2 c+\frac{5 d x}{2}\right )-63 \sin \left (3 c+\frac{5 d x}{2}\right )+45 \sin \left (3 c+\frac{7 d x}{2}\right )+5 \sin \left (4 c+\frac{9 d x}{2}\right )+315 \sin \left (\frac{d x}{2}\right )\right ) \sec ^9\left (\frac{1}{2} (c+d x)\right )}{16128 a^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + a*Sec[c + d*x])^5,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^9*(315*Sin[(d*x)/2] - 315*Sin[c + (d*x)/2] + 273*Sin[c + (3*d*x)/2] - 147*Sin[2*c +
(3*d*x)/2] + 117*Sin[2*c + (5*d*x)/2] - 63*Sin[3*c + (5*d*x)/2] + 45*Sin[3*c + (7*d*x)/2] + 5*Sin[4*c + (9*d*
x)/2]))/(16128*a^5*d)

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Maple [A]  time = 0.035, size = 58, normalized size = 0.4 \begin{align*}{\frac{1}{16\,d{a}^{5}} \left ( -{\frac{1}{9} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9}}+{\frac{2}{7} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\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)^2/(a+a*sec(d*x+c))^5,x)

[Out]

1/16/d/a^5*(-1/9*tan(1/2*d*x+1/2*c)^9+2/7*tan(1/2*d*x+1/2*c)^7-2/3*tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.05075, size = 117, normalized size = 0.82 \begin{align*} \frac{\frac{63 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{42 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{18 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{1008 \, a^{5} 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))^5,x, algorithm="maxima")

[Out]

1/1008*(63*sin(d*x + c)/(cos(d*x + c) + 1) - 42*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 18*sin(d*x + c)^7/(cos(d
*x + c) + 1)^7 - 7*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/(a^5*d)

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Fricas [A]  time = 1.61194, size = 312, normalized size = 2.18 \begin{align*} \frac{{\left (5 \, \cos \left (d x + c\right )^{4} + 25 \, \cos \left (d x + c\right )^{3} + 21 \, \cos \left (d x + c\right )^{2} + 10 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{63 \,{\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} 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))^5,x, algorithm="fricas")

[Out]

1/63*(5*cos(d*x + c)^4 + 25*cos(d*x + c)^3 + 21*cos(d*x + c)^2 + 10*cos(d*x + c) + 2)*sin(d*x + c)/(a^5*d*cos(
d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c)
+ a^5*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 ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec{\left (c + d x \right )} + 1}\, dx}{a^{5}} \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))**5,x)

[Out]

Integral(sec(c + d*x)**2/(sec(c + d*x)**5 + 5*sec(c + d*x)**4 + 10*sec(c + d*x)**3 + 10*sec(c + d*x)**2 + 5*se
c(c + d*x) + 1), x)/a**5

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Giac [A]  time = 1.4709, size = 80, normalized size = 0.56 \begin{align*} -\frac{7 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 18 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 42 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 63 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{1008 \, a^{5} 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))^5,x, algorithm="giac")

[Out]

-1/1008*(7*tan(1/2*d*x + 1/2*c)^9 - 18*tan(1/2*d*x + 1/2*c)^7 + 42*tan(1/2*d*x + 1/2*c)^3 - 63*tan(1/2*d*x + 1
/2*c))/(a^5*d)