3.83 \(\int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx\)

Optimal. Leaf size=159 \[ \frac {\tan (c+d x)}{9 d \left (a^5 \sec (c+d x)+a^5\right )}-\frac {8 \tan (c+d x)}{63 a d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {\tan (c+d x)}{21 a^2 d (a \sec (c+d x)+a)^3}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}+\frac {5 \tan (c+d x) \sec ^3(c+d x)}{63 a d (a \sec (c+d x)+a)^4} \]

[Out]

-1/9*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^5+5/63*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^4+1/21*tan
(d*x+c)/a^2/d/(a+a*sec(d*x+c))^3-8/63*tan(d*x+c)/a/d/(a^2+a^2*sec(d*x+c))^2+1/9*tan(d*x+c)/d/(a^5+a^5*sec(d*x+
c))

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Rubi [A]  time = 0.21, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3811, 3810, 3799, 4000, 3794} \[ \frac {\tan (c+d x)}{9 d \left (a^5 \sec (c+d x)+a^5\right )}-\frac {8 \tan (c+d x)}{63 a d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {\tan (c+d x)}{21 a^2 d (a \sec (c+d x)+a)^3}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}+\frac {5 \tan (c+d x) \sec ^3(c+d x)}{63 a d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + a*Sec[c + d*x])^5,x]

[Out]

-(Sec[c + d*x]^4*Tan[c + d*x])/(9*d*(a + a*Sec[c + d*x])^5) + (5*Sec[c + d*x]^3*Tan[c + d*x])/(63*a*d*(a + a*S
ec[c + d*x])^4) + Tan[c + d*x]/(21*a^2*d*(a + a*Sec[c + d*x])^3) - (8*Tan[c + d*x])/(63*a*d*(a^2 + a^2*Sec[c +
d*x])^2) + Tan[c + d*x]/(9*d*(a^5 + a^5*Sec[c + d*x]))

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]

Rule 3799

Int[csc[(e_.) + (f_.)*(x_)]^3*(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[1/(a^2*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)
]

Rule 3810

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(b*d*C
ot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[(d*(m + 1))/(b*(2*m +
1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && E
qQ[a^2 - b^2, 0] && EqQ[m + n, 0] && LtQ[m, -2^(-1)] && IntegerQ[2*m]

Rule 3811

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[
e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*(2*m + 1)), x] + Dist[m/(a*(2*m + 1)), Int[(a + b*Csc[e
+ f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1
, 0] && LtQ[m, -2^(-1)]

Rule 4000

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*B*m + A*b*
(m + 1))/(a*b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x
] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^4} \, dx}{9 a}\\ &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {5 \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx}{21 a^2}\\ &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \sec ^3(c+d x) \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 {\int \frac {\sec (c+d x) (-3 a+5 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{21 a^4}\\ &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \sec ^3(c+d x) \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 {8 \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{9 a^4}\\ &=-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \sec ^3(c+d x) \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 {8 \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {\tan (c+d x)}{9 d \left (a^5+a^5 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 97, normalized size = 0.61 \[ \frac {\sec \left (\frac {c}{2}\right ) \left (-63 \sin \left (c+\frac {d x}{2}\right )+84 \sin \left (c+\frac {3 d x}{2}\right )+36 \sin \left (2 c+\frac {5 d x}{2}\right )+9 \sin \left (3 c+\frac {7 d x}{2}\right )+\sin \left (4 c+\frac {9 d x}{2}\right )+63 \sin \left (\frac {d x}{2}\right )\right ) \sec ^9\left (\frac {1}{2} (c+d x)\right )}{8064 a^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + a*Sec[c + d*x])^5,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^9*(63*Sin[(d*x)/2] - 63*Sin[c + (d*x)/2] + 84*Sin[c + (3*d*x)/2] + 36*Sin[2*c + (5*
d*x)/2] + 9*Sin[3*c + (7*d*x)/2] + Sin[4*c + (9*d*x)/2]))/(8064*a^5*d)

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fricas [A]  time = 0.67, size = 123, normalized size = 0.77 \[ \frac {{\left (2 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 21 \, \cos \left (d x + c\right )^{2} + 25 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{63 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sec(d*x+c))^5,x, algorithm="fricas")

[Out]

1/63*(2*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 21*cos(d*x + c)^2 + 25*cos(d*x + c) + 5)*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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giac [A]  time = 1.43, size = 59, normalized size = 0.37 \[ -\frac {7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 42 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 63 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{1008 \, a^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(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)

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maple [A]  time = 0.41, size = 58, normalized size = 0.36 \[ \frac {-\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}-\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(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 = 0.60, size = 87, normalized size = 0.55 \[ \frac {\frac {63 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {42 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {18 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{1008 \, a^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(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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mupad [B]  time = 0.72, size = 58, normalized size = 0.36 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+42\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+63\right )}{1008\,a^5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + a/cos(c + d*x))^5),x)

[Out]

(tan(c/2 + (d*x)/2)*(42*tan(c/2 + (d*x)/2)^2 - 18*tan(c/2 + (d*x)/2)^6 - 7*tan(c/2 + (d*x)/2)^8 + 63))/(1008*a
^5*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+a*sec(d*x+c))**5,x)

[Out]

Integral(sec(c + d*x)**4/(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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