\(\int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx\) [83]

Optimal result

Integrand size = 21, antiderivative size = 159 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {5 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {\tan (c+d x)}{21 a^2 d (a+a \sec (c+d x))^3}-\frac {8 \tan (c+d x)}{63 a d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {\tan (c+d x)}{9 d \left (a^5+a^5 \sec (c+d x)\right )} \]

[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))

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3896, 3895, 3884, 4085, 3879} \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\tan (c+d x)}{9 d \left (a^5 \sec (c+d x)+a^5\right )}-\frac {8 \tan (c+d x)}{63 a d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {\tan (c+d x)}{21 a^2 d (a \sec (c+d x)+a)^3}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}+\frac {5 \tan (c+d x) \sec ^3(c+d x)}{63 a d (a \sec (c+d x)+a)^4} \]

[In]

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

[Out]

-1/9*(Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^5) + (5*Sec[c + d*x]^3*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) - (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 3879

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 3884

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 3895

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b*d*Co
t[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 3896

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 4085

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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.40 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {(65+130 \cos (c+d x)+46 \cos (2 (c+d x))+10 \cos (3 (c+d x))+\cos (4 (c+d x))) \sin (c+d x)}{252 a^5 d (1+\cos (c+d x))^5} \]

[In]

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

[Out]

((65 + 130*Cos[c + d*x] + 46*Cos[2*(c + d*x)] + 10*Cos[3*(c + d*x)] + Cos[4*(c + d*x)])*Sin[c + d*x])/(252*a^5
*d*(1 + Cos[c + d*x])^5)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36

method	result	size
parallelrisch	\(-\frac {\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{7}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{144 d \,a^{5}}\)	\(57\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\)	\(58\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\)	\(58\)
risch	\(\frac {4 i \left (63 \,{\mathrm e}^{5 i \left (d x +c \right )}+63 \,{\mathrm e}^{4 i \left (d x +c \right )}+84 \,{\mathrm e}^{3 i \left (d x +c \right )}+36 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{63 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}\)	\(80\)
norman	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{48 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{16 a d}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{112 a d}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{1008 a d}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{336 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{336 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{144 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{4}}\)	\(171\)

[In]

int(sec(d*x+c)^4/(a+a*sec(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

-1/144*(tan(1/2*d*x+1/2*c)^8+18/7*tan(1/2*d*x+1/2*c)^6-6*tan(1/2*d*x+1/2*c)^2-9)*tan(1/2*d*x+1/2*c)/d/a^5

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.77 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {{\left (2 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 21 \, \cos \left (d x + c\right )^{2} + 25 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{63 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \]

[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)

Sympy [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.55 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\frac {63 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {42 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {18 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{1008 \, a^{5} d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.37 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 42 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 63 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{1008 \, a^{5} d} \]

[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)

Mupad [B] (verification not implemented)

Time = 13.62 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.36 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+42\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+63\right )}{1008\,a^5\,d} \]

[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)