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

Optimal. Leaf size=120 \[ \frac {\tan (c+d x)}{5 d \left (a^4 \sec (c+d x)+a^4\right )}-\frac {8 \tan (c+d x)}{35 d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {3 \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

[Out]

1/7*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+3/35*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3-8/35*tan(d*x+c)/d/(a^2
+a^2*sec(d*x+c))^2+1/5*tan(d*x+c)/d/(a^4+a^4*sec(d*x+c))

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Rubi [A]  time = 0.17, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3810, 3799, 4000, 3794} \[ \frac {\tan (c+d x)}{5 d \left (a^4 \sec (c+d x)+a^4\right )}-\frac {8 \tan (c+d x)}{35 d \left (a^2 \sec (c+d x)+a^2\right )^2}+\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {3 \tan (c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*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)
- (8*Tan[c + d*x])/(35*d*(a^2 + a^2*Sec[c + d*x])^2) + Tan[c + d*x]/(5*d*(a^4 + a^4*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 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))^4} \, dx &=\frac {\sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {3 \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx}{7 a}\\ &=\frac {\sec ^3(c+d x) \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 {3 \int \frac {\sec (c+d x) (-3 a+5 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{35 a^3}\\ &=\frac {\sec ^3(c+d x) \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 {8 \tan (c+d x)}{35 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {\int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{5 a^3}\\ &=\frac {\sec ^3(c+d x) \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 {8 \tan (c+d x)}{35 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {\tan (c+d x)}{5 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 69, normalized size = 0.58 \[ \frac {\left (35 \sin \left (\frac {1}{2} (c+d x)\right )+21 \sin \left (\frac {3}{2} (c+d x)\right )+7 \sin \left (\frac {5}{2} (c+d x)\right )+\sin \left (\frac {7}{2} (c+d x)\right )\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right )}{1120 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[(c + d*x)/2]^7*(35*Sin[(c + d*x)/2] + 21*Sin[(3*(c + d*x))/2] + 7*Sin[(5*(c + d*x))/2] + Sin[(7*(c + d*x)
)/2]))/(1120*a^4*d)

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fricas [A]  time = 1.43, size = 99, normalized size = 0.82 \[ \frac {{\left (2 \, \cos \left (d x + c\right )^{3} + 8 \, \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right ) + 12\right )} \sin \left (d x + c\right )}{35 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(2*cos(d*x + c)^3 + 8*cos(d*x + c)^2 + 13*cos(d*x + c) + 12)*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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giac [A]  time = 2.87, size = 59, normalized size = 0.49 \[ \frac {5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{280 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.41, size = 56, normalized size = 0.47 \[ \frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(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 = 1.02, size = 87, normalized size = 0.72 \[ \frac {\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{280 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(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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mupad [B]  time = 0.67, size = 58, normalized size = 0.48 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+35\right )}{280\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(35*tan(c/2 + (d*x)/2)^2 + 21*tan(c/2 + (d*x)/2)^4 + 5*tan(c/2 + (d*x)/2)^6 + 35))/(280*a^
4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)**4/(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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