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

Optimal. Leaf size=132 \[ \frac {2 i \sec (c+d x)}{35 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {2 i \sec (c+d x)}{35 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {3 i \sec (c+d x)}{35 a d (a+i a \tan (c+d x))^3}+\frac {i \sec (c+d x)}{7 d (a+i a \tan (c+d x))^4} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3502, 3488} \[ \frac {2 i \sec (c+d x)}{35 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {2 i \sec (c+d x)}{35 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {3 i \sec (c+d x)}{35 a d (a+i a \tan (c+d x))^3}+\frac {i \sec (c+d x)}{7 d (a+i a \tan (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3488

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(a*f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 73, normalized size = 0.55 \[ \frac {i \sec ^4(c+d x) (7 i \sin (c+d x)+15 i \sin (3 (c+d x))+28 \cos (c+d x)+20 \cos (3 (c+d x)))}{140 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/140)*Sec[c + d*x]^4*(28*Cos[c + d*x] + 20*Cos[3*(c + d*x)] + (7*I)*Sin[c + d*x] + (15*I)*Sin[3*(c + d*x)])
)/(a^4*d*(-I + Tan[c + d*x])^4)

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fricas [A]  time = 0.54, size = 52, normalized size = 0.39 \[ \frac {{\left (35 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 35 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 21 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{280 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/280*(35*I*e^(6*I*d*x + 6*I*c) + 35*I*e^(4*I*d*x + 4*I*c) + 21*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-7*I*d*x - 7*I
*c)/(a^4*d)

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giac [A]  time = 1.95, size = 99, normalized size = 0.75 \[ \frac {2 \, {\left (35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 105 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 210 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 210 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 147 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 49 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12\right )}}{35 \, a^{4} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(35*tan(1/2*d*x + 1/2*c)^6 - 105*I*tan(1/2*d*x + 1/2*c)^5 - 210*tan(1/2*d*x + 1/2*c)^4 + 210*I*tan(1/2*d*
x + 1/2*c)^3 + 147*tan(1/2*d*x + 1/2*c)^2 - 49*I*tan(1/2*d*x + 1/2*c) - 12)/(a^4*d*(tan(1/2*d*x + 1/2*c) - I)^
7)

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maple [A]  time = 0.23, size = 123, normalized size = 0.93 \[ \frac {\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}+\frac {72}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {16}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{7}}-\frac {16 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {6 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {12}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {8 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{6}}}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a+I*a*tan(d*x+c))^4,x)

[Out]

2/d/a^4*(1/(tan(1/2*d*x+1/2*c)-I)+36/5/(tan(1/2*d*x+1/2*c)-I)^5-8/7/(tan(1/2*d*x+1/2*c)-I)^7-8*I/(tan(1/2*d*x+
1/2*c)-I)^4+3*I/(tan(1/2*d*x+1/2*c)-I)^2-6/(tan(1/2*d*x+1/2*c)-I)^3+4*I/(tan(1/2*d*x+1/2*c)-I)^6)

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maxima [A]  time = 0.44, size = 91, normalized size = 0.69 \[ \frac {5 i \, \cos \left (7 \, d x + 7 \, c\right ) + 21 i \, \cos \left (5 \, d x + 5 \, c\right ) + 35 i \, \cos \left (3 \, d x + 3 \, c\right ) + 35 i \, \cos \left (d x + c\right ) + 5 \, \sin \left (7 \, d x + 7 \, c\right ) + 21 \, \sin \left (5 \, d x + 5 \, c\right ) + 35 \, \sin \left (3 \, d x + 3 \, c\right ) + 35 \, \sin \left (d x + c\right )}{280 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/280*(5*I*cos(7*d*x + 7*c) + 21*I*cos(5*d*x + 5*c) + 35*I*cos(3*d*x + 3*c) + 35*I*cos(d*x + c) + 5*sin(7*d*x
+ 7*c) + 21*sin(5*d*x + 5*c) + 35*sin(3*d*x + 3*c) + 35*sin(d*x + c))/(a^4*d)

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mupad [B]  time = 3.74, size = 64, normalized size = 0.48 \[ \frac {\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{8}+\frac {{\mathrm {e}}^{-c\,3{}\mathrm {i}-d\,x\,3{}\mathrm {i}}\,1{}\mathrm {i}}{8}+\frac {{\mathrm {e}}^{-c\,5{}\mathrm {i}-d\,x\,5{}\mathrm {i}}\,3{}\mathrm {i}}{40}+\frac {{\mathrm {e}}^{-c\,7{}\mathrm {i}-d\,x\,7{}\mathrm {i}}\,1{}\mathrm {i}}{56}}{a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((exp(- c*1i - d*x*1i)*1i)/8 + (exp(- c*3i - d*x*3i)*1i)/8 + (exp(- c*5i - d*x*5i)*3i)/40 + (exp(- c*7i - d*x*
7i)*1i)/56)/(a^4*d)

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sympy [A]  time = 3.90, size = 354, normalized size = 2.68 \[ \begin {cases} \frac {2 \tan ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{35 a^{4} d \tan ^{4}{\left (c + d x \right )} - 140 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 210 a^{4} d \tan ^{2}{\left (c + d x \right )} + 140 i a^{4} d \tan {\left (c + d x \right )} + 35 a^{4} d} - \frac {8 i \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{35 a^{4} d \tan ^{4}{\left (c + d x \right )} - 140 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 210 a^{4} d \tan ^{2}{\left (c + d x \right )} + 140 i a^{4} d \tan {\left (c + d x \right )} + 35 a^{4} d} - \frac {13 \tan {\left (c + d x \right )} \sec {\left (c + d x \right )}}{35 a^{4} d \tan ^{4}{\left (c + d x \right )} - 140 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 210 a^{4} d \tan ^{2}{\left (c + d x \right )} + 140 i a^{4} d \tan {\left (c + d x \right )} + 35 a^{4} d} + \frac {12 i \sec {\left (c + d x \right )}}{35 a^{4} d \tan ^{4}{\left (c + d x \right )} - 140 i a^{4} d \tan ^{3}{\left (c + d x \right )} - 210 a^{4} d \tan ^{2}{\left (c + d x \right )} + 140 i a^{4} d \tan {\left (c + d x \right )} + 35 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sec {\relax (c )}}{\left (i a \tan {\relax (c )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*tan(c + d*x)**3*sec(c + d*x)/(35*a**4*d*tan(c + d*x)**4 - 140*I*a**4*d*tan(c + d*x)**3 - 210*a**4
*d*tan(c + d*x)**2 + 140*I*a**4*d*tan(c + d*x) + 35*a**4*d) - 8*I*tan(c + d*x)**2*sec(c + d*x)/(35*a**4*d*tan(
c + d*x)**4 - 140*I*a**4*d*tan(c + d*x)**3 - 210*a**4*d*tan(c + d*x)**2 + 140*I*a**4*d*tan(c + d*x) + 35*a**4*
d) - 13*tan(c + d*x)*sec(c + d*x)/(35*a**4*d*tan(c + d*x)**4 - 140*I*a**4*d*tan(c + d*x)**3 - 210*a**4*d*tan(c
+ d*x)**2 + 140*I*a**4*d*tan(c + d*x) + 35*a**4*d) + 12*I*sec(c + d*x)/(35*a**4*d*tan(c + d*x)**4 - 140*I*a**
4*d*tan(c + d*x)**3 - 210*a**4*d*tan(c + d*x)**2 + 140*I*a**4*d*tan(c + d*x) + 35*a**4*d), Ne(d, 0)), (x*sec(c
)/(I*a*tan(c) + a)**4, True))

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