3.149 \(\int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=55 \[ \frac {i (a-i a \tan (c+d x))^6}{3 a^{10} d}-\frac {i (a-i a \tan (c+d x))^7}{7 a^{11} d} \]

[Out]

1/3*I*(a-I*a*tan(d*x+c))^6/a^10/d-1/7*I*(a-I*a*tan(d*x+c))^7/a^11/d

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Rubi [A]  time = 0.05, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3487, 43} \[ \frac {i (a-i a \tan (c+d x))^6}{3 a^{10} d}-\frac {i (a-i a \tan (c+d x))^7}{7 a^{11} d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^12/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I/3)*(a - I*a*Tan[c + d*x])^6)/(a^10*d) - ((I/7)*(a - I*a*Tan[c + d*x])^7)/(a^11*d)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx &=-\frac {i \operatorname {Subst}\left (\int (a-x)^5 (a+x) \, dx,x,i a \tan (c+d x)\right )}{a^{11} d}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (2 a (a-x)^5-(a-x)^6\right ) \, dx,x,i a \tan (c+d x)\right )}{a^{11} d}\\ &=\frac {i (a-i a \tan (c+d x))^6}{3 a^{10} d}-\frac {i (a-i a \tan (c+d x))^7}{7 a^{11} d}\\ \end {align*}

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Mathematica [B]  time = 0.53, size = 127, normalized size = 2.31 \[ \frac {\sec (c) \sec ^7(c+d x) (-35 \sin (2 c+d x)+21 \sin (2 c+3 d x)-21 \sin (4 c+3 d x)+14 \sin (4 c+5 d x)+2 \sin (6 c+7 d x)-35 i \cos (2 c+d x)-21 i \cos (2 c+3 d x)-21 i \cos (4 c+3 d x)+35 \sin (d x)-35 i \cos (d x))}{84 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^12/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(Sec[c]*Sec[c + d*x]^7*((-35*I)*Cos[d*x] - (35*I)*Cos[2*c + d*x] - (21*I)*Cos[2*c + 3*d*x] - (21*I)*Cos[4*c +
3*d*x] + 35*Sin[d*x] - 35*Sin[2*c + d*x] + 21*Sin[2*c + 3*d*x] - 21*Sin[4*c + 3*d*x] + 14*Sin[4*c + 5*d*x] + 2
*Sin[6*c + 7*d*x]))/(84*a^4*d)

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fricas [B]  time = 0.66, size = 127, normalized size = 2.31 \[ \frac {448 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 64 i}{21 \, {\left (a^{4} d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, a^{4} d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/21*(448*I*e^(2*I*d*x + 2*I*c) + 64*I)/(a^4*d*e^(14*I*d*x + 14*I*c) + 7*a^4*d*e^(12*I*d*x + 12*I*c) + 21*a^4*
d*e^(10*I*d*x + 10*I*c) + 35*a^4*d*e^(8*I*d*x + 8*I*c) + 35*a^4*d*e^(6*I*d*x + 6*I*c) + 21*a^4*d*e^(4*I*d*x +
4*I*c) + 7*a^4*d*e^(2*I*d*x + 2*I*c) + a^4*d)

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giac [A]  time = 1.97, size = 67, normalized size = 1.22 \[ \frac {3 \, \tan \left (d x + c\right )^{7} + 14 i \, \tan \left (d x + c\right )^{6} - 21 \, \tan \left (d x + c\right )^{5} - 35 \, \tan \left (d x + c\right )^{3} - 42 i \, \tan \left (d x + c\right )^{2} + 21 \, \tan \left (d x + c\right )}{21 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/21*(3*tan(d*x + c)^7 + 14*I*tan(d*x + c)^6 - 21*tan(d*x + c)^5 - 35*tan(d*x + c)^3 - 42*I*tan(d*x + c)^2 + 2
1*tan(d*x + c))/(a^4*d)

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maple [A]  time = 0.39, size = 67, normalized size = 1.22 \[ \frac {\tan \left (d x +c \right )+\frac {\left (\tan ^{7}\left (d x +c \right )\right )}{7}+\frac {2 i \left (\tan ^{6}\left (d x +c \right )\right )}{3}-\left (\tan ^{5}\left (d x +c \right )\right )-\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{3}-2 i \left (\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^4,x)

[Out]

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

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maxima [A]  time = 0.52, size = 67, normalized size = 1.22 \[ \frac {3 \, \tan \left (d x + c\right )^{7} + 14 i \, \tan \left (d x + c\right )^{6} - 21 \, \tan \left (d x + c\right )^{5} - 35 \, \tan \left (d x + c\right )^{3} - 42 i \, \tan \left (d x + c\right )^{2} + 21 \, \tan \left (d x + c\right )}{21 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/21*(3*tan(d*x + c)^7 + 14*I*tan(d*x + c)^6 - 21*tan(d*x + c)^5 - 35*tan(d*x + c)^3 - 42*I*tan(d*x + c)^2 + 2
1*tan(d*x + c))/(a^4*d)

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mupad [B]  time = 3.34, size = 113, normalized size = 2.05 \[ \frac {\sin \left (c+d\,x\right )\,\left (21\,{\cos \left (c+d\,x\right )}^6-{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )\,42{}\mathrm {i}-35\,{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2-21\,{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^4+\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^5\,14{}\mathrm {i}+3\,{\sin \left (c+d\,x\right )}^6\right )}{21\,a^4\,d\,{\cos \left (c+d\,x\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^12*(a + a*tan(c + d*x)*1i)^4),x)

[Out]

(sin(c + d*x)*(cos(c + d*x)*sin(c + d*x)^5*14i - cos(c + d*x)^5*sin(c + d*x)*42i + 21*cos(c + d*x)^6 + 3*sin(c
+ d*x)^6 - 21*cos(c + d*x)^2*sin(c + d*x)^4 - 35*cos(c + d*x)^4*sin(c + d*x)^2))/(21*a^4*d*cos(c + d*x)^7)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{12}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**12/(a+I*a*tan(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)**12/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I*tan(c + d*x) + 1),
x)/a**4

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