3.130 \(\int \frac{\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=109 \[ -\frac{i (a-i a \tan (c+d x))^{10}}{10 a^{13} d}+\frac{2 i (a-i a \tan (c+d x))^9}{3 a^{12} d}-\frac{3 i (a-i a \tan (c+d x))^8}{2 a^{11} d}+\frac{8 i (a-i a \tan (c+d x))^7}{7 a^{10} d} \]

[Out]

(((8*I)/7)*(a - I*a*Tan[c + d*x])^7)/(a^10*d) - (((3*I)/2)*(a - I*a*Tan[c + d*x])^8)/(a^11*d) + (((2*I)/3)*(a
- I*a*Tan[c + d*x])^9)/(a^12*d) - ((I/10)*(a - I*a*Tan[c + d*x])^10)/(a^13*d)

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Rubi [A]  time = 0.0688227, antiderivative size = 109, normalized size of antiderivative = 1., 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} \[ -\frac{i (a-i a \tan (c+d x))^{10}}{10 a^{13} d}+\frac{2 i (a-i a \tan (c+d x))^9}{3 a^{12} d}-\frac{3 i (a-i a \tan (c+d x))^8}{2 a^{11} d}+\frac{8 i (a-i a \tan (c+d x))^7}{7 a^{10} d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((8*I)/7)*(a - I*a*Tan[c + d*x])^7)/(a^10*d) - (((3*I)/2)*(a - I*a*Tan[c + d*x])^8)/(a^11*d) + (((2*I)/3)*(a
- I*a*Tan[c + d*x])^9)/(a^12*d) - ((I/10)*(a - I*a*Tan[c + d*x])^10)/(a^13*d)

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]

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])

Rubi steps

\begin{align*} \int \frac{\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^6 (a+x)^3 \, dx,x,i a \tan (c+d x)\right )}{a^{13} d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (8 a^3 (a-x)^6-12 a^2 (a-x)^7+6 a (a-x)^8-(a-x)^9\right ) \, dx,x,i a \tan (c+d x)\right )}{a^{13} d}\\ &=\frac{8 i (a-i a \tan (c+d x))^7}{7 a^{10} d}-\frac{3 i (a-i a \tan (c+d x))^8}{2 a^{11} d}+\frac{2 i (a-i a \tan (c+d x))^9}{3 a^{12} d}-\frac{i (a-i a \tan (c+d x))^{10}}{10 a^{13} d}\\ \end{align*}

Mathematica [A]  time = 0.682843, size = 117, normalized size = 1.07 \[ \frac{\sec (c) \sec ^{10}(c+d x) (105 \sin (c+2 d x)-105 \sin (3 c+2 d x)+120 \sin (3 c+4 d x)+45 \sin (5 c+6 d x)+10 \sin (7 c+8 d x)+\sin (9 c+10 d x)-105 i \cos (c+2 d x)-105 i \cos (3 c+2 d x)-126 \sin (c)-126 i \cos (c))}{840 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c]*Sec[c + d*x]^10*((-126*I)*Cos[c] - (105*I)*Cos[c + 2*d*x] - (105*I)*Cos[3*c + 2*d*x] - 126*Sin[c] + 10
5*Sin[c + 2*d*x] - 105*Sin[3*c + 2*d*x] + 120*Sin[3*c + 4*d*x] + 45*Sin[5*c + 6*d*x] + 10*Sin[7*c + 8*d*x] + S
in[9*c + 10*d*x]))/(840*a^3*d)

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Maple [A]  time = 0.084, size = 89, normalized size = 0.8 \begin{align*}{\frac{1}{d{a}^{3}} \left ( \tan \left ( dx+c \right ) +{\frac{i}{10}} \left ( \tan \left ( dx+c \right ) \right ) ^{10}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{9}}{3}}-{\frac{8\, \left ( \tan \left ( dx+c \right ) \right ) ^{7}}{7}}-i \left ( \tan \left ( dx+c \right ) \right ) ^{6}-{\frac{6\, \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5}}-2\,i \left ( \tan \left ( dx+c \right ) \right ) ^{4}-{\frac{3\,i}{2}} \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x)

[Out]

1/d/a^3*(tan(d*x+c)+1/10*I*tan(d*x+c)^10-1/3*tan(d*x+c)^9-8/7*tan(d*x+c)^7-I*tan(d*x+c)^6-6/5*tan(d*x+c)^5-2*I
*tan(d*x+c)^4-3/2*I*tan(d*x+c)^2)

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Maxima [A]  time = 1.0175, size = 117, normalized size = 1.07 \begin{align*} \frac{42 i \, \tan \left (d x + c\right )^{10} - 140 \, \tan \left (d x + c\right )^{9} - 480 \, \tan \left (d x + c\right )^{7} - 420 i \, \tan \left (d x + c\right )^{6} - 504 \, \tan \left (d x + c\right )^{5} - 840 i \, \tan \left (d x + c\right )^{4} - 630 i \, \tan \left (d x + c\right )^{2} + 420 \, \tan \left (d x + c\right )}{420 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/420*(42*I*tan(d*x + c)^10 - 140*tan(d*x + c)^9 - 480*tan(d*x + c)^7 - 420*I*tan(d*x + c)^6 - 504*tan(d*x + c
)^5 - 840*I*tan(d*x + c)^4 - 630*I*tan(d*x + c)^2 + 420*tan(d*x + c))/(a^3*d)

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Fricas [B]  time = 2.61, size = 587, normalized size = 5.39 \begin{align*} \frac{15360 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 5760 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 1280 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 128 i}{105 \,{\left (a^{3} d e^{\left (20 i \, d x + 20 i \, c\right )} + 10 \, a^{3} d e^{\left (18 i \, d x + 18 i \, c\right )} + 45 \, a^{3} d e^{\left (16 i \, d x + 16 i \, c\right )} + 120 \, a^{3} d e^{\left (14 i \, d x + 14 i \, c\right )} + 210 \, a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 252 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 210 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 120 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 45 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/105*(15360*I*e^(6*I*d*x + 6*I*c) + 5760*I*e^(4*I*d*x + 4*I*c) + 1280*I*e^(2*I*d*x + 2*I*c) + 128*I)/(a^3*d*e
^(20*I*d*x + 20*I*c) + 10*a^3*d*e^(18*I*d*x + 18*I*c) + 45*a^3*d*e^(16*I*d*x + 16*I*c) + 120*a^3*d*e^(14*I*d*x
+ 14*I*c) + 210*a^3*d*e^(12*I*d*x + 12*I*c) + 252*a^3*d*e^(10*I*d*x + 10*I*c) + 210*a^3*d*e^(8*I*d*x + 8*I*c)
+ 120*a^3*d*e^(6*I*d*x + 6*I*c) + 45*a^3*d*e^(4*I*d*x + 4*I*c) + 10*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**14/(a+I*a*tan(d*x+c))**3,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.19416, size = 117, normalized size = 1.07 \begin{align*} -\frac{-21 i \, \tan \left (d x + c\right )^{10} + 70 \, \tan \left (d x + c\right )^{9} + 240 \, \tan \left (d x + c\right )^{7} + 210 i \, \tan \left (d x + c\right )^{6} + 252 \, \tan \left (d x + c\right )^{5} + 420 i \, \tan \left (d x + c\right )^{4} + 315 i \, \tan \left (d x + c\right )^{2} - 210 \, \tan \left (d x + c\right )}{210 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/210*(-21*I*tan(d*x + c)^10 + 70*tan(d*x + c)^9 + 240*tan(d*x + c)^7 + 210*I*tan(d*x + c)^6 + 252*tan(d*x +
c)^5 + 420*I*tan(d*x + c)^4 + 315*I*tan(d*x + c)^2 - 210*tan(d*x + c))/(a^3*d)