3.132 \(\int \frac{\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=55 \[ \frac{2 i (a-i a \tan (c+d x))^5}{5 a^8 d}-\frac{i (a-i a \tan (c+d x))^6}{6 a^9 d} \]

[Out]

(((2*I)/5)*(a - I*a*Tan[c + d*x])^5)/(a^8*d) - ((I/6)*(a - I*a*Tan[c + d*x])^6)/(a^9*d)

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Rubi [A]  time = 0.0473699, antiderivative size = 55, 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} \[ \frac{2 i (a-i a \tan (c+d x))^5}{5 a^8 d}-\frac{i (a-i a \tan (c+d x))^6}{6 a^9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((2*I)/5)*(a - I*a*Tan[c + d*x])^5)/(a^8*d) - ((I/6)*(a - I*a*Tan[c + d*x])^6)/(a^9*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 ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^4 (a+x) \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (2 a (a-x)^4-(a-x)^5\right ) \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=\frac{2 i (a-i a \tan (c+d x))^5}{5 a^8 d}-\frac{i (a-i a \tan (c+d x))^6}{6 a^9 d}\\ \end{align*}

Mathematica [A]  time = 0.435988, size = 97, normalized size = 1.76 \[ \frac{\sec (c) \sec ^6(c+d x) (15 \sin (c+2 d x)-15 \sin (3 c+2 d x)+12 \sin (3 c+4 d x)+2 \sin (5 c+6 d x)-15 i \cos (c+2 d x)-15 i \cos (3 c+2 d x)-20 \sin (c)-20 i \cos (c))}{60 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c]*Sec[c + d*x]^6*((-20*I)*Cos[c] - (15*I)*Cos[c + 2*d*x] - (15*I)*Cos[3*c + 2*d*x] - 20*Sin[c] + 15*Sin[
c + 2*d*x] - 15*Sin[3*c + 2*d*x] + 12*Sin[3*c + 4*d*x] + 2*Sin[5*c + 6*d*x]))/(60*a^3*d)

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Maple [A]  time = 0.078, size = 68, normalized size = 1.2 \begin{align*}{\frac{1}{d{a}^{3}} \left ( \tan \left ( dx+c \right ) +{\frac{i}{6}} \left ( \tan \left ( dx+c \right ) \right ) ^{6}-{\frac{3\, \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5}}-{\frac{i}{2}} \left ( \tan \left ( dx+c \right ) \right ) ^{4}-{\frac{2\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}-{\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)^10/(a+I*a*tan(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.05913, size = 90, normalized size = 1.64 \begin{align*} \frac{5 i \, \tan \left (d x + c\right )^{6} - 18 \, \tan \left (d x + c\right )^{5} - 15 i \, \tan \left (d x + c\right )^{4} - 20 \, \tan \left (d x + c\right )^{3} - 45 i \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right )}{30 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/30*(5*I*tan(d*x + c)^6 - 18*tan(d*x + c)^5 - 15*I*tan(d*x + c)^4 - 20*tan(d*x + c)^3 - 45*I*tan(d*x + c)^2 +
30*tan(d*x + c))/(a^3*d)

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Fricas [B]  time = 2.35389, size = 316, normalized size = 5.75 \begin{align*} \frac{192 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 32 i}{15 \,{\left (a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, 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)^10/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(192*I*e^(2*I*d*x + 2*I*c) + 32*I)/(a^3*d*e^(12*I*d*x + 12*I*c) + 6*a^3*d*e^(10*I*d*x + 10*I*c) + 15*a^3*
d*e^(8*I*d*x + 8*I*c) + 20*a^3*d*e^(6*I*d*x + 6*I*c) + 15*a^3*d*e^(4*I*d*x + 4*I*c) + 6*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)**10/(a+I*a*tan(d*x+c))**3,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.18569, size = 90, normalized size = 1.64 \begin{align*} -\frac{-5 i \, \tan \left (d x + c\right )^{6} + 18 \, \tan \left (d x + c\right )^{5} + 15 i \, \tan \left (d x + c\right )^{4} + 20 \, \tan \left (d x + c\right )^{3} + 45 i \, \tan \left (d x + c\right )^{2} - 30 \, \tan \left (d x + c\right )}{30 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/30*(-5*I*tan(d*x + c)^6 + 18*tan(d*x + c)^5 + 15*I*tan(d*x + c)^4 + 20*tan(d*x + c)^3 + 45*I*tan(d*x + c)^2
- 30*tan(d*x + c))/(a^3*d)