3.115 \(\int \frac{\sec ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=55 \[ \frac{i (a-i a \tan (c+d x))^4}{2 a^6 d}-\frac{i (a-i a \tan (c+d x))^5}{5 a^7 d} \]

[Out]

((I/2)*(a - I*a*Tan[c + d*x])^4)/(a^6*d) - ((I/5)*(a - I*a*Tan[c + d*x])^5)/(a^7*d)

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Rubi [A]  time = 0.0505928, 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} \[ \frac{i (a-i a \tan (c+d x))^4}{2 a^6 d}-\frac{i (a-i a \tan (c+d x))^5}{5 a^7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^2,x]

[Out]

((I/2)*(a - I*a*Tan[c + d*x])^4)/(a^6*d) - ((I/5)*(a - I*a*Tan[c + d*x])^5)/(a^7*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 ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^3 (a+x) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (2 a (a-x)^3-(a-x)^4\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=\frac{i (a-i a \tan (c+d x))^4}{2 a^6 d}-\frac{i (a-i a \tan (c+d x))^5}{5 a^7 d}\\ \end{align*}

Mathematica [A]  time = 0.328308, size = 77, normalized size = 1.4 \[ \frac{\sec (c) \sec ^5(c+d x) (-5 \sin (2 c+d x)+5 \sin (2 c+3 d x)+\sin (4 c+5 d x)-5 i \cos (2 c+d x)+5 \sin (d x)-5 i \cos (d x))}{20 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.072, size = 47, normalized size = 0.9 \begin{align*}{\frac{1}{{a}^{2}d} \left ( \tan \left ( dx+c \right ) -{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5}}-{\frac{i}{2}} \left ( \tan \left ( dx+c \right ) \right ) ^{4}-i \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)^8/(a+I*a*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.15832, size = 63, normalized size = 1.15 \begin{align*} -\frac{2 \, \tan \left (d x + c\right )^{5} + 5 i \, \tan \left (d x + c\right )^{4} + 10 i \, \tan \left (d x + c\right )^{2} - 10 \, \tan \left (d x + c\right )}{10 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/10*(2*tan(d*x + c)^5 + 5*I*tan(d*x + c)^4 + 10*I*tan(d*x + c)^2 - 10*tan(d*x + c))/(a^2*d)

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Fricas [B]  time = 2.12755, size = 267, normalized size = 4.85 \begin{align*} \frac{40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i}{5 \,{\left (a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/5*(40*I*e^(2*I*d*x + 2*I*c) + 8*I)/(a^2*d*e^(10*I*d*x + 10*I*c) + 5*a^2*d*e^(8*I*d*x + 8*I*c) + 10*a^2*d*e^(
6*I*d*x + 6*I*c) + 10*a^2*d*e^(4*I*d*x + 4*I*c) + 5*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*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)**8/(a+I*a*tan(d*x+c))**2,x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.15277, size = 63, normalized size = 1.15 \begin{align*} -\frac{2 \, \tan \left (d x + c\right )^{5} + 5 i \, \tan \left (d x + c\right )^{4} + 10 i \, \tan \left (d x + c\right )^{2} - 10 \, \tan \left (d x + c\right )}{10 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/10*(2*tan(d*x + c)^5 + 5*I*tan(d*x + c)^4 + 10*I*tan(d*x + c)^2 - 10*tan(d*x + c))/(a^2*d)