3.114 \(\int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=82 \[ \frac {i (a-i a \tan (c+d x))^7}{7 a^9 d}-\frac {2 i (a-i a \tan (c+d x))^6}{3 a^8 d}+\frac {4 i (a-i a \tan (c+d x))^5}{5 a^7 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 82, 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} \[ \frac {i (a-i a \tan (c+d x))^7}{7 a^9 d}-\frac {2 i (a-i a \tan (c+d x))^6}{3 a^8 d}+\frac {4 i (a-i a \tan (c+d x))^5}{5 a^7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(((4*I)/5)*(a - I*a*Tan[c + d*x])^5)/(a^7*d) - (((2*I)/3)*(a - I*a*Tan[c + d*x])^6)/(a^8*d) + ((I/7)*(a - I*a*
Tan[c + d*x])^7)/(a^9*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 ^{10}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac {i \operatorname {Subst}\left (\int (a-x)^4 (a+x)^2 \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (4 a^2 (a-x)^4-4 a (a-x)^5+(a-x)^6\right ) \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=\frac {4 i (a-i a \tan (c+d x))^5}{5 a^7 d}-\frac {2 i (a-i a \tan (c+d x))^6}{3 a^8 d}+\frac {i (a-i a \tan (c+d x))^7}{7 a^9 d}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 90, normalized size = 1.10 \[ \frac {\sec (c) \sec ^7(c+d x) (-35 \sin (2 c+d x)+42 \sin (2 c+3 d x)+14 \sin (4 c+5 d x)+2 \sin (6 c+7 d x)-35 i \cos (2 c+d x)+35 \sin (d x)-35 i \cos (d x))}{210 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(Sec[c]*Sec[c + d*x]^7*((-35*I)*Cos[d*x] - (35*I)*Cos[2*c + d*x] + 35*Sin[d*x] - 35*Sin[2*c + d*x] + 42*Sin[2*
c + 3*d*x] + 14*Sin[4*c + 5*d*x] + 2*Sin[6*c + 7*d*x]))/(210*a^2*d)

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fricas [B]  time = 0.50, size = 138, normalized size = 1.68 \[ \frac {2688 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 896 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 128 i}{105 \, {\left (a^{2} d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, a^{2} d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/105*(2688*I*e^(4*I*d*x + 4*I*c) + 896*I*e^(2*I*d*x + 2*I*c) + 128*I)/(a^2*d*e^(14*I*d*x + 14*I*c) + 7*a^2*d*
e^(12*I*d*x + 12*I*c) + 21*a^2*d*e^(10*I*d*x + 10*I*c) + 35*a^2*d*e^(8*I*d*x + 8*I*c) + 35*a^2*d*e^(6*I*d*x +
6*I*c) + 21*a^2*d*e^(4*I*d*x + 4*I*c) + 7*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

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giac [A]  time = 1.03, size = 77, normalized size = 0.94 \[ -\frac {15 \, \tan \left (d x + c\right )^{7} + 35 i \, \tan \left (d x + c\right )^{6} + 21 \, \tan \left (d x + c\right )^{5} + 105 i \, \tan \left (d x + c\right )^{4} - 35 \, \tan \left (d x + c\right )^{3} + 105 i \, \tan \left (d x + c\right )^{2} - 105 \, \tan \left (d x + c\right )}{105 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/105*(15*tan(d*x + c)^7 + 35*I*tan(d*x + c)^6 + 21*tan(d*x + c)^5 + 105*I*tan(d*x + c)^4 - 35*tan(d*x + c)^3
 + 105*I*tan(d*x + c)^2 - 105*tan(d*x + c))/(a^2*d)

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maple [A]  time = 0.41, size = 78, normalized size = 0.95 \[ \frac {\tan \left (d x +c \right )-\frac {\left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {i \left (\tan ^{6}\left (d x +c \right )\right )}{3}-\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-i \left (\tan ^{4}\left (d x +c \right )\right )+\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-i \left (\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.37, size = 77, normalized size = 0.94 \[ -\frac {15 \, \tan \left (d x + c\right )^{7} + 35 i \, \tan \left (d x + c\right )^{6} + 21 \, \tan \left (d x + c\right )^{5} + 105 i \, \tan \left (d x + c\right )^{4} - 35 \, \tan \left (d x + c\right )^{3} + 105 i \, \tan \left (d x + c\right )^{2} - 105 \, \tan \left (d x + c\right )}{105 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/105*(15*tan(d*x + c)^7 + 35*I*tan(d*x + c)^6 + 21*tan(d*x + c)^5 + 105*I*tan(d*x + c)^4 - 35*tan(d*x + c)^3
 + 105*I*tan(d*x + c)^2 - 105*tan(d*x + c))/(a^2*d)

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mupad [B]  time = 3.46, size = 93, normalized size = 1.13 \[ \frac {{\cos \left (c+d\,x\right )}^7\,35{}\mathrm {i}+64\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^6+32\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^4+24\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^2-\cos \left (c+d\,x\right )\,35{}\mathrm {i}-15\,\sin \left (c+d\,x\right )}{105\,a^2\,d\,{\cos \left (c+d\,x\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^10*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

(24*cos(c + d*x)^2*sin(c + d*x) - 15*sin(c + d*x) - cos(c + d*x)*35i + 32*cos(c + d*x)^4*sin(c + d*x) + 64*cos
(c + d*x)^6*sin(c + d*x) + cos(c + d*x)^7*35i)/(105*a^2*d*cos(c + d*x)^7)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\sec ^{10}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**10/(a+I*a*tan(d*x+c))**2,x)

[Out]

-Integral(sec(c + d*x)**10/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x)/a**2

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