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

Optimal. Leaf size=107 \[ -\frac {i (a-i a \tan (c+d x))^8}{8 a^9 d}+\frac {6 i (a-i a \tan (c+d x))^7}{7 a^8 d}-\frac {2 i (a-i a \tan (c+d x))^6}{a^7 d}+\frac {8 i (a-i a \tan (c+d x))^5}{5 a^6 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 107, 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} \[ -\frac {i (a-i a \tan (c+d x))^8}{8 a^9 d}+\frac {6 i (a-i a \tan (c+d x))^7}{7 a^8 d}-\frac {2 i (a-i a \tan (c+d x))^6}{a^7 d}+\frac {8 i (a-i a \tan (c+d x))^5}{5 a^6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((8*I)/5)*(a - I*a*Tan[c + d*x])^5)/(a^6*d) - ((2*I)*(a - I*a*Tan[c + d*x])^6)/(a^7*d) + (((6*I)/7)*(a - I*a*
Tan[c + d*x])^7)/(a^8*d) - ((I/8)*(a - I*a*Tan[c + d*x])^8)/(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)} \, dx &=-\frac {i \operatorname {Subst}\left (\int (a-x)^4 (a+x)^3 \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (8 a^3 (a-x)^4-12 a^2 (a-x)^5+6 a (a-x)^6-(a-x)^7\right ) \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=\frac {8 i (a-i a \tan (c+d x))^5}{5 a^6 d}-\frac {2 i (a-i a \tan (c+d x))^6}{a^7 d}+\frac {6 i (a-i a \tan (c+d x))^7}{7 a^8 d}-\frac {i (a-i a \tan (c+d x))^8}{8 a^9 d}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 71, normalized size = 0.66 \[ \frac {\sec (c) \sec ^8(c+d x) (56 \sin (c+2 d x)+28 \sin (3 c+4 d x)+8 \sin (5 c+6 d x)+\sin (7 c+8 d x)-35 \sin (c)-35 i \cos (c))}{280 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c]*Sec[c + d*x]^8*((-35*I)*Cos[c] - 35*Sin[c] + 56*Sin[c + 2*d*x] + 28*Sin[3*c + 4*d*x] + 8*Sin[5*c + 6*d
*x] + Sin[7*c + 8*d*x]))/(280*a*d)

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fricas [A]  time = 0.63, size = 146, normalized size = 1.36 \[ \frac {1792 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 896 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 256 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 32 i}{35 \, {\left (a d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, a d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, a d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, a d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(1792*I*e^(6*I*d*x + 6*I*c) + 896*I*e^(4*I*d*x + 4*I*c) + 256*I*e^(2*I*d*x + 2*I*c) + 32*I)/(a*d*e^(16*I*
d*x + 16*I*c) + 8*a*d*e^(14*I*d*x + 14*I*c) + 28*a*d*e^(12*I*d*x + 12*I*c) + 56*a*d*e^(10*I*d*x + 10*I*c) + 70
*a*d*e^(8*I*d*x + 8*I*c) + 56*a*d*e^(6*I*d*x + 6*I*c) + 28*a*d*e^(4*I*d*x + 4*I*c) + 8*a*d*e^(2*I*d*x + 2*I*c)
+ a*d)

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giac [A]  time = 0.69, size = 87, normalized size = 0.81 \[ -\frac {35 i \, \tan \left (d x + c\right )^{8} - 40 \, \tan \left (d x + c\right )^{7} + 140 i \, \tan \left (d x + c\right )^{6} - 168 \, \tan \left (d x + c\right )^{5} + 210 i \, \tan \left (d x + c\right )^{4} - 280 \, \tan \left (d x + c\right )^{3} + 140 i \, \tan \left (d x + c\right )^{2} - 280 \, \tan \left (d x + c\right )}{280 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/280*(35*I*tan(d*x + c)^8 - 40*tan(d*x + c)^7 + 140*I*tan(d*x + c)^6 - 168*tan(d*x + c)^5 + 210*I*tan(d*x +
c)^4 - 280*tan(d*x + c)^3 + 140*I*tan(d*x + c)^2 - 280*tan(d*x + c))/(a*d)

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maple [A]  time = 0.39, size = 87, normalized size = 0.81 \[ \frac {\tan \left (d x +c \right )-\frac {i \left (\tan ^{8}\left (d x +c \right )\right )}{8}+\frac {\left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {i \left (\tan ^{6}\left (d x +c \right )\right )}{2}+\frac {3 \left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {3 i \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\tan ^{3}\left (d x +c \right )-\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2}}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 87, normalized size = 0.81 \[ \frac {-105 i \, \tan \left (d x + c\right )^{8} + 120 \, \tan \left (d x + c\right )^{7} - 420 i \, \tan \left (d x + c\right )^{6} + 504 \, \tan \left (d x + c\right )^{5} - 630 i \, \tan \left (d x + c\right )^{4} + 840 \, \tan \left (d x + c\right )^{3} - 420 i \, \tan \left (d x + c\right )^{2} + 840 \, \tan \left (d x + c\right )}{840 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(-105*I*tan(d*x + c)^8 + 120*tan(d*x + c)^7 - 420*I*tan(d*x + c)^6 + 504*tan(d*x + c)^5 - 630*I*tan(d*x
+ c)^4 + 840*tan(d*x + c)^3 - 420*I*tan(d*x + c)^2 + 840*tan(d*x + c))/(a*d)

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mupad [B]  time = 3.53, size = 92, normalized size = 0.86 \[ \frac {{\cos \left (c+d\,x\right )}^8\,35{}\mathrm {i}+128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^7+64\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^5+48\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^3+40\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )-35{}\mathrm {i}}{280\,a\,d\,{\cos \left (c+d\,x\right )}^8} \]

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)),x)

[Out]

(40*cos(c + d*x)*sin(c + d*x) + 48*cos(c + d*x)^3*sin(c + d*x) + 64*cos(c + d*x)^5*sin(c + d*x) + 128*cos(c +
d*x)^7*sin(c + d*x) + cos(c + d*x)^8*35i - 35i)/(280*a*d*cos(c + d*x)^8)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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