3.2 \(\int \sec ^8(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=75 \[ \frac {a \tan ^7(c+d x)}{7 d}+\frac {3 a \tan ^5(c+d x)}{5 d}+\frac {a \tan ^3(c+d x)}{d}+\frac {a \tan (c+d x)}{d}+\frac {i a \sec ^8(c+d x)}{8 d} \]

[Out]

1/8*I*a*sec(d*x+c)^8/d+a*tan(d*x+c)/d+a*tan(d*x+c)^3/d+3/5*a*tan(d*x+c)^5/d+1/7*a*tan(d*x+c)^7/d

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Rubi [A]  time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3486, 3767} \[ \frac {a \tan ^7(c+d x)}{7 d}+\frac {3 a \tan ^5(c+d x)}{5 d}+\frac {a \tan ^3(c+d x)}{d}+\frac {a \tan (c+d x)}{d}+\frac {i a \sec ^8(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/8)*a*Sec[c + d*x]^8)/d + (a*Tan[c + d*x])/d + (a*Tan[c + d*x]^3)/d + (3*a*Tan[c + d*x]^5)/(5*d) + (a*Tan[c
+ d*x]^7)/(7*d)

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \sec ^8(c+d x) (a+i a \tan (c+d x)) \, dx &=\frac {i a \sec ^8(c+d x)}{8 d}+a \int \sec ^8(c+d x) \, dx\\ &=\frac {i a \sec ^8(c+d x)}{8 d}-\frac {a \operatorname {Subst}\left (\int \left (1+3 x^2+3 x^4+x^6\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac {i a \sec ^8(c+d x)}{8 d}+\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{d}+\frac {3 a \tan ^5(c+d x)}{5 d}+\frac {a \tan ^7(c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 63, normalized size = 0.84 \[ \frac {a \left (\frac {1}{7} \tan ^7(c+d x)+\frac {3}{5} \tan ^5(c+d x)+\tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac {i a \sec ^8(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/8)*a*Sec[c + d*x]^8)/d + (a*(Tan[c + d*x] + Tan[c + d*x]^3 + (3*Tan[c + d*x]^5)/5 + Tan[c + d*x]^7/7))/d

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fricas [B]  time = 0.69, size = 153, normalized size = 2.04 \[ \frac {2240 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 1792 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 896 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 256 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 32 i \, a}{35 \, {\left (d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(2240*I*a*e^(8*I*d*x + 8*I*c) + 1792*I*a*e^(6*I*d*x + 6*I*c) + 896*I*a*e^(4*I*d*x + 4*I*c) + 256*I*a*e^(2
*I*d*x + 2*I*c) + 32*I*a)/(d*e^(16*I*d*x + 16*I*c) + 8*d*e^(14*I*d*x + 14*I*c) + 28*d*e^(12*I*d*x + 12*I*c) +
56*d*e^(10*I*d*x + 10*I*c) + 70*d*e^(8*I*d*x + 8*I*c) + 56*d*e^(6*I*d*x + 6*I*c) + 28*d*e^(4*I*d*x + 4*I*c) +
8*d*e^(2*I*d*x + 2*I*c) + d)

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giac [A]  time = 1.00, size = 92, normalized size = 1.23 \[ -\frac {-35 i \, a \tan \left (d x + c\right )^{8} - 40 \, a \tan \left (d x + c\right )^{7} - 140 i \, a \tan \left (d x + c\right )^{6} - 168 \, a \tan \left (d x + c\right )^{5} - 210 i \, a \tan \left (d x + c\right )^{4} - 280 \, a \tan \left (d x + c\right )^{3} - 140 i \, a \tan \left (d x + c\right )^{2} - 280 \, a \tan \left (d x + c\right )}{280 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.44, size = 59, normalized size = 0.79 \[ \frac {\frac {i a}{8 \cos \left (d x +c \right )^{8}}-a \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8*(a+I*a*tan(d*x+c)),x)

[Out]

1/d*(1/8*I*a/cos(d*x+c)^8-a*(-16/35-1/7*sec(d*x+c)^6-6/35*sec(d*x+c)^4-8/35*sec(d*x+c)^2)*tan(d*x+c))

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maxima [A]  time = 0.57, size = 92, normalized size = 1.23 \[ \frac {35 i \, a \tan \left (d x + c\right )^{8} + 40 \, a \tan \left (d x + c\right )^{7} + 140 i \, a \tan \left (d x + c\right )^{6} + 168 \, a \tan \left (d x + c\right )^{5} + 210 i \, a \tan \left (d x + c\right )^{4} + 280 \, a \tan \left (d x + c\right )^{3} + 140 i \, a \tan \left (d x + c\right )^{2} + 280 \, a \tan \left (d x + c\right )}{280 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.27, size = 149, normalized size = 1.99 \[ \frac {a\,\sin \left (c+d\,x\right )\,\left (280\,{\cos \left (c+d\,x\right )}^7+{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )\,140{}\mathrm {i}+280\,{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^2+{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^3\,210{}\mathrm {i}+168\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^4+{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^5\,140{}\mathrm {i}+40\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^6+{\sin \left (c+d\,x\right )}^7\,35{}\mathrm {i}\right )}{280\,d\,{\cos \left (c+d\,x\right )}^8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)/cos(c + d*x)^8,x)

[Out]

(a*sin(c + d*x)*(40*cos(c + d*x)*sin(c + d*x)^6 + cos(c + d*x)^6*sin(c + d*x)*140i + 280*cos(c + d*x)^7 + sin(
c + d*x)^7*35i + cos(c + d*x)^2*sin(c + d*x)^5*140i + 168*cos(c + d*x)^3*sin(c + d*x)^4 + cos(c + d*x)^4*sin(c
+ d*x)^3*210i + 280*cos(c + d*x)^5*sin(c + d*x)^2))/(280*d*cos(c + d*x)^8)

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sympy [A]  time = 7.06, size = 68, normalized size = 0.91 \[ \begin {cases} \frac {a \left (\frac {\tan ^{7}{\left (c + d x \right )}}{7} + \frac {3 \tan ^{5}{\left (c + d x \right )}}{5} + \tan ^{3}{\left (c + d x \right )} + \tan {\left (c + d x \right )}\right ) + \frac {i a \sec ^{8}{\left (c + d x \right )}}{8}}{d} & \text {for}\: d \neq 0 \\x \left (i a \tan {\relax (c )} + a\right ) \sec ^{8}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8*(a+I*a*tan(d*x+c)),x)

[Out]

Piecewise(((a*(tan(c + d*x)**7/7 + 3*tan(c + d*x)**5/5 + tan(c + d*x)**3 + tan(c + d*x)) + I*a*sec(c + d*x)**8
/8)/d, Ne(d, 0)), (x*(I*a*tan(c) + a)*sec(c)**8, True))

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