3.11 \(\int \sec ^7(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=98 \[ \frac {i a \sec ^7(c+d x)}{7 d}+\frac {5 a \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a \tan (c+d x) \sec ^5(c+d x)}{6 d}+\frac {5 a \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {5 a \tan (c+d x) \sec (c+d x)}{16 d} \]

[Out]

5/16*a*arctanh(sin(d*x+c))/d+1/7*I*a*sec(d*x+c)^7/d+5/16*a*sec(d*x+c)*tan(d*x+c)/d+5/24*a*sec(d*x+c)^3*tan(d*x
+c)/d+1/6*a*sec(d*x+c)^5*tan(d*x+c)/d

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Rubi [A]  time = 0.06, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3486, 3768, 3770} \[ \frac {i a \sec ^7(c+d x)}{7 d}+\frac {5 a \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a \tan (c+d x) \sec ^5(c+d x)}{6 d}+\frac {5 a \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {5 a \tan (c+d x) \sec (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7*(a + I*a*Tan[c + d*x]),x]

[Out]

(5*a*ArcTanh[Sin[c + d*x]])/(16*d) + ((I/7)*a*Sec[c + d*x]^7)/d + (5*a*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (5*
a*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + (a*Sec[c + d*x]^5*Tan[c + d*x])/(6*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \sec ^7(c+d x) (a+i a \tan (c+d x)) \, dx &=\frac {i a \sec ^7(c+d x)}{7 d}+a \int \sec ^7(c+d x) \, dx\\ &=\frac {i a \sec ^7(c+d x)}{7 d}+\frac {a \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{6} (5 a) \int \sec ^5(c+d x) \, dx\\ &=\frac {i a \sec ^7(c+d x)}{7 d}+\frac {5 a \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {a \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{8} (5 a) \int \sec ^3(c+d x) \, dx\\ &=\frac {i a \sec ^7(c+d x)}{7 d}+\frac {5 a \sec (c+d x) \tan (c+d x)}{16 d}+\frac {5 a \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {a \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{16} (5 a) \int \sec (c+d x) \, dx\\ &=\frac {5 a \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {i a \sec ^7(c+d x)}{7 d}+\frac {5 a \sec (c+d x) \tan (c+d x)}{16 d}+\frac {5 a \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {a \sec ^5(c+d x) \tan (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 61, normalized size = 0.62 \[ \frac {a \left (3360 \tanh ^{-1}(\sin (c+d x))+(1981 \sin (2 (c+d x))+700 \sin (4 (c+d x))+105 \sin (6 (c+d x))+1536 i) \sec ^7(c+d x)\right )}{10752 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7*(a + I*a*Tan[c + d*x]),x]

[Out]

(a*(3360*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]^7*(1536*I + 1981*Sin[2*(c + d*x)] + 700*Sin[4*(c + d*x)] + 105*S
in[6*(c + d*x)])))/(10752*d)

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fricas [B]  time = 0.54, size = 372, normalized size = 3.80 \[ \frac {-210 i \, a e^{\left (13 i \, d x + 13 i \, c\right )} - 1400 i \, a e^{\left (11 i \, d x + 11 i \, c\right )} - 3962 i \, a e^{\left (9 i \, d x + 9 i \, c\right )} + 6144 i \, a e^{\left (7 i \, d x + 7 i \, c\right )} + 3962 i \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 1400 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} + 210 i \, a e^{\left (i \, d x + i \, c\right )} + 105 \, {\left (a e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, a e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, a e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 105 \, {\left (a e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, a e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, a e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{336 \, {\left (d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/336*(-210*I*a*e^(13*I*d*x + 13*I*c) - 1400*I*a*e^(11*I*d*x + 11*I*c) - 3962*I*a*e^(9*I*d*x + 9*I*c) + 6144*I
*a*e^(7*I*d*x + 7*I*c) + 3962*I*a*e^(5*I*d*x + 5*I*c) + 1400*I*a*e^(3*I*d*x + 3*I*c) + 210*I*a*e^(I*d*x + I*c)
+ 105*(a*e^(14*I*d*x + 14*I*c) + 7*a*e^(12*I*d*x + 12*I*c) + 21*a*e^(10*I*d*x + 10*I*c) + 35*a*e^(8*I*d*x + 8
*I*c) + 35*a*e^(6*I*d*x + 6*I*c) + 21*a*e^(4*I*d*x + 4*I*c) + 7*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(I*d*x + I*c)
+ I) - 105*(a*e^(14*I*d*x + 14*I*c) + 7*a*e^(12*I*d*x + 12*I*c) + 21*a*e^(10*I*d*x + 10*I*c) + 35*a*e^(8*I*d*
x + 8*I*c) + 35*a*e^(6*I*d*x + 6*I*c) + 21*a*e^(4*I*d*x + 4*I*c) + 7*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(I*d*x +
I*c) - I))/(d*e^(14*I*d*x + 14*I*c) + 7*d*e^(12*I*d*x + 12*I*c) + 21*d*e^(10*I*d*x + 10*I*c) + 35*d*e^(8*I*d*
x + 8*I*c) + 35*d*e^(6*I*d*x + 6*I*c) + 21*d*e^(4*I*d*x + 4*I*c) + 7*d*e^(2*I*d*x + 2*I*c) + d)

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giac [B]  time = 2.25, size = 181, normalized size = 1.85 \[ \frac {105 \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 105 \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + \frac {2 \, {\left (231 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} - 336 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{12} - 196 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 595 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1680 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 595 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1008 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 196 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 231 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 i \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{7}}}{336 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

1/336*(105*a*log(tan(1/2*d*x + 1/2*c) + 1) - 105*a*log(tan(1/2*d*x + 1/2*c) - 1) + 2*(231*a*tan(1/2*d*x + 1/2*
c)^13 - 336*I*a*tan(1/2*d*x + 1/2*c)^12 - 196*a*tan(1/2*d*x + 1/2*c)^11 + 595*a*tan(1/2*d*x + 1/2*c)^9 - 1680*
I*a*tan(1/2*d*x + 1/2*c)^8 - 595*a*tan(1/2*d*x + 1/2*c)^5 - 1008*I*a*tan(1/2*d*x + 1/2*c)^4 + 196*a*tan(1/2*d*
x + 1/2*c)^3 - 231*a*tan(1/2*d*x + 1/2*c) - 48*I*a)/(tan(1/2*d*x + 1/2*c)^2 - 1)^7)/d

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maple [A]  time = 0.45, size = 95, normalized size = 0.97 \[ \frac {i a}{7 d \cos \left (d x +c \right )^{7}}+\frac {a \left (\sec ^{5}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{6 d}+\frac {5 a \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{24 d}+\frac {5 a \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {5 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*(a+I*a*tan(d*x+c)),x)

[Out]

1/7*I/d*a/cos(d*x+c)^7+1/6*a*sec(d*x+c)^5*tan(d*x+c)/d+5/24*a*sec(d*x+c)^3*tan(d*x+c)/d+5/16*a*sec(d*x+c)*tan(
d*x+c)/d+5/16/d*a*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.35, size = 106, normalized size = 1.08 \[ -\frac {7 \, a {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {96 i \, a}{\cos \left (d x + c\right )^{7}}}{672 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/672*(7*a*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 +
3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 96*I*a/cos(d*x + c)^7)/d

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mupad [B]  time = 7.37, size = 247, normalized size = 2.52 \[ \frac {5\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}-\frac {-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{8}+2{}\mathrm {i}\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{6}-\frac {85\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+10{}\mathrm {i}\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {85\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{24}+6{}\mathrm {i}\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {11\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {a\,2{}\mathrm {i}}{7}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)/cos(c + d*x)^7,x)

[Out]

(5*a*atanh(tan(c/2 + (d*x)/2)))/(8*d) - ((a*2i)/7 + (11*a*tan(c/2 + (d*x)/2))/8 - (7*a*tan(c/2 + (d*x)/2)^3)/6
+ a*tan(c/2 + (d*x)/2)^4*6i + (85*a*tan(c/2 + (d*x)/2)^5)/24 + a*tan(c/2 + (d*x)/2)^8*10i - (85*a*tan(c/2 + (
d*x)/2)^9)/24 + (7*a*tan(c/2 + (d*x)/2)^11)/6 + a*tan(c/2 + (d*x)/2)^12*2i - (11*a*tan(c/2 + (d*x)/2)^13)/8)/(
d*(7*tan(c/2 + (d*x)/2)^2 - 21*tan(c/2 + (d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 - 35*tan(c/2 + (d*x)/2)^8 + 21*t
an(c/2 + (d*x)/2)^10 - 7*tan(c/2 + (d*x)/2)^12 + tan(c/2 + (d*x)/2)^14 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int \left (- i \sec ^{7}{\left (c + d x \right )}\right )\, dx + \int \tan {\left (c + d x \right )} \sec ^{7}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7*(a+I*a*tan(d*x+c)),x)

[Out]

I*a*(Integral(-I*sec(c + d*x)**7, x) + Integral(tan(c + d*x)*sec(c + d*x)**7, x))

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