3.1.100 \(\int \frac {\sec ^8(c+d x)}{a+i a \tan (c+d x)} \, dx\) [100]

Optimal. Leaf size=80 \[ \frac {i (a-i a \tan (c+d x))^4}{a^5 d}-\frac {4 i (a-i a \tan (c+d x))^5}{5 a^6 d}+\frac {i (a-i a \tan (c+d x))^6}{6 a^7 d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 80, 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 = {3568, 45} \begin {gather*} \frac {i (a-i a \tan (c+d x))^6}{6 a^7 d}-\frac {4 i (a-i a \tan (c+d x))^5}{5 a^6 d}+\frac {i (a-i a \tan (c+d x))^4}{a^5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

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 3568

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

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Mathematica [A]
time = 0.40, size = 60, normalized size = 0.75 \begin {gather*} \frac {\sec (c) \sec ^6(c+d x) (-10 i \cos (c)-10 \sin (c)+15 \sin (c+2 d x)+6 \sin (3 c+4 d x)+\sin (5 c+6 d x))}{60 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c]*Sec[c + d*x]^6*((-10*I)*Cos[c] - 10*Sin[c] + 15*Sin[c + 2*d*x] + 6*Sin[3*c + 4*d*x] + Sin[5*c + 6*d*x]
))/(60*a*d)

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Maple [A]
time = 0.25, size = 72, normalized size = 0.90

method	result	size
risch	\(\frac {16 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{15 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\)	\(47\)
derivativedivides	\(-\frac {i \left (\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{2}+\frac {i \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {2 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+i \tan \left (d x +c \right )\right )}{d a}\)	\(72\)
default	\(-\frac {i \left (\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{2}+\frac {i \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {2 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+i \tan \left (d x +c \right )\right )}{d a}\)	\(72\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 67, normalized size = 0.84 \begin {gather*} \frac {-5 i \, \tan \left (d x + c\right )^{6} + 6 \, \tan \left (d x + c\right )^{5} - 15 i \, \tan \left (d x + c\right )^{4} + 20 \, \tan \left (d x + c\right )^{3} - 15 i \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right )}{30 \, a d} \end {gather*}

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]

1/30*(-5*I*tan(d*x + c)^6 + 6*tan(d*x + c)^5 - 15*I*tan(d*x + c)^4 + 20*tan(d*x + c)^3 - 15*I*tan(d*x + c)^2 +
30*tan(d*x + c))/(a*d)

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Fricas [A]
time = 0.37, size = 109, normalized size = 1.36 \begin {gather*} -\frac {16 \, {\left (-15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{15 \, {\left (a d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \end {gather*}

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]

-16/15*(-15*I*e^(4*I*d*x + 4*I*c) - 6*I*e^(2*I*d*x + 2*I*c) - I)/(a*d*e^(12*I*d*x + 12*I*c) + 6*a*d*e^(10*I*d*
x + 10*I*c) + 15*a*d*e^(8*I*d*x + 8*I*c) + 20*a*d*e^(6*I*d*x + 6*I*c) + 15*a*d*e^(4*I*d*x + 4*I*c) + 6*a*d*e^(
2*I*d*x + 2*I*c) + a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {\sec ^{8}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \end {gather*}

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]

-I*Integral(sec(c + d*x)**8/(tan(c + d*x) - I), x)/a

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Giac [A]
time = 0.57, size = 67, normalized size = 0.84 \begin {gather*} -\frac {5 i \, \tan \left (d x + c\right )^{6} - 6 \, \tan \left (d x + c\right )^{5} + 15 i \, \tan \left (d x + c\right )^{4} - 20 \, \tan \left (d x + c\right )^{3} + 15 i \, \tan \left (d x + c\right )^{2} - 30 \, \tan \left (d x + c\right )}{30 \, a d} \end {gather*}

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]

-1/30*(5*I*tan(d*x + c)^6 - 6*tan(d*x + c)^5 + 15*I*tan(d*x + c)^4 - 20*tan(d*x + c)^3 + 15*I*tan(d*x + c)^2 -
30*tan(d*x + c))/(a*d)

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Mupad [B]
time = 3.32, size = 114, normalized size = 1.42 \begin {gather*} \frac {\sin \left (c+d\,x\right )\,\left (30\,{\cos \left (c+d\,x\right )}^5-{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,15{}\mathrm {i}+20\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,15{}\mathrm {i}+6\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^5\,5{}\mathrm {i}\right )}{30\,a\,d\,{\cos \left (c+d\,x\right )}^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(c + d*x)*(6*cos(c + d*x)*sin(c + d*x)^4 - cos(c + d*x)^4*sin(c + d*x)*15i + 30*cos(c + d*x)^5 - sin(c + d
*x)^5*5i - cos(c + d*x)^2*sin(c + d*x)^3*15i + 20*cos(c + d*x)^3*sin(c + d*x)^2))/(30*a*d*cos(c + d*x)^6)

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