3.2.68 \(\int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\) [168]

Optimal. Leaf size=116 \[ -\frac {8 x}{a^8}-\frac {8 i \log (\cos (c+d x))}{a^8 d}+\frac {\tan (c+d x)}{a^8 d}+\frac {16 i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2}+\frac {24 i}{d \left (a^8+i a^8 \tan (c+d x)\right )} \]

[Out]

-8*x/a^8-8*I*ln(cos(d*x+c))/a^8/d+tan(d*x+c)/a^8/d+16/3*I/a^5/d/(a+I*a*tan(d*x+c))^3-16*I/d/(a^4+I*a^4*tan(d*x
+c))^2+24*I/d/(a^8+I*a^8*tan(d*x+c))

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Rubi [A]
time = 0.05, antiderivative size = 116, 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 {\tan (c+d x)}{a^8 d}+\frac {24 i}{d \left (a^8+i a^8 \tan (c+d x)\right )}-\frac {8 i \log (\cos (c+d x))}{a^8 d}-\frac {8 x}{a^8}+\frac {16 i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(-8*x)/a^8 - ((8*I)*Log[Cos[c + d*x]])/(a^8*d) + Tan[c + d*x]/(a^8*d) + ((16*I)/3)/(a^5*d*(a + I*a*Tan[c + d*x
])^3) - (16*I)/(d*(a^4 + I*a^4*Tan[c + d*x])^2) + (24*I)/(d*(a^8 + I*a^8*Tan[c + d*x]))

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 ^{10}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=-\frac {i \text {Subst}\left (\int \frac {(a-x)^4}{(a+x)^4} \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac {i \text {Subst}\left (\int \left (1+\frac {16 a^4}{(a+x)^4}-\frac {32 a^3}{(a+x)^3}+\frac {24 a^2}{(a+x)^2}-\frac {8 a}{a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac {8 x}{a^8}-\frac {8 i \log (\cos (c+d x))}{a^8 d}+\frac {\tan (c+d x)}{a^8 d}+\frac {16 i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2}+\frac {24 i}{d \left (a^8+i a^8 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(397\) vs. \(2(116)=232\).
time = 1.12, size = 397, normalized size = 3.42 \begin {gather*} \frac {\sec (c) \sec ^9(c+d x) (-\cos (5 (c+d x))-i \sin (5 (c+d x))) (-12 i \cos (c)-10 i \cos (3 c+2 d x)+12 d x \cos (3 c+2 d x)+2 i \cos (3 c+4 d x)+12 d x \cos (3 c+4 d x)-i \cos (5 c+4 d x)+12 d x \cos (5 c+4 d x)+\cos (c+2 d x) (-7 i+12 d x+12 i \log (\cos (c+d x)))+12 i \cos (3 c+2 d x) \log (\cos (c+d x))+12 i \cos (3 c+4 d x) \log (\cos (c+d x))+12 i \cos (5 c+4 d x) \log (\cos (c+d x))+11 \sin (c+2 d x)+12 i d x \sin (c+2 d x)-12 \log (\cos (c+d x)) \sin (c+2 d x)+14 \sin (3 c+2 d x)+12 i d x \sin (3 c+2 d x)-12 \log (\cos (c+d x)) \sin (3 c+2 d x)-4 \sin (3 c+4 d x)+12 i d x \sin (3 c+4 d x)-12 \log (\cos (c+d x)) \sin (3 c+4 d x)-\sin (5 c+4 d x)+12 i d x \sin (5 c+4 d x)-12 \log (\cos (c+d x)) \sin (5 c+4 d x))}{6 a^8 d (-i+\tan (c+d x))^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(Sec[c]*Sec[c + d*x]^9*(-Cos[5*(c + d*x)] - I*Sin[5*(c + d*x)])*((-12*I)*Cos[c] - (10*I)*Cos[3*c + 2*d*x] + 12
*d*x*Cos[3*c + 2*d*x] + (2*I)*Cos[3*c + 4*d*x] + 12*d*x*Cos[3*c + 4*d*x] - I*Cos[5*c + 4*d*x] + 12*d*x*Cos[5*c
+ 4*d*x] + Cos[c + 2*d*x]*(-7*I + 12*d*x + (12*I)*Log[Cos[c + d*x]]) + (12*I)*Cos[3*c + 2*d*x]*Log[Cos[c + d*
x]] + (12*I)*Cos[3*c + 4*d*x]*Log[Cos[c + d*x]] + (12*I)*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]] + 11*Sin[c + 2*d*x
] + (12*I)*d*x*Sin[c + 2*d*x] - 12*Log[Cos[c + d*x]]*Sin[c + 2*d*x] + 14*Sin[3*c + 2*d*x] + (12*I)*d*x*Sin[3*c
+ 2*d*x] - 12*Log[Cos[c + d*x]]*Sin[3*c + 2*d*x] - 4*Sin[3*c + 4*d*x] + (12*I)*d*x*Sin[3*c + 4*d*x] - 12*Log[
Cos[c + d*x]]*Sin[3*c + 4*d*x] - Sin[5*c + 4*d*x] + (12*I)*d*x*Sin[5*c + 4*d*x] - 12*Log[Cos[c + d*x]]*Sin[5*c
+ 4*d*x]))/(6*a^8*d*(-I + Tan[c + d*x])^8)

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Maple [A]
time = 0.33, size = 68, normalized size = 0.59

method	result	size
derivativedivides	\(\frac {\tan \left (d x +c \right )+8 i \ln \left (\tan \left (d x +c \right )-i\right )-\frac {16}{3 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {24}{\tan \left (d x +c \right )-i}+\frac {16 i}{\left (\tan \left (d x +c \right )-i\right )^{2}}}{d \,a^{8}}\)	\(68\)
default	\(\frac {\tan \left (d x +c \right )+8 i \ln \left (\tan \left (d x +c \right )-i\right )-\frac {16}{3 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {24}{\tan \left (d x +c \right )-i}+\frac {16 i}{\left (\tan \left (d x +c \right )-i\right )^{2}}}{d \,a^{8}}\)	\(68\)
risch	\(\frac {6 i {\mathrm e}^{-2 i \left (d x +c \right )}}{a^{8} d}-\frac {2 i {\mathrm e}^{-4 i \left (d x +c \right )}}{a^{8} d}+\frac {2 i {\mathrm e}^{-6 i \left (d x +c \right )}}{3 a^{8} d}-\frac {16 x}{a^{8}}-\frac {16 c}{a^{8} d}+\frac {2 i}{d \,a^{8} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {8 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{8} d}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x,method=_RETURNVERBOSE)

[Out]

1/d/a^8*(tan(d*x+c)+8*I*ln(tan(d*x+c)-I)-16/3/(tan(d*x+c)-I)^3+24/(tan(d*x+c)-I)+16*I/(tan(d*x+c)-I)^2)

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Maxima [A]
time = 0.31, size = 191, normalized size = 1.65 \begin {gather*} \frac {\frac {8 \, {\left (9 \, \tan \left (d x + c\right )^{6} - 48 i \, \tan \left (d x + c\right )^{5} - 107 \, \tan \left (d x + c\right )^{4} + 128 i \, \tan \left (d x + c\right )^{3} + 87 \, \tan \left (d x + c\right )^{2} - 32 i \, \tan \left (d x + c\right ) - 5\right )}}{a^{8} \tan \left (d x + c\right )^{7} - 7 i \, a^{8} \tan \left (d x + c\right )^{6} - 21 \, a^{8} \tan \left (d x + c\right )^{5} + 35 i \, a^{8} \tan \left (d x + c\right )^{4} + 35 \, a^{8} \tan \left (d x + c\right )^{3} - 21 i \, a^{8} \tan \left (d x + c\right )^{2} - 7 \, a^{8} \tan \left (d x + c\right ) + i \, a^{8}} + \frac {24 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{8}} + \frac {3 \, \tan \left (d x + c\right )}{a^{8}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

1/3*(8*(9*tan(d*x + c)^6 - 48*I*tan(d*x + c)^5 - 107*tan(d*x + c)^4 + 128*I*tan(d*x + c)^3 + 87*tan(d*x + c)^2
- 32*I*tan(d*x + c) - 5)/(a^8*tan(d*x + c)^7 - 7*I*a^8*tan(d*x + c)^6 - 21*a^8*tan(d*x + c)^5 + 35*I*a^8*tan(
d*x + c)^4 + 35*a^8*tan(d*x + c)^3 - 21*I*a^8*tan(d*x + c)^2 - 7*a^8*tan(d*x + c) + I*a^8) + 24*I*log(I*tan(d*
x + c) + 1)/a^8 + 3*tan(d*x + c)/a^8)/d

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Fricas [A]
time = 0.40, size = 124, normalized size = 1.07 \begin {gather*} -\frac {2 \, {\left (24 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} + 12 \, {\left (2 \, d x - i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 12 \, {\left (i \, e^{\left (8 i \, d x + 8 i \, c\right )} + i \, e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{3 \, {\left (a^{8} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{8} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

-2/3*(24*d*x*e^(8*I*d*x + 8*I*c) + 12*(2*d*x - I)*e^(6*I*d*x + 6*I*c) + 12*(I*e^(8*I*d*x + 8*I*c) + I*e^(6*I*d
*x + 6*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 6*I*e^(4*I*d*x + 4*I*c) + 2*I*e^(2*I*d*x + 2*I*c) - I)/(a^8*d*e^(8
*I*d*x + 8*I*c) + a^8*d*e^(6*I*d*x + 6*I*c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sec ^{10}{\left (c + d x \right )}}{\tan ^{8}{\left (c + d x \right )} - 8 i \tan ^{7}{\left (c + d x \right )} - 28 \tan ^{6}{\left (c + d x \right )} + 56 i \tan ^{5}{\left (c + d x \right )} + 70 \tan ^{4}{\left (c + d x \right )} - 56 i \tan ^{3}{\left (c + d x \right )} - 28 \tan ^{2}{\left (c + d x \right )} + 8 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**10/(a+I*a*tan(d*x+c))**8,x)

[Out]

Integral(sec(c + d*x)**10/(tan(c + d*x)**8 - 8*I*tan(c + d*x)**7 - 28*tan(c + d*x)**6 + 56*I*tan(c + d*x)**5 +
70*tan(c + d*x)**4 - 56*I*tan(c + d*x)**3 - 28*tan(c + d*x)**2 + 8*I*tan(c + d*x) + 1), x)/a**8

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Giac [A]
time = 1.67, size = 199, normalized size = 1.72 \begin {gather*} -\frac {2 \, {\left (\frac {60 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{8}} - \frac {120 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{8}} + \frac {60 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{8}} - \frac {15 \, {\left (4 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{8}} + \frac {2 \, {\left (147 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 942 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2445 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3460 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2445 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 942 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 147 i\right )}}{a^{8} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{6}}\right )}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

-2/15*(60*I*log(tan(1/2*d*x + 1/2*c) + 1)/a^8 - 120*I*log(tan(1/2*d*x + 1/2*c) - I)/a^8 + 60*I*log(tan(1/2*d*x
+ 1/2*c) - 1)/a^8 - 15*(4*I*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) - 4*I)/((tan(1/2*d*x + 1/2*c)^2 - 1
)*a^8) + 2*(147*I*tan(1/2*d*x + 1/2*c)^6 + 942*tan(1/2*d*x + 1/2*c)^5 - 2445*I*tan(1/2*d*x + 1/2*c)^4 - 3460*t
an(1/2*d*x + 1/2*c)^3 + 2445*I*tan(1/2*d*x + 1/2*c)^2 + 942*tan(1/2*d*x + 1/2*c) - 147*I)/(a^8*(tan(1/2*d*x +
1/2*c) - I)^6))/d

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Mupad [B]
time = 3.50, size = 104, normalized size = 0.90 \begin {gather*} \frac {\mathrm {tan}\left (c+d\,x\right )}{a^8\,d}-\frac {\frac {32\,\mathrm {tan}\left (c+d\,x\right )}{a^8}-\frac {40{}\mathrm {i}}{3\,a^8}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,24{}\mathrm {i}}{a^8}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,8{}\mathrm {i}}{a^8\,d} \end {gather*}

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)^8),x)

[Out]

(log(tan(c + d*x) - 1i)*8i)/(a^8*d) - ((32*tan(c + d*x))/a^8 - 40i/(3*a^8) + (tan(c + d*x)^2*24i)/a^8)/(d*(tan
(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) + tan(c + d*x)/(a^8*d)

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