3.2.15 \(\int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [115]

Optimal. Leaf size=55 \[ \frac {i (a-i a \tan (c+d x))^4}{2 a^6 d}-\frac {i (a-i a \tan (c+d x))^5}{5 a^7 d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 55, 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))^4}{2 a^6 d}-\frac {i (a-i a \tan (c+d x))^5}{5 a^7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

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Mathematica [A]
time = 0.45, size = 77, normalized size = 1.40 \begin {gather*} \frac {\sec (c) \sec ^5(c+d x) (-5 i \cos (d x)-5 i \cos (2 c+d x)+5 \sin (d x)-5 \sin (2 c+d x)+5 \sin (2 c+3 d x)+\sin (4 c+5 d x))}{20 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(Sec[c]*Sec[c + d*x]^5*((-5*I)*Cos[d*x] - (5*I)*Cos[2*c + d*x] + 5*Sin[d*x] - 5*Sin[2*c + d*x] + 5*Sin[2*c + 3
*d*x] + Sin[4*c + 5*d*x]))/(20*a^2*d)

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Maple [A]
time = 0.26, size = 47, normalized size = 0.85

method	result	size
risch	\(\frac {8 i \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{5 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}\)	\(36\)
derivativedivides	\(\frac {\tan \left (d x +c \right )-\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{2}-i \left (\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}\)	\(47\)
default	\(\frac {\tan \left (d x +c \right )-\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{2}-i \left (\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 47, normalized size = 0.85 \begin {gather*} -\frac {2 \, \tan \left (d x + c\right )^{5} + 5 i \, \tan \left (d x + c\right )^{4} + 10 i \, \tan \left (d x + c\right )^{2} - 10 \, \tan \left (d x + c\right )}{10 \, a^{2} 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))^2,x, algorithm="maxima")

[Out]

-1/10*(2*tan(d*x + c)^5 + 5*I*tan(d*x + c)^4 + 10*I*tan(d*x + c)^2 - 10*tan(d*x + c))/(a^2*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (43) = 86\).
time = 0.37, size = 97, normalized size = 1.76 \begin {gather*} -\frac {8 \, {\left (-5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{5 \, {\left (a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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))^2,x, algorithm="fricas")

[Out]

-8/5*(-5*I*e^(2*I*d*x + 2*I*c) - I)/(a^2*d*e^(10*I*d*x + 10*I*c) + 5*a^2*d*e^(8*I*d*x + 8*I*c) + 10*a^2*d*e^(6
*I*d*x + 6*I*c) + 10*a^2*d*e^(4*I*d*x + 4*I*c) + 5*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\sec ^{8}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \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))**2,x)

[Out]

-Integral(sec(c + d*x)**8/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x)/a**2

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Giac [A]
time = 0.57, size = 47, normalized size = 0.85 \begin {gather*} -\frac {2 \, \tan \left (d x + c\right )^{5} + 5 i \, \tan \left (d x + c\right )^{4} + 10 i \, \tan \left (d x + c\right )^{2} - 10 \, \tan \left (d x + c\right )}{10 \, a^{2} 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))^2,x, algorithm="giac")

[Out]

-1/10*(2*tan(d*x + c)^5 + 5*I*tan(d*x + c)^4 + 10*I*tan(d*x + c)^2 - 10*tan(d*x + c))/(a^2*d)

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Mupad [B]
time = 3.36, size = 77, normalized size = 1.40 \begin {gather*} -\frac {\sin \left (c+d\,x\right )\,\left (-10\,{\cos \left (c+d\,x\right )}^4+{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )\,10{}\mathrm {i}+\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^3\,5{}\mathrm {i}+2\,{\sin \left (c+d\,x\right )}^4\right )}{10\,a^2\,d\,{\cos \left (c+d\,x\right )}^5} \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)^2),x)

[Out]

-(sin(c + d*x)*(cos(c + d*x)*sin(c + d*x)^3*5i + cos(c + d*x)^3*sin(c + d*x)*10i - 10*cos(c + d*x)^4 + 2*sin(c
+ d*x)^4))/(10*a^2*d*cos(c + d*x)^5)

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