3.79 \(\int \sec ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx\)

Optimal. Leaf size=55 \[ \frac{i (a+i a \tan (c+d x))^{11}}{11 a^3 d}-\frac{i (a+i a \tan (c+d x))^{10}}{5 a^2 d} \]

[Out]

((-I/5)*(a + I*a*Tan[c + d*x])^10)/(a^2*d) + ((I/11)*(a + I*a*Tan[c + d*x])^11)/(a^3*d)

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Rubi [A]  time = 0.0454868, antiderivative size = 55, normalized size of antiderivative = 1., 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 = {3487, 43} \[ \frac{i (a+i a \tan (c+d x))^{11}}{11 a^3 d}-\frac{i (a+i a \tan (c+d x))^{10}}{5 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^8,x]

[Out]

((-I/5)*(a + I*a*Tan[c + d*x])^10)/(a^2*d) + ((I/11)*(a + I*a*Tan[c + d*x])^11)/(a^3*d)

Rule 3487

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]

Rule 43

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])

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x) (a+x)^9 \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (2 a (a+x)^9-(a+x)^{10}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{i (a+i a \tan (c+d x))^{10}}{5 a^2 d}+\frac{i (a+i a \tan (c+d x))^{11}}{11 a^3 d}\\ \end{align*}

Mathematica [B]  time = 4.20502, size = 223, normalized size = 4.05 \[ \frac{a^8 \sec (c) \sec ^{11}(c+d x) (-462 \sin (2 c+d x)+330 \sin (2 c+3 d x)-330 \sin (4 c+3 d x)+165 \sin (4 c+5 d x)-165 \sin (6 c+5 d x)+55 \sin (6 c+7 d x)-55 \sin (8 c+7 d x)+22 \sin (8 c+9 d x)+2 \sin (10 c+11 d x)+462 i \cos (2 c+d x)+330 i \cos (2 c+3 d x)+330 i \cos (4 c+3 d x)+165 i \cos (4 c+5 d x)+165 i \cos (6 c+5 d x)+55 i \cos (6 c+7 d x)+55 i \cos (8 c+7 d x)+462 \sin (d x)+462 i \cos (d x))}{220 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^8,x]

[Out]

(a^8*Sec[c]*Sec[c + d*x]^11*((462*I)*Cos[d*x] + (462*I)*Cos[2*c + d*x] + (330*I)*Cos[2*c + 3*d*x] + (330*I)*Co
s[4*c + 3*d*x] + (165*I)*Cos[4*c + 5*d*x] + (165*I)*Cos[6*c + 5*d*x] + (55*I)*Cos[6*c + 7*d*x] + (55*I)*Cos[8*
c + 7*d*x] + 462*Sin[d*x] - 462*Sin[2*c + d*x] + 330*Sin[2*c + 3*d*x] - 330*Sin[4*c + 3*d*x] + 165*Sin[4*c + 5
*d*x] - 165*Sin[6*c + 5*d*x] + 55*Sin[6*c + 7*d*x] - 55*Sin[8*c + 7*d*x] + 22*Sin[8*c + 9*d*x] + 2*Sin[10*c +
11*d*x]))/(220*d)

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Maple [B]  time = 0.095, size = 339, normalized size = 6.2 \begin{align*}{\frac{1}{d} \left ({a}^{8} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{11\, \left ( \cos \left ( dx+c \right ) \right ) ^{11}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{99\, \left ( \cos \left ( dx+c \right ) \right ) ^{9}}} \right ) -56\,i{a}^{8} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{12\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \right ) -28\,{a}^{8} \left ( 1/9\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{9}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{63\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}} \right ) +56\,i{a}^{8} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{24\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}} \right ) +70\,{a}^{8} \left ( 1/7\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \right ) -8\,i{a}^{8} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{10\, \left ( \cos \left ( dx+c \right ) \right ) ^{10}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{40\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}}} \right ) -28\,{a}^{8} \left ( 1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+2/15\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{2\,i{a}^{8}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{a}^{8} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x)

[Out]

1/d*(a^8*(1/11*sin(d*x+c)^9/cos(d*x+c)^11+2/99*sin(d*x+c)^9/cos(d*x+c)^9)-56*I*a^8*(1/6*sin(d*x+c)^4/cos(d*x+c
)^6+1/12*sin(d*x+c)^4/cos(d*x+c)^4)-28*a^8*(1/9*sin(d*x+c)^7/cos(d*x+c)^9+2/63*sin(d*x+c)^7/cos(d*x+c)^7)+56*I
*a^8*(1/8*sin(d*x+c)^6/cos(d*x+c)^8+1/24*sin(d*x+c)^6/cos(d*x+c)^6)+70*a^8*(1/7*sin(d*x+c)^5/cos(d*x+c)^7+2/35
*sin(d*x+c)^5/cos(d*x+c)^5)-8*I*a^8*(1/10*sin(d*x+c)^8/cos(d*x+c)^10+1/40*sin(d*x+c)^8/cos(d*x+c)^8)-28*a^8*(1
/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)+2*I*a^8/cos(d*x+c)^4-a^8*(-2/3-1/3*sec(d*x+c)^2)*
tan(d*x+c))

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Maxima [B]  time = 1.34338, size = 181, normalized size = 3.29 \begin{align*} \frac{45 \, a^{8} \tan \left (d x + c\right )^{11} - 396 i \, a^{8} \tan \left (d x + c\right )^{10} - 1485 \, a^{8} \tan \left (d x + c\right )^{9} + 2970 i \, a^{8} \tan \left (d x + c\right )^{8} + 2970 \, a^{8} \tan \left (d x + c\right )^{7} + 4158 \, a^{8} \tan \left (d x + c\right )^{5} - 5940 i \, a^{8} \tan \left (d x + c\right )^{4} - 4455 \, a^{8} \tan \left (d x + c\right )^{3} + 1980 i \, a^{8} \tan \left (d x + c\right )^{2} + 495 \, a^{8} \tan \left (d x + c\right )}{495 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

1/495*(45*a^8*tan(d*x + c)^11 - 396*I*a^8*tan(d*x + c)^10 - 1485*a^8*tan(d*x + c)^9 + 2970*I*a^8*tan(d*x + c)^
8 + 2970*a^8*tan(d*x + c)^7 + 4158*a^8*tan(d*x + c)^5 - 5940*I*a^8*tan(d*x + c)^4 - 4455*a^8*tan(d*x + c)^3 +
1980*I*a^8*tan(d*x + c)^2 + 495*a^8*tan(d*x + c))/d

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Fricas [B]  time = 1.54089, size = 890, normalized size = 16.18 \begin{align*} \frac{56320 i \, a^{8} e^{\left (18 i \, d x + 18 i \, c\right )} + 168960 i \, a^{8} e^{\left (16 i \, d x + 16 i \, c\right )} + 337920 i \, a^{8} e^{\left (14 i \, d x + 14 i \, c\right )} + 473088 i \, a^{8} e^{\left (12 i \, d x + 12 i \, c\right )} + 473088 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} + 337920 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} + 168960 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 56320 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 11264 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + 1024 i \, a^{8}}{55 \,{\left (d e^{\left (22 i \, d x + 22 i \, c\right )} + 11 \, d e^{\left (20 i \, d x + 20 i \, c\right )} + 55 \, d e^{\left (18 i \, d x + 18 i \, c\right )} + 165 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 330 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 462 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 462 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 330 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 165 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 55 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 11 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/55*(56320*I*a^8*e^(18*I*d*x + 18*I*c) + 168960*I*a^8*e^(16*I*d*x + 16*I*c) + 337920*I*a^8*e^(14*I*d*x + 14*I
*c) + 473088*I*a^8*e^(12*I*d*x + 12*I*c) + 473088*I*a^8*e^(10*I*d*x + 10*I*c) + 337920*I*a^8*e^(8*I*d*x + 8*I*
c) + 168960*I*a^8*e^(6*I*d*x + 6*I*c) + 56320*I*a^8*e^(4*I*d*x + 4*I*c) + 11264*I*a^8*e^(2*I*d*x + 2*I*c) + 10
24*I*a^8)/(d*e^(22*I*d*x + 22*I*c) + 11*d*e^(20*I*d*x + 20*I*c) + 55*d*e^(18*I*d*x + 18*I*c) + 165*d*e^(16*I*d
*x + 16*I*c) + 330*d*e^(14*I*d*x + 14*I*c) + 462*d*e^(12*I*d*x + 12*I*c) + 462*d*e^(10*I*d*x + 10*I*c) + 330*d
*e^(8*I*d*x + 8*I*c) + 165*d*e^(6*I*d*x + 6*I*c) + 55*d*e^(4*I*d*x + 4*I*c) + 11*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+I*a*tan(d*x+c))**8,x)

[Out]

Timed out

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Giac [B]  time = 1.81662, size = 181, normalized size = 3.29 \begin{align*} \frac{5 \, a^{8} \tan \left (d x + c\right )^{11} - 44 i \, a^{8} \tan \left (d x + c\right )^{10} - 165 \, a^{8} \tan \left (d x + c\right )^{9} + 330 i \, a^{8} \tan \left (d x + c\right )^{8} + 330 \, a^{8} \tan \left (d x + c\right )^{7} + 462 \, a^{8} \tan \left (d x + c\right )^{5} - 660 i \, a^{8} \tan \left (d x + c\right )^{4} - 495 \, a^{8} \tan \left (d x + c\right )^{3} + 220 i \, a^{8} \tan \left (d x + c\right )^{2} + 55 \, a^{8} \tan \left (d x + c\right )}{55 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

1/55*(5*a^8*tan(d*x + c)^11 - 44*I*a^8*tan(d*x + c)^10 - 165*a^8*tan(d*x + c)^9 + 330*I*a^8*tan(d*x + c)^8 + 3
30*a^8*tan(d*x + c)^7 + 462*a^8*tan(d*x + c)^5 - 660*I*a^8*tan(d*x + c)^4 - 495*a^8*tan(d*x + c)^3 + 220*I*a^8
*tan(d*x + c)^2 + 55*a^8*tan(d*x + c))/d