3.36 \(\int \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=109 \[ \frac{i (a+i a \tan (c+d x))^{10}}{10 a^7 d}-\frac{2 i (a+i a \tan (c+d x))^9}{3 a^6 d}+\frac{3 i (a+i a \tan (c+d x))^8}{2 a^5 d}-\frac{8 i (a+i a \tan (c+d x))^7}{7 a^4 d} \]

[Out]

(((-8*I)/7)*(a + I*a*Tan[c + d*x])^7)/(a^4*d) + (((3*I)/2)*(a + I*a*Tan[c + d*x])^8)/(a^5*d) - (((2*I)/3)*(a +
I*a*Tan[c + d*x])^9)/(a^6*d) + ((I/10)*(a + I*a*Tan[c + d*x])^10)/(a^7*d)

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Rubi [A]  time = 0.0637078, antiderivative size = 109, 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} \[ \frac{i (a+i a \tan (c+d x))^{10}}{10 a^7 d}-\frac{2 i (a+i a \tan (c+d x))^9}{3 a^6 d}+\frac{3 i (a+i a \tan (c+d x))^8}{2 a^5 d}-\frac{8 i (a+i a \tan (c+d x))^7}{7 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((-8*I)/7)*(a + I*a*Tan[c + d*x])^7)/(a^4*d) + (((3*I)/2)*(a + I*a*Tan[c + d*x])^8)/(a^5*d) - (((2*I)/3)*(a +
I*a*Tan[c + d*x])^9)/(a^6*d) + ((I/10)*(a + I*a*Tan[c + d*x])^10)/(a^7*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 ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x)^3 (a+x)^6 \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{i \operatorname{Subst}\left (\int \left (8 a^3 (a+x)^6-12 a^2 (a+x)^7+6 a (a+x)^8-(a+x)^9\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac{8 i (a+i a \tan (c+d x))^7}{7 a^4 d}+\frac{3 i (a+i a \tan (c+d x))^8}{2 a^5 d}-\frac{2 i (a+i a \tan (c+d x))^9}{3 a^6 d}+\frac{i (a+i a \tan (c+d x))^{10}}{10 a^7 d}\\ \end{align*}

Mathematica [A]  time = 1.79137, size = 117, normalized size = 1.07 \[ \frac{a^3 \sec (c) \sec ^{10}(c+d x) (105 \sin (c+2 d x)-105 \sin (3 c+2 d x)+120 \sin (3 c+4 d x)+45 \sin (5 c+6 d x)+10 \sin (7 c+8 d x)+\sin (9 c+10 d x)+105 i \cos (c+2 d x)+105 i \cos (3 c+2 d x)-126 \sin (c)+126 i \cos (c))}{840 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*Sec[c]*Sec[c + d*x]^10*((126*I)*Cos[c] + (105*I)*Cos[c + 2*d*x] + (105*I)*Cos[3*c + 2*d*x] - 126*Sin[c] +
105*Sin[c + 2*d*x] - 105*Sin[3*c + 2*d*x] + 120*Sin[3*c + 4*d*x] + 45*Sin[5*c + 6*d*x] + 10*Sin[7*c + 8*d*x]
+ Sin[9*c + 10*d*x]))/(840*d)

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Maple [B]  time = 0.065, size = 220, normalized size = 2. \begin{align*}{\frac{1}{d} \left ( -i{a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{10\, \left ( \cos \left ( dx+c \right ) \right ) ^{10}}}+{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{40\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{20\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{40\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \right ) -3\,{a}^{3} \left ( 1/9\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{9}}}+2/21\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{16\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{315\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{{\frac{3\,i}{8}}{a}^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}-{a}^{3} \left ( -{\frac{16}{35}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{35}} \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)^8*(a+I*a*tan(d*x+c))^3,x)

[Out]

1/d*(-I*a^3*(1/10*sin(d*x+c)^4/cos(d*x+c)^10+3/40*sin(d*x+c)^4/cos(d*x+c)^8+1/20*sin(d*x+c)^4/cos(d*x+c)^6+1/4
0*sin(d*x+c)^4/cos(d*x+c)^4)-3*a^3*(1/9*sin(d*x+c)^3/cos(d*x+c)^9+2/21*sin(d*x+c)^3/cos(d*x+c)^7+8/105*sin(d*x
+c)^3/cos(d*x+c)^5+16/315*sin(d*x+c)^3/cos(d*x+c)^3)+3/8*I*a^3/cos(d*x+c)^8-a^3*(-16/35-1/7*sec(d*x+c)^6-6/35*
sec(d*x+c)^4-8/35*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.09396, size = 146, normalized size = 1.34 \begin{align*} \frac{-84 i \, a^{3} \tan \left (d x + c\right )^{10} - 280 \, a^{3} \tan \left (d x + c\right )^{9} - 960 \, a^{3} \tan \left (d x + c\right )^{7} + 840 i \, a^{3} \tan \left (d x + c\right )^{6} - 1008 \, a^{3} \tan \left (d x + c\right )^{5} + 1680 i \, a^{3} \tan \left (d x + c\right )^{4} + 1260 i \, a^{3} \tan \left (d x + c\right )^{2} + 840 \, a^{3} \tan \left (d x + c\right )}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/840*(-84*I*a^3*tan(d*x + c)^10 - 280*a^3*tan(d*x + c)^9 - 960*a^3*tan(d*x + c)^7 + 840*I*a^3*tan(d*x + c)^6
- 1008*a^3*tan(d*x + c)^5 + 1680*I*a^3*tan(d*x + c)^4 + 1260*I*a^3*tan(d*x + c)^2 + 840*a^3*tan(d*x + c))/d

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Fricas [B]  time = 1.2306, size = 693, normalized size = 6.36 \begin{align*} \frac{26880 i \, a^{3} e^{\left (12 i \, d x + 12 i \, c\right )} + 32256 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + 26880 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 15360 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 5760 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 1280 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 128 i \, a^{3}}{105 \,{\left (d e^{\left (20 i \, d x + 20 i \, c\right )} + 10 \, d e^{\left (18 i \, d x + 18 i \, c\right )} + 45 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 120 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 210 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 252 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 210 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 120 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 45 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \, 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)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/105*(26880*I*a^3*e^(12*I*d*x + 12*I*c) + 32256*I*a^3*e^(10*I*d*x + 10*I*c) + 26880*I*a^3*e^(8*I*d*x + 8*I*c)
+ 15360*I*a^3*e^(6*I*d*x + 6*I*c) + 5760*I*a^3*e^(4*I*d*x + 4*I*c) + 1280*I*a^3*e^(2*I*d*x + 2*I*c) + 128*I*a
^3)/(d*e^(20*I*d*x + 20*I*c) + 10*d*e^(18*I*d*x + 18*I*c) + 45*d*e^(16*I*d*x + 16*I*c) + 120*d*e^(14*I*d*x + 1
4*I*c) + 210*d*e^(12*I*d*x + 12*I*c) + 252*d*e^(10*I*d*x + 10*I*c) + 210*d*e^(8*I*d*x + 8*I*c) + 120*d*e^(6*I*
d*x + 6*I*c) + 45*d*e^(4*I*d*x + 4*I*c) + 10*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)**8*(a+I*a*tan(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25443, size = 146, normalized size = 1.34 \begin{align*} -\frac{21 i \, a^{3} \tan \left (d x + c\right )^{10} + 70 \, a^{3} \tan \left (d x + c\right )^{9} + 240 \, a^{3} \tan \left (d x + c\right )^{7} - 210 i \, a^{3} \tan \left (d x + c\right )^{6} + 252 \, a^{3} \tan \left (d x + c\right )^{5} - 420 i \, a^{3} \tan \left (d x + c\right )^{4} - 315 i \, a^{3} \tan \left (d x + c\right )^{2} - 210 \, a^{3} \tan \left (d x + c\right )}{210 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/210*(21*I*a^3*tan(d*x + c)^10 + 70*a^3*tan(d*x + c)^9 + 240*a^3*tan(d*x + c)^7 - 210*I*a^3*tan(d*x + c)^6 +
252*a^3*tan(d*x + c)^5 - 420*I*a^3*tan(d*x + c)^4 - 315*I*a^3*tan(d*x + c)^2 - 210*a^3*tan(d*x + c))/d