3.37 \(\int \sec ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=82 \[ -\frac{i (a+i a \tan (c+d x))^8}{8 a^5 d}+\frac{4 i (a+i a \tan (c+d x))^7}{7 a^4 d}-\frac{2 i (a+i a \tan (c+d x))^6}{3 a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.0563161, antiderivative size = 82, 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} \[ -\frac{i (a+i a \tan (c+d x))^8}{8 a^5 d}+\frac{4 i (a+i a \tan (c+d x))^7}{7 a^4 d}-\frac{2 i (a+i a \tan (c+d x))^6}{3 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^3,x]

[Out]

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

Mathematica [A]  time = 1.20586, size = 106, normalized size = 1.29 \[ \frac{a^3 \sec (c) \sec ^8(c+d x) (28 \sin (c+2 d x)-28 \sin (3 c+2 d x)+28 \sin (3 c+4 d x)+8 \sin (5 c+6 d x)+\sin (7 c+8 d x)+28 i \cos (c+2 d x)+28 i \cos (3 c+2 d x)-35 \sin (c)+35 i \cos (c))}{168 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*Sec[c]*Sec[c + d*x]^8*((35*I)*Cos[c] + (28*I)*Cos[c + 2*d*x] + (28*I)*Cos[3*c + 2*d*x] - 35*Sin[c] + 28*S
in[c + 2*d*x] - 28*Sin[3*c + 2*d*x] + 28*Sin[3*c + 4*d*x] + 8*Sin[5*c + 6*d*x] + Sin[7*c + 8*d*x]))/(168*d)

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Maple [B]  time = 0.061, size = 174, normalized size = 2.1 \begin{align*}{\frac{1}{d} \left ( -i{a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{12\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{24\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \right ) -3\,{a}^{3} \left ( 1/7\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{{\frac{i}{2}}{a}^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}-{a}^{3} \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \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)^6*(a+I*a*tan(d*x+c))^3,x)

[Out]

1/d*(-I*a^3*(1/8*sin(d*x+c)^4/cos(d*x+c)^8+1/12*sin(d*x+c)^4/cos(d*x+c)^6+1/24*sin(d*x+c)^4/cos(d*x+c)^4)-3*a^
3*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3)+1/2*I*a^3/cos
(d*x+c)^6-a^3*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 1.13342, size = 146, normalized size = 1.78 \begin{align*} \frac{-105 i \, a^{3} \tan \left (d x + c\right )^{8} - 360 \, a^{3} \tan \left (d x + c\right )^{7} + 140 i \, a^{3} \tan \left (d x + c\right )^{6} - 840 \, a^{3} \tan \left (d x + c\right )^{5} + 1050 i \, a^{3} \tan \left (d x + c\right )^{4} - 280 \, a^{3} \tan \left (d x + c\right )^{3} + 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)^6*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/840*(-105*I*a^3*tan(d*x + c)^8 - 360*a^3*tan(d*x + c)^7 + 140*I*a^3*tan(d*x + c)^6 - 840*a^3*tan(d*x + c)^5
+ 1050*I*a^3*tan(d*x + c)^4 - 280*a^3*tan(d*x + c)^3 + 1260*I*a^3*tan(d*x + c)^2 + 840*a^3*tan(d*x + c))/d

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Fricas [B]  time = 1.03985, size = 547, normalized size = 6.67 \begin{align*} \frac{1792 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + 2240 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 1792 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 896 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 256 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 32 i \, a^{3}}{21 \,{\left (d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, 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)^6*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/21*(1792*I*a^3*e^(10*I*d*x + 10*I*c) + 2240*I*a^3*e^(8*I*d*x + 8*I*c) + 1792*I*a^3*e^(6*I*d*x + 6*I*c) + 896
*I*a^3*e^(4*I*d*x + 4*I*c) + 256*I*a^3*e^(2*I*d*x + 2*I*c) + 32*I*a^3)/(d*e^(16*I*d*x + 16*I*c) + 8*d*e^(14*I*
d*x + 14*I*c) + 28*d*e^(12*I*d*x + 12*I*c) + 56*d*e^(10*I*d*x + 10*I*c) + 70*d*e^(8*I*d*x + 8*I*c) + 56*d*e^(6
*I*d*x + 6*I*c) + 28*d*e^(4*I*d*x + 4*I*c) + 8*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int - 3 \tan ^{2}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int 3 i \tan{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int - i \tan ^{3}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \sec ^{6}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6*(a+I*a*tan(d*x+c))**3,x)

[Out]

a**3*(Integral(-3*tan(c + d*x)**2*sec(c + d*x)**6, x) + Integral(3*I*tan(c + d*x)*sec(c + d*x)**6, x) + Integr
al(-I*tan(c + d*x)**3*sec(c + d*x)**6, x) + Integral(sec(c + d*x)**6, x))

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Giac [A]  time = 1.2548, size = 146, normalized size = 1.78 \begin{align*} -\frac{21 i \, a^{3} \tan \left (d x + c\right )^{8} + 72 \, a^{3} \tan \left (d x + c\right )^{7} - 28 i \, a^{3} \tan \left (d x + c\right )^{6} + 168 \, a^{3} \tan \left (d x + c\right )^{5} - 210 i \, a^{3} \tan \left (d x + c\right )^{4} + 56 \, a^{3} \tan \left (d x + c\right )^{3} - 252 i \, a^{3} \tan \left (d x + c\right )^{2} - 168 \, a^{3} \tan \left (d x + c\right )}{168 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/168*(21*I*a^3*tan(d*x + c)^8 + 72*a^3*tan(d*x + c)^7 - 28*I*a^3*tan(d*x + c)^6 + 168*a^3*tan(d*x + c)^5 - 2
10*I*a^3*tan(d*x + c)^4 + 56*a^3*tan(d*x + c)^3 - 252*I*a^3*tan(d*x + c)^2 - 168*a^3*tan(d*x + c))/d