3.61 \(\int \sec ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx\)

Optimal. Leaf size=55 \[ \frac{i (a+i a \tan (c+d x))^8}{8 a^3 d}-\frac{2 i (a+i a \tan (c+d x))^7}{7 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0429323, 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} \[ \frac{i (a+i a \tan (c+d x))^8}{8 a^3 d}-\frac{2 i (a+i a \tan (c+d x))^7}{7 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^5,x]

[Out]

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

Mathematica [B]  time = 1.6968, size = 143, normalized size = 2.6 \[ \frac{a^5 \sec (c) \sec ^8(c+d x) (28 \sin (c+2 d x)-28 \sin (3 c+2 d x)+14 \sin (3 c+4 d x)-14 \sin (5 c+4 d x)+8 \sin (5 c+6 d x)+\sin (7 c+8 d x)+28 i \cos (c+2 d x)+28 i \cos (3 c+2 d x)+14 i \cos (3 c+4 d x)+14 i \cos (5 c+4 d x)-35 \sin (c)+35 i \cos (c))}{56 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*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] + (14*I)*Cos[3*c +
4*d*x] + (14*I)*Cos[5*c + 4*d*x] - 35*Sin[c] + 28*Sin[c + 2*d*x] - 28*Sin[3*c + 2*d*x] + 14*Sin[3*c + 4*d*x]
- 14*Sin[5*c + 4*d*x] + 8*Sin[5*c + 6*d*x] + Sin[7*c + 8*d*x]))/(56*d)

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Maple [B]  time = 0.078, size = 213, normalized size = 3.9 \begin{align*}{\frac{1}{d} \left ( i{a}^{5} \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 ) +5\,{a}^{5} \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 ) -10\,i{a}^{5} \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 ) -10\,{a}^{5} \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{{\frac{5\,i}{4}}{a}^{5}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{a}^{5} \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))^5,x)

[Out]

1/d*(I*a^5*(1/8*sin(d*x+c)^6/cos(d*x+c)^8+1/24*sin(d*x+c)^6/cos(d*x+c)^6)+5*a^5*(1/7*sin(d*x+c)^5/cos(d*x+c)^7
+2/35*sin(d*x+c)^5/cos(d*x+c)^5)-10*I*a^5*(1/6*sin(d*x+c)^4/cos(d*x+c)^6+1/12*sin(d*x+c)^4/cos(d*x+c)^4)-10*a^
5*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)+5/4*I*a^5/cos(d*x+c)^4-a^5*(-2/3-1/3*sec(d*x+
c)^2)*tan(d*x+c))

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Maxima [B]  time = 1.13411, size = 146, normalized size = 2.65 \begin{align*} \frac{21 i \, a^{5} \tan \left (d x + c\right )^{8} + 120 \, a^{5} \tan \left (d x + c\right )^{7} - 252 i \, a^{5} \tan \left (d x + c\right )^{6} - 168 \, a^{5} \tan \left (d x + c\right )^{5} - 210 i \, a^{5} \tan \left (d x + c\right )^{4} - 504 \, a^{5} \tan \left (d x + c\right )^{3} + 420 i \, a^{5} \tan \left (d x + c\right )^{2} + 168 \, a^{5} \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)^4*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

1/168*(21*I*a^5*tan(d*x + c)^8 + 120*a^5*tan(d*x + c)^7 - 252*I*a^5*tan(d*x + c)^6 - 168*a^5*tan(d*x + c)^5 -
210*I*a^5*tan(d*x + c)^4 - 504*a^5*tan(d*x + c)^3 + 420*I*a^5*tan(d*x + c)^2 + 168*a^5*tan(d*x + c))/d

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Fricas [B]  time = 1.09312, size = 591, normalized size = 10.75 \begin{align*} \frac{896 i \, a^{5} e^{\left (12 i \, d x + 12 i \, c\right )} + 1792 i \, a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} + 2240 i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 1792 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 896 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 256 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + 32 i \, a^{5}}{7 \,{\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)^4*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

1/7*(896*I*a^5*e^(12*I*d*x + 12*I*c) + 1792*I*a^5*e^(10*I*d*x + 10*I*c) + 2240*I*a^5*e^(8*I*d*x + 8*I*c) + 179
2*I*a^5*e^(6*I*d*x + 6*I*c) + 896*I*a^5*e^(4*I*d*x + 4*I*c) + 256*I*a^5*e^(2*I*d*x + 2*I*c) + 32*I*a^5)/(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^{5} \left (\int - 10 \tan ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 5 \tan ^{4}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 5 i \tan{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int - 10 i \tan ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int i \tan ^{5}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \sec ^{4}{\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)**4*(a+I*a*tan(d*x+c))**5,x)

[Out]

a**5*(Integral(-10*tan(c + d*x)**2*sec(c + d*x)**4, x) + Integral(5*tan(c + d*x)**4*sec(c + d*x)**4, x) + Inte
gral(5*I*tan(c + d*x)*sec(c + d*x)**4, x) + Integral(-10*I*tan(c + d*x)**3*sec(c + d*x)**4, x) + Integral(I*ta
n(c + d*x)**5*sec(c + d*x)**4, x) + Integral(sec(c + d*x)**4, x))

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Giac [B]  time = 1.53702, size = 146, normalized size = 2.65 \begin{align*} -\frac{-7 i \, a^{5} \tan \left (d x + c\right )^{8} - 40 \, a^{5} \tan \left (d x + c\right )^{7} + 84 i \, a^{5} \tan \left (d x + c\right )^{6} + 56 \, a^{5} \tan \left (d x + c\right )^{5} + 70 i \, a^{5} \tan \left (d x + c\right )^{4} + 168 \, a^{5} \tan \left (d x + c\right )^{3} - 140 i \, a^{5} \tan \left (d x + c\right )^{2} - 56 \, a^{5} \tan \left (d x + c\right )}{56 \, 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))^5,x, algorithm="giac")

[Out]

-1/56*(-7*I*a^5*tan(d*x + c)^8 - 40*a^5*tan(d*x + c)^7 + 84*I*a^5*tan(d*x + c)^6 + 56*a^5*tan(d*x + c)^5 + 70*
I*a^5*tan(d*x + c)^4 + 168*a^5*tan(d*x + c)^3 - 140*I*a^5*tan(d*x + c)^2 - 56*a^5*tan(d*x + c))/d