3.21 \(\int \sec ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=55 \[ \frac {i (a+i a \tan (c+d x))^5}{5 a^3 d}-\frac {i (a+i a \tan (c+d x))^4}{2 a^2 d} \]

[Out]

-1/2*I*(a+I*a*tan(d*x+c))^4/a^2/d+1/5*I*(a+I*a*tan(d*x+c))^5/a^3/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 = {3487, 43} \[ \frac {i (a+i a \tan (c+d x))^5}{5 a^3 d}-\frac {i (a+i a \tan (c+d x))^4}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.50, size = 77, normalized size = 1.40 \[ \frac {a^2 \sec (c) \sec ^5(c+d x) (-5 \sin (2 c+d x)+5 \sin (2 c+3 d x)+\sin (4 c+5 d x)+5 i \cos (2 c+d x)+5 \sin (d x)+5 i \cos (d x))}{20 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*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*d)

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fricas [B]  time = 0.51, size = 113, normalized size = 2.05 \[ \frac {80 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 80 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 40 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a^{2}}{5 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/5*(80*I*a^2*e^(6*I*d*x + 6*I*c) + 80*I*a^2*e^(4*I*d*x + 4*I*c) + 40*I*a^2*e^(2*I*d*x + 2*I*c) + 8*I*a^2)/(d*
e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^
(2*I*d*x + 2*I*c) + d)

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giac [A]  time = 0.95, size = 56, normalized size = 1.02 \[ -\frac {2 \, a^{2} \tan \left (d x + c\right )^{5} - 5 i \, a^{2} \tan \left (d x + c\right )^{4} - 10 i \, a^{2} \tan \left (d x + c\right )^{2} - 10 \, a^{2} \tan \left (d x + c\right )}{10 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.42, size = 85, normalized size = 1.55 \[ \frac {-a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{2}}{2 \cos \left (d x +c \right )^{4}}-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x)

[Out]

1/d*(-a^2*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)+1/2*I*a^2/cos(d*x+c)^4-a^2*(-2/3-1/3*
sec(d*x+c)^2)*tan(d*x+c))

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maxima [A]  time = 0.41, size = 56, normalized size = 1.02 \[ -\frac {6 \, a^{2} \tan \left (d x + c\right )^{5} - 15 i \, a^{2} \tan \left (d x + c\right )^{4} - 30 i \, a^{2} \tan \left (d x + c\right )^{2} - 30 \, a^{2} \tan \left (d x + c\right )}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/30*(6*a^2*tan(d*x + c)^5 - 15*I*a^2*tan(d*x + c)^4 - 30*I*a^2*tan(d*x + c)^2 - 30*a^2*tan(d*x + c))/d

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mupad [B]  time = 3.22, size = 56, normalized size = 1.02 \[ \frac {-\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}}{2}+a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+a^2\,\mathrm {tan}\left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^2/cos(c + d*x)^4,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \tan ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 2 i \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int \left (- \sec ^{4}{\left (c + d x \right )}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+I*a*tan(d*x+c))**2,x)

[Out]

-a**2*(Integral(tan(c + d*x)**2*sec(c + d*x)**4, x) + Integral(-2*I*tan(c + d*x)*sec(c + d*x)**4, x) + Integra
l(-sec(c + d*x)**4, x))

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