3.38 \(\int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=55 \[ \frac {i (a+i a \tan (c+d x))^6}{6 a^3 d}-\frac {2 i (a+i a \tan (c+d x))^5}{5 a^2 d} \]

[Out]

-2/5*I*(a+I*a*tan(d*x+c))^5/a^2/d+1/6*I*(a+I*a*tan(d*x+c))^6/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} \[ \frac {i (a+i a \tan (c+d x))^6}{6 a^3 d}-\frac {2 i (a+i a \tan (c+d x))^5}{5 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^3,x]

[Out]

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

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Mathematica [A]  time = 1.30, size = 97, normalized size = 1.76 \[ \frac {a^3 \sec (c) \sec ^6(c+d x) (15 \sin (c+2 d x)-15 \sin (3 c+2 d x)+12 \sin (3 c+4 d x)+2 \sin (5 c+6 d x)+15 i \cos (c+2 d x)+15 i \cos (3 c+2 d x)-20 \sin (c)+20 i \cos (c))}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*Sec[c]*Sec[c + d*x]^6*((20*I)*Cos[c] + (15*I)*Cos[c + 2*d*x] + (15*I)*Cos[3*c + 2*d*x] - 20*Sin[c] + 15*S
in[c + 2*d*x] - 15*Sin[3*c + 2*d*x] + 12*Sin[3*c + 4*d*x] + 2*Sin[5*c + 6*d*x]))/(60*d)

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fricas [B]  time = 0.59, size = 139, normalized size = 2.53 \[ \frac {480 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 640 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 480 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 192 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 32 i \, a^{3}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(480*I*a^3*e^(8*I*d*x + 8*I*c) + 640*I*a^3*e^(6*I*d*x + 6*I*c) + 480*I*a^3*e^(4*I*d*x + 4*I*c) + 192*I*a^
3*e^(2*I*d*x + 2*I*c) + 32*I*a^3)/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I
*c) + 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*d*e^(2*I*d*x + 2*I*c) + d)

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giac [A]  time = 1.62, size = 82, normalized size = 1.49 \[ -\frac {5 i \, a^{3} \tan \left (d x + c\right )^{6} + 18 \, a^{3} \tan \left (d x + c\right )^{5} - 15 i \, a^{3} \tan \left (d x + c\right )^{4} + 20 \, a^{3} \tan \left (d x + c\right )^{3} - 45 i \, a^{3} \tan \left (d x + c\right )^{2} - 30 \, a^{3} \tan \left (d x + c\right )}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/30*(5*I*a^3*tan(d*x + c)^6 + 18*a^3*tan(d*x + c)^5 - 15*I*a^3*tan(d*x + c)^4 + 20*a^3*tan(d*x + c)^3 - 45*I
*a^3*tan(d*x + c)^2 - 30*a^3*tan(d*x + c))/d

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maple [B]  time = 0.44, size = 128, normalized size = 2.33 \[ \frac {-i a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )-3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {3 i a^{3}}{4 \cos \left (d x +c \right )^{4}}-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.53, size = 82, normalized size = 1.49 \[ \frac {-10 i \, a^{3} \tan \left (d x + c\right )^{6} - 36 \, a^{3} \tan \left (d x + c\right )^{5} + 30 i \, a^{3} \tan \left (d x + c\right )^{4} - 40 \, a^{3} \tan \left (d x + c\right )^{3} + 90 i \, a^{3} \tan \left (d x + c\right )^{2} + 60 \, a^{3} \tan \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(-10*I*a^3*tan(d*x + c)^6 - 36*a^3*tan(d*x + c)^5 + 30*I*a^3*tan(d*x + c)^4 - 40*a^3*tan(d*x + c)^3 + 90*
I*a^3*tan(d*x + c)^2 + 60*a^3*tan(d*x + c))/d

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mupad [B]  time = 3.26, size = 114, normalized size = 2.07 \[ -\frac {a^3\,\sin \left (c+d\,x\right )\,\left (-30\,{\cos \left (c+d\,x\right )}^5-{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,45{}\mathrm {i}+20\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,15{}\mathrm {i}+18\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^5\,5{}\mathrm {i}\right )}{30\,d\,{\cos \left (c+d\,x\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^3/cos(c + d*x)^4,x)

[Out]

-(a^3*sin(c + d*x)*(18*cos(c + d*x)*sin(c + d*x)^4 - cos(c + d*x)^4*sin(c + d*x)*45i - 30*cos(c + d*x)^5 + sin
(c + d*x)^5*5i - cos(c + d*x)^2*sin(c + d*x)^3*15i + 20*cos(c + d*x)^3*sin(c + d*x)^2))/(30*d*cos(c + d*x)^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i a^{3} \left (\int i \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 3 \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int \tan ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 3 i \tan ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+I*a*tan(d*x+c))**3,x)

[Out]

-I*a**3*(Integral(I*sec(c + d*x)**4, x) + Integral(-3*tan(c + d*x)*sec(c + d*x)**4, x) + Integral(tan(c + d*x)
**3*sec(c + d*x)**4, x) + Integral(-3*I*tan(c + d*x)**2*sec(c + d*x)**4, x))

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