3.101 \(\int \frac {\sec ^6(c+d x)}{a+i a \tan (c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.05, 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} \[ \frac {2 i (a-i a \tan (c+d x))^3}{3 a^4 d}-\frac {i (a-i a \tan (c+d x))^4}{4 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.21, size = 49, normalized size = 0.89 \[ \frac {\sec (c) \sec ^4(c+d x) (4 \sin (c+2 d x)+\sin (3 c+4 d x)-3 \sin (c)-3 i \cos (c))}{12 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c]*Sec[c + d*x]^4*((-3*I)*Cos[c] - 3*Sin[c] + 4*Sin[c + 2*d*x] + Sin[3*c + 4*d*x]))/(12*a*d)

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fricas [A]  time = 0.59, size = 72, normalized size = 1.31 \[ \frac {16 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i}{3 \, {\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(16*I*e^(2*I*d*x + 2*I*c) + 4*I)/(a*d*e^(8*I*d*x + 8*I*c) + 4*a*d*e^(6*I*d*x + 6*I*c) + 6*a*d*e^(4*I*d*x +
4*I*c) + 4*a*d*e^(2*I*d*x + 2*I*c) + a*d)

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giac [A]  time = 0.56, size = 47, normalized size = 0.85 \[ -\frac {3 i \, \tan \left (d x + c\right )^{4} - 4 \, \tan \left (d x + c\right )^{3} + 6 i \, \tan \left (d x + c\right )^{2} - 12 \, \tan \left (d x + c\right )}{12 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*I*tan(d*x + c)^4 - 4*tan(d*x + c)^3 + 6*I*tan(d*x + c)^2 - 12*tan(d*x + c))/(a*d)

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maple [A]  time = 0.38, size = 47, normalized size = 0.85 \[ \frac {\tan \left (d x +c \right )-\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2}}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(a+I*a*tan(d*x+c)),x)

[Out]

1/d/a*(tan(d*x+c)-1/4*I*tan(d*x+c)^4+1/3*tan(d*x+c)^3-1/2*I*tan(d*x+c)^2)

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maxima [A]  time = 0.55, size = 47, normalized size = 0.85 \[ \frac {-3 i \, \tan \left (d x + c\right )^{4} + 4 \, \tan \left (d x + c\right )^{3} - 6 i \, \tan \left (d x + c\right )^{2} + 12 \, \tan \left (d x + c\right )}{12 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(-3*I*tan(d*x + c)^4 + 4*tan(d*x + c)^3 - 6*I*tan(d*x + c)^2 + 12*tan(d*x + c))/(a*d)

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mupad [B]  time = 3.34, size = 77, normalized size = 1.40 \[ \frac {\sin \left (c+d\,x\right )\,\left (12\,{\cos \left (c+d\,x\right )}^3-{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,6{}\mathrm {i}+4\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^2-{\sin \left (c+d\,x\right )}^3\,3{}\mathrm {i}\right )}{12\,a\,d\,{\cos \left (c+d\,x\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)),x)

[Out]

(sin(c + d*x)*(4*cos(c + d*x)*sin(c + d*x)^2 - cos(c + d*x)^2*sin(c + d*x)*6i + 12*cos(c + d*x)^3 - sin(c + d*
x)^3*3i))/(12*a*d*cos(c + d*x)^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \int \frac {\sec ^{6}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*Integral(sec(c + d*x)**6/(tan(c + d*x) - I), x)/a

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