3.62 \(\int \sec ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx\)

Optimal. Leaf size=27 \[ -\frac {i (a+i a \tan (c+d x))^6}{6 a d} \]

[Out]

-1/6*I*(a+I*a*tan(d*x+c))^6/a/d

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Rubi [A]  time = 0.04, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3487, 32} \[ -\frac {i (a+i a \tan (c+d x))^6}{6 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + I*a*Tan[c + d*x])^5,x]

[Out]

((-I/6)*(a + I*a*Tan[c + d*x])^6)/(a*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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

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Mathematica [B]  time = 1.91, size = 134, normalized size = 4.96 \[ \frac {a^5 \sec (c) \sec ^6(c+d x) (15 \sin (c+2 d x)-15 \sin (3 c+2 d x)+6 \sin (3 c+4 d x)-6 \sin (5 c+4 d x)+2 \sin (5 c+6 d x)+15 i \cos (c+2 d x)+15 i \cos (3 c+2 d x)+6 i \cos (3 c+4 d x)+6 i \cos (5 c+4 d x)-20 \sin (c)+20 i \cos (c))}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + I*a*Tan[c + d*x])^5,x]

[Out]

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

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fricas [B]  time = 0.58, size = 153, normalized size = 5.67 \[ \frac {192 i \, a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} + 480 i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 640 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 480 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 192 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + 32 i \, a^{5}}{3 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

1/3*(192*I*a^5*e^(10*I*d*x + 10*I*c) + 480*I*a^5*e^(8*I*d*x + 8*I*c) + 640*I*a^5*e^(6*I*d*x + 6*I*c) + 480*I*a
^5*e^(4*I*d*x + 4*I*c) + 192*I*a^5*e^(2*I*d*x + 2*I*c) + 32*I*a^5)/(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 [B]  time = 1.92, size = 82, normalized size = 3.04 \[ -\frac {-i \, a^{5} \tan \left (d x + c\right )^{6} - 6 \, a^{5} \tan \left (d x + c\right )^{5} + 15 i \, a^{5} \tan \left (d x + c\right )^{4} + 20 \, a^{5} \tan \left (d x + c\right )^{3} - 15 i \, a^{5} \tan \left (d x + c\right )^{2} - 6 \, a^{5} \tan \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.44, size = 115, normalized size = 4.26 \[ \frac {\frac {i a^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{6 \cos \left (d x +c \right )^{6}}+\frac {a^{5} \left (\sin ^{5}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{5}}-\frac {5 i a^{5} \left (\sin ^{4}\left (d x +c \right )\right )}{2 \cos \left (d x +c \right )^{4}}-\frac {10 a^{5} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {5 i a^{5}}{2 \cos \left (d x +c \right )^{2}}+a^{5} \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x)

[Out]

1/d*(1/6*I*a^5*sin(d*x+c)^6/cos(d*x+c)^6+a^5*sin(d*x+c)^5/cos(d*x+c)^5-5/2*I*a^5*sin(d*x+c)^4/cos(d*x+c)^4-10/
3*a^5*sin(d*x+c)^3/cos(d*x+c)^3+5/2*I*a^5/cos(d*x+c)^2+a^5*tan(d*x+c))

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maxima [A]  time = 0.41, size = 21, normalized size = 0.78 \[ -\frac {i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6}}{6 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/6*I*(I*a*tan(d*x + c) + a)^6/(a*d)

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mupad [B]  time = 3.26, size = 114, normalized size = 4.22 \[ \frac {a^5\,\sin \left (c+d\,x\right )\,\left (6\,{\cos \left (c+d\,x\right )}^5+{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,15{}\mathrm {i}-20\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,15{}\mathrm {i}+6\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^5\,1{}\mathrm {i}\right )}{6\,d\,{\cos \left (c+d\,x\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^5/cos(c + d*x)^2,x)

[Out]

(a^5*sin(c + d*x)*(6*cos(c + d*x)*sin(c + d*x)^4 + cos(c + d*x)^4*sin(c + d*x)*15i + 6*cos(c + d*x)^5 + sin(c
+ d*x)^5*1i - cos(c + d*x)^2*sin(c + d*x)^3*15i - 20*cos(c + d*x)^3*sin(c + d*x)^2))/(6*d*cos(c + d*x)^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a^{5} \left (\int \left (- i \sec ^{2}{\left (c + d x \right )}\right )\, dx + \int 5 \tan {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \left (- 10 \tan ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\right )\, dx + \int \tan ^{5}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 10 i \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \left (- 5 i \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+I*a*tan(d*x+c))**5,x)

[Out]

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

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