3.185 \(\int \frac {\sec ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=34 \[ \frac {i \tan ^4(c+d x) (-\cot (c+d x)+i)^4}{4 a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3088, 848, 37} \[ \frac {i \tan ^4(c+d x) (-\cot (c+d x)+i)^4}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^5 (i a+a x)^3} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\left (-\frac {i}{a}+\frac {x}{a}\right )^3}{x^5} \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {i (i-\cot (c+d x))^4 \tan ^4(c+d x)}{4 a^3 d}\\ \end {align*}

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Mathematica [B]  time = 0.47, size = 90, normalized size = 2.65 \[ -\frac {i \sec (c) \sec ^4(c+d x) (2 i \sin (c+2 d x)-2 i \sin (3 c+2 d x)+i \sin (3 c+4 d x)+2 \cos (c+2 d x)+2 \cos (3 c+2 d x)-3 i \sin (c)+3 \cos (c))}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]

[Out]

((-1/4*I)*Sec[c]*Sec[c + d*x]^4*(3*Cos[c] + 2*Cos[c + 2*d*x] + 2*Cos[3*c + 2*d*x] - (3*I)*Sin[c] + (2*I)*Sin[c
+ 2*d*x] - (2*I)*Sin[3*c + 2*d*x] + I*Sin[3*c + 4*d*x]))/(a^3*d)

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fricas [B]  time = 1.42, size = 69, normalized size = 2.03 \[ \frac {4 i}{a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.41, size = 47, normalized size = 1.38 \[ -\frac {-i \, \tan \left (d x + c\right )^{4} + 4 \, \tan \left (d x + c\right )^{3} + 6 i \, \tan \left (d x + c\right )^{2} - 4 \, \tan \left (d x + c\right )}{4 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.30, size = 47, normalized size = 1.38 \[ \frac {\tan \left (d x +c \right )+\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{4}-\left (\tan ^{3}\left (d x +c \right )\right )-\frac {3 i \left (\tan ^{2}\left (d x +c \right )\right )}{2}}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x)

[Out]

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

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maxima [B]  time = 0.38, size = 240, normalized size = 7.06 \[ \frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 i \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {8 i \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {7 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 i \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {\sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{{\left (a^{3} - \frac {4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*I*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 7*sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3 + 8*I*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 7*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 3*I*sin(d*x + c)^6/
(cos(d*x + c) + 1)^6 - sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/((a^3 - 4*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
+ 6*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^8/(
cos(d*x + c) + 1)^8)*d)

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mupad [B]  time = 0.88, size = 55, normalized size = 1.62 \[ \frac {{\sin \left (c+d\,x\right )}^2\,1{}\mathrm {i}-\frac {{\sin \left (2\,c+2\,d\,x\right )}^2\,7{}\mathrm {i}}{4}+\sin \left (4\,c+4\,d\,x\right )}{4\,a^3\,d\,{\left ({\sin \left (c+d\,x\right )}^2-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a*cos(c + d*x) + a*sin(c + d*x)*1i)^3),x)

[Out]

(sin(4*c + 4*d*x) - (sin(2*c + 2*d*x)^2*7i)/4 + sin(c + d*x)^2*1i)/(4*a^3*d*(sin(c + d*x)^2 - 1)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a*cos(d*x+c)+I*a*sin(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**5/(-I*sin(c + d*x)**3 - 3*sin(c + d*x)**2*cos(c + d*x) + 3*I*sin(c + d*x)*cos(c + d*x)*
*2 + cos(c + d*x)**3), x)/a**3

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