3.162 \(\int \frac {\sec ^7(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx\)

Optimal. Leaf size=70 \[ \frac {\tan ^5(c+d x)}{5 a d}+\frac {2 \tan ^3(c+d x)}{3 a d}+\frac {\tan (c+d x)}{a d}-\frac {i \sec ^6(c+d x)}{6 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3092, 3090, 3767, 2606, 30} \[ \frac {\tan ^5(c+d x)}{5 a d}+\frac {2 \tan ^3(c+d x)}{3 a d}+\frac {\tan (c+d x)}{a d}-\frac {i \sec ^6(c+d x)}{6 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7/(a*Cos[c + d*x] + I*a*Sin[c + d*x]),x]

[Out]

((-I/6)*Sec[c + d*x]^6)/(a*d) + Tan[c + d*x]/(a*d) + (2*Tan[c + d*x]^3)/(3*a*d) + Tan[c + d*x]^5/(5*a*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 3092

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[a^n*b^n, Int[Cos[c + d*x]^m/(b*Cos[c + d*x] + a*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, m},
x] && EqQ[a^2 + b^2, 0] && ILtQ[n, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {\sec ^7(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx &=-\frac {i \int \sec ^7(c+d x) (i a \cos (c+d x)+a \sin (c+d x)) \, dx}{a^2}\\ &=-\frac {i \int \left (i a \sec ^6(c+d x)+a \sec ^6(c+d x) \tan (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {i \int \sec ^6(c+d x) \tan (c+d x) \, dx}{a}+\frac {\int \sec ^6(c+d x) \, dx}{a}\\ &=-\frac {i \operatorname {Subst}\left (\int x^5 \, dx,x,\sec (c+d x)\right )}{a d}-\frac {\operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{a d}\\ &=-\frac {i \sec ^6(c+d x)}{6 a d}+\frac {\tan (c+d x)}{a d}+\frac {2 \tan ^3(c+d x)}{3 a d}+\frac {\tan ^5(c+d x)}{5 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.37, size = 67, normalized size = 0.96 \[ -\frac {i \sec (c) \sec ^6(c+d x) (10 \cos (c)-i (-15 \sin (c+2 d x)-6 \sin (3 c+4 d x)-\sin (5 c+6 d x)+10 \sin (c)))}{60 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7/(a*Cos[c + d*x] + I*a*Sin[c + d*x]),x]

[Out]

((-1/60*I)*Sec[c]*Sec[c + d*x]^6*(10*Cos[c] - I*(10*Sin[c] - 15*Sin[c + 2*d*x] - 6*Sin[3*c + 4*d*x] - Sin[5*c
+ 6*d*x])))/(a*d)

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 109, normalized size = 1.56 \[ \frac {240 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 96 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i}{15 \, {\left (a d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a*cos(d*x+c)+I*a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/15*(240*I*e^(4*I*d*x + 4*I*c) + 96*I*e^(2*I*d*x + 2*I*c) + 16*I)/(a*d*e^(12*I*d*x + 12*I*c) + 6*a*d*e^(10*I*
d*x + 10*I*c) + 15*a*d*e^(8*I*d*x + 8*I*c) + 20*a*d*e^(6*I*d*x + 6*I*c) + 15*a*d*e^(4*I*d*x + 4*I*c) + 6*a*d*e
^(2*I*d*x + 2*I*c) + a*d)

________________________________________________________________________________________

giac [A]  time = 0.22, size = 67, normalized size = 0.96 \[ -\frac {5 i \, \tan \left (d x + c\right )^{6} - 6 \, \tan \left (d x + c\right )^{5} + 15 i \, \tan \left (d x + c\right )^{4} - 20 \, \tan \left (d x + c\right )^{3} + 15 i \, \tan \left (d x + c\right )^{2} - 30 \, \tan \left (d x + c\right )}{30 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a*cos(d*x+c)+I*a*sin(d*x+c)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.23, size = 68, normalized size = 0.97 \[ \frac {\tan \left (d x +c \right )-\frac {i \left (\tan ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{2}+\frac {2 \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2}}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7/(a*cos(d*x+c)+I*a*sin(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.36, size = 313, normalized size = 4.47 \[ \frac {2 \, {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {15 i \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {78 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {50 i \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {78 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {35 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {15 i \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {15 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}\right )}}{15 \, {\left (a - \frac {6 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {20 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {6 \, a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7/(a*cos(d*x+c)+I*a*sin(d*x+c)),x, algorithm="maxima")

[Out]

2/15*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 15*I*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 35*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 78*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 50*I*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 78*sin(d*
x + c)^7/(cos(d*x + c) + 1)^7 + 35*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 15*I*sin(d*x + c)^10/(cos(d*x + c) +
1)^10 - 15*sin(d*x + c)^11/(cos(d*x + c) + 1)^11)/((a - 6*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a*sin(d*x
+ c)^4/(cos(d*x + c) + 1)^4 - 20*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a*sin(d*x + c)^8/(cos(d*x + c) +
1)^8 - 6*a*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + a*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*d)

________________________________________________________________________________________

mupad [B]  time = 1.89, size = 139, normalized size = 1.99 \[ -\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,15{}\mathrm {i}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+78\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,50{}\mathrm {i}-78\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,15{}\mathrm {i}-15\right )}{15\,a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^7*(a*cos(c + d*x) + a*sin(c + d*x)*1i)),x)

[Out]

-(2*tan(c/2 + (d*x)/2)*(tan(c/2 + (d*x)/2)*15i + 35*tan(c/2 + (d*x)/2)^2 - 78*tan(c/2 + (d*x)/2)^4 + tan(c/2 +
(d*x)/2)^5*50i + 78*tan(c/2 + (d*x)/2)^6 - 35*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^9*15i + 15*tan(c/2 +
(d*x)/2)^10 - 15))/(15*a*d*(tan(c/2 + (d*x)/2)^2 - 1)^6)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7/(a*cos(d*x+c)+I*a*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________