\(\int \frac {\sec ^7(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx\) [162]

Optimal result

Integrand size = 31, antiderivative size = 70 \[ \int \frac {\sec ^7(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {i \sec ^6(c+d x)}{6 a d}+\frac {\tan (c+d x)}{a d}+\frac {2 \tan ^3(c+d x)}{3 a d}+\frac {\tan ^5(c+d x)}{5 a d} \]

[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] (verified)

Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3171, 3169, 3852, 2686, 30} \[ \int \frac {\sec ^7(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {\tan ^5(c+d x)}{5 a d}+\frac {2 \tan ^3(c+d x)}{3 a d}+\frac {\tan (c+d x)}{a d}-\frac {i \sec ^6(c+d x)}{6 a d} \]

[In]

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

[Out]

((-1/6*I)*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 2686

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 3169

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 3171

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 3852

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*} \text {integral}& = -\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 \text {Subst}\left (\int x^5 \, dx,x,\sec (c+d x)\right )}{a d}-\frac {\text {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] (verified)

Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^7(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {i \tan (c+d x) \left (30 i+15 \tan (c+d x)+20 i \tan ^2(c+d x)+15 \tan ^3(c+d x)+6 i \tan ^4(c+d x)+5 \tan ^5(c+d x)\right )}{30 a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {16 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{15 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\)	\(47\)
derivativedivides	\(-\frac {i \left (\frac {\tan \left (d x +c \right )^{6}}{6}+\frac {\tan \left (d x +c \right )^{4}}{2}+\frac {i \tan \left (d x +c \right )^{5}}{5}+\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {2 i \tan \left (d x +c \right )^{3}}{3}+i \tan \left (d x +c \right )\right )}{d a}\)	\(72\)
default	\(-\frac {i \left (\frac {\tan \left (d x +c \right )^{6}}{6}+\frac {\tan \left (d x +c \right )^{4}}{2}+\frac {i \tan \left (d x +c \right )^{5}}{5}+\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {2 i \tan \left (d x +c \right )^{3}}{3}+i \tan \left (d x +c \right )\right )}{d a}\)	\(72\)
norman	\(\frac {\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {52 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 a d}-\frac {52 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 a d}+\frac {14 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 a d}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{a d}-\frac {2 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a d}-\frac {20 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 a d}-\frac {2 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{6}}\)	\(190\)

[In]

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

[Out]

16/15*I*(15*exp(4*I*(d*x+c))+6*exp(2*I*(d*x+c))+1)/d/a/(exp(2*I*(d*x+c))+1)^6

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.56 \[ \int \frac {\sec ^7(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {16 \, {\left (-15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{15 \, {\left (a d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

[In]

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

[Out]

-16/15*(-15*I*e^(4*I*d*x + 4*I*c) - 6*I*e^(2*I*d*x + 2*I*c) - 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)

Sympy [F]

\[ \int \frac {\sec ^7(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {\int \frac {\sec ^{7}{\left (c + d x \right )}}{i \sin {\left (c + d x \right )} + \cos {\left (c + d x \right )}}\, dx}{a} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (62) = 124\).

Time = 0.22 (sec) , antiderivative size = 313, normalized size of antiderivative = 4.47 \[ \int \frac {\sec ^7(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {2 \, {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {15 i \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {78 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {50 i \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {78 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {35 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {15 i \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {15 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}\right )}}{15 \, {\left (a - \frac {6 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {20 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {6 \, a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^7(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {5 i \, \tan \left (d x + c\right )^{6} - 6 \, \tan \left (d x + c\right )^{5} + 15 i \, \tan \left (d x + c\right )^{4} - 20 \, \tan \left (d x + c\right )^{3} + 15 i \, \tan \left (d x + c\right )^{2} - 30 \, \tan \left (d x + c\right )}{30 \, a d} \]

[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)

Mupad [B] (verification not implemented)

Time = 24.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.99 \[ \int \frac {\sec ^7(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,15{}\mathrm {i}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+78\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,50{}\mathrm {i}-78\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,15{}\mathrm {i}-15\right )}{15\,a\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}^6} \]

[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)