\(\int \sec ^7(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [286]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F(-1)]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 147 \[ \int \sec ^7(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {256 i a^4 \sec ^7(c+d x)}{3003 d (a+i a \tan (c+d x))^{7/2}}+\frac {64 i a^3 \sec ^7(c+d x)}{429 d (a+i a \tan (c+d x))^{5/2}}+\frac {24 i a^2 \sec ^7(c+d x)}{143 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^7(c+d x)}{13 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

2/13*I*a*sec(d*x+c)^7/d/(a+I*a*tan(d*x+c))^(1/2)+256/3003*I*a^4*sec(d*x+c)^7/d/(a+I*a*tan(d*x+c))^(7/2)+64/429
*I*a^3*sec(d*x+c)^7/d/(a+I*a*tan(d*x+c))^(5/2)+24/143*I*a^2*sec(d*x+c)^7/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3575, 3574} \[ \int \sec ^7(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {256 i a^4 \sec ^7(c+d x)}{3003 d (a+i a \tan (c+d x))^{7/2}}+\frac {64 i a^3 \sec ^7(c+d x)}{429 d (a+i a \tan (c+d x))^{5/2}}+\frac {24 i a^2 \sec ^7(c+d x)}{143 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^7(c+d x)}{13 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

Int[Sec[c + d*x]^7*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((256*I)/3003)*a^4*Sec[c + d*x]^7)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((64*I)/429)*a^3*Sec[c + d*x]^7)/(d*(a
 + I*a*Tan[c + d*x])^(5/2)) + (((24*I)/143)*a^2*Sec[c + d*x]^7)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((2*I)/13)
*a*Sec[c + d*x]^7)/(d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3574

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(
d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rule 3575

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a \sec ^7(c+d x)}{13 d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{13} (12 a) \int \frac {\sec ^7(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {24 i a^2 \sec ^7(c+d x)}{143 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^7(c+d x)}{13 d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{143} \left (96 a^2\right ) \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx \\ & = \frac {64 i a^3 \sec ^7(c+d x)}{429 d (a+i a \tan (c+d x))^{5/2}}+\frac {24 i a^2 \sec ^7(c+d x)}{143 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^7(c+d x)}{13 d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{429} \left (128 a^3\right ) \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx \\ & = \frac {256 i a^4 \sec ^7(c+d x)}{3003 d (a+i a \tan (c+d x))^{7/2}}+\frac {64 i a^3 \sec ^7(c+d x)}{429 d (a+i a \tan (c+d x))^{5/2}}+\frac {24 i a^2 \sec ^7(c+d x)}{143 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^7(c+d x)}{13 d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.65 \[ \int \sec ^7(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2 \sec ^6(c+d x) (390 \cos (c+d x)+445 \cos (3 (c+d x))+7 i (26 \sin (c+d x)+59 \sin (3 (c+d x)))) (i \cos (4 (c+d x))+\sin (4 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{3003 d} \]

[In]

Integrate[Sec[c + d*x]^7*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(2*Sec[c + d*x]^6*(390*Cos[c + d*x] + 445*Cos[3*(c + d*x)] + (7*I)*(26*Sin[c + d*x] + 59*Sin[3*(c + d*x)]))*(I
*Cos[4*(c + d*x)] + Sin[4*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(3003*d)

Maple [A] (verified)

Time = 8.63 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.79

method result size
default \(\frac {2 \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (1024 i \cos \left (d x +c \right )+1024 \sin \left (d x +c \right )-128 i \sec \left (d x +c \right )+384 \sec \left (d x +c \right ) \tan \left (d x +c \right )-40 i \left (\sec ^{3}\left (d x +c \right )\right )+280 \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )-21 i \left (\sec ^{5}\left (d x +c \right )\right )+231 \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )\right )}{3003 d}\) \(116\)

[In]

int(sec(d*x+c)^7*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3003/d*(a*(1+I*tan(d*x+c)))^(1/2)*(1024*I*cos(d*x+c)+1024*sin(d*x+c)-128*I*sec(d*x+c)+384*sec(d*x+c)*tan(d*x
+c)-40*I*sec(d*x+c)^3+280*tan(d*x+c)*sec(d*x+c)^3-21*I*sec(d*x+c)^5+231*tan(d*x+c)*sec(d*x+c)^5)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90 \[ \int \sec ^7(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {128 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-429 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 286 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 104 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i\right )}}{3003 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^7*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-128/3003*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-429*I*e^(6*I*d*x + 6*I*c) - 286*I*e^(4*I*d*x + 4*I*c) -
104*I*e^(2*I*d*x + 2*I*c) - 16*I)/(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)

Sympy [F]

\[ \int \sec ^7(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \sec ^{7}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**7*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*sec(c + d*x)**7, x)

Maxima [F(-1)]

Timed out. \[ \int \sec ^7(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^7*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

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

[In]

integrate(sec(d*x+c)^7*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*tan(d*x + c) + a)*sec(d*x + c)^7, x)

Mupad [B] (verification not implemented)

Time = 9.18 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.97 \[ \int \sec ^7(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{7\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}+\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,384{}\mathrm {i}}{11\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{13\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6} \]

[In]

int((a + a*tan(c + d*x)*1i)^(1/2)/cos(c + d*x)^7,x)

[Out]

(exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*128i)/(7*d*(exp
(c*2i + d*x*2i) + 1)^3) - (exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) +
 1))^(1/2)*128i)/(3*d*(exp(c*2i + d*x*2i) + 1)^4) + (exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)
*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*384i)/(11*d*(exp(c*2i + d*x*2i) + 1)^5) - (exp(- c*1i - d*x*1i)*(a - (a*(
exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*128i)/(13*d*(exp(c*2i + d*x*2i) + 1)^6)