\(\int \sec ^8(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [306]

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

Optimal result

Integrand size = 26, antiderivative size = 117 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {16 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d}+\frac {8 i (a+i a \tan (c+d x))^{15/2}}{5 a^5 d}-\frac {12 i (a+i a \tan (c+d x))^{17/2}}{17 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{19/2}}{19 a^7 d} \]

[Out]

-16/13*I*(a+I*a*tan(d*x+c))^(13/2)/a^4/d+8/5*I*(a+I*a*tan(d*x+c))^(15/2)/a^5/d-12/17*I*(a+I*a*tan(d*x+c))^(17/
2)/a^6/d+2/19*I*(a+I*a*tan(d*x+c))^(19/2)/a^7/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 45} \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {2 i (a+i a \tan (c+d x))^{19/2}}{19 a^7 d}-\frac {12 i (a+i a \tan (c+d x))^{17/2}}{17 a^6 d}+\frac {8 i (a+i a \tan (c+d x))^{15/2}}{5 a^5 d}-\frac {16 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d} \]

[In]

Int[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((-16*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^4*d) + (((8*I)/5)*(a + I*a*Tan[c + d*x])^(15/2))/(a^5*d) - (((
12*I)/17)*(a + I*a*Tan[c + d*x])^(17/2))/(a^6*d) + (((2*I)/19)*(a + I*a*Tan[c + d*x])^(19/2))/(a^7*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

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*} \text {integral}& = -\frac {i \text {Subst}\left (\int (a-x)^3 (a+x)^{11/2} \, dx,x,i a \tan (c+d x)\right )}{a^7 d} \\ & = -\frac {i \text {Subst}\left (\int \left (8 a^3 (a+x)^{11/2}-12 a^2 (a+x)^{13/2}+6 a (a+x)^{15/2}-(a+x)^{17/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d} \\ & = -\frac {16 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d}+\frac {8 i (a+i a \tan (c+d x))^{15/2}}{5 a^5 d}-\frac {12 i (a+i a \tan (c+d x))^{17/2}}{17 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{19/2}}{19 a^7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 a^2 (-i+\tan (c+d x))^6 \sqrt {a+i a \tan (c+d x)} \left (-2429 i-5291 \tan (c+d x)+4095 i \tan ^2(c+d x)+1105 \tan ^3(c+d x)\right )}{20995 d} \]

[In]

Integrate[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(-2*a^2*(-I + Tan[c + d*x])^6*Sqrt[a + I*a*Tan[c + d*x]]*(-2429*I - 5291*Tan[c + d*x] + (4095*I)*Tan[c + d*x]^
2 + 1105*Tan[c + d*x]^3))/(20995*d)

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.70

\[\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {19}{2}}}{19}-\frac {6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {17}{2}}}{17}+\frac {4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {15}{2}}}{5}-\frac {8 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}\right )}{d \,a^{7}}\]

[In]

int(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

2*I/d/a^7*(1/19*(a+I*a*tan(d*x+c))^(19/2)-6/17*a*(a+I*a*tan(d*x+c))^(17/2)+4/5*a^2*(a+I*a*tan(d*x+c))^(15/2)-8
/13*a^3*(a+I*a*tan(d*x+c))^(13/2))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (85) = 170\).

Time = 0.29 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.62 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {1024 \, \sqrt {2} {\left (16 i \, a^{2} e^{\left (19 i \, d x + 19 i \, c\right )} + 152 i \, a^{2} e^{\left (17 i \, d x + 17 i \, c\right )} + 646 i \, a^{2} e^{\left (15 i \, d x + 15 i \, c\right )} + 1615 i \, a^{2} e^{\left (13 i \, d x + 13 i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{20995 \, {\left (d e^{\left (18 i \, d x + 18 i \, c\right )} + 9 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 36 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 84 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 126 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 126 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 84 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 9 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1024/20995*sqrt(2)*(16*I*a^2*e^(19*I*d*x + 19*I*c) + 152*I*a^2*e^(17*I*d*x + 17*I*c) + 646*I*a^2*e^(15*I*d*x
+ 15*I*c) + 1615*I*a^2*e^(13*I*d*x + 13*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(18*I*d*x + 18*I*c) + 9*d
*e^(16*I*d*x + 16*I*c) + 36*d*e^(14*I*d*x + 14*I*c) + 84*d*e^(12*I*d*x + 12*I*c) + 126*d*e^(10*I*d*x + 10*I*c)
 + 126*d*e^(8*I*d*x + 8*I*c) + 84*d*e^(6*I*d*x + 6*I*c) + 36*d*e^(4*I*d*x + 4*I*c) + 9*d*e^(2*I*d*x + 2*I*c) +
 d)

Sympy [F(-1)]

Timed out. \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**8*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.65 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {2 i \, {\left (1105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {19}{2}} - 7410 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {17}{2}} a + 16796 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {15}{2}} a^{2} - 12920 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a^{3}\right )}}{20995 \, a^{7} d} \]

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

2/20995*I*(1105*(I*a*tan(d*x + c) + a)^(19/2) - 7410*(I*a*tan(d*x + c) + a)^(17/2)*a + 16796*(I*a*tan(d*x + c)
 + a)^(15/2)*a^2 - 12920*(I*a*tan(d*x + c) + a)^(13/2)*a^3)/(a^7*d)

Giac [F]

\[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{8} \,d x } \]

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)*sec(d*x + c)^8, x)

Mupad [B] (verification not implemented)

Time = 16.99 (sec) , antiderivative size = 626, normalized size of antiderivative = 5.35 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {a^2\,\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}}\,16384{}\mathrm {i}}{20995\,d}-\frac {a^2\,\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}}\,8192{}\mathrm {i}}{20995\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {a^2\,\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}}\,6144{}\mathrm {i}}{20995\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {a^2\,\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}}\,1024{}\mathrm {i}}{4199\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {a^2\,\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}}\,536576{}\mathrm {i}}{4199\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {a^2\,\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}}\,10484736{}\mathrm {i}}{20995\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}+\frac {a^2\,\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}}\,17262592{}\mathrm {i}}{20995\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}-\frac {a^2\,\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}}\,1129472{}\mathrm {i}}{1615\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7}+\frac {a^2\,\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}}\,98304{}\mathrm {i}}{323\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^8}-\frac {a^2\,\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}}\,1024{}\mathrm {i}}{19\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^9} \]

[In]

int((a + a*tan(c + d*x)*1i)^(5/2)/cos(c + d*x)^8,x)

[Out]

(a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*536576i)/(4199*d*(exp(c*2i + d*x
*2i) + 1)^4) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*8192i)/(20995*d*(
exp(c*2i + d*x*2i) + 1)) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*6144i
)/(20995*d*(exp(c*2i + d*x*2i) + 1)^2) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1
))^(1/2)*1024i)/(4199*d*(exp(c*2i + d*x*2i) + 1)^3) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i
+ d*x*2i) + 1))^(1/2)*16384i)/(20995*d) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) +
1))^(1/2)*10484736i)/(20995*d*(exp(c*2i + d*x*2i) + 1)^5) + (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp
(c*2i + d*x*2i) + 1))^(1/2)*17262592i)/(20995*d*(exp(c*2i + d*x*2i) + 1)^6) - (a^2*(a - (a*(exp(c*2i + d*x*2i)
*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*1129472i)/(1615*d*(exp(c*2i + d*x*2i) + 1)^7) + (a^2*(a - (a*(ex
p(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*98304i)/(323*d*(exp(c*2i + d*x*2i) + 1)^8) - (a^
2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*1024i)/(19*d*(exp(c*2i + d*x*2i) +
1)^9)