\(\int \sec ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [307]

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {8 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d}+\frac {8 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d}-\frac {2 i (a+i a \tan (c+d x))^{15/2}}{15 a^5 d} \]

[Out]

-8/11*I*(a+I*a*tan(d*x+c))^(11/2)/a^3/d+8/13*I*(a+I*a*tan(d*x+c))^(13/2)/a^4/d-2/15*I*(a+I*a*tan(d*x+c))^(15/2
)/a^5/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 88, 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} \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{15/2}}{15 a^5 d}+\frac {8 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d}-\frac {8 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d} \]

[In]

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

[Out]

(((-8*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^3*d) + (((8*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^4*d) - (((
2*I)/15)*(a + I*a*Tan[c + d*x])^(15/2))/(a^5*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)^2 (a+x)^{9/2} \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {i \text {Subst}\left (\int \left (4 a^2 (a+x)^{9/2}-4 a (a+x)^{11/2}+(a+x)^{13/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {8 i (a+i a \tan (c+d x))^{11/2}}{11 a^3 d}+\frac {8 i (a+i a \tan (c+d x))^{13/2}}{13 a^4 d}-\frac {2 i (a+i a \tan (c+d x))^{15/2}}{15 a^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.69 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 a^2 (-i+\tan (c+d x))^5 \sqrt {a+i a \tan (c+d x)} \left (-263+374 i \tan (c+d x)+143 \tan ^2(c+d x)\right )}{2145 d} \]

[In]

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

[Out]

(-2*a^2*(-I + Tan[c + d*x])^5*Sqrt[a + I*a*Tan[c + d*x]]*(-263 + (374*I)*Tan[c + d*x] + 143*Tan[c + d*x]^2))/(
2145*d)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72

\[\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {15}{2}}}{15}+\frac {4 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}-\frac {4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}\right )}{d \,a^{5}}\]

[In]

int(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

2*I/d/a^5*(-1/15*(a+I*a*tan(d*x+c))^(15/2)+4/13*a*(a+I*a*tan(d*x+c))^(13/2)-4/11*a^2*(a+I*a*tan(d*x+c))^(11/2)
)

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. 152 vs. \(2 (64) = 128\).

Time = 0.25 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.73 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {256 \, \sqrt {2} {\left (8 i \, a^{2} e^{\left (15 i \, d x + 15 i \, c\right )} + 60 i \, a^{2} e^{\left (13 i \, d x + 13 i \, c\right )} + 195 i \, a^{2} e^{\left (11 i \, d x + 11 i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{2145 \, {\left (d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-256/2145*sqrt(2)*(8*I*a^2*e^(15*I*d*x + 15*I*c) + 60*I*a^2*e^(13*I*d*x + 13*I*c) + 195*I*a^2*e^(11*I*d*x + 11
*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(14*I*d*x + 14*I*c) + 7*d*e^(12*I*d*x + 12*I*c) + 21*d*e^(10*I*d
*x + 10*I*c) + 35*d*e^(8*I*d*x + 8*I*c) + 35*d*e^(6*I*d*x + 6*I*c) + 21*d*e^(4*I*d*x + 4*I*c) + 7*d*e^(2*I*d*x
+ 2*I*c) + d)

Sympy [F(-1)]

Timed out. \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**6*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.66 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 i \, {\left (143 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {15}{2}} - 660 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a + 780 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{2}\right )}}{2145 \, a^{5} d} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-2/2145*I*(143*(I*a*tan(d*x + c) + a)^(15/2) - 660*(I*a*tan(d*x + c) + a)^(13/2)*a + 780*(I*a*tan(d*x + c) + a
)^(11/2)*a^2)/(a^5*d)

Giac [F]

\[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{6} \,d x } \]

[In]

integrate(sec(d*x+c)^6*(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)^6, x)

Mupad [B] (verification not implemented)

Time = 12.64 (sec) , antiderivative size = 498, normalized size of antiderivative = 5.66 \[ \int \sec ^6(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}}\,2048{}\mathrm {i}}{2145\,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}}\,1024{}\mathrm {i}}{2145\,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}}\,256{}\mathrm {i}}{715\,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}}\,18176{}\mathrm {i}}{429\,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}}\,52736{}\mathrm {i}}{429\,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}}\,103936{}\mathrm {i}}{715\,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}}\,15616{}\mathrm {i}}{195\,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}}\,256{}\mathrm {i}}{15\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7} \]

[In]

int((a + a*tan(c + d*x)*1i)^(5/2)/cos(c + d*x)^6,x)

[Out]

(a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*18176i)/(429*d*(exp(c*2i + d*x*2
i) + 1)^3) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*1024i)/(2145*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)*256i)/(7
15*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)*2048i)/(2145*d) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*52736i)/(42
9*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
)*103936i)/(715*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)*15616i)/(195*d*(exp(c*2i + d*x*2i) + 1)^6) + (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(ex
p(c*2i + d*x*2i) + 1))^(1/2)*256i)/(15*d*(exp(c*2i + d*x*2i) + 1)^7)