\(\int \sec ^8(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [293]

Optimal result

Integrand size = 26, antiderivative size = 117 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {16 i (a+i a \tan (c+d x))^{11/2}}{11 a^4 d}+\frac {24 i (a+i a \tan (c+d x))^{13/2}}{13 a^5 d}-\frac {4 i (a+i a \tan (c+d x))^{15/2}}{5 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{17/2}}{17 a^7 d} \]

[Out]

-16/11*I*(a+I*a*tan(d*x+c))^(11/2)/a^4/d+24/13*I*(a+I*a*tan(d*x+c))^(13/2)/a^5/d-4/5*I*(a+I*a*tan(d*x+c))^(15/
2)/a^6/d+2/17*I*(a+I*a*tan(d*x+c))^(17/2)/a^7/d

Rubi [A] (verified)

Time = 0.10 (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} \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 i (a+i a \tan (c+d x))^{17/2}}{17 a^7 d}-\frac {4 i (a+i a \tan (c+d x))^{15/2}}{5 a^6 d}+\frac {24 i (a+i a \tan (c+d x))^{13/2}}{13 a^5 d}-\frac {16 i (a+i a \tan (c+d x))^{11/2}}{11 a^4 d} \]

[In]

Int[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((-16*I)/11)*(a + I*a*Tan[c + d*x])^(11/2))/(a^4*d) + (((24*I)/13)*(a + I*a*Tan[c + d*x])^(13/2))/(a^5*d) - (
((4*I)/5)*(a + I*a*Tan[c + d*x])^(15/2))/(a^6*d) + (((2*I)/17)*(a + I*a*Tan[c + d*x])^(17/2))/(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)^{9/2} \, dx,x,i a \tan (c+d x)\right )}{a^7 d} \\ & = -\frac {i \text {Subst}\left (\int \left (8 a^3 (a+x)^{9/2}-12 a^2 (a+x)^{11/2}+6 a (a+x)^{13/2}-(a+x)^{15/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d} \\ & = -\frac {16 i (a+i a \tan (c+d x))^{11/2}}{11 a^4 d}+\frac {24 i (a+i a \tan (c+d x))^{13/2}}{13 a^5 d}-\frac {4 i (a+i a \tan (c+d x))^{15/2}}{5 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{17/2}}{17 a^7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 a (1+i \tan (c+d x))^5 \sqrt {a+i a \tan (c+d x)} \left (-1767 i-3641 \tan (c+d x)+2717 i \tan ^2(c+d x)+715 \tan ^3(c+d x)\right )}{12155 d} \]

[In]

Integrate[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(2*a*(1 + I*Tan[c + d*x])^5*Sqrt[a + I*a*Tan[c + d*x]]*(-1767*I - 3641*Tan[c + d*x] + (2717*I)*Tan[c + d*x]^2
+ 715*Tan[c + d*x]^3))/(12155*d)

Maple [A] (verified)

Time = 1.25 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {17}{2}}}{17}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {15}{2}}}{5}+\frac {12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}-\frac {8 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}\right )}{d \,a^{7}}\)	\(82\)
default	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {17}{2}}}{17}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {15}{2}}}{5}+\frac {12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}-\frac {8 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}\right )}{d \,a^{7}}\)	\(82\)

[In]

int(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2*I/d/a^7*(1/17*(a+I*a*tan(d*x+c))^(17/2)-2/5*a*(a+I*a*tan(d*x+c))^(15/2)+12/13*a^2*(a+I*a*tan(d*x+c))^(13/2)-
8/11*a^3*(a+I*a*tan(d*x+c))^(11/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.45 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {512 \, \sqrt {2} {\left (16 i \, a e^{\left (17 i \, d x + 17 i \, c\right )} + 136 i \, a e^{\left (15 i \, d x + 15 i \, c\right )} + 510 i \, a e^{\left (13 i \, d x + 13 i \, c\right )} + 1105 i \, a e^{\left (11 i \, d x + 11 i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12155 \, {\left (d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-512/12155*sqrt(2)*(16*I*a*e^(17*I*d*x + 17*I*c) + 136*I*a*e^(15*I*d*x + 15*I*c) + 510*I*a*e^(13*I*d*x + 13*I*
c) + 1105*I*a*e^(11*I*d*x + 11*I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(16*I*d*x + 16*I*c) + 8*d*e^(14*I*
d*x + 14*I*c) + 28*d*e^(12*I*d*x + 12*I*c) + 56*d*e^(10*I*d*x + 10*I*c) + 70*d*e^(8*I*d*x + 8*I*c) + 56*d*e^(6
*I*d*x + 6*I*c) + 28*d*e^(4*I*d*x + 4*I*c) + 8*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F(-1)]

Timed out. \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**8*(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.65 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 i \, {\left (715 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {17}{2}} - 4862 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {15}{2}} a + 11220 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} a^{2} - 8840 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{3}\right )}}{12155 \, a^{7} d} \]

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

2/12155*I*(715*(I*a*tan(d*x + c) + a)^(17/2) - 4862*(I*a*tan(d*x + c) + a)^(15/2)*a + 11220*(I*a*tan(d*x + c)
+ a)^(13/2)*a^2 - 8840*(I*a*tan(d*x + c) + a)^(11/2)*a^3)/(a^7*d)

Giac [F]

\[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{8} \,d x } \]

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)*sec(d*x + c)^8, x)

Mupad [B] (verification not implemented)

Time = 17.49 (sec) , antiderivative size = 544, normalized size of antiderivative = 4.65 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {a\,\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}}{12155\,d}-\frac {a\,\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}}\,4096{}\mathrm {i}}{12155\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {a\,\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}}\,3072{}\mathrm {i}}{12155\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {a\,\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}}\,512{}\mathrm {i}}{2431\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {a\,\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}}\,155136{}\mathrm {i}}{2431\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {a\,\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}}\,2413568{}\mathrm {i}}{12155\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}+\frac {a\,\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}}\,270336{}\mathrm {i}}{1105\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6}-\frac {a\,\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}}\,11776{}\mathrm {i}}{85\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^7}+\frac {a\,\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}}\,512{}\mathrm {i}}{17\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^8} \]

[In]

int((a + a*tan(c + d*x)*1i)^(3/2)/cos(c + d*x)^8,x)

[Out]

(a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*155136i)/(2431*d*(exp(c*2i + d*x*2
i) + 1)^4) - (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*4096i)/(12155*d*(exp(
c*2i + d*x*2i) + 1)) - (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*3072i)/(121
55*d*(exp(c*2i + d*x*2i) + 1)^2) - (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)
*512i)/(2431*d*(exp(c*2i + d*x*2i) + 1)^3) - (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) +
1))^(1/2)*8192i)/(12155*d) - (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*2413
568i)/(12155*d*(exp(c*2i + d*x*2i) + 1)^5) + (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) +
1))^(1/2)*270336i)/(1105*d*(exp(c*2i + d*x*2i) + 1)^6) - (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2
i + d*x*2i) + 1))^(1/2)*11776i)/(85*d*(exp(c*2i + d*x*2i) + 1)^7) + (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i
)/(exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/(17*d*(exp(c*2i + d*x*2i) + 1)^8)