\(\int \sec ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [294]

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {8 i (a+i a \tan (c+d x))^{9/2}}{9 a^3 d}+\frac {8 i (a+i a \tan (c+d x))^{11/2}}{11 a^4 d}-\frac {2 i (a+i a \tan (c+d x))^{13/2}}{13 a^5 d} \]

[Out]

-8/9*I*(a+I*a*tan(d*x+c))^(9/2)/a^3/d+8/11*I*(a+I*a*tan(d*x+c))^(11/2)/a^4/d-2/13*I*(a+I*a*tan(d*x+c))^(13/2)/
a^5/d

Rubi [A] (verified)

Time = 0.09 (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} \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{13/2}}{13 a^5 d}+\frac {8 i (a+i a \tan (c+d x))^{11/2}}{11 a^4 d}-\frac {8 i (a+i a \tan (c+d x))^{9/2}}{9 a^3 d} \]

[In]

Int[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.69 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 i a (-i+\tan (c+d x))^4 \sqrt {a+i a \tan (c+d x)} \left (-203+270 i \tan (c+d x)+99 \tan ^2(c+d x)\right )}{1287 d} \]

[In]

Integrate[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((2*I)/1287)*a*(-I + Tan[c + d*x])^4*Sqrt[a + I*a*Tan[c + d*x]]*(-203 + (270*I)*Tan[c + d*x] + 99*Tan[c + d*x
]^2))/d

Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}+\frac {4 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9}\right )}{d \,a^{5}}\)	\(63\)
default	\(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}+\frac {4 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9}\right )}{d \,a^{5}}\)	\(63\)

[In]

int(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2*I/d/a^5*(-1/13*(a+I*a*tan(d*x+c))^(13/2)+4/11*a*(a+I*a*tan(d*x+c))^(11/2)-4/9*a^2*(a+I*a*tan(d*x+c))^(9/2))

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

Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.52 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {128 \, \sqrt {2} {\left (8 i \, a e^{\left (13 i \, d x + 13 i \, c\right )} + 52 i \, a e^{\left (11 i \, d x + 11 i \, c\right )} + 143 i \, a e^{\left (9 i \, d x + 9 i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{1287 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-128/1287*sqrt(2)*(8*I*a*e^(13*I*d*x + 13*I*c) + 52*I*a*e^(11*I*d*x + 11*I*c) + 143*I*a*e^(9*I*d*x + 9*I*c))*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(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]

\[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \sec ^{6}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**6*(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)*sec(c + d*x)**6, x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.66 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2 i \, {\left (99 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 468 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a + 572 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{2}\right )}}{1287 \, a^{5} d} \]

[In]

integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-2/1287*I*(99*(I*a*tan(d*x + c) + a)^(13/2) - 468*(I*a*tan(d*x + c) + a)^(11/2)*a + 572*(I*a*tan(d*x + c) + a)
^(9/2)*a^2)/(a^5*d)

Giac [F]

\[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{6} \,d x } \]

[In]

integrate(sec(d*x+c)^6*(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)^6, x)

Mupad [B] (verification not implemented)

Time = 8.31 (sec) , antiderivative size = 420, normalized size of antiderivative = 4.77 \[ \int \sec ^6(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}}\,1024{}\mathrm {i}}{1287\,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}}\,512{}\mathrm {i}}{1287\,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}}\,128{}\mathrm {i}}{429\,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}}\,27136{}\mathrm {i}}{1287\,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}}\,58624{}\mathrm {i}}{1287\,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}}\,5120{}\mathrm {i}}{143\,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}}\,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)^(3/2)/cos(c + d*x)^6,x)

[Out]

(a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*27136i)/(1287*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)*512i)/(1287*d*(exp(c*2
i + 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)*128i)/(429*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)*1024i
)/(1287*d) - (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*58624i)/(1287*d*(exp(
c*2i + d*x*2i) + 1)^4) + (a*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*5120i)/(1
43*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)
*128i)/(13*d*(exp(c*2i + d*x*2i) + 1)^6)