\(\int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [313]

Optimal result

Integrand size = 26, antiderivative size = 147 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {256 i a^4 \sec ^3(c+d x)}{315 d (a+i a \tan (c+d x))^{3/2}}+\frac {64 i a^3 \sec ^3(c+d x)}{105 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i a^2 \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \]

[Out]

64/105*I*a^3*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)+8/21*I*a^2*sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/d+256/31
5*I*a^4*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(3/2)+2/9*I*a*sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2)/d

Rubi [A] (verified)

Time = 0.32 (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} \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {256 i a^4 \sec ^3(c+d x)}{315 d (a+i a \tan (c+d x))^{3/2}}+\frac {64 i a^3 \sec ^3(c+d x)}{105 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i a^2 \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \]

[In]

Int[Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((256*I)/315)*a^4*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((64*I)/105)*a^3*Sec[c + d*x]^3)/(d*Sqr
t[a + I*a*Tan[c + d*x]]) + (((8*I)/21)*a^2*Sec[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d + (((2*I)/9)*a*Sec[c +
d*x]^3*(a + I*a*Tan[c + d*x])^(3/2))/d

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 ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac {1}{3} (4 a) \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx \\ & = \frac {8 i a^2 \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac {1}{21} \left (32 a^2\right ) \int \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx \\ & = \frac {64 i a^3 \sec ^3(c+d x)}{105 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i a^2 \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d}+\frac {1}{105} \left (128 a^3\right ) \int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {256 i a^4 \sec ^3(c+d x)}{315 d (a+i a \tan (c+d x))^{3/2}}+\frac {64 i a^3 \sec ^3(c+d x)}{105 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i a^2 \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {2 i a \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.70 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {2 a^2 \sec ^3(c+d x) (i \cos (2 c)+\sin (2 c)) (77+242 \cos (2 (c+d x))+89 i \sec (c+d x) \sin (3 (c+d x))+54 i \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{315 d (\cos (d x)+i \sin (d x))^2} \]

[In]

Integrate[Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(2*a^2*Sec[c + d*x]^3*(I*Cos[2*c] + Sin[2*c])*(77 + 242*Cos[2*(c + d*x)] + (89*I)*Sec[c + d*x]*Sin[3*(c + d*x)
] + (54*I)*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*(Cos[d*x] + I*Sin[d*x])^2)

Maple [A] (verified)

Time = 41.25 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.52

method	result	size
default	\(\frac {2 i \left (-\tan \left (d x +c \right )+i\right )^{2} a^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (256 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+128 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-256 \left (\cos ^{3}\left (d x +c \right )\right )+96 i \sin \left (d x +c \right )-128 \left (\cos ^{2}\left (d x +c \right )\right )-130 i \tan \left (d x +c \right )+32 \cos \left (d x +c \right )-35 i \tan \left (d x +c \right ) \sec \left (d x +c \right )-226-95 \sec \left (d x +c \right )+35 \left (\sec ^{2}\left (d x +c \right )\right )\right )}{315 d \left (4 \left (\cos ^{3}\left (d x +c \right )\right )+2 \left (\cos ^{2}\left (d x +c \right )\right )+4 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-3 \cos \left (d x +c \right )+2 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-1-i \sin \left (d x +c \right )\right )}\)	\(224\)

[In]

int(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/315*I/d*(-tan(d*x+c)+I)^2*a^2*(a*(1+I*tan(d*x+c)))^(1/2)/(4*cos(d*x+c)^3+2*cos(d*x+c)^2+4*I*cos(d*x+c)^2*sin
(d*x+c)-3*cos(d*x+c)+2*I*cos(d*x+c)*sin(d*x+c)-1-I*sin(d*x+c))*(256*I*cos(d*x+c)^2*sin(d*x+c)+128*I*cos(d*x+c)
*sin(d*x+c)-256*cos(d*x+c)^3+96*I*sin(d*x+c)-128*cos(d*x+c)^2-130*I*tan(d*x+c)+32*cos(d*x+c)-35*I*tan(d*x+c)*s
ec(d*x+c)-226-95*sec(d*x+c)+35*sec(d*x+c)^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.82 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {32 \, \sqrt {2} {\left (-105 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 126 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 72 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-32/315*sqrt(2)*(-105*I*a^2*e^(6*I*d*x + 6*I*c) - 126*I*a^2*e^(4*I*d*x + 4*I*c) - 72*I*a^2*e^(2*I*d*x + 2*I*c)
- 16*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d
*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**3*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 624 vs. \(2 (115) = 230\).

Time = 243.69 (sec) , antiderivative size = 624, normalized size of antiderivative = 4.24 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {32 \, {\left (105 i \, \sqrt {2} a^{2} \cos \left (6 \, d x + 6 \, c\right ) + 126 i \, \sqrt {2} a^{2} \cos \left (4 \, d x + 4 \, c\right ) + 72 i \, \sqrt {2} a^{2} \cos \left (2 \, d x + 2 \, c\right ) - 105 \, \sqrt {2} a^{2} \sin \left (6 \, d x + 6 \, c\right ) - 126 \, \sqrt {2} a^{2} \sin \left (4 \, d x + 4 \, c\right ) - 72 \, \sqrt {2} a^{2} \sin \left (2 \, d x + 2 \, c\right ) + 16 i \, \sqrt {2} a^{2}\right )} \sqrt {a}}{315 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} {\left ({\left (2 \, \cos \left (2 \, d x + 2 \, c\right )^{3} + {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \sin \left (2 \, d x + 2 \, c\right )^{3} + {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 5 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + {\left (2 i \, \cos \left (2 \, d x + 2 \, c\right )^{3} + {\left (2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \cos \left (4 \, d x + 4 \, c\right ) + 5 i \, \cos \left (2 \, d x + 2 \, c\right )^{2} - {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 2 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right ) + 4 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )} d} \]

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

32/315*(105*I*sqrt(2)*a^2*cos(6*d*x + 6*c) + 126*I*sqrt(2)*a^2*cos(4*d*x + 4*c) + 72*I*sqrt(2)*a^2*cos(2*d*x +
2*c) - 105*sqrt(2)*a^2*sin(6*d*x + 6*c) - 126*sqrt(2)*a^2*sin(4*d*x + 4*c) - 72*sqrt(2)*a^2*sin(2*d*x + 2*c)
+ 16*I*sqrt(2)*a^2)*sqrt(a)/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((2*cos(
2*d*x + 2*c)^3 + (2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 2*I*sin(2*d*x + 2*c)^3 + (cos(2*d*x + 2*c)^2 +
sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 5*cos(2*d*x + 2*c)^2 + (I*cos(2*d*x + 2*c)^2 +
I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin(4*d*x + 4*c) + 2*(I*cos(2*d*x + 2*c)^2 + 2*I*cos(2*d*x +
2*c) + I)*sin(2*d*x + 2*c) + 4*cos(2*d*x + 2*c) + 1)*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))
+ (2*I*cos(2*d*x + 2*c)^3 + (2*I*cos(2*d*x + 2*c) + I)*sin(2*d*x + 2*c)^2 - 2*sin(2*d*x + 2*c)^3 + (I*cos(2*d
*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*cos(4*d*x + 4*c) + 5*I*cos(2*d*x + 2*c)^2 - (co
s(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c) - 2*(cos(2*d*x + 2*c)^2 + 2*c
os(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c) + 4*I*cos(2*d*x + 2*c) + I)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c) + 1)))*d)

Giac [F]

\[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(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)^3, x)

Mupad [B] (verification not implemented)

Time = 8.78 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.05 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{3\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,96{}\mathrm {i}}{5\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,96{}\mathrm {i}}{7\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {a^2\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{9\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4} \]

[In]

int((a + a*tan(c + d*x)*1i)^(5/2)/cos(c + d*x)^3,x)

[Out]

(a^2*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)*32i)/(3*d*(
exp(c*2i + d*x*2i) + 1)) - (a^2*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)*96i)/(5*d*(exp(c*2i + d*x*2i) + 1)^2) + (a^2*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)*96i)/(7*d*(exp(c*2i + d*x*2i) + 1)^3) - (a^2*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)*32i)/(9*d*(exp(c*2i + d*x*2i) + 1)^4
)