\(\int \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [288]

Optimal result

Integrand size = 26, antiderivative size = 73 \[ \int \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {8 i a^2 \sec ^3(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^3(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

2/5*I*a*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)+8/15*I*a^2*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3575, 3574} \[ \int \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {8 i a^2 \sec ^3(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^3(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

Int[Sec[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((8*I)/15)*a^2*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((2*I)/5)*a*Sec[c + d*x]^3)/(d*Sqrt[a + I*
a*Tan[c + d*x]])

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)}{5 d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{5} (4 a) \int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {8 i a^2 \sec ^3(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i a \sec ^3(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86 \[ \int \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 \sec (c+d x) (\cos (2 (c+d x))-i \sin (2 (c+d x))) (-7 i+3 \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{15 d} \]

[In]

Integrate[Sec[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(-2*Sec[c + d*x]*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)])*(-7*I + 3*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(
15*d)

Maple [A] (verified)

Time = 7.95 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {2 \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (8 i \cos \left (d x +c \right )+8 \sin \left (d x +c \right )-i \sec \left (d x +c \right )+3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d}\)	\(62\)

[In]

int(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/15/d*(a*(1+I*tan(d*x+c)))^(1/2)*(8*I*cos(d*x+c)+8*sin(d*x+c)-I*sec(d*x+c)+3*sec(d*x+c)*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.85 \[ \int \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {8 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-8/15*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-5*I*e^(2*I*d*x + 2*I*c) - 2*I)/(d*e^(4*I*d*x + 4*I*c) + 2*d*
e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

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

[In]

integrate(sec(d*x+c)**3*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*sec(c + d*x)**3, x)

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. 222 vs. \(2 (57) = 114\).

Time = 21.46 (sec) , antiderivative size = 222, normalized size of antiderivative = 3.04 \[ \int \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {8 \, {\left (5 i \, \sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) - 5 \, \sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) + 2 i \, \sqrt {2}\right )} \sqrt {a}}{15 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} {\left ({\left (\cos \left (4 \, d x + 4 \, c\right ) + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + i \, \sin \left (4 \, d x + 4 \, c\right ) + 2 i \, \sin \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + {\left (i \, \cos \left (4 \, d x + 4 \, c\right ) + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) - \sin \left (4 \, d x + 4 \, c\right ) - 2 \, \sin \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )} d} \]

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

8/15*(5*I*sqrt(2)*cos(2*d*x + 2*c) - 5*sqrt(2)*sin(2*d*x + 2*c) + 2*I*sqrt(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)*((cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + I*sin(4*d*x + 4*c
) + 2*I*sin(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (I*cos(4*d*x + 4*c) +
2*I*cos(2*d*x + 2*c) - sin(4*d*x + 4*c) - 2*sin(2*d*x + 2*c) + I)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c) + 1)))*d)

Giac [F]

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

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*tan(d*x + c) + a)*sec(d*x + c)^3, x)

Mupad [B] (verification not implemented)

Time = 6.65 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.21 \[ \int \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {8\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,5{}\mathrm {i}+2{}\mathrm {i}\right )\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}}{15\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]

[In]

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

[Out]

(8*exp(- c*1i - d*x*1i)*(exp(c*2i + d*x*2i)*5i + 2i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*
2i) + 1))^(1/2))/(15*d*(exp(c*2i + d*x*2i) + 1)^2)