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

Optimal. Leaf size=73 \[ \frac{8 i a^2 \sec ^3(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}+\frac{2 i a \sec ^3(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

(((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]])

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Rubi [A]  time = 0.107429, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3494, 3493} \[ \frac{8 i a^2 \sec ^3(c+d x)}{15 d (a+i a \tan (c+d x))^{3/2}}+\frac{2 i a \sec ^3(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[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 3494

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]

Rule 3493

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]

Rubi steps

\begin{align*} \int \sec ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx &=\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]  time = 0.217766, size = 63, normalized size = 0.86 \[ -\frac{2 (3 \tan (c+d x)-7 i) \sec (c+d x) \sqrt{a+i a \tan (c+d x)} (\cos (2 (c+d x))-i \sin (2 (c+d x)))}{15 d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.317, size = 87, normalized size = 1.2 \begin{align*}{\frac{16\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+16\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -2\,i\cos \left ( dx+c \right ) +6\,\sin \left ( dx+c \right ) }{15\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 116.376, size = 304, normalized size = 4.16 \begin{align*} -\frac{{\left (-600 i \, \sqrt{2} \cos \left (2 \, d x + 2 \, c\right ) + 600 \, \sqrt{2} \sin \left (2 \, d x + 2 \, c\right ) - 240 i \, \sqrt{2}\right )} \sqrt{a}}{{\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 (225 \, \cos \left (4 \, d x + 4 \, c\right ) + 450 \, \cos \left (2 \, d x + 2 \, c\right ) + 225 i \, \sin \left (4 \, d x + 4 \, c\right ) + 450 i \, \sin \left (2 \, d x + 2 \, c\right ) + 225\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 (-225 i \, \cos \left (4 \, d x + 4 \, c\right ) - 450 i \, \cos \left (2 \, d x + 2 \, c\right ) + 225 \, \sin \left (4 \, d x + 4 \, c\right ) + 450 \, \sin \left (2 \, d x + 2 \, c\right ) - 225 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-600*I*sqrt(2)*cos(2*d*x + 2*c) + 600*sqrt(2)*sin(2*d*x + 2*c) - 240*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)*((225*cos(4*d*x + 4*c) + 450*cos(2*d*x + 2*c) + 225*I*si
n(4*d*x + 4*c) + 450*I*sin(2*d*x + 2*c) + 225)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (-22
5*I*cos(4*d*x + 4*c) - 450*I*cos(2*d*x + 2*c) + 225*sin(4*d*x + 4*c) + 450*sin(2*d*x + 2*c) - 225*I)*sin(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*d)

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Fricas [A]  time = 2.33966, size = 227, normalized size = 3.11 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i\right )} e^{\left (i \, d x + i \, c\right )}}{15 \,{\left (d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(40*I*e^(2*I*d*x + 2*I*c) + 16*I)*e^(I*d*x + I*c)/(d*e^(5*I*d*x
+ 5*I*c) + 2*d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )} \sec ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(I*tan(c + d*x) + 1))*sec(c + d*x)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)