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]

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)

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Rubi [A]  time = 0.11, antiderivative size = 73, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.28, 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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fricas [A]  time = 0.51, size = 62, normalized size = 0.85 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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)/(d*e^(4*I*d*x + 4*I*c) + 2*d*
e^(2*I*d*x + 2*I*c) + d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i \, a \tan \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3}\,{d x} \]

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)

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maple [A]  time = 1.19, size = 87, normalized size = 1.19 \[ \frac {2 \left (8 i \left (\cos ^{3}\left (d x +c \right )\right )+8 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-i \cos \left (d x +c \right )+3 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{15 d \cos \left (d x +c \right )^{2}} \]

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 = 33.00, size = 225, normalized size = 3.08 \[ -\frac {{\left (-600 i \, \sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) + 600 \, \sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) - 240 i \, \sqrt {2}\right )} \sqrt {a}}{{\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} {\left ({\left (225 \, \cos \left (4 \, d x + 4 \, c\right ) + 450 \, \cos \left (2 \, d x + 2 \, c\right ) + 225 i \, \sin \left (4 \, d x + 4 \, c\right ) + 450 i \, \sin \left (2 \, d x + 2 \, c\right ) + 225\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) - {\left (-225 i \, \cos \left (4 \, d x + 4 \, c\right ) - 450 i \, \cos \left (2 \, d x + 2 \, c\right ) + 225 \, \sin \left (4 \, d x + 4 \, c\right ) + 450 \, \sin \left (2 \, d x + 2 \, c\right ) - 225 i\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )} d} \]

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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mupad [B]  time = 5.95, size = 88, normalized size = 1.21 \[ \frac {8\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,5{}\mathrm {i}+2{}\mathrm {i}\right )\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}}{15\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \sec ^{3}{\left (c + d x \right )}\, dx \]

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

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