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

Optimal. Leaf size=69 \[ \frac {8 i a^2 \sec (c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3494, 3493} \[ \frac {8 i a^2 \sec (c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec (c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.32, size = 57, normalized size = 0.83 \[ -\frac {2 a (\cos (c)-i \sin (c)) (\tan (c+d x)-5 i) (\cos (d x)-i \sin (d x)) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(Cos[c] - I*Sin[c])*(Cos[d*x] - I*Sin[d*x])*(-5*I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(3*d)

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fricas [A]  time = 0.61, size = 53, normalized size = 0.77 \[ \frac {\sqrt {2} {\left (12 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(2)*(12*I*a*e^(2*I*d*x + 2*I*c) + 8*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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), x)

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maple [A]  time = 1.02, size = 71, normalized size = 1.03 \[ \frac {2 \left (4 i \left (\cos ^{2}\left (d x +c \right )\right )+4 \cos \left (d x +c \right ) \sin \left (d x +c \right )+i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{3 d \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d*(4*I*cos(d*x+c)^2+4*cos(d*x+c)*sin(d*x+c)+I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)*a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.49, size = 98, normalized size = 1.42 \[ \frac {2\,a\,\sqrt {\frac {a\,\left (2\,{\cos \left (c+d\,x\right )}^2+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (c+d\,x\right )}^2}}\,\left ({\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,8{}\mathrm {i}+{\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}^2\,2{}\mathrm {i}+\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )-5{}\mathrm {i}\right )}{3\,d\,{\cos \left (c+d\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*((a*(sin(2*c + 2*d*x)*1i + 2*cos(c + d*x)^2))/(2*cos(c + d*x)^2))^(1/2)*(sin(c + d*x) + sin(3*c + 3*d*x)
+ cos(c/2 + (d*x)/2)^2*8i + cos((3*c)/2 + (3*d*x)/2)^2*2i - 5i))/(3*d*cos(c + d*x)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \sec {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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), x)

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