3.4.2 \(\int \sec (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [302]

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

________________________________________________________________________________________

Rubi [A]
time = 0.05, 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 = {3575, 3574} \begin {gather*} \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 {gather*}

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 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*} \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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.37, size = 57, normalized size = 0.83 \begin {gather*} -\frac {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} \end {gather*}

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)

________________________________________________________________________________________

Maple [A]
time = 0.83, size = 71, normalized size = 1.03

method	result	size
default	\(\frac {2 \left (4 i \left (\cos ^{2}\left (d x +c \right )\right )+4 \sin \left (d x +c \right ) \cos \left (d x +c \right )+i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{3 d \cos \left (d x +c \right )}\)	\(71\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 53, normalized size = 0.77 \begin {gather*} -\frac {4 \, \sqrt {2} {\left (-3 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 2 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 )}} \end {gather*}

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]

-4/3*sqrt(2)*(-3*I*a*e^(2*I*d*x + 2*I*c) - 2*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(2*I*d*x + 2*I*c) + d
)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [B]
time = 4.49, size = 98, normalized size = 1.42 \begin {gather*} \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} \end {gather*}

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)

________________________________________________________________________________________