3.970 \(\int (a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=27 \[ \frac{2 i a (c-i c \tan (e+f x))^{5/2}}{5 f} \]

[Out]

(((2*I)/5)*a*(c - I*c*Tan[e + f*x])^(5/2))/f

________________________________________________________________________________________

Rubi [A]  time = 0.100823, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3522, 3487, 32} \[ \frac{2 i a (c-i c \tan (e+f x))^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(((2*I)/5)*a*(c - I*c*Tan[e + f*x])^(5/2))/f

Rule 3522

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int (a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx &=(a c) \int \sec ^2(e+f x) (c-i c \tan (e+f x))^{3/2} \, dx\\ &=\frac{(i a) \operatorname{Subst}\left (\int (c+x)^{3/2} \, dx,x,-i c \tan (e+f x)\right )}{f}\\ &=\frac{2 i a (c-i c \tan (e+f x))^{5/2}}{5 f}\\ \end{align*}

Mathematica [B]  time = 1.37617, size = 70, normalized size = 2.59 \[ \frac{2 a c^2 \sec ^2(e+f x) (\cos (f x)-i \sin (f x)) \sqrt{c-i c \tan (e+f x)} (\sin (2 e+f x)+i \cos (2 e+f x))}{5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(2*a*c^2*Sec[e + f*x]^2*(Cos[f*x] - I*Sin[f*x])*(I*Cos[2*e + f*x] + Sin[2*e + f*x])*Sqrt[c - I*c*Tan[e + f*x]]
)/(5*f)

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 22, normalized size = 0.8 \begin{align*}{\frac{{\frac{2\,i}{5}}a}{f} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

2/5*I*a*(c-I*c*tan(f*x+e))^(5/2)/f

________________________________________________________________________________________

Maxima [A]  time = 1.28123, size = 26, normalized size = 0.96 \begin{align*} \frac{2 i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} a}{5 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

2/5*I*(-I*c*tan(f*x + e) + c)^(5/2)*a/f

________________________________________________________________________________________

Fricas [B]  time = 1.42088, size = 147, normalized size = 5.44 \begin{align*} \frac{8 i \, \sqrt{2} a c^{2} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{5 \,{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)*(-I*c*tan(f*x + e) + c)^(5/2), x)