3.10.64 \(\int (a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx\) [964]

Optimal. Leaf size=27 \[ \frac {2 i a (c-i c \tan (e+f x))^{3/2}}{3 f} \]

[Out]

2/3*I*a*(c-I*c*tan(f*x+e))^(3/2)/f

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Rubi [A]
time = 0.07, antiderivative size = 27, normalized size of antiderivative = 1.00, 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 = {3603, 3568, 32} \begin {gather*} \frac {2 i a (c-i c \tan (e+f x))^{3/2}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

(((2*I)/3)*a*(c - I*c*Tan[e + f*x])^(3/2))/f

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]

Rule 3568

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 3603

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]))

Rubi steps

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

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Mathematica [A]
time = 0.55, size = 54, normalized size = 2.00 \begin {gather*} \frac {2 a c \sec (e+f x) (i \cos (e)+\sin (e)) (\cos (f x)-i \sin (f x)) \sqrt {c-i c \tan (e+f x)}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2),x]

[Out]

(2*a*c*Sec[e + f*x]*(I*Cos[e] + Sin[e])*(Cos[f*x] - I*Sin[f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(3*f)

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Maple [A]
time = 0.16, size = 22, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {2 i a \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}\)	\(22\)
default	\(\frac {2 i a \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}\)	\(22\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/3*I*a*(c-I*c*tan(f*x+e))^(3/2)/f

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Maxima [A]
time = 0.28, size = 20, normalized size = 0.74 \begin {gather*} \frac {2 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a}{3 \, f} \end {gather*}

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))^(3/2),x, algorithm="maxima")

[Out]

2/3*I*(-I*c*tan(f*x + e) + c)^(3/2)*a/f

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
time = 0.88, size = 41, normalized size = 1.52 \begin {gather*} \frac {4 i \, \sqrt {2} a c \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

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))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 1.75, size = 44, normalized size = 1.63 \begin {gather*} \begin {cases} \frac {2 i a \left (- i c \tan {\left (e + f x \right )} + c\right )^{\frac {3}{2}}}{3 f} & \text {for}\: f \neq 0 \\x \left (i a \tan {\left (e \right )} + a\right ) \left (- i c \tan {\left (e \right )} + c\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \end {gather*}

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))**(3/2),x)

[Out]

Piecewise((2*I*a*(-I*c*tan(e + f*x) + c)**(3/2)/(3*f), Ne(f, 0)), (x*(I*a*tan(e) + a)*(-I*c*tan(e) + c)**(3/2)
, True))

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Giac [A]
time = 0.46, size = 20, normalized size = 0.74 \begin {gather*} \frac {2 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a}{3 \, f} \end {gather*}

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))^(3/2),x, algorithm="giac")

[Out]

2/3*I*(-I*c*tan(f*x + e) + c)^(3/2)*a/f

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Mupad [B]
time = 0.20, size = 47, normalized size = 1.74 \begin {gather*} \frac {\sqrt {2}\,a\,c\,\sqrt {\frac {c}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,4{}\mathrm {i}}{3\,\left (f+f\,{\mathrm {e}}^{e\,2{}\mathrm {i}}\,{\mathrm {e}}^{f\,x\,2{}\mathrm {i}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)*(c - c*tan(e + f*x)*1i)^(3/2),x)

[Out]

(2^(1/2)*a*c*(c/(exp(e*2i + f*x*2i) + 1))^(1/2)*4i)/(3*(f + f*exp(e*2i)*exp(f*x*2i)))

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