3.10.82 \(\int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [982]

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

[Out]

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

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Rubi [A]
time = 0.08, 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}{3 f (c-i c \tan (e+f x))^{3/2}} \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)/(f*(c - I*c*Tan[e + f*x])^(3/2))

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]

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 \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^{3/2}} \, dx &=(a c) \int \frac {\sec ^2(e+f x)}{(c-i c \tan (e+f x))^{5/2}} \, dx\\ &=\frac {(i a) \text {Subst}\left (\int \frac {1}{(c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{f}\\ &=-\frac {2 i a}{3 f (c-i c \tan (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(72\) vs. \(2(27)=54\).
time = 0.80, size = 72, normalized size = 2.67 \begin {gather*} \frac {2 a \cos ^2(e+f x) (\cos (f x)-i \sin (f x)) (-i \cos (2 e+3 f x)+\sin (2 e+3 f x)) \sqrt {c-i c \tan (e+f x)}}{3 c^2 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*Cos[e + f*x]^2*(Cos[f*x] - I*Sin[f*x])*((-I)*Cos[2*e + 3*f*x] + Sin[2*e + 3*f*x])*Sqrt[c - I*c*Tan[e + f*
x]])/(3*c^2*f)

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Maple [A]
time = 0.20, size = 22, normalized size = 0.81

method	result	size
derivativedivides	\(-\frac {2 i a}{3 f \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\)	\(22\)
default	\(-\frac {2 i a}{3 f \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\)	\(22\)
risch	\(-\frac {i a \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {2}}{6 c \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(42\)

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

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Maxima [A]
time = 0.30, size = 20, normalized size = 0.74 \begin {gather*} -\frac {2 i \, a}{3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} 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*a/((-I*c*tan(f*x + e) + c)^(3/2)*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. 59 vs. \(2 (20) = 40\).
time = 1.55, size = 59, normalized size = 2.19 \begin {gather*} \frac {\sqrt {2} {\left (-i \, a e^{\left (4 i \, f x + 4 i \, e\right )} - 2 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{6 \, c^{2} 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="fricas")

[Out]

1/6*sqrt(2)*(-I*a*e^(4*I*f*x + 4*I*e) - 2*I*a*e^(2*I*f*x + 2*I*e) - I*a)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^
2*f)

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Sympy [A]
time = 7.32, size = 46, normalized size = 1.70 \begin {gather*} \begin {cases} - \frac {2 i a}{3 f \left (- i c \tan {\left (e + f x \right )} + c\right )^{\frac {3}{2}}} & \text {for}\: f \neq 0 \\\frac {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/(3*f*(-I*c*tan(e + f*x) + c)**(3/2)), Ne(f, 0)), (x*(I*a*tan(e) + a)/(-I*c*tan(e) + c)**(3/2
), True))

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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((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 5.09, size = 93, normalized size = 3.44 \begin {gather*} \frac {a\,\sqrt {-\frac {c\,\left (-2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}}\,\left (-{\cos \left (e+f\,x\right )}^2\,4{}\mathrm {i}-{\cos \left (2\,e+2\,f\,x\right )}^2\,2{}\mathrm {i}+2\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )+2{}\mathrm {i}\right )}{6\,c^2\,f} \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]

(a*(-(c*(sin(2*e + 2*f*x)*1i - 2*cos(e + f*x)^2))/(2*cos(e + f*x)^2))^(1/2)*(2*sin(2*e + 2*f*x) + sin(4*e + 4*
f*x) - cos(2*e + 2*f*x)^2*2i - cos(e + f*x)^2*4i + 2i))/(6*c^2*f)

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