3.11.3 \(\int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx\) [1003]

Optimal. Leaf size=43 \[ \frac {i (c-i c \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {3604, 37} \begin {gather*} \frac {i (c-i c \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 3604

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

Rubi steps

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

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Mathematica [A]
time = 1.20, size = 69, normalized size = 1.60 \begin {gather*} \frac {c (1-i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{3 a f (-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(1 - I*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(3*a*f*(-I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [A]
time = 0.45, size = 64, normalized size = 1.49

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan ^{2}\left (f x +e \right )\right )}{3 f \,a^{2} \left (-\tan \left (f x +e \right )+i\right )^{3}}\)	\(64\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan ^{2}\left (f x +e \right )\right )}{3 f \,a^{2} \left (-\tan \left (f x +e \right )+i\right )^{3}}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.54, size = 37, normalized size = 0.86 \begin {gather*} \frac {{\left (i \, c \cos \left (3 \, f x + 3 \, e\right ) + c \sin \left (3 \, f x + 3 \, e\right )\right )} \sqrt {c}}{3 \, a^{\frac {3}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(I*c*cos(3*f*x + 3*e) + c*sin(3*f*x + 3*e))*sqrt(c)/(a^(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. 71 vs. \(2 (33) = 66\).
time = 1.55, size = 71, normalized size = 1.65 \begin {gather*} \frac {{\left (i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-3 i \, f x - 3 i \, e\right )}}{3 \, a^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [B]
time = 5.52, size = 62, normalized size = 1.44 \begin {gather*} -\frac {c\,\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,\sqrt {-c\,\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}}{3\,a\,f\,\left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\sqrt {a\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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