3.10.65 \(\int \frac {(c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx\) [965]

Optimal. Leaf size=95 \[ -\frac {i c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a f}+\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{a f (c+i c \tan (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3603, 3568, 43, 65, 212} \begin {gather*} \frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{a f (c+i c \tan (e+f x))}-\frac {i c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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

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Mathematica [A]
time = 0.91, size = 114, normalized size = 1.20 \begin {gather*} \frac {(-i c \cos (e+f x)-c \sin (e+f x)) \left (\sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (\cos (e+f x)+i \sin (e+f x))-2 \cos (e+f x) \sqrt {c-i c \tan (e+f x)}\right )}{2 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.32, size = 77, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {2 i c^{2} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{2 c +2 i c \tan \left (f x +e \right )}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\right )}{f a}\)	\(77\)
default	\(\frac {2 i c^{2} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{2 c +2 i c \tan \left (f x +e \right )}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\right )}{f a}\)	\(77\)

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)),x,method=_RETURNVERBOSE)

[Out]

2*I/f/a*c^2*(1/4*(c-I*c*tan(f*x+e))^(1/2)/(1/2*c+1/2*I*c*tan(f*x+e))-1/4*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*ta
n(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))

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Maxima [A]
time = 0.51, size = 110, normalized size = 1.16 \begin {gather*} \frac {i \, {\left (\frac {\sqrt {2} c^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a} - \frac {4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{3}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a - 2 \, a c}\right )}}{4 \, c 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)),x, algorithm="maxima")

[Out]

1/4*I*(sqrt(2)*c^(5/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f
*x + e) + c)))/a - 4*sqrt(-I*c*tan(f*x + e) + c)*c^3/((-I*c*tan(f*x + e) + c)*a - 2*a*c))/(c*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. 267 vs. \(2 (77) = 154\).
time = 0.71, size = 267, normalized size = 2.81 \begin {gather*} -\frac {{\left (\sqrt {2} a f \sqrt {-\frac {c^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, c^{2} + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{3}}{a^{2} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - \sqrt {2} a f \sqrt {-\frac {c^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, c^{2} - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{3}}{a^{2} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + 2 \, \sqrt {2} {\left (-i \, c e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a 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)),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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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)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.29, size = 83, normalized size = 0.87 \begin {gather*} \frac {\sqrt {2}\,{\left (-c\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{2\,a\,f}+\frac {c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{a\,f\,\left (c+c\,\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),x)

[Out]

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

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