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

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

[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]))

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Rubi [A]  time = 0.179456, antiderivative size = 95, normalized size of antiderivative = 1., 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 = {3522, 3487, 47, 63, 206} \[ \frac{i c^2 \sqrt{c-i c \tan (e+f x)}}{a f (c+i c \tan (e+f x))}-\frac{i c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{\sqrt{2} a f} \]

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 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 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 206

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

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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 1.31201, size = 114, normalized size = 1.2 \[ \frac{(-c \sin (e+f x)-i c \cos (e+f x)) \left (\sqrt{2} \sqrt{c} (\cos (e+f x)+i \sin (e+f x)) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )-2 \cos (e+f x) \sqrt{c-i c \tan (e+f x)}\right )}{2 a f} \]

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 = 0.033, size = 77, normalized size = 0.8 \begin{align*}{\frac{2\,i{c}^{2}}{fa} \left ( -{\frac{1}{-2\,c-2\,ic\tan \left ( fx+e \right ) }\sqrt{c-ic\tan \left ( fx+e \right ) }}-{\frac{\sqrt{2}}{4}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}} \right ) } \end{align*}

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)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 1.40429, size = 651, normalized size = 6.85 \begin{align*} \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{{\left (-2 i \, c^{2} + 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{{\left (-2 i \, c^{2} - 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}{\left (2 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + 2 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{align*}

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 + 2*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sq
rt(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 - 2*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqr
t(-c^3/(a^2*f^2)))*e^(-I*f*x - I*e)/(a*f)) + sqrt(2)*(2*I*c*e^(2*I*f*x + 2*I*e) + 2*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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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]

Exception raised: AttributeError

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

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)