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

Optimal. Leaf size=146 \[ \frac{i c^3 \sqrt{c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac{i c^2 \sqrt{c-i c \tan (e+f x)}}{8 a^2 f (c+i c \tan (e+f x))}-\frac{i c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a^2 f} \]

[Out]

((-I/8)*c^(3/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^2*f) + ((I/2)*c^3*Sqrt[c - I
*c*Tan[e + f*x]])/(a^2*f*(c + I*c*Tan[e + f*x])^2) - ((I/8)*c^2*Sqrt[c - I*c*Tan[e + f*x]])/(a^2*f*(c + I*c*Ta
n[e + f*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.191614, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3522, 3487, 47, 51, 63, 206} \[ \frac{i c^3 \sqrt{c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac{i c^2 \sqrt{c-i c \tan (e+f x)}}{8 a^2 f (c+i c \tan (e+f x))}-\frac{i c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/8)*c^(3/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^2*f) + ((I/2)*c^3*Sqrt[c - I
*c*Tan[e + f*x]])/(a^2*f*(c + I*c*Tan[e + f*x])^2) - ((I/8)*c^2*Sqrt[c - I*c*Tan[e + f*x]])/(a^2*f*(c + I*c*Ta
n[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 51

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] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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))^2} \, dx &=\frac{\int \cos ^4(e+f x) (c-i c \tan (e+f x))^{7/2} \, dx}{a^2 c^2}\\ &=\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+x}}{(c-x)^3} \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=\frac{i c^3 \sqrt{c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac{\left (i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^2 \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{4 a^2 f}\\ &=\frac{i c^3 \sqrt{c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac{i c^2 \sqrt{c-i c \tan (e+f x)}}{8 a^2 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 )}{16 a^2 f}\\ &=\frac{i c^3 \sqrt{c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac{i c^2 \sqrt{c-i c \tan (e+f x)}}{8 a^2 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 )}{8 a^2 f}\\ &=-\frac{i c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a^2 f}+\frac{i c^3 \sqrt{c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac{i c^2 \sqrt{c-i c \tan (e+f x)}}{8 a^2 f (c+i c \tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 2.0655, size = 136, normalized size = 0.93 \[ \frac{c (\cos (2 (e+f x))-i \sin (2 (e+f x))) \left (\sqrt{c-i c \tan (e+f x)} (\sin (2 (e+f x))+3 i \cos (2 (e+f x))+3 i)+\sqrt{2} \sqrt{c} (\sin (2 (e+f x))-i \cos (2 (e+f x))) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )\right )}{16 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)])*(Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c
])]*((-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x)]) + (3*I + (3*I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x)])*Sqrt[c - I*
c*Tan[e + f*x]]))/(16*a^2*f)

________________________________________________________________________________________

Maple [A]  time = 0.033, size = 98, normalized size = 0.7 \begin{align*}{\frac{-2\,i{c}^{3}}{f{a}^{2}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}} \left ( -{\frac{1}{16\,c} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{1}{8}\sqrt{c-ic\tan \left ( fx+e \right ) }} \right ) }+{\frac{\sqrt{2}}{32}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){c}^{-{\frac{3}{2}}}} \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))^2,x)

[Out]

-2*I/f/a^2*c^3*((-1/16/c*(c-I*c*tan(f*x+e))^(3/2)-1/8*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^2+1/32/c^(
3/2)*2^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))

________________________________________________________________________________________

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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.36099, size = 765, normalized size = 5.24 \begin{align*} -\frac{{\left (\sqrt{\frac{1}{2}} a^{2} f \sqrt{-\frac{c^{3}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{c^{3}}{a^{4} f^{2}}} + i \, c^{2}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2} f}\right ) - \sqrt{\frac{1}{2}} a^{2} f \sqrt{-\frac{c^{3}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{c^{3}}{a^{4} f^{2}}} - i \, c^{2}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2} f}\right ) - \sqrt{2}{\left (i \, c e^{\left (4 i \, f x + 4 i \, e\right )} + 3 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 (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

-1/16*(sqrt(1/2)*a^2*f*sqrt(-c^3/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/4*(sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x
+ 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c^3/(a^4*f^2)) + I*c^2)*e^(-I*f*x - I*e)/(a^2*f)) -
sqrt(1/2)*a^2*f*sqrt(-c^3/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/4*(sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e)
+ a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c^3/(a^4*f^2)) - I*c^2)*e^(-I*f*x - I*e)/(a^2*f)) - sqrt(2)*
(I*c*e^(4*I*f*x + 4*I*e) + 3*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^(-4*I*f*x -
4*I*e)/(a^2*f)

________________________________________________________________________________________

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))**2,x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

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}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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