3.985 \(\int \frac{1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=242 \[ \frac{105 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{256 \sqrt{2} a^3 c^{3/2} f}-\frac{105 i}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \]

[Out]

(((105*I)/256)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*c^(3/2)*f) - ((35*I)/128)/(
a^3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I/6)/(a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(3/2)) + ((3*
I)/16)/(a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)) + ((21*I)/64)/(a^3*f*(1 + I*Tan[e + f*x])*(
c - I*c*Tan[e + f*x])^(3/2)) - ((105*I)/256)/(a^3*c*f*Sqrt[c - I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.244592, antiderivative size = 242, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {3522, 3487, 51, 63, 206} \[ \frac{105 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{256 \sqrt{2} a^3 c^{3/2} f}-\frac{105 i}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

(((105*I)/256)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*c^(3/2)*f) - ((35*I)/128)/(
a^3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I/6)/(a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(3/2)) + ((3*
I)/16)/(a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)) + ((21*I)/64)/(a^3*f*(1 + I*Tan[e + f*x])*(
c - I*c*Tan[e + f*x])^(3/2)) - ((105*I)/256)/(a^3*c*f*Sqrt[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 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{1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac{\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{3/2} \, dx}{a^3 c^3}\\ &=\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^4 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{\left (3 i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^3 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{4 a^3 f}\\ &=\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{\left (21 i c^2\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^2 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^3 f}\\ &=\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{(105 i c) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 f}\\ &=-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{(105 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{256 a^3 f}\\ &=-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{105 i}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{(105 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{512 a^3 c f}\\ &=-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{105 i}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}+\frac{(105 i) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{256 a^3 c f}\\ &=\frac{105 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{256 \sqrt{2} a^3 c^{3/2} f}-\frac{35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac{3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{105 i}{256 a^3 c f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.70186, size = 160, normalized size = 0.66 \[ \frac{\sqrt{c-i c \tan (e+f x)} (\cos (e+f x)-i \sin (e+f x)) \left (258 \sin (e+f x)+282 \sin (3 (e+f x))+24 \sin (5 (e+f x))+172 i \cos (e+f x)-166 i \cos (3 (e+f x))-8 i \cos (5 (e+f x))+315 i e^{i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )\right )}{1536 a^3 c^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(3/2)),x]

[Out]

((Cos[e + f*x] - I*Sin[e + f*x])*((315*I)*E^(I*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2
*I)*(e + f*x))]] + (172*I)*Cos[e + f*x] - (166*I)*Cos[3*(e + f*x)] - (8*I)*Cos[5*(e + f*x)] + 258*Sin[e + f*x]
+ 282*Sin[3*(e + f*x)] + 24*Sin[5*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(1536*a^3*c^2*f)

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Maple [A]  time = 0.046, size = 159, normalized size = 0.7 \begin{align*}{\frac{2\,i{c}^{4}}{f{a}^{3}} \left ( -{\frac{1}{16\,{c}^{5}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ({\frac{41}{32} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{35\,c}{6} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{55\,{c}^{2}}{8}\sqrt{c-ic\tan \left ( fx+e \right ) }} \right ) }-{\frac{105\,\sqrt{2}}{64}{\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 ) }-{\frac{1}{8\,{c}^{5}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{1}{48\,{c}^{4}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f/a^3*c^4*(-1/16/c^5*((41/32*(c-I*c*tan(f*x+e))^(5/2)-35/6*(c-I*c*tan(f*x+e))^(3/2)*c+55/8*c^2*(c-I*c*tan(
f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3-105/64*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/
2)))-1/8/c^5/(c-I*c*tan(f*x+e))^(1/2)-1/48/c^4/(c-I*c*tan(f*x+e))^(3/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51662, size = 994, normalized size = 4.11 \begin{align*} \frac{{\left (315 i \, \sqrt{\frac{1}{2}} a^{3} c^{2} f \sqrt{\frac{1}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (13440 i \, a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + 13440 i \, a^{3} c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{6} c^{3} f^{2}}} + 13440 i\right )} e^{\left (-i \, f x - i \, e\right )}}{16384 \, a^{3} c f}\right ) - 315 i \, \sqrt{\frac{1}{2}} a^{3} c^{2} f \sqrt{\frac{1}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (-13440 i \, a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} - 13440 i \, a^{3} c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{6} c^{3} f^{2}}} + 13440 i\right )} e^{\left (-i \, f x - i \, e\right )}}{16384 \, a^{3} c f}\right ) + \sqrt{2} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (-16 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 224 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 43 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 215 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 58 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1536 \, a^{3} c^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/1536*(315*I*sqrt(1/2)*a^3*c^2*f*sqrt(1/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I*e)*log(1/16384*(sqrt(2)*sqrt(1/2)*(13
440*I*a^3*c*f*e^(2*I*f*x + 2*I*e) + 13440*I*a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c^3*f^2)) +
13440*I)*e^(-I*f*x - I*e)/(a^3*c*f)) - 315*I*sqrt(1/2)*a^3*c^2*f*sqrt(1/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I*e)*lo
g(1/16384*(sqrt(2)*sqrt(1/2)*(-13440*I*a^3*c*f*e^(2*I*f*x + 2*I*e) - 13440*I*a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I
*e) + 1))*sqrt(1/(a^6*c^3*f^2)) + 13440*I)*e^(-I*f*x - I*e)/(a^3*c*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) +
1))*(-16*I*e^(10*I*f*x + 10*I*e) - 224*I*e^(8*I*f*x + 8*I*e) - 43*I*e^(6*I*f*x + 6*I*e) + 215*I*e^(4*I*f*x +
4*I*e) + 58*I*e^(2*I*f*x + 2*I*e) + 8*I))*e^(-6*I*f*x - 6*I*e)/(a^3*c^2*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(1/(a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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