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

Optimal. Leaf size=199 \[ \frac{35 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{64 \sqrt{2} a^2 c^{3/2} f}-\frac{35 i}{64 a^2 c f \sqrt{c-i c \tan (e+f x)}}-\frac{35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \]

[Out]

(((35*I)/64)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^2*c^(3/2)*f) - ((35*I)/96)/(a^2
*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I/4)/(a^2*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)) + ((7*I)/
16)/(a^2*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) - ((35*I)/64)/(a^2*c*f*Sqrt[c - I*c*Tan[e + f*x]
])

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Rubi [A]  time = 0.211571, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{35 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{64 \sqrt{2} a^2 c^{3/2} f}-\frac{35 i}{64 a^2 c f \sqrt{c-i c \tan (e+f x)}}-\frac{35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((35*I)/64)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^2*c^(3/2)*f) - ((35*I)/96)/(a^2
*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I/4)/(a^2*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)) + ((7*I)/
16)/(a^2*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) - ((35*I)/64)/(a^2*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))^2 (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac{\int \cos ^4(e+f x) \sqrt{c-i c \tan (e+f x)} \, dx}{a^2 c^2}\\ &=\frac{\left (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 )}{a^2 f}\\ &=\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{\left (7 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 )}{8 a^2 f}\\ &=\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{(35 i c) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^2 f}\\ &=-\frac{35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac{(35 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{64 a^2 f}\\ &=-\frac{35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{35 i}{64 a^2 c f \sqrt{c-i c \tan (e+f x)}}+\frac{(35 i) \operatorname{Subst}\left (\int \frac{1}{(c-x) \sqrt{c+x}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^2 c f}\\ &=-\frac{35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{35 i}{64 a^2 c f \sqrt{c-i c \tan (e+f x)}}+\frac{(35 i) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,\sqrt{c-i c \tan (e+f x)}\right )}{64 a^2 c f}\\ &=\frac{35 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{64 \sqrt{2} a^2 c^{3/2} f}-\frac{35 i}{96 a^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac{7 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac{35 i}{64 a^2 c f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.99822, size = 145, normalized size = 0.73 \[ -\frac{i e^{-4 i (e+f x)} \left (-45 e^{2 i (e+f x)}+41 e^{4 i (e+f x)}+88 e^{6 i (e+f x)}+8 e^{8 i (e+f x)}-105 e^{4 i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )-6\right ) \sqrt{c-i c \tan (e+f x)}}{384 a^2 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/384)*(-6 - 45*E^((2*I)*(e + f*x)) + 41*E^((4*I)*(e + f*x)) + 88*E^((6*I)*(e + f*x)) + 8*E^((8*I)*(e + f*x
)) - 105*E^((4*I)*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(e + f*x))]])*Sqrt[c - I*
c*Tan[e + f*x]])/(a^2*c^2*E^((4*I)*(e + f*x))*f)

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Maple [A]  time = 0.04, size = 140, normalized size = 0.7 \begin{align*}{\frac{-2\,i{c}^{3}}{f{a}^{2}} \left ({\frac{1}{16\,{c}^{4}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{2}} \left ({\frac{11}{8} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{13\,c}{4}\sqrt{c-ic\tan \left ( fx+e \right ) }} \right ) }-{\frac{35\,\sqrt{2}}{16}{\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{3}{16\,{c}^{4}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}+{\frac{1}{24\,{c}^{3}} \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))^2/(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

-2*I/f/a^2*c^3*(1/16/c^4*((11/8*(c-I*c*tan(f*x+e))^(3/2)-13/4*c*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^
2-35/16*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))+3/16/c^4/(c-I*c*tan(f*x+e))^(1/
2)+1/24/c^3/(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))^2/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.53078, size = 938, normalized size = 4.71 \begin{align*} \frac{{\left (105 i \, \sqrt{\frac{1}{2}} a^{2} c^{2} f \sqrt{\frac{1}{a^{4} c^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (1120 i \, a^{2} c f e^{\left (2 i \, f x + 2 i \, e\right )} + 1120 i \, a^{2} c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{4} c^{3} f^{2}}} + 1120 i\right )} e^{\left (-i \, f x - i \, e\right )}}{1024 \, a^{2} c f}\right ) - 105 i \, \sqrt{\frac{1}{2}} a^{2} c^{2} f \sqrt{\frac{1}{a^{4} c^{3} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (-1120 i \, a^{2} c f e^{\left (2 i \, f x + 2 i \, e\right )} - 1120 i \, a^{2} c f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{4} c^{3} f^{2}}} + 1120 i\right )} e^{\left (-i \, f x - i \, e\right )}}{1024 \, a^{2} c f}\right ) + \sqrt{2} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (-8 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 88 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 41 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 45 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} 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))^2/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/384*(105*I*sqrt(1/2)*a^2*c^2*f*sqrt(1/(a^4*c^3*f^2))*e^(4*I*f*x + 4*I*e)*log(1/1024*(sqrt(2)*sqrt(1/2)*(1120
*I*a^2*c*f*e^(2*I*f*x + 2*I*e) + 1120*I*a^2*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c^3*f^2)) + 112
0*I)*e^(-I*f*x - I*e)/(a^2*c*f)) - 105*I*sqrt(1/2)*a^2*c^2*f*sqrt(1/(a^4*c^3*f^2))*e^(4*I*f*x + 4*I*e)*log(1/1
024*(sqrt(2)*sqrt(1/2)*(-1120*I*a^2*c*f*e^(2*I*f*x + 2*I*e) - 1120*I*a^2*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)
)*sqrt(1/(a^4*c^3*f^2)) + 1120*I)*e^(-I*f*x - I*e)/(a^2*c*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-8*
I*e^(8*I*f*x + 8*I*e) - 88*I*e^(6*I*f*x + 6*I*e) - 41*I*e^(4*I*f*x + 4*I*e) + 45*I*e^(2*I*f*x + 2*I*e) + 6*I))
*e^(-4*I*f*x - 4*I*e)/(a^2*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))**2/(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 )}^{2}{\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))^2/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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