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

Optimal. Leaf size=210 \[ -\frac{35 i}{128 a^3 f \sqrt{c-i c \tan (e+f x)}}+\frac{35 i}{192 a^3 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{35 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^3 \sqrt{c} f} \]

[Out]

(((35*I)/128)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*Sqrt[c]*f) - ((35*I)/128)/(a
^3*f*Sqrt[c - I*c*Tan[e + f*x]]) + (I/6)/(a^3*f*(1 + I*Tan[e + f*x])^3*Sqrt[c - I*c*Tan[e + f*x]]) + ((7*I)/48
)/(a^3*f*(1 + I*Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]]) + ((35*I)/192)/(a^3*f*(1 + I*Tan[e + f*x])*Sqrt[c
- I*c*Tan[e + f*x]])

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Rubi [A]  time = 0.215194, antiderivative size = 210, 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} \[ -\frac{35 i}{128 a^3 f \sqrt{c-i c \tan (e+f x)}}+\frac{35 i}{192 a^3 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{35 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^3 \sqrt{c} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((35*I)/128)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*Sqrt[c]*f) - ((35*I)/128)/(a
^3*f*Sqrt[c - I*c*Tan[e + f*x]]) + (I/6)/(a^3*f*(1 + I*Tan[e + f*x])^3*Sqrt[c - I*c*Tan[e + f*x]]) + ((7*I)/48
)/(a^3*f*(1 + I*Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]]) + ((35*I)/192)/(a^3*f*(1 + I*Tan[e + f*x])*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 \sqrt{c-i c \tan (e+f x)}} \, dx &=\frac{\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{5/2} \, dx}{a^3 c^3}\\ &=\frac{\left (i c^4\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^4 (c+x)^{3/2}} \, 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 \sqrt{c-i c \tan (e+f x)}}+\frac{\left (7 i c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^3 (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{12 a^3 f}\\ &=\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{\left (35 i c^2\right ) \operatorname{Subst}\left (\int \frac{1}{(c-x)^2 (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{96 a^3 f}\\ &=\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{35 i}{192 a^3 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}+\frac{(35 i c) \operatorname{Subst}\left (\int \frac{1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 f}\\ &=-\frac{35 i}{128 a^3 f \sqrt{c-i c \tan (e+f x)}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{35 i}{192 a^3 f (1+i \tan (e+f x)) \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 )}{256 a^3 f}\\ &=-\frac{35 i}{128 a^3 f \sqrt{c-i c \tan (e+f x)}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{35 i}{192 a^3 f (1+i \tan (e+f x)) \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 )}{128 a^3 f}\\ &=\frac{35 i \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{128 \sqrt{2} a^3 \sqrt{c} f}-\frac{35 i}{128 a^3 f \sqrt{c-i c \tan (e+f x)}}+\frac{i}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt{c-i c \tan (e+f x)}}+\frac{7 i}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt{c-i c \tan (e+f x)}}+\frac{35 i}{192 a^3 f (1+i \tan (e+f x)) \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 2.68418, size = 146, normalized size = 0.7 \[ -\frac{i \sec ^2(e+f x) \sqrt{c-i c \tan (e+f x)} \left (-7 i \sin (2 (e+f x))-56 i \sin (4 (e+f x))+85 \cos (2 (e+f x))-40 \cos (4 (e+f x))+105 e^{2 i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )+125\right )}{768 a^3 c f (\tan (e+f x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/768)*Sec[e + f*x]^2*(125 + 105*E^((2*I)*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I
)*(e + f*x))]] + 85*Cos[2*(e + f*x)] - 40*Cos[4*(e + f*x)] - (7*I)*Sin[2*(e + f*x)] - (56*I)*Sin[4*(e + f*x)])
*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*c*f*(-I + Tan[e + f*x])^2)

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Maple [A]  time = 0.087, size = 140, normalized size = 0.7 \begin{align*}{\frac{2\,i{c}^{4}}{f{a}^{3}} \left ( -{\frac{1}{16\,{c}^{4}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ({\frac{19}{16} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{17\,c}{3} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{29\,{c}^{2}}{4}\sqrt{c-ic\tan \left ( fx+e \right ) }} \right ) }-{\frac{35\,\sqrt{2}}{32}{\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}{16\,{c}^{4}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f/a^3*c^4*(-1/16/c^4*((19/16*(c-I*c*tan(f*x+e))^(5/2)-17/3*(c-I*c*tan(f*x+e))^(3/2)*c+29/4*c^2*(c-I*c*tan(
f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3-35/32*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2
)))-1/16/c^4/(c-I*c*tan(f*x+e))^(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(1/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.49773, size = 906, normalized size = 4.31 \begin{align*} \frac{{\left (105 i \, \sqrt{\frac{1}{2}} a^{3} c f \sqrt{\frac{1}{a^{6} c f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (2240 i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + 2240 i \, a^{3} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{6} c f^{2}}} + 2240 i\right )} e^{\left (-i \, f x - i \, e\right )}}{4096 \, a^{3} f}\right ) - 105 i \, \sqrt{\frac{1}{2}} a^{3} c f \sqrt{\frac{1}{a^{6} c f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (-2240 i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - 2240 i \, a^{3} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{1}{a^{6} c f^{2}}} + 2240 i\right )} e^{\left (-i \, f x - i \, e\right )}}{4096 \, a^{3} f}\right ) + \sqrt{2} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (-48 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 39 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 125 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 46 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 )}}{768 \, a^{3} c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(105*I*sqrt(1/2)*a^3*c*f*sqrt(1/(a^6*c*f^2))*e^(6*I*f*x + 6*I*e)*log(1/4096*(sqrt(2)*sqrt(1/2)*(2240*I*a
^3*f*e^(2*I*f*x + 2*I*e) + 2240*I*a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c*f^2)) + 2240*I)*e^(-I
*f*x - I*e)/(a^3*f)) - 105*I*sqrt(1/2)*a^3*c*f*sqrt(1/(a^6*c*f^2))*e^(6*I*f*x + 6*I*e)*log(1/4096*(sqrt(2)*sqr
t(1/2)*(-2240*I*a^3*f*e^(2*I*f*x + 2*I*e) - 2240*I*a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c*f^2)
) + 2240*I)*e^(-I*f*x - I*e)/(a^3*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-48*I*e^(8*I*f*x + 8*I*e) +
39*I*e^(6*I*f*x + 6*I*e) + 125*I*e^(4*I*f*x + 4*I*e) + 46*I*e^(2*I*f*x + 2*I*e) + 8*I))*e^(-6*I*f*x - 6*I*e)/
(a^3*c*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/(c-I*c*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**3,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} \sqrt{-i \, c \tan \left (f x + e\right ) + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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