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

Optimal. Leaf size=167 \[ -\frac {15 i}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {5 i}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {15 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^2 \sqrt {c} f} \]

[Out]

15/64*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^2/f*2^(1/2)/c^(1/2)-15/32*I/a^2/f/(c-I*c*tan(f
*x+e))^(1/2)+1/4*I/a^2/f/(c-I*c*tan(f*x+e))^(1/2)/(1+I*tan(f*x+e))^2+5/16*I/a^2/f/(c-I*c*tan(f*x+e))^(1/2)/(1+
I*tan(f*x+e))

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Rubi [A]  time = 0.20, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ -\frac {15 i}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {5 i}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {15 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^2 \sqrt {c} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((15*I)/32)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^2*Sqrt[c]*f) - ((15*I)/32)/(a^2
*f*Sqrt[c - I*c*Tan[e + f*x]]) + (I/4)/(a^2*f*(1 + I*Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]]) + ((5*I)/16)/
(a^2*f*(1 + I*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*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])

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 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]))

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {\int \cos ^4(e+f x) (c-i c \tan (e+f x))^{3/2} \, dx}{a^2 c^2}\\ &=\frac {\left (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 )}{a^2 f}\\ &=\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {\left (5 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 )}{8 a^2 f}\\ &=\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(15 i c) \operatorname {Subst}\left (\int \frac {1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^2 f}\\ &=-\frac {15 i}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(15 i) \operatorname {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{64 a^2 f}\\ &=-\frac {15 i}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}+\frac {(15 i) \operatorname {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{32 a^2 f}\\ &=\frac {15 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^2 \sqrt {c} f}-\frac {15 i}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 2.47, size = 141, normalized size = 0.84 \[ -\frac {i e^{-4 i (e+f x)} \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \left (-11 e^{2 i (e+f x)}-e^{4 i (e+f x)}+8 e^{6 i (e+f x)}-15 e^{4 i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )-2\right )}{32 \sqrt {2} a^2 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/32*I)*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*(-2 - 11*E^((2*I)*(e + f*x)) - E^((4*I)*(e + f*x)) + 8*E^((6*I)*(
e + f*x)) - 15*E^((4*I)*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(e + f*x))]]))/(Sqr
t[2]*a^2*c*E^((4*I)*(e + f*x))*f)

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fricas [B]  time = 0.48, size = 295, normalized size = 1.77 \[ \frac {{\left (15 i \, \sqrt {\frac {1}{2}} a^{2} c f \sqrt {\frac {1}{a^{4} c f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (240 i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 240 i \, a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c f^{2}}} + 240 i\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{2} f}\right ) - 15 i \, \sqrt {\frac {1}{2}} a^{2} c f \sqrt {\frac {1}{a^{4} c f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-240 i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 240 i \, a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c f^{2}}} + 240 i\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-8 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 11 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{64 \, a^{2} c f} \]

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))^2,x, algorithm="fricas")

[Out]

1/64*(15*I*sqrt(1/2)*a^2*c*f*sqrt(1/(a^4*c*f^2))*e^(4*I*f*x + 4*I*e)*log(1/256*(sqrt(2)*sqrt(1/2)*(240*I*a^2*f
*e^(2*I*f*x + 2*I*e) + 240*I*a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c*f^2)) + 240*I)*e^(-I*f*x -
I*e)/(a^2*f)) - 15*I*sqrt(1/2)*a^2*c*f*sqrt(1/(a^4*c*f^2))*e^(4*I*f*x + 4*I*e)*log(1/256*(sqrt(2)*sqrt(1/2)*(
-240*I*a^2*f*e^(2*I*f*x + 2*I*e) - 240*I*a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c*f^2)) + 240*I)
*e^(-I*f*x - I*e)/(a^2*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-8*I*e^(6*I*f*x + 6*I*e) + I*e^(4*I*f*
x + 4*I*e) + 11*I*e^(2*I*f*x + 2*I*e) + 2*I))*e^(-4*I*f*x - 4*I*e)/(a^2*c*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\,{d x} \]

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))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.40, size = 121, normalized size = 0.72 \[ -\frac {2 i c^{3} \left (\frac {\frac {\frac {7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8}-\frac {9 \sqrt {c -i c \tan \left (f x +e \right )}\, c}{4}}{\left (-c -i c \tan \left (f x +e \right )\right )^{2}}-\frac {15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}}{8 c^{3}}+\frac {1}{8 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,a^{2}} \]

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))^2,x)

[Out]

-2*I/f/a^2*c^3*(1/8/c^3*((7/8*(c-I*c*tan(f*x+e))^(3/2)-9/4*(c-I*c*tan(f*x+e))^(1/2)*c)/(-c-I*c*tan(f*x+e))^2-1
5/16*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))+1/8/c^3/(c-I*c*tan(f*x+e))^(1/2))

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maxima [A]  time = 0.83, size = 169, normalized size = 1.01 \[ -\frac {i \, {\left (\frac {4 \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c - 50 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{2} + 32 \, c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} c + 4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{2} c^{2}} + \frac {15 \, \sqrt {2} \sqrt {c} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}}\right )}}{128 \, c f} \]

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))^2,x, algorithm="maxima")

[Out]

-1/128*I*(4*(15*(-I*c*tan(f*x + e) + c)^2*c - 50*(-I*c*tan(f*x + e) + c)*c^2 + 32*c^3)/((-I*c*tan(f*x + e) + c
)^(5/2)*a^2 - 4*(-I*c*tan(f*x + e) + c)^(3/2)*a^2*c + 4*sqrt(-I*c*tan(f*x + e) + c)*a^2*c^2) + 15*sqrt(2)*sqrt
(c)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/a^2)
/(c*f)

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mupad [B]  time = 5.00, size = 156, normalized size = 0.93 \[ -\frac {\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,15{}\mathrm {i}}{32\,a^2\,f}+\frac {c^2\,1{}\mathrm {i}}{a^2\,f}-\frac {c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,25{}\mathrm {i}}{16\,a^2\,f}}{-4\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}+{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}+4\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,15{}\mathrm {i}}{64\,a^2\,\sqrt {-c}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (((c - c*tan(e + f*x)*1i)^2*15i)/(32*a^2*f) + (c^2*1i)/(a^2*f) - (c*(c - c*tan(e + f*x)*1i)*25i)/(16*a^2*f))
/((c - c*tan(e + f*x)*1i)^(5/2) - 4*c*(c - c*tan(e + f*x)*1i)^(3/2) + 4*c^2*(c - c*tan(e + f*x)*1i)^(1/2)) - (
2^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*15i)/(64*a^2*(-c)^(1/2)*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {1}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{2}} \]

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))**2,x)

[Out]

-Integral(1/(sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - 2*I*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) - sqrt
(-I*c*tan(e + f*x) + c)), x)/a**2

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