3.10.60 \(\int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx\) [960]

Optimal. Leaf size=138 \[ \frac {3 i \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f}+\frac {i \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {3 i \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))} \]

[Out]

3/32*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^(1/2)/a^2/f*2^(1/2)+1/4*I*(c-I*c*tan(f*x+e))^(1
/2)/a^2/f/(1+I*tan(f*x+e))^2+3/16*I*(c-I*c*tan(f*x+e))^(1/2)/a^2/f/(1+I*tan(f*x+e))

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Rubi [A]
time = 0.13, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3603, 3568, 44, 65, 212} \begin {gather*} \frac {3 i \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac {i \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {3 i \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3568

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 3603

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

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Mathematica [A]
time = 1.10, size = 136, normalized size = 0.99 \begin {gather*} \frac {(i \cos (2 (e+f x))+\sin (2 (e+f x))) \left (3 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+(7+7 \cos (2 (e+f x))+3 i \sin (2 (e+f x))) \sqrt {c-i c \tan (e+f x)}\right )}{32 a^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I*Cos[2*(e + f*x)] + Sin[2*(e + f*x)])*(3*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c
])]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) + (7 + 7*Cos[2*(e + f*x)] + (3*I)*Sin[2*(e + f*x)])*Sqrt[c - I*c*T
an[e + f*x]]))/(32*a^2*f)

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Maple [A]
time = 0.32, size = 117, normalized size = 0.85

method	result	size
derivativedivides	\(-\frac {2 i c^{3} \left (-\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{8 c \left (c +i c \tan \left (f x +e \right )\right )^{2}}-\frac {3 \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{4 c \left (c +i c \tan \left (f x +e \right )\right )}+\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )}{8 c}\right )}{f \,a^{2}}\)	\(117\)
default	\(-\frac {2 i c^{3} \left (-\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{8 c \left (c +i c \tan \left (f x +e \right )\right )^{2}}-\frac {3 \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{4 c \left (c +i c \tan \left (f x +e \right )\right )}+\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )}{8 c}\right )}{f \,a^{2}}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 159, normalized size = 1.15 \begin {gather*} -\frac {i \, {\left (\frac {3 \, \sqrt {2} c^{\frac {3}{2}} \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}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{2} - 10 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}}\right )}}{64 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*I*(3*sqrt(2)*c^(3/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*t
an(f*x + e) + c)))/a^2 + 4*(3*(-I*c*tan(f*x + e) + c)^(3/2)*c^2 - 10*sqrt(-I*c*tan(f*x + e) + c)*c^3)/((-I*c*t
an(f*x + e) + c)^2*a^2 - 4*(-I*c*tan(f*x + e) + c)*a^2*c + 4*a^2*c^2))/(c*f)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (108) = 216\).
time = 1.27, size = 287, normalized size = 2.08 \begin {gather*} \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {3 \, {\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}{a^{4} f^{2}}} + i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {3 \, {\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}{a^{4} f^{2}}} - i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (5 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 7 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 )}}{32 \, a^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(3*sqrt(1/2)*a^2*f*sqrt(-c/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(3/8*(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/(a^4*f^2)) + I*c)*e^(-I*f*x - I*e)/(a^2*f)) - 3*sqrt
(1/2)*a^2*f*sqrt(-c/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-3/8*(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/(a^4*f^2)) - I*c)*e^(-I*f*x - I*e)/(a^2*f)) + sqrt(2)*sqrt(c/(e
^(2*I*f*x + 2*I*e) + 1))*(5*I*e^(4*I*f*x + 4*I*e) + 7*I*e^(2*I*f*x + 2*I*e) + 2*I))*e^(-4*I*f*x - 4*I*e)/(a^2*
f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.85, size = 132, normalized size = 0.96 \begin {gather*} \frac {\frac {c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,5{}\mathrm {i}}{8\,a^2\,f}-\frac {c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,3{}\mathrm {i}}{16\,a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}+\frac {\sqrt {2}\,\sqrt {-c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,3{}\mathrm {i}}{32\,a^2\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c^2*(c - c*tan(e + f*x)*1i)^(1/2)*5i)/(8*a^2*f) - (c*(c - c*tan(e + f*x)*1i)^(3/2)*3i)/(16*a^2*f))/((c - c*t
an(e + f*x)*1i)^2 - 4*c*(c - c*tan(e + f*x)*1i) + 4*c^2) + (2^(1/2)*(-c)^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*
x)*1i)^(1/2))/(2*(-c)^(1/2)))*3i)/(32*a^2*f)

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