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

Optimal. Leaf size=181 \[ \frac {5 i \sqrt {c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac {5 i \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {i \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {5 i \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^3 f} \]

[Out]

5/128*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^(1/2)/a^3/f*2^(1/2)+1/6*I*(c-I*c*tan(f*x+e))^(
1/2)/a^3/f/(1+I*tan(f*x+e))^3+5/48*I*(c-I*c*tan(f*x+e))^(1/2)/a^3/f/(1+I*tan(f*x+e))^2+5/64*I*(c-I*c*tan(f*x+e
))^(1/2)/a^3/f/(1+I*tan(f*x+e))

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Rubi [A]  time = 0.21, antiderivative size = 181, 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} \[ \frac {5 i \sqrt {c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac {5 i \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {i \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {5 i \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((5*I)/64)*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*f) + ((I/6)*Sqrt[c - I
*c*Tan[e + f*x]])/(a^3*f*(1 + I*Tan[e + f*x])^3) + (((5*I)/48)*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*f*(1 + I*Tan[e
+ f*x])^2) + (((5*I)/64)*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*f*(1 + I*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 {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx &=\frac {\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{7/2} \, dx}{a^3 c^3}\\ &=\frac {\left (i c^4\right ) \operatorname {Subst}\left (\int \frac {1}{(c-x)^4 \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac {i \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {\left (5 i c^3\right ) \operatorname {Subst}\left (\int \frac {1}{(c-x)^3 \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{12 a^3 f}\\ &=\frac {i \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {5 i \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {\left (5 i c^2\right ) \operatorname {Subst}\left (\int \frac {1}{(c-x)^2 \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^3 f}\\ &=\frac {i \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {5 i \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {5 i \sqrt {c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac {(5 i c) \operatorname {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 f}\\ &=\frac {i \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {5 i \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {5 i \sqrt {c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}+\frac {(5 i c) \operatorname {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{64 a^3 f}\\ &=\frac {5 i \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{64 \sqrt {2} a^3 f}+\frac {i \sqrt {c-i c \tan (e+f x)}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {5 i \sqrt {c-i c \tan (e+f x)}}{48 a^3 f (1+i \tan (e+f x))^2}+\frac {5 i \sqrt {c-i c \tan (e+f x)}}{64 a^3 f (1+i \tan (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 2.11, size = 149, normalized size = 0.82 \[ \frac {(\sin (3 (e+f x))+i \cos (3 (e+f x))) \left (\sqrt {c-i c \tan (e+f x)} \left (93 \cos (e+f x)+41 \cos (3 (e+f x))+100 i \sin (e+f x) \cos ^2(e+f x)\right )+15 \sqrt {2} \sqrt {c} (\cos (3 (e+f x))+i \sin (3 (e+f x))) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )\right )}{384 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*Cos[3*(e + f*x)] + Sin[3*(e + f*x)])*(15*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[
c])]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) + (93*Cos[e + f*x] + 41*Cos[3*(e + f*x)] + (100*I)*Cos[e + f*x]^2
*Sin[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]]))/(384*a^3*f)

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fricas [B]  time = 0.49, size = 286, normalized size = 1.58 \[ \frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {5 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{6} f^{2}}} + i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{3} f}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {5 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{6} f^{2}}} - i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{32 \, a^{3} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (33 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 59 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 34 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{384 \, a^{3} f} \]

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))^3,x, algorithm="fricas")

[Out]

1/384*(15*sqrt(1/2)*a^3*f*sqrt(-c/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(5/32*(sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x
+ 2*I*e) + a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c/(a^6*f^2)) + I*c)*e^(-I*f*x - I*e)/(a^3*f)) - 15*
sqrt(1/2)*a^3*f*sqrt(-c/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-5/32*(sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e)
+ a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c/(a^6*f^2)) - I*c)*e^(-I*f*x - I*e)/(a^3*f)) + sqrt(2)*sqrt
(c/(e^(2*I*f*x + 2*I*e) + 1))*(33*I*e^(6*I*f*x + 6*I*e) + 59*I*e^(4*I*f*x + 4*I*e) + 34*I*e^(2*I*f*x + 2*I*e)
+ 8*I))*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-i \, c \tan \left (f x + e\right ) + c}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}\,{d x} \]

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))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.36, size = 162, normalized size = 0.90 \[ \frac {2 i c^{4} \left (-\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{12 c \left (-c -i c \tan \left (f x +e \right )\right )^{3}}-\frac {5 \left (-\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{8 c \left (-c -i c \tan \left (f x +e \right )\right )^{2}}-\frac {3 \left (-\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{4 c \left (-c -i c \tan \left (f x +e \right )\right )}+\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )}{8 c}\right )}{12 c}\right )}{f \,a^{3}} \]

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))^3,x)

[Out]

2*I/f/a^3*c^4*(-1/12*(c-I*c*tan(f*x+e))^(1/2)/c/(-c-I*c*tan(f*x+e))^3-5/12/c*(-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 = 1.16, size = 192, normalized size = 1.06 \[ -\frac {i \, {\left (\frac {4 \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c^{2} - 80 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{3} + 132 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c^{2} - 8 \, a^{3} c^{3}} + \frac {15 \, \sqrt {2} c^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}}\right )}}{768 \, c f} \]

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))^3,x, algorithm="maxima")

[Out]

-1/768*I*(4*(15*(-I*c*tan(f*x + e) + c)^(5/2)*c^2 - 80*(-I*c*tan(f*x + e) + c)^(3/2)*c^3 + 132*sqrt(-I*c*tan(f
*x + e) + c)*c^4)/((-I*c*tan(f*x + e) + c)^3*a^3 - 6*(-I*c*tan(f*x + e) + c)^2*a^3*c + 12*(-I*c*tan(f*x + e) +
c)*a^3*c^2 - 8*a^3*c^3) + 15*sqrt(2)*c^(3/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sq
rt(c) + sqrt(-I*c*tan(f*x + e) + c)))/a^3)/(c*f)

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mupad [B]  time = 4.98, size = 179, normalized size = 0.99 \[ \frac {\frac {c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,11{}\mathrm {i}}{16\,a^3\,f}-\frac {c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,5{}\mathrm {i}}{12\,a^3\,f}+\frac {c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,5{}\mathrm {i}}{64\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+8\,c^3}+\frac {\sqrt {2}\,\sqrt {-c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,5{}\mathrm {i}}{128\,a^3\,f} \]

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)^3,x)

[Out]

((c^3*(c - c*tan(e + f*x)*1i)^(1/2)*11i)/(16*a^3*f) - (c^2*(c - c*tan(e + f*x)*1i)^(3/2)*5i)/(12*a^3*f) + (c*(
c - c*tan(e + f*x)*1i)^(5/2)*5i)/(64*a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^2 - 12*c^2*(c - c*tan(e + f*x)*1i) -
(c - c*tan(e + f*x)*1i)^3 + 8*c^3) + (2^(1/2)*(-c)^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)
^(1/2)))*5i)/(128*a^3*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {i \int \frac {\sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \]

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))**3,x)

[Out]

I*Integral(sqrt(-I*c*tan(e + f*x) + c)/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x)/a**3

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