3.10.67 \(\int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx\) [967]

Optimal. Leaf size=193 \[ -\frac {i c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^3 f}+\frac {i c^4 \sqrt {c-i c \tan (e+f x)}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{24 a^3 f (c+i c \tan (e+f x))^2}-\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (c+i c \tan (e+f x))} \]

[Out]

-1/64*I*c^(3/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/f*2^(1/2)+1/3*I*c^4*(c-I*c*tan(f*x+e
))^(1/2)/a^3/f/(c+I*c*tan(f*x+e))^3-1/24*I*c^3*(c-I*c*tan(f*x+e))^(1/2)/a^3/f/(c+I*c*tan(f*x+e))^2-1/32*I*c^2*
(c-I*c*tan(f*x+e))^(1/2)/a^3/f/(c+I*c*tan(f*x+e))

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Rubi [A]
time = 0.15, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3603, 3568, 43, 44, 65, 212} \begin {gather*} -\frac {i c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^3 f}+\frac {i c^4 \sqrt {c-i c \tan (e+f x)}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{24 a^3 f (c+i c \tan (e+f x))^2}-\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (c+i c \tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/32*I)*c^(3/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*f) + ((I/3)*c^4*Sqrt[c
- I*c*Tan[e + f*x]])/(a^3*f*(c + I*c*Tan[e + f*x])^3) - ((I/24)*c^3*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*f*(c + I*
c*Tan[e + f*x])^2) - ((I/32)*c^2*Sqrt[c - I*c*Tan[e + f*x]])/(a^3*f*(c + I*c*Tan[e + f*x]))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx &=\frac {\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{9/2} \, dx}{a^3 c^3}\\ &=\frac {\left (i c^4\right ) \text {Subst}\left (\int \frac {\sqrt {c+x}}{(c-x)^4} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac {i c^4 \sqrt {c-i c \tan (e+f x)}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac {\left (i c^4\right ) \text {Subst}\left (\int \frac {1}{(c-x)^3 \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{6 a^3 f}\\ &=\frac {i c^4 \sqrt {c-i c \tan (e+f x)}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{24 a^3 f (c+i c \tan (e+f x))^2}-\frac {\left (i c^3\right ) \text {Subst}\left (\int \frac {1}{(c-x)^2 \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{16 a^3 f}\\ &=\frac {i c^4 \sqrt {c-i c \tan (e+f x)}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{24 a^3 f (c+i c \tan (e+f x))^2}-\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (c+i c \tan (e+f x))}-\frac {\left (i c^2\right ) \text {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{64 a^3 f}\\ &=\frac {i c^4 \sqrt {c-i c \tan (e+f x)}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{24 a^3 f (c+i c \tan (e+f x))^2}-\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (c+i c \tan (e+f x))}-\frac {\left (i c^2\right ) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{32 a^3 f}\\ &=-\frac {i c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^3 f}+\frac {i c^4 \sqrt {c-i c \tan (e+f x)}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{24 a^3 f (c+i c \tan (e+f x))^2}-\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (c+i c \tan (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 1.88, size = 152, normalized size = 0.79 \begin {gather*} \frac {c (\cos (3 (e+f x))-i \sin (3 (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 ) (-i \cos (3 (e+f x))+\sin (3 (e+f x)))+\left (39 i \cos (e+f x)+11 i \cos (3 (e+f x))+20 \cos ^2(e+f x) \sin (e+f x)\right ) \sqrt {c-i c \tan (e+f x)}\right )}{192 a^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(Cos[3*(e + f*x)] - I*Sin[3*(e + f*x)])*(3*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt
[c])]*((-I)*Cos[3*(e + f*x)] + Sin[3*(e + f*x)]) + ((39*I)*Cos[e + f*x] + (11*I)*Cos[3*(e + f*x)] + 20*Cos[e +
f*x]^2*Sin[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]]))/(192*a^3*f)

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Maple [A]
time = 0.33, size = 116, normalized size = 0.60

method	result	size
derivativedivides	\(\frac {2 i c^{4} \left (\frac {-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{64 c^{2}}+\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{12 c}+\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{16}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{128 c^{\frac {5}{2}}}\right )}{f \,a^{3}}\)	\(116\)
default	\(\frac {2 i c^{4} \left (\frac {-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{64 c^{2}}+\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{12 c}+\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{16}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{128 c^{\frac {5}{2}}}\right )}{f \,a^{3}}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f/a^3*c^4*(8*(-1/512/c^2*(c-I*c*tan(f*x+e))^(5/2)+1/96/c*(c-I*c*tan(f*x+e))^(3/2)+1/128*(c-I*c*tan(f*x+e))
^(1/2))/(c+I*c*tan(f*x+e))^3-1/128/c^(5/2)*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 = 0.50, size = 200, normalized size = 1.04 \begin {gather*} \frac {i \, {\left (\frac {3 \, \sqrt {2} c^{\frac {5}{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}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c^{3} - 16 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{4} - 12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{5}\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}}\right )}}{384 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*I*(3*sqrt(2)*c^(5/2)*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^3 + 4*(3*(-I*c*tan(f*x + e) + c)^(5/2)*c^3 - 16*(-I*c*tan(f*x + e) + c)^(3/2)*c^4 - 12*sq
rt(-I*c*tan(f*x + e) + c)*c^5)/((-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*t
an(f*x + e) + c)*a^3*c^2 - 8*a^3*c^3))/(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. 317 vs. \(2 (155) = 310\).
time = 0.86, size = 317, normalized size = 1.64 \begin {gather*} -\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{3}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {{\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^{3}}{a^{6} f^{2}}} + i \, c^{2}\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{3} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{3}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\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^{3}}{a^{6} f^{2}}} - i \, c^{2}\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{3} f}\right ) - \sqrt {2} {\left (3 i \, c e^{\left (6 i \, f x + 6 i \, e\right )} + 17 i \, c e^{\left (4 i \, f x + 4 i \, e\right )} + 22 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{192 \, a^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(3*sqrt(1/2)*a^3*f*sqrt(-c^3/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-1/16*(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^3/(a^6*f^2)) + I*c^2)*e^(-I*f*x - I*e)/(a^3*f)
) - 3*sqrt(1/2)*a^3*f*sqrt(-c^3/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/16*(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^3/(a^6*f^2)) - I*c^2)*e^(-I*f*x - I*e)/(a^3*f)) - s
qrt(2)*(3*I*c*e^(6*I*f*x + 6*I*e) + 17*I*c*e^(4*I*f*x + 4*I*e) + 22*I*c*e^(2*I*f*x + 2*I*e) + 8*I*c)*sqrt(c/(e
^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 4.94, size = 181, normalized size = 0.94 \begin {gather*} \frac {\frac {c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{8\,a^3\,f}+\frac {c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,1{}\mathrm {i}}{6\,a^3\,f}-\frac {c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,1{}\mathrm {i}}{32\,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}\,{\left (-c\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{64\,a^3\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c^4*(c - c*tan(e + f*x)*1i)^(1/2)*1i)/(8*a^3*f) + (c^3*(c - c*tan(e + f*x)*1i)^(3/2)*1i)/(6*a^3*f) - (c^2*(c
- c*tan(e + f*x)*1i)^(5/2)*1i)/(32*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)^(3/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^
(1/2)))*1i)/(64*a^3*f)

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