3.10.66 \(\int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx\) [966]

Optimal. Leaf size=146 \[ -\frac {i c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a^2 f}+\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{8 a^2 f (c+i c \tan (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, 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 )}{8 \sqrt {2} a^2 f}+\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{8 a^2 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])^2,x]

[Out]

((-1/8*I)*c^(3/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^2*f) + ((I/2)*c^3*Sqrt[c -
I*c*Tan[e + f*x]])/(a^2*f*(c + I*c*Tan[e + f*x])^2) - ((I/8)*c^2*Sqrt[c - I*c*Tan[e + f*x]])/(a^2*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))^2} \, dx &=\frac {\int \cos ^4(e+f x) (c-i c \tan (e+f x))^{7/2} \, dx}{a^2 c^2}\\ &=\frac {\left (i c^3\right ) \text {Subst}\left (\int \frac {\sqrt {c+x}}{(c-x)^3} \, dx,x,-i c \tan (e+f x)\right )}{a^2 f}\\ &=\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{2 a^2 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 )}{4 a^2 f}\\ &=\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{8 a^2 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 )}{16 a^2 f}\\ &=\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{8 a^2 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 )}{8 a^2 f}\\ &=-\frac {i c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a^2 f}+\frac {i c^3 \sqrt {c-i c \tan (e+f x)}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{8 a^2 f (c+i c \tan (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 1.41, size = 136, normalized size = 0.93 \begin {gather*} \frac {c (\cos (2 (e+f x))-i \sin (2 (e+f x))) \left (\sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (-i \cos (2 (e+f x))+\sin (2 (e+f x)))+(3 i+3 i \cos (2 (e+f x))+\sin (2 (e+f x))) \sqrt {c-i c \tan (e+f x)}\right )}{16 a^2 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])^2,x]

[Out]

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

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Maple [A]
time = 0.32, size = 97, normalized size = 0.66

method	result	size
derivativedivides	\(-\frac {2 i c^{3} \left (-\frac {4 \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{64 c}+\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{32}\right )}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 c^{\frac {3}{2}}}\right )}{f \,a^{2}}\)	\(97\)
default	\(-\frac {2 i c^{3} \left (-\frac {4 \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{64 c}+\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{32}\right )}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 c^{\frac {3}{2}}}\right )}{f \,a^{2}}\)	\(97\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

-2*I/f/a^2*c^3*(-4*(1/64/c*(c-I*c*tan(f*x+e))^(3/2)+1/32*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^2+1/32/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 = 0.54, size = 157, normalized size = 1.08 \begin {gather*} \frac {i \, {\left (\frac {\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^{2}} + \frac {4 \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{3} + 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{4}\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 )}}{32 \, 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))^2,x, algorithm="maxima")

[Out]

1/32*I*(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*tan(
f*x + e) + c)))/a^2 + 4*((-I*c*tan(f*x + e) + c)^(3/2)*c^3 + 2*sqrt(-I*c*tan(f*x + e) + c)*c^4)/((-I*c*tan(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. 303 vs. \(2 (116) = 232\).
time = 0.76, size = 303, normalized size = 2.08 \begin {gather*} -\frac {{\left (\sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c^{3}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {{\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^{3}}{a^{4} f^{2}}} + i \, c^{2}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2} f}\right ) - \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c^{3}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\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^{3}}{a^{4} f^{2}}} - i \, c^{2}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{2} f}\right ) - \sqrt {2} {\left (i \, c e^{\left (4 i \, f x + 4 i \, e\right )} + 3 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

-1/16*(sqrt(1/2)*a^2*f*sqrt(-c^3/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/4*(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^3/(a^4*f^2)) + I*c^2)*e^(-I*f*x - I*e)/(a^2*f)) -
sqrt(1/2)*a^2*f*sqrt(-c^3/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/4*(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^3/(a^4*f^2)) - I*c^2)*e^(-I*f*x - I*e)/(a^2*f)) - sqrt(2)*
(I*c*e^(4*I*f*x + 4*I*e) + 3*I*c*e^(2*I*f*x + 2*I*e) + 2*I*c)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*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 {c \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 + \int \left (- \frac {i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\right )\, dx}{a^{2}} \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))**2,x)

[Out]

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

[Out]

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

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Mupad [B]
time = 4.89, size = 134, normalized size = 0.92 \begin {gather*} \frac {\frac {c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{4\,a^2\,f}+\frac {c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,1{}\mathrm {i}}{8\,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}\,{\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}}{16\,a^2\,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)^2,x)

[Out]

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

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