3.11.2 \(\int \frac {(c-i c \tan (e+f x))^{3/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx\) [1002]

Optimal. Leaf size=106 \[ \frac {2 i c^{3/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a} f}+\frac {2 i c \sqrt {c-i c \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}} \]

[Out]

2*I*c^(3/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))/f/a^(1/2)+2*I*c*(c-I*c*t
an(f*x+e))^(1/2)/f/(a+I*a*tan(f*x+e))^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3604, 49, 65, 223, 209} \begin {gather*} \frac {2 i c^{3/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a} f}+\frac {2 i c \sqrt {c-i c \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 49

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 209

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3604

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[a*(c/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [A]
time = 1.93, size = 119, normalized size = 1.12 \begin {gather*} \frac {c^2 (\cos (f x)+i \sin (f x)) (i \cos (f x)+\sin (f x)) (1+\text {ArcTan}(\cos (e+f x)+i \sin (e+f x)) \sec (e+f x)-i \tan (e+f x))}{f \sqrt {\frac {c}{2+2 \cos (2 (e+f x))+2 i \sin (2 (e+f x))}} \sqrt {a+i a \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(Cos[f*x] + I*Sin[f*x])*(I*Cos[f*x] + Sin[f*x])*(1 + ArcTan[Cos[e + f*x] + I*Sin[e + f*x]]*Sec[e + f*x] -
I*Tan[e + f*x]))/(f*Sqrt[c/(2 + 2*Cos[2*(e + f*x)] + (2*I)*Sin[2*(e + f*x)])]*Sqrt[a + I*a*Tan[e + f*x]])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (84 ) = 168\).
time = 0.38, size = 269, normalized size = 2.54

method	result	size
derivativedivides	\(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -2 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )+2 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-2 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{f a \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (-\tan \left (f x +e \right )+i\right )^{2} \sqrt {a c}}\)	\(269\)
default	\(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -2 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )+2 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-2 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{f a \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (-\tan \left (f x +e \right )+i\right )^{2} \sqrt {a c}}\)	\(269\)

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))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.62, size = 139, normalized size = 1.31 \begin {gather*} -\frac {{\left (2 i \, c \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 2 i \, c \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 4 i \, c \cos \left (f x + e\right ) + c \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - c \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 4 \, c \sin \left (f x + e\right )\right )} \sqrt {c}}{2 \, \sqrt {a} 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))^(1/2),x, algorithm="maxima")

[Out]

-1/2*(2*I*c*arctan2(cos(f*x + e), sin(f*x + e) + 1) + 2*I*c*arctan2(cos(f*x + e), -sin(f*x + e) + 1) - 4*I*c*c
os(f*x + e) + c*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - c*log(cos(f*x + e)^2 + sin(f*x + e
)^2 - 2*sin(f*x + e) + 1) - 4*c*sin(f*x + e))*sqrt(c)/(sqrt(a)*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. 357 vs. \(2 (82) = 164\).
time = 1.85, size = 357, normalized size = 3.37 \begin {gather*} -\frac {{\left (a f \sqrt {\frac {c^{3}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (c e^{\left (3 i \, f x + 3 i \, e\right )} + c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a f\right )} \sqrt {\frac {c^{3}}{a f^{2}}}\right )}}{c e^{\left (2 i \, f x + 2 i \, e\right )} + c}\right ) - a f \sqrt {\frac {c^{3}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (c e^{\left (3 i \, f x + 3 i \, e\right )} + c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (-i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a f\right )} \sqrt {\frac {c^{3}}{a f^{2}}}\right )}}{c e^{\left (2 i \, f x + 2 i \, e\right )} + c}\right ) + 4 \, {\left (-i \, c e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a 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))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}}}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}\, dx \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))**(1/2),x)

[Out]

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

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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))^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \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)^(1/2),x)

[Out]

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

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