\(\int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [1018]

Optimal result

Integrand size = 35, antiderivative size = 106 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 i a^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3604, 49, 65, 223, 209} \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 i a^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

((2*I)*a^(3/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(Sqrt[c]*f)
- ((2*I)*a*Sqrt[a + I*a*Tan[e + f*x]])/(f*Sqrt[c - I*c*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*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {a^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 a \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}}+\frac {(2 i a) \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 )}{f} \\ & = -\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}}+\frac {(2 i a) \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 )}{f} \\ & = \frac {2 i a^{3/2} \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 {c} f}-\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 i a^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (84 ) = 168\).

Time = 0.88 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.52

method	result	size
derivativedivides	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -2 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )-2 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+2 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{f c \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan \left (f x +e \right )+i\right )^{2} \sqrt {a c}}\)	\(267\)
default	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -2 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )-2 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+2 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{f c \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan \left (f x +e \right )+i\right )^{2} \sqrt {a c}}\)	\(267\)

[In]

int((a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (78) = 156\).

Time = 0.26 (sec) , antiderivative size = 321, normalized size of antiderivative = 3.03 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {c f \sqrt {\frac {a^{3}}{c f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a e^{\left (3 i \, f x + 3 i \, e\right )} + a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c f\right )} \sqrt {\frac {a^{3}}{c f^{2}}}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right ) - c f \sqrt {\frac {a^{3}}{c f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a e^{\left (3 i \, f x + 3 i \, e\right )} + a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (-i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c f\right )} \sqrt {\frac {a^{3}}{c f^{2}}}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right ) + 4 \, {\left (i \, a e^{\left (3 i \, f x + 3 i \, e\right )} + i \, a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{2 \, c f} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate((a+I*a*tan(f*x+e))**(3/2)/(c-I*c*tan(f*x+e))**(1/2),x)

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (78) = 156\).

Time = 0.68 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.01 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {{\left (-2 i \, a \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - 2 i \, a \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), -\sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 4 i \, a \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + a \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - a \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - 4 \, a \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{2 \, \sqrt {c} f} \]

[In]

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

[Out]

-1/2*(-2*I*a*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x + 2*e), c
os(2*f*x + 2*e))) + 1) - 2*I*a*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 4*I*a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + a*log
(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - a*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e
), cos(2*f*x + 2*e))) + 1) - 4*a*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(sqrt(c)*f)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \]

[In]

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

[Out]

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