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

Optimal. Leaf size=106 \[ \frac{i a \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{f}-\frac{2 i a^{3/2} \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.14093, antiderivative size = 106, normalized size of antiderivative = 1., 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 = {3523, 50, 63, 217, 203} \[ \frac{i a \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{f}-\frac{2 i a^{3/2} \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

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

Rule 3523

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]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 217

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 203

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

Rubi steps

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

Mathematica [A]  time = 2.45676, size = 129, normalized size = 1.22 \[ \frac{a c e^{-\frac{1}{2} i (4 e+f x)} \left (\sin \left (\frac{3 e}{2}\right )-i \cos \left (\frac{3 e}{2}\right )\right ) \sqrt{a+i a \tan (e+f x)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-i \sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (-\sec (e+f x)+2 \tan ^{-1}\left (e^{i (e+f x)}\right )\right )}{\sqrt{2} f \sqrt{\frac{c}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c*(2*ArcTan[E^(I*(e + f*x))] - Sec[e + f*x])*((-I)*Cos[(3*e)/2] + Sin[(3*e)/2])*(Cos[(e + f*x)/2] - I*Sin[(
e + f*x)/2])*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*E^((I/2)*(4*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f)

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Maple [A]  time = 0.07, size = 122, normalized size = 1.2 \begin{align*}{\frac{a}{f}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) } \left ( ac\ln \left ({ \left ( ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}} \right ) +i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}

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))^(3/2),x)

[Out]

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

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Maxima [B]  time = 1.93842, size = 609, normalized size = 5.75 \begin{align*} \text{result too large to display} \end{align*}

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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.57339, size = 748, normalized size = 7.06 \begin{align*} \frac{8 i \, a \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}} e^{\left (i \, f x + i \, e\right )} - 2 \, \sqrt{\frac{a^{3} c}{f^{2}}} f \log \left (\frac{2 \,{\left (4 \,{\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\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}} e^{\left (i \, f x + i \, e\right )} + \sqrt{\frac{a^{3} c}{f^{2}}}{\left (2 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, f\right )}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right ) + 2 \, \sqrt{\frac{a^{3} c}{f^{2}}} f \log \left (\frac{2 \,{\left (4 \,{\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\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}} e^{\left (i \, f x + i \, e\right )} + \sqrt{\frac{a^{3} c}{f^{2}}}{\left (-2 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, f\right )}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right )}{4 \, f} \end{align*}

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))^(3/2),x, algorithm="fricas")

[Out]

1/4*(8*I*a*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) - 2*sqrt(a^3*c/
f^2)*f*log(2*(4*(a*e^(2*I*f*x + 2*I*e) + a)*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) + sqrt(a^3*c/f^2)*(2*I*f*e^(2*I*f*x + 2*I*e) - 2*I*f))/(a*e^(2*I*f*x + 2*I*e) + a)) + 2*sqrt
(a^3*c/f^2)*f*log(2*(4*(a*e^(2*I*f*x + 2*I*e) + a)*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) + sqrt(a^3*c/f^2)*(-2*I*f*e^(2*I*f*x + 2*I*e) + 2*I*f))/(a*e^(2*I*f*x + 2*I*e) + a)))
/f

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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} \end{align*}

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))^(3/2),x, algorithm="giac")

[Out]

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