3.992 \(\int (a+i a \tan (e+f x))^{5/2} \sqrt{c-i c \tan (e+f x)} \, dx\)

Optimal. Leaf size=154 \[ -\frac{3 i a^{5/2} \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}+\frac{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f} \]

[Out]

((-3*I)*a^(5/2)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/f +
(((3*I)/2)*a^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/f + ((I/2)*a*(a + I*a*Tan[e + f*x])^(3/
2)*Sqrt[c - I*c*Tan[e + f*x]])/f

________________________________________________________________________________________

Rubi [A]  time = 0.159651, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{3 i a^{5/2} \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}+\frac{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

((-3*I)*a^(5/2)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/f +
(((3*I)/2)*a^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/f + ((I/2)*a*(a + I*a*Tan[e + f*x])^(3/
2)*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))^{5/2} \sqrt{c-i c \tan (e+f x)} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{\left (3 a^2 c\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+i a x}}{\sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{\left (3 a^3 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 )}{2 f}\\ &=\frac{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{\left (3 i a^2 c\right ) \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{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{\left (3 i a^2 c\right ) \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{3 i a^{5/2} \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{3 i a^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}+\frac{i a (a+i a \tan (e+f x))^{3/2} \sqrt{c-i c \tan (e+f x)}}{2 f}\\ \end{align*}

Mathematica [A]  time = 3.3508, size = 87, normalized size = 0.56 \[ \frac{a^2 c (\tan (e+f x)+i) \sqrt{a+i a \tan (e+f x)} \left (i \tan (e+f x)-6 \cos (e+f x) \tan ^{-1}\left (e^{i (e+f x)}\right )+4\right )}{2 f \sqrt{c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

(a^2*c*(4 - 6*ArcTan[E^(I*(e + f*x))]*Cos[e + f*x] + I*Tan[e + f*x])*(I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f
*x]])/(2*f*Sqrt[c - I*c*Tan[e + f*x]])

________________________________________________________________________________________

Maple [A]  time = 0.087, size = 154, normalized size = 1. \begin{align*}{\frac{{a}^{2}}{2\,f}\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) } \left ( 4\,i\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}-\tan \left ( fx+e \right ) \sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}+3\,ac\ln \left ({\frac{ac\tan \left ( fx+e \right ) +\sqrt{ac \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }\sqrt{ac}}{\sqrt{ac}}} \right ) \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))^(5/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 2.21158, size = 900, normalized size = 5.84 \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))^(5/2),x, algorithm="maxima")

[Out]

(20*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*a^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*
f*x + 2*e))) + 20*I*a^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*I*a^2*sin(1/2*arctan2(sin(2*
f*x + 2*e), cos(2*f*x + 2*e))) - (6*a^2*cos(4*f*x + 4*e) + 12*a^2*cos(2*f*x + 2*e) + 6*I*a^2*sin(4*f*x + 4*e)
+ 12*I*a^2*sin(2*f*x + 2*e) + 6*a^2)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arc
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (6*a^2*cos(4*f*x + 4*e) + 12*a^2*cos(2*f*x + 2*e) + 6*I*a^2*s
in(4*f*x + 4*e) + 12*I*a^2*sin(2*f*x + 2*e) + 6*a^2)*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) + (-3*I*a^2*cos(4*f*x + 4*e) - 6*I*a^2*cos(2*f
*x + 2*e) + 3*a^2*sin(4*f*x + 4*e) + 6*a^2*sin(2*f*x + 2*e) - 3*I*a^2)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), c
os(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) + (3*I*a^2*cos(4*f*x + 4*e) + 6*I*a^2*cos(2*f*x + 2*e) - 3*a^2*sin(4*f*x + 4*e) -
6*a^2*sin(2*f*x + 2*e) + 3*I*a^2)*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))*sqrt(a
)*sqrt(c)/(f*(-4*I*cos(4*f*x + 4*e) - 8*I*cos(2*f*x + 2*e) + 4*sin(4*f*x + 4*e) + 8*sin(2*f*x + 2*e) - 4*I))

________________________________________________________________________________________

Fricas [B]  time = 1.57612, size = 919, normalized size = 5.97 \begin{align*} \frac{2 \,{\left (10 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, a^{2}\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 )} - 3 \, \sqrt{\frac{a^{5} c}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{8 \,{\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\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^{5} c}{f^{2}}}{\left (4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, f\right )}}{a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}}\right ) + 3 \, \sqrt{\frac{a^{5} c}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{8 \,{\left (a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}\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^{5} c}{f^{2}}}{\left (-4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, f\right )}}{a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}}\right )}{4 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \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))^(5/2),x, algorithm="fricas")

[Out]

1/4*(2*(10*I*a^2*e^(2*I*f*x + 2*I*e) + 6*I*a^2)*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) - 3*sqrt(a^5*c/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log((8*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*s
qrt(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^5*c/f^2)*(4*I*f*e^
(2*I*f*x + 2*I*e) - 4*I*f))/(a^2*e^(2*I*f*x + 2*I*e) + a^2)) + 3*sqrt(a^5*c/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*l
og((8*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*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^5*c/f^2)*(-4*I*f*e^(2*I*f*x + 2*I*e) + 4*I*f))/(a^2*e^(2*I*f*x + 2*I*e) + a^2)))/(f*e^(2*
I*f*x + 2*I*e) + f)

________________________________________________________________________________________

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))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(5/2)*sqrt(-I*c*tan(f*x + e) + c), x)