3.1009 \(\int \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.154225, 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} \[ -\frac{3 i \sqrt{a} c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c-i c \tan (e+f x)}}\right )}{f}-\frac{3 i c^2 \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)}}{2 f}-\frac{i c \sqrt{a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 2.74184, size = 155, normalized size = 1.01 \[ -\frac{i c e^{-i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \left (\frac{c}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (e^{i (e+f x)} \left (5+3 e^{2 i (e+f x)}\right )+3 \left (1+e^{2 i (e+f x)}\right )^2 \tan ^{-1}\left (e^{i (e+f x)}\right )\right ) \sqrt{a+i a \tan (e+f x)}}{f \sqrt{\sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*c*(c/(1 + E^((2*I)*(e + f*x))))^(3/2)*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*(E^(I*(e + f*x))*(
5 + 3*E^((2*I)*(e + f*x))) + 3*(1 + E^((2*I)*(e + f*x)))^2*ArcTan[E^(I*(e + f*x))])*Sqrt[a + I*a*Tan[e + f*x]]
)/(E^(I*(e + f*x))*f*Sqrt[Sec[e + f*x]])

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Maple [A]  time = 0.072, size = 153, normalized size = 1. \begin{align*} -{\frac{{c}^{2}}{2\,f}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \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((a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

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

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Maxima [B]  time = 2.07983, size = 608, normalized size = 3.95 \begin{align*} -\frac{{\left (12 \, c^{2} \cos \left (3 \, f x + 3 \, e\right ) + 20 \, c^{2} \cos \left (f x + e\right ) + 12 i \, c^{2} \sin \left (3 \, f x + 3 \, e\right ) + 20 i \, c^{2} \sin \left (f x + e\right ) +{\left (6 \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 12 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + 6 i \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 12 i \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + 6 \, c^{2}\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) +{\left (6 \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 12 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + 6 i \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 12 i \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + 6 \, c^{2}\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) -{\left (-3 i \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) - 6 i \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + 3 \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 6 \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) - 3 i \, c^{2}\right )} \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 ) -{\left (3 i \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 6 i \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) - 3 \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) - 6 \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + 3 i \, c^{2}\right )} \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 )\right )} \sqrt{a} \sqrt{c}}{f{\left (-4 i \, \cos \left (4 \, f x + 4 \, e\right ) - 8 i \, \cos \left (2 \, f x + 2 \, e\right ) + 4 \, \sin \left (4 \, f x + 4 \, e\right ) + 8 \, \sin \left (2 \, f x + 2 \, e\right ) - 4 i\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-(12*c^2*cos(3*f*x + 3*e) + 20*c^2*cos(f*x + e) + 12*I*c^2*sin(3*f*x + 3*e) + 20*I*c^2*sin(f*x + e) + (6*c^2*c
os(4*f*x + 4*e) + 12*c^2*cos(2*f*x + 2*e) + 6*I*c^2*sin(4*f*x + 4*e) + 12*I*c^2*sin(2*f*x + 2*e) + 6*c^2)*arct
an2(cos(f*x + e), sin(f*x + e) + 1) + (6*c^2*cos(4*f*x + 4*e) + 12*c^2*cos(2*f*x + 2*e) + 6*I*c^2*sin(4*f*x +
4*e) + 12*I*c^2*sin(2*f*x + 2*e) + 6*c^2)*arctan2(cos(f*x + e), -sin(f*x + e) + 1) - (-3*I*c^2*cos(4*f*x + 4*e
) - 6*I*c^2*cos(2*f*x + 2*e) + 3*c^2*sin(4*f*x + 4*e) + 6*c^2*sin(2*f*x + 2*e) - 3*I*c^2)*log(cos(f*x + e)^2 +
sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - (3*I*c^2*cos(4*f*x + 4*e) + 6*I*c^2*cos(2*f*x + 2*e) - 3*c^2*sin(4*f*x
+ 4*e) - 6*c^2*sin(2*f*x + 2*e) + 3*I*c^2)*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + 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))

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Fricas [B]  time = 1.51091, size = 921, normalized size = 5.98 \begin{align*} \frac{2 \,{\left (-6 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 10 i \, c^{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 c^{5}}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{8 \,{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{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 c^{5}}{f^{2}}}{\left (4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, f\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}}\right ) + 3 \, \sqrt{\frac{a c^{5}}{f^{2}}}{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac{8 \,{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{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 c^{5}}{f^{2}}}{\left (-4 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, f\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{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((a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/4*(2*(-6*I*c^2*e^(2*I*f*x + 2*I*e) - 10*I*c^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*c^5/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log((8*(c^2*e^(2*I*f*x + 2*I*e) + c^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*c^5/f^2)*(4*I*f*e
^(2*I*f*x + 2*I*e) - 4*I*f))/(c^2*e^(2*I*f*x + 2*I*e) + c^2)) + 3*sqrt(a*c^5/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*
log((8*(c^2*e^(2*I*f*x + 2*I*e) + c^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*c^5/f^2)*(-4*I*f*e^(2*I*f*x + 2*I*e) + 4*I*f))/(c^2*e^(2*I*f*x + 2*I*e) + c^2)))/(f*e^(2
*I*f*x + 2*I*e) + 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((a+I*a*tan(f*x+e))**(1/2)*(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (f x + e\right ) + a}{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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