3.10.92 \(\int (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)} \, dx\) [992]

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

[Out]

-3*I*a^(5/2)*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))*c^(1/2)/f+3/2*I*a^2*(a+
I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/f+1/2*I*a*(c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2)/f

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Rubi [A]
time = 0.12, antiderivative size = 154, normalized size of antiderivative = 1.00, 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 = {3604, 52, 65, 223, 209} \begin {gather*} -\frac {3 i a^{5/2} \sqrt {c} \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 )}{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 {gather*}

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 52

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 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 (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)} \, dx &=\frac {(a c) \text {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 ) \text {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 ) \text {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 ) \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 {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 ) \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 {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*}

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Mathematica [A]
time = 2.87, size = 87, normalized size = 0.56 \begin {gather*} \frac {a^2 c \left (4-6 \text {ArcTan}\left (e^{i (e+f x)}\right ) \cos (e+f x)+i \tan (e+f x)\right ) (i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}{2 f \sqrt {c-i c \tan (e+f x)}} \end {gather*}

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]])

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Maple [A]
time = 0.44, size = 154, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a^{2} \left (4 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}-\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+3 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )\right )}{2 f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\)	\(154\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a^{2} \left (4 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}-\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+3 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )\right )}{2 f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\)	\(154\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 710 vs. \(2 (118) = 236\).
time = 0.60, size = 710, normalized size = 4.61 \begin {gather*} \frac {{\left (20 \, a^{2} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 12 \, a^{2} \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 ) + 20 i \, a^{2} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 12 i \, a^{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 ) - 6 \, {\left (a^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, a^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 i \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) + a^{2}\right )} \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 ) - 6 \, {\left (a^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, a^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 i \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) + a^{2}\right )} \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 ) - 3 \, {\left (i \, a^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) - a^{2} \sin \left (4 \, f x + 4 \, e\right ) - 2 \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) + i \, a^{2}\right )} \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 ) - 3 \, {\left (-i \, a^{2} \cos \left (4 \, f x + 4 \, e\right ) - 2 i \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) + a^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) - i \, a^{2}\right )} \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 )\right )} \sqrt {a} \sqrt {c}}{-4 \, f {\left (i \, \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, \cos \left (2 \, f x + 2 \, e\right ) - \sin \left (4 \, f x + 4 \, e\right ) - 2 \, \sin \left (2 \, f x + 2 \, e\right ) + i\right )}} \end {gather*}

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) + 2*a^2*cos(2*f*x + 2*e) + I*a^2*sin(4*f*x + 4*e) + 2
*I*a^2*sin(2*f*x + 2*e) + a^2)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 6*(a^2*cos(4*f*x + 4*e) + 2*a^2*cos(2*f*x + 2*e) + I*a^2*sin(4*f*x
+ 4*e) + 2*I*a^2*sin(2*f*x + 2*e) + 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) + 2*I*a^2*cos(2*f*x + 2*e) - a
^2*sin(4*f*x + 4*e) - 2*a^2*sin(2*f*x + 2*e) + I*a^2)*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) - 3*(-I*a^2*cos(4*f*x + 4*e) - 2*I*a^2*cos(2*f*x + 2*e) + a^2*sin(4*f*x + 4*e) + 2*a^2*sin(2*f*x +
2*e) - I*a^2)*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))*sqrt(a)*sqrt(c)/(f*(-4*I*c
os(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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (118) = 236\).
time = 1.07, size = 380, normalized size = 2.47 \begin {gather*} \frac {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 {4 \, {\left (2 \, {\left (a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} 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}} - \sqrt {\frac {a^{5} c}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\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 {4 \, {\left (2 \, {\left (a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} 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}} - \sqrt {\frac {a^{5} c}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2}}\right ) - 4 \, {\left (-5 i \, a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} - 3 i \, a^{2} 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}}}{4 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

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*(3*sqrt(a^5*c/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log(4*(2*(a^2*e^(3*I*f*x + 3*I*e) + a^2*e^(I*f*x + I*e))*sq
rt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - sqrt(a^5*c/f^2)*(I*f*e^(2*I*f*x + 2*I*e) -
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)*log(4*(2*(a^2*e^(3*I*f
*x + 3*I*e) + a^2*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)) - sqrt(
a^5*c/f^2)*(-I*f*e^(2*I*f*x + 2*I*e) + I*f))/(a^2*e^(2*I*f*x + 2*I*e) + a^2)) - 4*(-5*I*a^2*e^(3*I*f*x + 3*I*e
) - 3*I*a^2*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)))/(f*e^(2*I*f*
x + 2*I*e) + f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \sqrt {- i c \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))**(1/2)*(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

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

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (118) = 236\).
time = 1.45, size = 325, normalized size = 2.11 \begin {gather*} \frac {15 \, {\left (a^{3} c - a^{2} c\right )} \sqrt {-a c} e^{\left (9 i \, f x + 9 i \, e\right )} + 74 \, {\left (a^{3} c - a^{2} c\right )} \sqrt {-a c} e^{\left (7 i \, f x + 7 i \, e\right )} + 132 \, {\left (a^{3} c - a^{2} c\right )} \sqrt {-a c} e^{\left (5 i \, f x + 5 i \, e\right )} + 102 \, {\left (a^{3} c - a^{2} c\right )} \sqrt {-a c} e^{\left (3 i \, f x + 3 i \, e\right )} + 29 \, {\left (a^{3} c - a^{2} c\right )} \sqrt {-a c} e^{\left (i \, f x + i \, e\right )}}{4 \, {\left ({\left (a - 1\right )} c f e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, {\left (a - 1\right )} c f e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, {\left (a - 1\right )} c f e^{\left (6 i \, f x + 6 i \, e\right )} + 10 \, {\left (a - 1\right )} c f e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, {\left (a - 1\right )} c f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (a - 1\right )} c f\right )}} - \frac {12 i \, a^{\frac {5}{2}} \sqrt {c} \arctan \left (e^{\left (i \, f x + i \, e\right )}\right ) - \frac {i \, {\left (5 \, a^{\frac {5}{2}} \sqrt {c} e^{\left (3 i \, f x + 3 i \, e\right )} + 7 \, a^{\frac {5}{2}} \sqrt {c} e^{\left (i \, f x + i \, e\right )}\right )}}{{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}^{2}}}{4 \, f} \end {gather*}

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]

1/4*(15*(a^3*c - a^2*c)*sqrt(-a*c)*e^(9*I*f*x + 9*I*e) + 74*(a^3*c - a^2*c)*sqrt(-a*c)*e^(7*I*f*x + 7*I*e) + 1
32*(a^3*c - a^2*c)*sqrt(-a*c)*e^(5*I*f*x + 5*I*e) + 102*(a^3*c - a^2*c)*sqrt(-a*c)*e^(3*I*f*x + 3*I*e) + 29*(a
^3*c - a^2*c)*sqrt(-a*c)*e^(I*f*x + I*e))/((a - 1)*c*f*e^(10*I*f*x + 10*I*e) + 5*(a - 1)*c*f*e^(8*I*f*x + 8*I*
e) + 10*(a - 1)*c*f*e^(6*I*f*x + 6*I*e) + 10*(a - 1)*c*f*e^(4*I*f*x + 4*I*e) + 5*(a - 1)*c*f*e^(2*I*f*x + 2*I*
e) + (a - 1)*c*f) - 1/4*(12*I*a^(5/2)*sqrt(c)*arctan(e^(I*f*x + I*e)) - I*(5*a^(5/2)*sqrt(c)*e^(3*I*f*x + 3*I*
e) + 7*a^(5/2)*sqrt(c)*e^(I*f*x + I*e))/(e^(2*I*f*x + 2*I*e) + 1)^2)/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,\sqrt {c-c\,\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((a + a*tan(e + f*x)*1i)^(5/2)*(c - c*tan(e + f*x)*1i)^(1/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^(5/2)*(c - c*tan(e + f*x)*1i)^(1/2), x)

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