3.1007 \(\int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=168 \[ -\frac {3 i a^{5/2} c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {3 a^2 c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 168, 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 = {3523, 38, 63, 217, 203} \[ -\frac {3 i a^{5/2} c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {3 a^2 c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/4)*a^(5/2)*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*a^2*c^2*Tan[e + f*x]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(8*f) + (a*c*Tan[e + f*x]*
(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/(4*f)

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

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 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])

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 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]

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int (a+i a x)^{3/2} (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac {\left (3 a^2 c^2\right ) \operatorname {Subst}\left (\int \sqrt {a+i a x} \sqrt {c-i c x} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {3 a^2 c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}+\frac {\left (3 a^3 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 )}{8 f}\\ &=\frac {3 a^2 c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}-\frac {\left (3 i a^2 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 )}{4 f}\\ &=\frac {3 a^2 c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}-\frac {\left (3 i a^2 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 )}{4 f}\\ &=-\frac {3 i a^{5/2} 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 )}{4 f}+\frac {3 a^2 c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f}\\ \end {align*}

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Mathematica [A]  time = 6.23, size = 110, normalized size = 0.65 \[ -\frac {a^2 c^3 (\tan (e+f x)+i) \sec ^3(e+f x) \sqrt {a+i a \tan (e+f x)} \left (11 i \sin (e+f x)+3 i \sin (3 (e+f x))+24 \cos ^4(e+f x) \tan ^{-1}\left (e^{i (e+f x)}\right )\right )}{32 f \sqrt {c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(a^2*c^3*Sec[e + f*x]^3*(24*ArcTan[E^(I*(e + f*x))]*Cos[e + f*x]^4 + (11*I)*Sin[e + f*x] + (3*I)*Sin[3*(
e + f*x)])*(I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]])/(f*Sqrt[c - I*c*Tan[e + f*x]])

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fricas [B]  time = 0.49, size = 505, normalized size = 3.01 \[ -\frac {3 \, \sqrt {\frac {a^{5} c^{5}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {32 \, {\left (a^{2} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a^{5} c^{5}}{f^{2}}} {\left (16 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - 16 i \, f\right )}}{4 \, {\left (a^{2} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2}\right )}}\right ) - 3 \, \sqrt {\frac {a^{5} c^{5}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {32 \, {\left (a^{2} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {a^{5} c^{5}}{f^{2}}} {\left (-16 i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + 16 i \, f\right )}}{4 \, {\left (a^{2} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2}\right )}}\right ) - 4 \, {\left (-3 i \, a^{2} c^{2} e^{\left (7 i \, f x + 7 i \, e\right )} - 11 i \, a^{2} c^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + 11 i \, a^{2} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + 3 i \, a^{2} c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{16 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(3*sqrt(a^5*c^5/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)*log
(1/4*(32*(a^2*c^2*e^(3*I*f*x + 3*I*e) + a^2*c^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^5/f^2)*(16*I*f*e^(2*I*f*x + 2*I*e) - 16*I*f))/(a^2*c^2*e^(2*I*f*x + 2*I*e)
+ a^2*c^2)) - 3*sqrt(a^5*c^5/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e)
+ f)*log(1/4*(32*(a^2*c^2*e^(3*I*f*x + 3*I*e) + a^2*c^2*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqr
t(c/(e^(2*I*f*x + 2*I*e) + 1)) + sqrt(a^5*c^5/f^2)*(-16*I*f*e^(2*I*f*x + 2*I*e) + 16*I*f))/(a^2*c^2*e^(2*I*f*x
+ 2*I*e) + a^2*c^2)) - 4*(-3*I*a^2*c^2*e^(7*I*f*x + 7*I*e) - 11*I*a^2*c^2*e^(5*I*f*x + 5*I*e) + 11*I*a^2*c^2*
e^(3*I*f*x + 3*I*e) + 3*I*a^2*c^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^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.25, size = 164, normalized size = 0.98 \[ \frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{2} c^{2} \left (2 \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+3 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right )+5 \tan \left (f x +e \right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{8 f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2*c^2*(2*tan(f*x+e)^3*(c*a*(1+tan(f*x+e)^2))^(
1/2)*(c*a)^(1/2)+3*a*c*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))+5*tan(f*x+e)*
(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a*(1+tan(f*x+e)^2))^(1/2)/(c*a)^(1/2)

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maxima [B]  time = 1.34, size = 1176, normalized size = 7.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(12*a^2*c^2*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 44*a^2*c^2*cos(5/2*arctan2(sin(2*f*x + 2*e
), cos(2*f*x + 2*e))) - 44*a^2*c^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 12*a^2*c^2*cos(1/2*a
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*I*a^2*c^2*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))
) + 44*I*a^2*c^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 44*I*a^2*c^2*sin(3/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))) - 12*I*a^2*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (6*a^2*c^2*c
os(8*f*x + 8*e) + 24*a^2*c^2*cos(6*f*x + 6*e) + 36*a^2*c^2*cos(4*f*x + 4*e) + 24*a^2*c^2*cos(2*f*x + 2*e) + 6*
I*a^2*c^2*sin(8*f*x + 8*e) + 24*I*a^2*c^2*sin(6*f*x + 6*e) + 36*I*a^2*c^2*sin(4*f*x + 4*e) + 24*I*a^2*c^2*sin(
2*f*x + 2*e) + 6*a^2*c^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) + (6*a^2*c^2*cos(8*f*x + 8*e) + 24*a^2*c^2*cos(6*f*x + 6*e) + 36*a^2*c^2*c
os(4*f*x + 4*e) + 24*a^2*c^2*cos(2*f*x + 2*e) + 6*I*a^2*c^2*sin(8*f*x + 8*e) + 24*I*a^2*c^2*sin(6*f*x + 6*e) +
36*I*a^2*c^2*sin(4*f*x + 4*e) + 24*I*a^2*c^2*sin(2*f*x + 2*e) + 6*a^2*c^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*c^2*cos(8*f
*x + 8*e) - 12*I*a^2*c^2*cos(6*f*x + 6*e) - 18*I*a^2*c^2*cos(4*f*x + 4*e) - 12*I*a^2*c^2*cos(2*f*x + 2*e) + 3*
a^2*c^2*sin(8*f*x + 8*e) + 12*a^2*c^2*sin(6*f*x + 6*e) + 18*a^2*c^2*sin(4*f*x + 4*e) + 12*a^2*c^2*sin(2*f*x +
2*e) - 3*I*a^2*c^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*c^2*cos(8*f
*x + 8*e) + 12*I*a^2*c^2*cos(6*f*x + 6*e) + 18*I*a^2*c^2*cos(4*f*x + 4*e) + 12*I*a^2*c^2*cos(2*f*x + 2*e) - 3*
a^2*c^2*sin(8*f*x + 8*e) - 12*a^2*c^2*sin(6*f*x + 6*e) - 18*a^2*c^2*sin(4*f*x + 4*e) - 12*a^2*c^2*sin(2*f*x +
2*e) + 3*I*a^2*c^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*(-
16*I*cos(8*f*x + 8*e) - 64*I*cos(6*f*x + 6*e) - 96*I*cos(4*f*x + 4*e) - 64*I*cos(2*f*x + 2*e) + 16*sin(8*f*x +
8*e) + 64*sin(6*f*x + 6*e) + 96*sin(4*f*x + 4*e) + 64*sin(2*f*x + 2*e) - 16*I))

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

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)^(5/2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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