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

Optimal. Leaf size=164 \[ -\frac {\sqrt {a} c^{3/2} (-B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}-\frac {c (-B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3588, 80, 50, 63, 217, 203} \[ -\frac {\sqrt {a} c^{3/2} (-B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}-\frac {c (-B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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])

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 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((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)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 6.16, size = 159, normalized size = 0.97 \[ \frac {c^2 e^{-i e} \left (\sin \left (\frac {e}{2}\right )-i \cos \left (\frac {e}{2}\right )\right ) \sqrt {a+i a \tan (e+f x)} \left (\cos \left (\frac {e}{2}+f x\right )-i \sin \left (\frac {e}{2}+f x\right )\right ) \left ((4 A+2 i B) \tan ^{-1}\left (e^{i (e+f x)}\right )+\sec (e+f x) (2 A+B \sec (e) \sin (f x) \sec (e+f x)+B \tan (e)+2 i B)\right )}{2 \sqrt {2} f \sqrt {\frac {c}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.99, size = 458, normalized size = 2.79 \[ -\frac {\sqrt {\frac {{\left (4 \, A^{2} + 4 i \, A B - B^{2}\right )} a c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-8 i \, A + 4 \, B\right )} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-8 i \, A + 4 \, B\right )} c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 2 \, \sqrt {\frac {{\left (4 \, A^{2} + 4 i \, A B - B^{2}\right )} a c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-2 i \, A + B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-2 i \, A + B\right )} c}\right ) - \sqrt {\frac {{\left (4 \, A^{2} + 4 i \, A B - B^{2}\right )} a c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-8 i \, A + 4 \, B\right )} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-8 i \, A + 4 \, B\right )} c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - 2 \, \sqrt {\frac {{\left (4 \, A^{2} + 4 i \, A B - B^{2}\right )} a c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-2 i \, A + B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-2 i \, A + B\right )} c}\right ) - 2 \, {\left ({\left (-4 i \, A + 2 \, B\right )} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-4 i \, A + 6 \, B\right )} c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/4*(sqrt((4*A^2 + 4*I*A*B - B^2)*a*c^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((-8*I*A + 4*B)*c*e^(3*I*f*x
+ 3*I*e) + (-8*I*A + 4*B)*c*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
)) + 2*sqrt((4*A^2 + 4*I*A*B - B^2)*a*c^3/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-2*I*A + B)*c*e^(2*I*f*x + 2*I*e
) + (-2*I*A + B)*c)) - sqrt((4*A^2 + 4*I*A*B - B^2)*a*c^3/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((-8*I*A + 4
*B)*c*e^(3*I*f*x + 3*I*e) + (-8*I*A + 4*B)*c*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)) - 2*sqrt((4*A^2 + 4*I*A*B - B^2)*a*c^3/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-2*I*A + B)*c*e
^(2*I*f*x + 2*I*e) + (-2*I*A + B)*c)) - 2*((-4*I*A + 2*B)*c*e^(3*I*f*x + 3*I*e) + (-4*I*A + 6*B)*c*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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giac [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))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.63, size = 223, normalized size = 1.36 \[ -\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, c \left (i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +2 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-2 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -2 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{2 f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x)

[Out]

-1/2/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*c*(I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)
*tan(f*x+e)-I*B*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))*c*a+2*I*A*(c*a*(1+ta
n(f*x+e)^2))^(1/2)*(c*a)^(1/2)-2*A*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))*a
*c-2*B*(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 = 1.03, size = 782, normalized size = 4.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-(16*(2*A + I*B)*c*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*(2*A + 3*I*B)*c*cos(1/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-32*I*A + 16*B)*c*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
- (-32*I*A + 48*B)*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (8*(2*A + I*B)*c*cos(4*f*x + 4*e)
+ 16*(2*A + I*B)*c*cos(2*f*x + 2*e) - (-16*I*A + 8*B)*c*sin(4*f*x + 4*e) - (-32*I*A + 16*B)*c*sin(2*f*x + 2*e)
+ 8*(2*A + I*B)*c)*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) + (8*(2*A + I*B)*c*cos(4*f*x + 4*e) + 16*(2*A + I*B)*c*cos(2*f*x + 2*e) - (-16*I
*A + 8*B)*c*sin(4*f*x + 4*e) - (-32*I*A + 16*B)*c*sin(2*f*x + 2*e) + 8*(2*A + I*B)*c)*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) - ((-8*I*A +
4*B)*c*cos(4*f*x + 4*e) + (-16*I*A + 8*B)*c*cos(2*f*x + 2*e) + 4*(2*A + I*B)*c*sin(4*f*x + 4*e) + 8*(2*A + I*B
)*c*sin(2*f*x + 2*e) + (-8*I*A + 4*B)*c)*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) -
((8*I*A - 4*B)*c*cos(4*f*x + 4*e) + (16*I*A - 8*B)*c*cos(2*f*x + 2*e) - 4*(2*A + I*B)*c*sin(4*f*x + 4*e) - 8*
(2*A + I*B)*c*sin(2*f*x + 2*e) + (8*I*A - 4*B)*c)*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(4*f*x + 4*e) - 32*I*cos(2*f*x + 2*e) + 16*sin(4*f*x + 4*e) + 32*sin(2*f
*x + 2*e) - 16*I))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(1/2)*(c - c*tan(e + f*x)*1i)^(3/2),x)

[Out]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(1/2)*(c - c*tan(e + f*x)*1i)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

Integral(sqrt(I*a*(tan(e + f*x) - I))*(-I*c*(tan(e + f*x) + I))**(3/2)*(A + B*tan(e + f*x)), x)

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