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

Optimal. Leaf size=160 \[ -\frac {a^{3/2} \sqrt {c} (B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a (B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f} \]

[Out]

-a^(3/2)*(2*I*A+B)*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+1/2*a*(
2*I*A+B)*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/f+1/2*B*(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 = 0.26, antiderivative size = 160, 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} \[ -\frac {a^{3/2} \sqrt {c} (B+2 i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a (B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^(3/2)*((2*I)*A + B)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x
]])])/f) + (a*((2*I)*A + B)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*f) + (B*(a + I*a*Tan[e +
f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]])/(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 (a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {\sqrt {a+i a x} (A+B x)}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {(a (2 A-i B) c) \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 {a (2 i A+B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {\left (a^2 (2 A-i B) 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 {a (2 i A+B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {(a (2 i A+B) c) \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 {a (2 i A+B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {(a (2 i A+B) c) \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 {a^{3/2} (2 i A+B) \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 {a (2 i A+B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {B (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 = 6.80, size = 220, normalized size = 1.38 \[ \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \left (\frac {\cos (e) (\tan (e)+i) \sqrt {\sec (e+f x)} \sqrt {c-i c \tan (e+f x)} (2 A+B \tan (e+f x)-2 i B)}{2 \cos (f x)+2 i \sin (f x)}-\frac {i c (2 A-i B) e^{-2 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right )}{\sqrt {\frac {c}{1+e^{2 i (e+f x)}}}}\right )}{f \sec ^{\frac {5}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.72, size = 458, normalized size = 2.86 \[ \frac {\sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (8 i \, A + 4 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (8 i \, A + 4 \, B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 2 \, \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (2 i \, A + B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A + B\right )} a}\right ) - \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {2 \, {\left ({\left ({\left (8 i \, A + 4 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (8 i \, A + 4 \, B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - 2 \, \sqrt {\frac {{\left (4 \, A^{2} - 4 i \, A B - B^{2}\right )} a^{3} c}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (2 i \, A + B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A + B\right )} a}\right ) + 2 \, {\left ({\left (4 i \, A + 6 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (4 i \, A + 2 \, B\right )} a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt((4*A^2 - 4*I*A*B - B^2)*a^3*c/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((8*I*A + 4*B)*a*e^(3*I*f*x +
3*I*e) + (8*I*A + 4*B)*a*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^3*c/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((2*I*A + B)*a*e^(2*I*f*x + 2*I*e) +
(2*I*A + B)*a)) - sqrt((4*A^2 - 4*I*A*B - B^2)*a^3*c/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log(2*(((8*I*A + 4*B)*a*
e^(3*I*f*x + 3*I*e) + (8*I*A + 4*B)*a*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^3*c/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((2*I*A + B)*a*e^(2*I*f*
x + 2*I*e) + (2*I*A + B)*a)) + 2*((4*I*A + 6*B)*a*e^(3*I*f*x + 3*I*e) + (4*I*A + 2*B)*a*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 [B]  time = 23.46, size = 565, normalized size = 3.53 \[ \frac {-3 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (8 i \, f x + 8 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) - 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) - 18 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) - 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) + 3 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (8 i \, f x + 8 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 18 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 12 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right ) + 10 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (7 i \, f x + 7 i \, e\right )} + 26 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (5 i \, f x + 5 i \, e\right )} + 22 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (3 i \, f x + 3 i \, e\right )} + 6 \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (i \, f x + i \, e\right )} - 3 i \, B a^{\frac {3}{2}} \sqrt {c} \log \left (e^{\left (i \, f x + i \, e\right )} + i\right ) + 3 i \, B a^{\frac {3}{2}} \sqrt {c} \log \left (e^{\left (i \, f x + i \, e\right )} - i\right )}{8 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} - \frac {i \, {\left ({\left (8 \, A a^{\frac {3}{2}} \sqrt {c} - i \, B a^{\frac {3}{2}} \sqrt {c}\right )} \arctan \left (e^{\left (i \, f x + i \, e\right )}\right ) - \frac {8 \, A a^{\frac {3}{2}} \sqrt {c} e^{\left (3 i \, f x + 3 i \, e\right )} - 7 i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (3 i \, f x + 3 i \, e\right )} + 8 \, A a^{\frac {3}{2}} \sqrt {c} e^{\left (i \, f x + i \, e\right )} - i \, B a^{\frac {3}{2}} \sqrt {c} e^{\left (i \, f x + i \, e\right )}}{{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}^{2}}\right )}}{4 \, f} \]

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))^(3/2)*(A+B*tan(f*x+e)),x, algorithm="giac")

[Out]

1/8*(-3*I*B*a^(3/2)*sqrt(c)*e^(8*I*f*x + 8*I*e)*log(e^(I*f*x + I*e) + I) - 12*I*B*a^(3/2)*sqrt(c)*e^(6*I*f*x +
6*I*e)*log(e^(I*f*x + I*e) + I) - 18*I*B*a^(3/2)*sqrt(c)*e^(4*I*f*x + 4*I*e)*log(e^(I*f*x + I*e) + I) - 12*I*
B*a^(3/2)*sqrt(c)*e^(2*I*f*x + 2*I*e)*log(e^(I*f*x + I*e) + I) + 3*I*B*a^(3/2)*sqrt(c)*e^(8*I*f*x + 8*I*e)*log
(e^(I*f*x + I*e) - I) + 12*I*B*a^(3/2)*sqrt(c)*e^(6*I*f*x + 6*I*e)*log(e^(I*f*x + I*e) - I) + 18*I*B*a^(3/2)*s
qrt(c)*e^(4*I*f*x + 4*I*e)*log(e^(I*f*x + I*e) - I) + 12*I*B*a^(3/2)*sqrt(c)*e^(2*I*f*x + 2*I*e)*log(e^(I*f*x
+ I*e) - I) + 10*B*a^(3/2)*sqrt(c)*e^(7*I*f*x + 7*I*e) + 26*B*a^(3/2)*sqrt(c)*e^(5*I*f*x + 5*I*e) + 22*B*a^(3/
2)*sqrt(c)*e^(3*I*f*x + 3*I*e) + 6*B*a^(3/2)*sqrt(c)*e^(I*f*x + I*e) - 3*I*B*a^(3/2)*sqrt(c)*log(e^(I*f*x + I*
e) + I) + 3*I*B*a^(3/2)*sqrt(c)*log(e^(I*f*x + I*e) - I))/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6
*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f) - 1/4*I*((8*A*a^(3/2)*sqrt(c) - I*B*a^(3/2)*sqrt(c))*arc
tan(e^(I*f*x + I*e)) - (8*A*a^(3/2)*sqrt(c)*e^(3*I*f*x + 3*I*e) - 7*I*B*a^(3/2)*sqrt(c)*e^(3*I*f*x + 3*I*e) +
8*A*a^(3/2)*sqrt(c)*e^(I*f*x + I*e) - I*B*a^(3/2)*sqrt(c)*e^(I*f*x + I*e))/(e^(2*I*f*x + 2*I*e) + 1)^2)/f

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maple [A]  time = 0.61, size = 223, normalized size = 1.39 \[ \frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (-i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+2 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+2 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +2 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{2 f \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*a*(-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+I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)+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 = 0.96, size = 779, normalized size = 4.87 \[ \text {result too large to display} \]

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))^(3/2)*(A+B*tan(f*x+e)),x, algorithm="maxima")

[Out]

(16*(2*A - 3*I*B)*a*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*(2*A - I*B)*a*cos(1/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-32*I*A - 48*B)*a*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) -
(-32*I*A - 16*B)*a*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (8*(2*A - I*B)*a*cos(4*f*x + 4*e) +
16*(2*A - I*B)*a*cos(2*f*x + 2*e) + (16*I*A + 8*B)*a*sin(4*f*x + 4*e) + (32*I*A + 16*B)*a*sin(2*f*x + 2*e) +
8*(2*A - I*B)*a)*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)*a*cos(4*f*x + 4*e) + 16*(2*A - I*B)*a*cos(2*f*x + 2*e) + (16*I*A +
8*B)*a*sin(4*f*x + 4*e) + (32*I*A + 16*B)*a*sin(2*f*x + 2*e) + 8*(2*A - I*B)*a)*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)*a
*cos(4*f*x + 4*e) + (16*I*A + 8*B)*a*cos(2*f*x + 2*e) - 4*(2*A - I*B)*a*sin(4*f*x + 4*e) - 8*(2*A - I*B)*a*sin
(2*f*x + 2*e) + (8*I*A + 4*B)*a)*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)*a*cos(4*f*x + 4*e) + (-16*I*A - 8*B)*a*cos(2*f*x + 2*e) + 4*(2*A - I*B)*a*sin(4*f*x + 4*e) + 8*(2*A -
I*B)*a*sin(2*f*x + 2*e) + (-8*I*A - 4*B)*a)*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 \[ \int \left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \]

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

[Out]

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

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

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))**(3/2)*(A+B*tan(f*x+e)),x)

[Out]

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

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