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

Optimal. Leaf size=169 \[ \frac {2 c^{3/2} (-2 B+i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a} f}+\frac {c (-2 B+i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 169, 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, 78, 50, 63, 217, 203} \[ \frac {2 c^{3/2} (-2 B+i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a} f}+\frac {c (-2 B+i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

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Mathematica [A]  time = 6.77, size = 161, normalized size = 0.95 \[ \frac {c^2 (\cos (f x)+i \sin (f x)) (\sin (f x)+i \cos (f x)) (A+B \tan (e+f x)) \left (\cos (e+f x) (\tan (e+f x)+i) (-2 i A+i B \tan (e+f x)+3 B)+2 (A+2 i B) \tan ^{-1}(\cos (e+f x)+i \sin (e+f x))\right )}{f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.78, size = 464, normalized size = 2.75 \[ \frac {{\left (a \sqrt {\frac {{\left (4 \, A^{2} + 16 i \, A B - 16 \, B^{2}\right )} c^{3}}{a f^{2}}} f e^{\left (i \, f x + i \, e\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-4 i \, A + 8 \, B\right )} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-4 i \, A + 8 \, B\right )} c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} - a f\right )} \sqrt {\frac {{\left (4 \, A^{2} + 16 i \, A B - 16 \, B^{2}\right )} c^{3}}{a f^{2}}}\right )}}{{\left (-i \, A + 2 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A + 2 \, B\right )} c}\right ) - a \sqrt {\frac {{\left (4 \, A^{2} + 16 i \, A B - 16 \, B^{2}\right )} c^{3}}{a f^{2}}} f e^{\left (i \, f x + i \, e\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-4 i \, A + 8 \, B\right )} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-4 i \, A + 8 \, B\right )} c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} - a f\right )} \sqrt {\frac {{\left (4 \, A^{2} + 16 i \, A B - 16 \, B^{2}\right )} c^{3}}{a f^{2}}}\right )}}{{\left (-i \, A + 2 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A + 2 \, B\right )} c}\right ) + 2 \, {\left ({\left (4 i \, A - 8 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (4 i \, A - 4 \, B\right )} c\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*sqrt((4*A^2 + 16*I*A*B - 16*B^2)*c^3/(a*f^2))*f*e^(I*f*x + I*e)*log(2*(((-4*I*A + 8*B)*c*e^(3*I*f*x + 3
*I*e) + (-4*I*A + 8*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))
+ (a*f*e^(2*I*f*x + 2*I*e) - a*f)*sqrt((4*A^2 + 16*I*A*B - 16*B^2)*c^3/(a*f^2)))/((-I*A + 2*B)*c*e^(2*I*f*x +
2*I*e) + (-I*A + 2*B)*c)) - a*sqrt((4*A^2 + 16*I*A*B - 16*B^2)*c^3/(a*f^2))*f*e^(I*f*x + I*e)*log(2*(((-4*I*A
+ 8*B)*c*e^(3*I*f*x + 3*I*e) + (-4*I*A + 8*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)) - (a*f*e^(2*I*f*x + 2*I*e) - a*f)*sqrt((4*A^2 + 16*I*A*B - 16*B^2)*c^3/(a*f^2)))/((-I*A
+ 2*B)*c*e^(2*I*f*x + 2*I*e) + (-I*A + 2*B)*c)) + 2*((4*I*A - 8*B)*c*e^(2*I*f*x + 2*I*e) + (4*I*A - 4*B)*c)*s
qrt(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)/(a*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.54, size = 499, normalized size = 2.95 \[ \frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (-2 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c +2 i A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c -A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c +2 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +4 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-4 B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c -B \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-2 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +2 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+3 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{f a \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (-\tan \left (f x +e \right )+i\right )^{2} \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c/a*(-2*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))*tan(f*x+e)^2*a*c+2*I*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))*tan(f*x+e)*a*c-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))*tan(f*x+e)^2*a*c+2*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))*a*c+4*I*B
*(c*a)^(1/2)*(c*a*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-4*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))*tan(f*x+e)*a*c-B*tan(f*x+e)^2*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-2*I*A*(c*a)^(1/2)*
(c*a*(1+tan(f*x+e)^2))^(1/2)+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
*A*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)+3*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a*(1+t
an(f*x+e)^2))^(1/2)/(-tan(f*x+e)+I)^2/(c*a)^(1/2)

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maxima [B]  time = 0.92, size = 904, normalized size = 5.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*(A + 2*I*B)*c*cos(2*f*x + 2*e) - (-4*I*A + 8*B)*c*sin(2*f*x + 2*e) + 4*(A + I*B)*c + (2*(A + 2*I*B)*c*cos(3
/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*(A + 2*I*B)*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) - (-2*I*A + 4*B)*c*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-2*I*A + 4*B)*c*sin(1/2*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*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) + (2*(A + 2*I*B)*c*cos(3/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))) + 2*(A + 2*I*B)*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-2*I*A + 4*B)*c*sin(
3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-2*I*A + 4*B)*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e))))*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), c
os(2*f*x + 2*e))) + 1) - ((-I*A + 2*B)*c*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (-I*A + 2*B)*c
*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (A + 2*I*B)*c*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*
f*x + 2*e))) + (A + 2*I*B)*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*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(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - ((I*A - 2*B)*c*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))
+ (I*A - 2*B)*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (A + 2*I*B)*c*sin(3/2*arctan2(sin(2*f*
x + 2*e), cos(2*f*x + 2*e))) - (A + 2*I*B)*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*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*s
in(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1))*sqrt(a)*sqrt(c)/((-2*I*a*cos(3/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))) - 2*I*a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*a*sin(3/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*a*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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