3.8.99 \(\int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [799]

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

[Out]

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

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Rubi [A]
time = 0.18, 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 = {3669, 79, 52, 65, 223, 209} \begin {gather*} \frac {2 a^{3/2} (2 B+i A) \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 )}{\sqrt {c} f}-\frac {a (2 B+i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c f}-\frac {(B+i A) (a+i a \tan (e+f x))^{3/2}}{f \sqrt {c-i c \tan (e+f x)}} \end {gather*}

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]

(2*a^(3/2)*(I*A + 2*B)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(Sqr
t[c]*f) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(3/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (a*(I*A + 2*B)*Sqrt[a + I*
a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(c*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 79

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] || L
tQ[p, n]))))

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 3669

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

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Mathematica [A]
time = 3.02, size = 190, normalized size = 1.12 \begin {gather*} \frac {2 a e^{-2 i (e+f x)} \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \left (e^{i (e+f x)} \left (A \left (1+e^{2 i (e+f x)}\right )-i B \left (2+e^{2 i (e+f x)}\right )\right )-(A-2 i B) \left (1+e^{2 i (e+f x)}\right ) \text {ArcTan}\left (e^{i (e+f x)}\right )\right ) (-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}{c f \sec ^{\frac {3}{2}}(e+f x)} \end {gather*}

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]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 496 vs. \(2 (140 ) = 280\).
time = 0.42, size = 497, normalized size = 2.94

method	result	size
derivativedivides	\(\frac {\left (2 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-2 i A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-2 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c -4 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-4 i B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )-B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )+A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c +2 i A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+2 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )+3 B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right ) \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a}{f c \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (i+\tan \left (f x +e \right )\right )^{2} \sqrt {a c}}\)	\(497\)
default	\(\frac {\left (2 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-2 i A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-2 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c -4 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-4 i B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )-B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )+A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {a c}}\right ) a c +2 i A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+2 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )+3 B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right ) \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a}{f c \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (i+\tan \left (f x +e \right )\right )^{2} \sqrt {a c}}\)	\(497\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(2*I*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^2-2*I*A*ln
((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)-A*ln((a*c*tan(f*x+e)+(a
*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^2-2*I*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*
c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c-4*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(
a*c)^(1/2))*a*c*tan(f*x+e)-4*I*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)-B*(a*c*(1+tan(f*x+e)^2))^
(1/2)*(a*c)^(1/2)*tan(f*x+e)^2+A*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c
+2*I*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+2*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)+3*B*(a
*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a/c/(a*c*(1+tan
(f*x+e)^2))^(1/2)/(I+tan(f*x+e))^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. 637 vs. \(2 (137) = 274\).
time = 0.64, size = 637, normalized size = 3.77 \begin {gather*} \frac {{\left (2 \, {\left ({\left (A - 2 i \, B\right )} a \cos \left (2 \, f x + 2 \, e\right ) - {\left (-i \, A - 2 \, B\right )} a \sin \left (2 \, f x + 2 \, e\right ) + {\left (A - 2 i \, B\right )} a\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 ) + 2 \, {\left ({\left (A - 2 i \, B\right )} a \cos \left (2 \, f x + 2 \, e\right ) - {\left (-i \, A - 2 \, B\right )} a \sin \left (2 \, f x + 2 \, e\right ) + {\left (A - 2 i \, B\right )} a\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 ) - 4 \, {\left ({\left (A - i \, B\right )} a \cos \left (2 \, f x + 2 \, e\right ) + {\left (i \, A + B\right )} a \sin \left (2 \, f x + 2 \, e\right ) + {\left (A - 2 i \, B\right )} a\right )} \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 ) + {\left ({\left (i \, A + 2 \, B\right )} a \cos \left (2 \, f x + 2 \, e\right ) - {\left (A - 2 i \, B\right )} a \sin \left (2 \, f x + 2 \, e\right ) + {\left (i \, A + 2 \, B\right )} a\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 ) + {\left ({\left (-i \, A - 2 \, B\right )} a \cos \left (2 \, f x + 2 \, e\right ) + {\left (A - 2 i \, B\right )} a \sin \left (2 \, f x + 2 \, e\right ) + {\left (-i \, A - 2 \, B\right )} a\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 ) - 4 \, {\left ({\left (i \, A + B\right )} a \cos \left (2 \, f x + 2 \, e\right ) - {\left (A - i \, B\right )} a \sin \left (2 \, f x + 2 \, e\right ) + {\left (i \, A + 2 \, B\right )} a\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 )\right )} \sqrt {a} \sqrt {c}}{-2 \, {\left (i \, c \cos \left (2 \, f x + 2 \, e\right ) - c \sin \left (2 \, f x + 2 \, e\right ) + i \, c\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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. 454 vs. \(2 (137) = 274\).
time = 4.43, size = 454, normalized size = 2.69 \begin {gather*} \frac {c \sqrt {\frac {{\left (A^{2} - 4 i \, A B - 4 \, B^{2}\right )} a^{3}}{c f^{2}}} f \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-i \, A - 2 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-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}} + {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )} \sqrt {\frac {{\left (A^{2} - 4 i \, A B - 4 \, B^{2}\right )} a^{3}}{c f^{2}}}\right )}}{{\left (i \, A + 2 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A + 2 \, B\right )} a}\right ) - c \sqrt {\frac {{\left (A^{2} - 4 i \, A B - 4 \, B^{2}\right )} a^{3}}{c f^{2}}} f \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-i \, A - 2 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-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}} - {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )} \sqrt {\frac {{\left (A^{2} - 4 i \, A B - 4 \, B^{2}\right )} a^{3}}{c f^{2}}}\right )}}{{\left (i \, A + 2 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A + 2 \, B\right )} a}\right ) - 4 \, {\left ({\left (i \, A + B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (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}}}{2 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (i a \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 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((a+I*a*tan(f*x+e))**(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(1/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\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 \end {gather*}

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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