3.9.9 \(\int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [809]

Optimal. Leaf size=227 \[ \frac {3 a^{5/2} (2 i A+3 B) \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 a^2 (2 i A+3 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f} \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 7, 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 {3 a^{5/2} (3 B+2 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 {3 a^2 (3 B+2 i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (3 B+2 i A) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^(5/2)*((2*I)*A + 3*B)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/
(Sqrt[c]*f) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (3*a^2*((2*I)*A + 3*B)
*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/(2*c*f) - (a*((2*I)*A + 3*B)*(a + I*a*Tan[e + f*x])^(3
/2)*Sqrt[c - I*c*Tan[e + f*x]])/(2*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))^{5/2} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(a+i a x)^{3/2} (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))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {(a (2 A-3 i B)) \text {Subst}\left (\int \frac {(a+i a x)^{3/2}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {\left (3 a^2 (2 A-3 i B)\right ) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 a^2 (2 i A+3 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {\left (3 a^3 (2 A-3 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 )}{2 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 a^2 (2 i A+3 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}+\frac {\left (3 a^2 (2 i A+3 B)\right ) \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))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 a^2 (2 i A+3 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}+\frac {\left (3 a^2 (2 i A+3 B)\right ) \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 {3 a^{5/2} (2 i A+3 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))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {3 a^2 (2 i A+3 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {a (2 i A+3 B) (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f}\\ \end {align*}

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Mathematica [A]
time = 4.72, size = 239, normalized size = 1.05 \begin {gather*} \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) \left (\frac {3 (2 i A+3 B) e^{-3 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \text {ArcTan}\left (e^{i (e+f x)}\right )}{\sqrt {\frac {c}{1+e^{2 i (e+f x)}}}}-\frac {\sqrt {\sec (e+f x)} (5 (2 A-3 i B)+(10 A-13 i B) \cos (2 (e+f x))+(-2 i A-5 B) \sin (2 (e+f x))) (i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{4 c}\right )}{f \sec ^{\frac {7}{2}}(e+f x) (A \cos (e+f x)+B \sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x])*((3*((2*I)*A + 3*B)*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e +
f*x)))]*ArcTan[E^(I*(e + f*x))])/(E^((3*I)*(e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]) - (Sqrt[Sec[e + f*x]
]*(5*(2*A - (3*I)*B) + (10*A - (13*I)*B)*Cos[2*(e + f*x)] + ((-2*I)*A - 5*B)*Sin[2*(e + f*x)])*(I + Tan[e + f*
x])*Sqrt[c - I*c*Tan[e + f*x]])/(4*c)))/(f*Sec[e + f*x]^(7/2)*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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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. 564 vs. \(2 (187 ) = 374\).
time = 0.42, size = 565, normalized size = 2.49

method	result	size
derivativedivides	\(\frac {i \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^{2} \left (6 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 \left (\tan ^{2}\left (f x +e \right )\right )+18 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 \tan \left (f x +e \right )+9 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 )+4 i 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 )-B \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )-6 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 -12 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 )-12 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )-2 A \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 )-9 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 -14 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}-19 B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )+10 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{2 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}}\)	\(565\)
default	\(\frac {i \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^{2} \left (6 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 \left (\tan ^{2}\left (f x +e \right )\right )+18 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 \tan \left (f x +e \right )+9 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 )+4 i 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 )-B \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )-6 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 -12 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 )-12 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )-2 A \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 )-9 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 -14 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}-19 B \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )+10 A \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{2 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}}\)	\(565\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^2/c*(6*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)^2+18*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)+9*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+4*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^2-B*(a*c)^(1/2)*(a*c*(1+
tan(f*x+e)^2))^(1/2)*tan(f*x+e)^3-6*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-12*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)-12*I*A*(
a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-2*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2-9
*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-14*I*B*(a*c)^(1/2)*(a*c*(1+ta
n(f*x+e)^2))^(1/2)-19*B*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)+10*A*(a*c*(1+tan(f*x+e)^2))^(1/2)*
(a*c)^(1/2))/(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. 1049 vs. \(2 (183) = 366\).
time = 0.73, size = 1049, normalized size = 4.62 \begin {gather*} -\frac {4 \, {\left (4 \, {\left (2 \, A - 7 i \, B\right )} a^{2} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 4 \, {\left (2 i \, A + 7 \, B\right )} a^{2} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 6 \, {\left ({\left (2 \, A - 3 i \, B\right )} a^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, {\left (2 \, A - 3 i \, B\right )} a^{2} \cos \left (2 \, f x + 2 \, e\right ) - {\left (-2 i \, A - 3 \, B\right )} a^{2} \sin \left (4 \, f x + 4 \, e\right ) - 2 \, {\left (-2 i \, A - 3 \, B\right )} a^{2} \sin \left (2 \, f x + 2 \, e\right ) + {\left (2 \, A - 3 i \, B\right )} a^{2}\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 ) - 6 \, {\left ({\left (2 \, A - 3 i \, B\right )} a^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, {\left (2 \, A - 3 i \, B\right )} a^{2} \cos \left (2 \, f x + 2 \, e\right ) - {\left (-2 i \, A - 3 \, B\right )} a^{2} \sin \left (4 \, f x + 4 \, e\right ) - 2 \, {\left (-2 i \, A - 3 \, B\right )} a^{2} \sin \left (2 \, f x + 2 \, e\right ) + {\left (2 \, A - 3 i \, B\right )} a^{2}\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 (4 \, {\left (A - i \, B\right )} a^{2} \cos \left (4 \, f x + 4 \, e\right ) + 8 \, {\left (A - i \, B\right )} a^{2} \cos \left (2 \, f x + 2 \, e\right ) + 4 \, {\left (i \, A + B\right )} a^{2} \sin \left (4 \, f x + 4 \, e\right ) + 8 \, {\left (i \, A + B\right )} a^{2} \sin \left (2 \, f x + 2 \, e\right ) + 3 \, {\left (2 \, A - 3 i \, B\right )} a^{2}\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 ) + 3 \, {\left ({\left (-2 i \, A - 3 \, B\right )} a^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, {\left (-2 i \, A - 3 \, B\right )} a^{2} \cos \left (2 \, f x + 2 \, e\right ) + {\left (2 \, A - 3 i \, B\right )} a^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 \, {\left (2 \, A - 3 i \, B\right )} a^{2} \sin \left (2 \, f x + 2 \, e\right ) + {\left (-2 i \, A - 3 \, B\right )} a^{2}\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 ) + 3 \, {\left ({\left (2 i \, A + 3 \, B\right )} a^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, {\left (2 i \, A + 3 \, B\right )} a^{2} \cos \left (2 \, f x + 2 \, e\right ) - {\left (2 \, A - 3 i \, B\right )} a^{2} \sin \left (4 \, f x + 4 \, e\right ) - 2 \, {\left (2 \, A - 3 i \, B\right )} a^{2} \sin \left (2 \, f x + 2 \, e\right ) + {\left (2 i \, A + 3 \, B\right )} a^{2}\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 (4 \, {\left (i \, A + B\right )} a^{2} \cos \left (4 \, f x + 4 \, e\right ) + 8 \, {\left (i \, A + B\right )} a^{2} \cos \left (2 \, f x + 2 \, e\right ) - 4 \, {\left (A - i \, B\right )} a^{2} \sin \left (4 \, f x + 4 \, e\right ) - 8 \, {\left (A - i \, B\right )} a^{2} \sin \left (2 \, f x + 2 \, e\right ) + 3 \, {\left (2 i \, A + 3 \, B\right )} a^{2}\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}}{-16 \, {\left (i \, c \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, c \cos \left (2 \, f x + 2 \, e\right ) - c \sin \left (4 \, f x + 4 \, e\right ) - 2 \, 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))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-4*(4*(2*A - 7*I*B)*a^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*(2*I*A + 7*B)*a^2*sin(3/2*arc
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 6*((2*A - 3*I*B)*a^2*cos(4*f*x + 4*e) + 2*(2*A - 3*I*B)*a^2*cos(2*
f*x + 2*e) - (-2*I*A - 3*B)*a^2*sin(4*f*x + 4*e) - 2*(-2*I*A - 3*B)*a^2*sin(2*f*x + 2*e) + (2*A - 3*I*B)*a^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*((2*A - 3*I*B)*a^2*cos(4*f*x + 4*e) + 2*(2*A - 3*I*B)*a^2*cos(2*f*x + 2*e) - (-2*I*A - 3*B)*a^2*
sin(4*f*x + 4*e) - 2*(-2*I*A - 3*B)*a^2*sin(2*f*x + 2*e) + (2*A - 3*I*B)*a^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) + 4*(4*(A - I*B)*a^2*
cos(4*f*x + 4*e) + 8*(A - I*B)*a^2*cos(2*f*x + 2*e) + 4*(I*A + B)*a^2*sin(4*f*x + 4*e) + 8*(I*A + B)*a^2*sin(2
*f*x + 2*e) + 3*(2*A - 3*I*B)*a^2)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 3*((-2*I*A - 3*B)*a^
2*cos(4*f*x + 4*e) + 2*(-2*I*A - 3*B)*a^2*cos(2*f*x + 2*e) + (2*A - 3*I*B)*a^2*sin(4*f*x + 4*e) + 2*(2*A - 3*I
*B)*a^2*sin(2*f*x + 2*e) + (-2*I*A - 3*B)*a^2)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + si
n(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*((2*I*A + 3*B)*a^2*cos(4*f*x + 4*e) + 2*(2*I*A + 3*B)*a^2*cos(2*f*x + 2*e) - (2*A - 3*I*B)*a^2*sin(4*
f*x + 4*e) - 2*(2*A - 3*I*B)*a^2*sin(2*f*x + 2*e) + (2*I*A + 3*B)*a^2)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), c
os(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*(4*(I*A + B)*a^2*cos(4*f*x + 4*e) + 8*(I*A + B)*a^2*cos(2*f*x + 2*e) - 4*(A -
I*B)*a^2*sin(4*f*x + 4*e) - 8*(A - I*B)*a^2*sin(2*f*x + 2*e) + 3*(2*I*A + 3*B)*a^2)*sin(1/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-16*I*c*cos(4*f*x + 4*e) - 32*I*c*cos(2*f*x + 2*e) + 16*c*sin(4
*f*x + 4*e) + 32*c*sin(2*f*x + 2*e) - 16*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. 550 vs. \(2 (183) = 366\).
time = 4.06, size = 550, normalized size = 2.42 \begin {gather*} \frac {3 \, \sqrt {\frac {{\left (4 \, A^{2} - 12 i \, A B - 9 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac {4 \, {\left (2 \, {\left ({\left (-2 i \, A - 3 \, B\right )} a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-2 i \, A - 3 \, B\right )} a^{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 {{\left (4 \, A^{2} - 12 i \, A B - 9 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )}\right )}}{{\left (-2 i \, A - 3 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-2 i \, A - 3 \, B\right )} a^{2}}\right ) - 3 \, \sqrt {\frac {{\left (4 \, A^{2} - 12 i \, A B - 9 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac {4 \, {\left (2 \, {\left ({\left (-2 i \, A - 3 \, B\right )} a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-2 i \, A - 3 \, B\right )} a^{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 {{\left (4 \, A^{2} - 12 i \, A B - 9 \, B^{2}\right )} a^{5}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )}\right )}}{{\left (-2 i \, A - 3 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-2 i \, A - 3 \, B\right )} a^{2}}\right ) - 4 \, {\left (4 \, {\left (i \, A + B\right )} a^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + 5 \, {\left (2 i \, A + 3 \, B\right )} a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + 3 \, {\left (2 i \, A + 3 \, B\right )} a^{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}}}{4 \, {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/4*(3*sqrt((4*A^2 - 12*I*A*B - 9*B^2)*a^5/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) + c*f)*log(4*(2*((-2*I*A - 3*B)*a
^2*e^(3*I*f*x + 3*I*e) + (-2*I*A - 3*B)*a^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((4*A^2 - 12*I*A*B - 9*B^2)*a^5/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) - c*f))/((-2*I*A -
3*B)*a^2*e^(2*I*f*x + 2*I*e) + (-2*I*A - 3*B)*a^2)) - 3*sqrt((4*A^2 - 12*I*A*B - 9*B^2)*a^5/(c*f^2))*(c*f*e^(2
*I*f*x + 2*I*e) + c*f)*log(4*(2*((-2*I*A - 3*B)*a^2*e^(3*I*f*x + 3*I*e) + (-2*I*A - 3*B)*a^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((4*A^2 - 12*I*A*B - 9*B^2)*a^5/(c*f
^2))*(c*f*e^(2*I*f*x + 2*I*e) - c*f))/((-2*I*A - 3*B)*a^2*e^(2*I*f*x + 2*I*e) + (-2*I*A - 3*B)*a^2)) - 4*(4*(I
*A + B)*a^2*e^(5*I*f*x + 5*I*e) + 5*(2*I*A + 3*B)*a^2*e^(3*I*f*x + 3*I*e) + 3*(2*I*A + 3*B)*a^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)))/(c*f*e^(2*I*f*x + 2*I*e) + 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 {5}{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))**(5/2)*(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))**(5/2)*(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))^(5/2)*(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)^(5/2)/sqrt(-I*c*tan(f*x + e) + c), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \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 )}^{5/2}}{\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)^(5/2))/(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)^(5/2))/(c - c*tan(e + f*x)*1i)^(1/2), x)

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