3.12.0 ∫ (a+ia tan(e+fx))3 ____________________________

Integrand size = 30, antiderivative size = 158 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {8 i a^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac {4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt {c+d \tan (e+f x)}} \]

-8*I*a^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(5/2)/f+4/3*a^3*(I*c-d)*(c-4*I*d)/(c-I*d)^2/d^2

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

Rubi [A] (veriﬁed)

Time = 0.53 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3634, 3672, 3618, 65, 214} \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {8 i a^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (c-i d)^{5/2}}+\frac {4 a^3 (-d+i c) (c-4 i d)}{3 d^2 f (c-i d)^2 \sqrt {c+d \tan (e+f x)}}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 d f (c-i d) (c+d \tan (e+f x))^{3/2}} \]

Int[(a + I*a*Tan[e + f*x])^3/(c + d*Tan[e + f*x])^(5/2),x]

((-8*I)*a^3*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(5/2)*f) + (2*(c + I*d)*(a^3 + I*a^3*T

an[e + f*x]))/(3*(c - I*d)*d*f*(c + d*Tan[e + f*x])^(3/2)) + (4*a^3*(I*c - d)*(c - (4*I)*d))/(3*(c - I*d)^2*d^

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,

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(

d/f), Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&

NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim

p[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x]

+ Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(

m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /;

FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && Lt

Q[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e

_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(A*b - a*B)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2

+ b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*

c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}-\frac {2 \int \frac {(a+i a \tan (e+f x)) \left (-a^2 (c+4 i d)+a^2 (i c+2 d) \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{3 d (i c+d)} \\ & = \frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac {4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt {c+d \tan (e+f x)}}-\frac {2 \int \frac {-6 a^3 (i c-d) d+6 a^3 (c+i d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 d (i c+d) \left (c^2+d^2\right )} \\ & = \frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac {4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\left (24 a^6 (c+i d) d\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\frac {x}{6 a^3 (c+i d)}} \left (36 a^6 (c+i d)^2 d^2-6 a^3 (i c-d) d x\right )} \, dx,x,6 a^3 (c+i d) d \tan (e+f x)\right )}{(c-i d)^2 f} \\ & = \frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac {4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\left (288 a^9 (c+i d)^2 d\right ) \text {Subst}\left (\int \frac {1}{36 a^6 c (i c-d) (c+i d) d+36 a^6 (c+i d)^2 d^2-36 a^6 (i c-d) (c+i d) d x^2} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{(c-i d)^2 f} \\ & = -\frac {8 i a^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{3 (c-i d) d f (c+d \tan (e+f x))^{3/2}}+\frac {4 a^3 (i c-d) (c-4 i d)}{3 (c-i d)^2 d^2 f \sqrt {c+d \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 3.03 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {2 a^3 \left (-\frac {12 i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2}}+\frac {(c+i d) \left (2 i c^2+9 c d-i d^2+3 d (i c+3 d) \tan (e+f x)\right )}{(c-i d)^2 d^2 (c+d \tan (e+f x))^{3/2}}\right )}{3 f} \]

Integrate[(a + I*a*Tan[e + f*x])^3/(c + d*Tan[e + f*x])^(5/2),x]

(2*a^3*(((-12*I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(c - I*d)^(5/2) + ((c + I*d)*((2*I)*c^2 + 9*

c*d - I*d^2 + 3*d*(I*c + 3*d)*Tan[e + f*x]))/((c - I*d)^2*d^2*(c + d*Tan[e + f*x])^(3/2))))/(3*f)

Maple [B] (veriﬁed)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2771 vs. \(2 (135 ) = 270\).

Time = 1.07 (sec) , antiderivative size = 2772, normalized size of antiderivative = 17.54


method	result	size

derivativedivides	\(\text {Expression too large to display}\)	\(2772\)
default	\(\text {Expression too large to display}\)	\(2772\)
parts	\(\text {Expression too large to display}\)	\(8982\)

int((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

4/f*a^3*d^3/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x

+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))+2/f*a^3*d^3/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*t

an(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))-2/f*a^3*d^3/(c^2+d^2)^(5/2)/(2*

(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1

/2))-6*I/f*a^3*d^2/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+

2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c+12*I/f*a^3*d^2/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arct

an((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-12*I/f*a^3*d^2/(c

^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2

*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+6*I/f*a^3*d^2/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(

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

(3/2)+2/f*a^3/d/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)*c^2+12*I/f*a^3/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*c^2-6*I/f*a

^3*d^2/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)+6/f*a^3*d/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x

+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^2-12/f*a^3*d/(c^2+d^2)^(5/2)/(2*

(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-

2*c)^(1/2))*c^2-4*I/f*a^3/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c

+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^2+4*I/f*a^3/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*a

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

^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+

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

))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^2+2*I/f*a^3/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2

)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c^2+4*I/f*a^3/(c^2+d

^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^

2+d^2)^(1/2)-2*c)^(1/2))*c^3-4*I/f*a^3/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e)

)^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^3-2*I/f*a^3/(c^2+d^2)^(5/2)/(2*(c^2+d^

2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c^

3-4*I/f*a^3*d^2/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+

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

(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)

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

^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))+4*I/f*a^3*d^2/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*

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

c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*c^4-8/f*a^3*d/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f

*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-4/f*a^3*d/(c^2+d^2)^2/(2*(c^2+d^2

)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c+4

/f*a^3*d/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)

+2*c)^(1/2)+(c^2+d^2)^(1/2))*c+8/f*a^3*d/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+

2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+12/f*a^3*d/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^

(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)

)*c^2-6/f*a^3*d/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c

)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c^2+2*I/f*a^3/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)*c-16/f*a^3*d/(c^2+d^2)^

2/(c+d*tan(f*x+e))^(1/2)*c-4/f*a^3*d^3/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e)

)^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))

Fricas [B] (veriﬁcation not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 963 vs. \(2 (126) = 252\).

Time = 0.32 (sec) , antiderivative size = 963, normalized size of antiderivative = 6.09 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]

integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

1/12*(3*sqrt(-64*I*a^6/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*((c^4*d^2 - 4*I*

c^3*d^3 - 6*c^2*d^4 + 4*I*c*d^5 + d^6)*f*e^(4*I*f*x + 4*I*e) + 2*(c^4*d^2 - 2*I*c^3*d^3 - 2*I*c*d^5 - d^6)*f*e

^(2*I*f*x + 2*I*e) + (c^4*d^2 + 2*c^2*d^4 + d^6)*f)*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c^5 + 5*c^4*d - 10*I

*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*((I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (I

*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))

+ 8*(a^3*c - I*a^3*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) - 3*sqrt(-64*I*a^6/((I*c^5 + 5*c^4*d - 1

0*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*((c^4*d^2 - 4*I*c^3*d^3 - 6*c^2*d^4 + 4*I*c*d^5 + d^6)*f*e^(

4*I*f*x + 4*I*e) + 2*(c^4*d^2 - 2*I*c^3*d^3 - 2*I*c*d^5 - d^6)*f*e^(2*I*f*x + 2*I*e) + (c^4*d^2 + 2*c^2*d^4 +

d^6)*f)*log(1/4*(8*a^3*c + sqrt(-64*I*a^6/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2

))*((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f*e^(2*I*f*x + 2*I*e) + (-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f)*sqrt(

((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)) + 8*(a^3*c - I*a^3*d)*e^(2*I*f*x + 2*I*e)

)*e^(-2*I*f*x - 2*I*e)/a^3) - 16*(-I*a^3*c^3 - 2*a^3*c^2*d - 7*I*a^3*c*d^2 + 4*a^3*d^3 + (-I*a^3*c^3 - 5*a^3*c

^2*d - I*a^3*c*d^2 - 5*a^3*d^3)*e^(4*I*f*x + 4*I*e) + (-2*I*a^3*c^3 - 7*a^3*c^2*d - 8*I*a^3*c*d^2 - a^3*d^3)*e

^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/((c^4*d^2 - 4*I

*c^3*d^3 - 6*c^2*d^4 + 4*I*c*d^5 + d^6)*f*e^(4*I*f*x + 4*I*e) + 2*(c^4*d^2 - 2*I*c^3*d^3 - 2*I*c*d^5 - d^6)*f*

e^(2*I*f*x + 2*I*e) + (c^4*d^2 + 2*c^2*d^4 + d^6)*f)

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx=- i a^{3} \left (\int \frac {i}{c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} + 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} + d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} + 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} + d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} + 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} + d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} + 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} + d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}\right )\, dx\right ) \]

integrate((a+I*a*tan(f*x+e))**3/(c+d*tan(f*x+e))**(5/2),x)

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

+ d*tan(e + f*x))*tan(e + f*x)**2), x) + Integral(-3*tan(e + f*x)/(c**2*sqrt(c + d*tan(e + f*x)) + 2*c*d*sqrt

(c + d*tan(e + f*x))*tan(e + f*x) + d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2), x) + Integral(tan(e + f*x)

**3/(c**2*sqrt(c + d*tan(e + f*x)) + 2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x) + d**2*sqrt(c + d*tan(e + f*x

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

e + f*x))*tan(e + f*x) + d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2), x))

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]

integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

Giac [B] (veriﬁcation not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (126) = 252\).

Time = 1.19 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.95 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {16 \, a^{3} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{{\left (-i \, c^{2} f - 2 \, c d f + i \, d^{2} f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 \, {\left (3 \, {\left (-i \, d \tan \left (f x + e\right ) - i \, c\right )} a^{3} c^{2} + i \, a^{3} c^{3} - 6 \, {\left (d \tan \left (f x + e\right ) + c\right )} a^{3} c d - a^{3} c^{2} d + 9 \, {\left (-i \, d \tan \left (f x + e\right ) - i \, c\right )} a^{3} d^{2} + i \, a^{3} c d^{2} - a^{3} d^{3}\right )}}{3 \, {\left (c^{2} d^{2} f - 2 i \, c d^{3} f - d^{4} f\right )} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \]

integrate((a+I*a*tan(f*x+e))^3/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

16*a^3*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-2*c + 2*sqrt(

c^2 + d^2)) - I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/((-I*c^2*f

- 2*c*d*f + I*d^2*f)*sqrt(-2*c + 2*sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) - 2/3*(3*(-I*d*tan(f*x

+ e) - I*c)*a^3*c^2 + I*a^3*c^3 - 6*(d*tan(f*x + e) + c)*a^3*c*d - a^3*c^2*d + 9*(-I*d*tan(f*x + e) - I*c)*a^3

*d^2 + I*a^3*c*d^2 - a^3*d^3)/((c^2*d^2*f - 2*I*c*d^3*f - d^4*f)*(d*tan(f*x + e) + c)^(3/2))

Mupad [B] (veriﬁcation not implemented)

Time = 9.84 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.61 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {\frac {\left (a^3\,c^2+a^3\,c\,d\,2{}\mathrm {i}-a^3\,d^2\right )\,2{}\mathrm {i}}{3\,d^2\,f\,\left (c-d\,1{}\mathrm {i}\right )}-\frac {a^3\,\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )\,\left (c^2-c\,d\,2{}\mathrm {i}+3\,d^2\right )\,2{}\mathrm {i}}{d^2\,f\,{\left (c-d\,1{}\mathrm {i}\right )}^2}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {a^3\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\left (2\,c^8\,f^2+8\,c^6\,d^2\,f^2+12\,c^4\,d^4\,f^2+8\,c^2\,d^6\,f^2+2\,d^8\,f^2\right )}{2\,f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{5/2}\,\left (f\,c^6+2{}\mathrm {i}\,f\,c^5\,d+f\,c^4\,d^2+4{}\mathrm {i}\,f\,c^3\,d^3-f\,c^2\,d^4+2{}\mathrm {i}\,f\,c\,d^5-f\,d^6\right )}\right )\,8{}\mathrm {i}}{f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{5/2}} \]

int((a + a*tan(e + f*x)*1i)^3/(c + d*tan(e + f*x))^(5/2),x)

(a^3*atan(((c + d*tan(e + f*x))^(1/2)*(2*c^8*f^2 + 2*d^8*f^2 + 8*c^2*d^6*f^2 + 12*c^4*d^4*f^2 + 8*c^6*d^2*f^2)

)/(2*f*(d*1i - c)^(5/2)*(c^6*f - d^6*f - c^2*d^4*f + c^3*d^3*f*4i + c^4*d^2*f + c*d^5*f*2i + c^5*d*f*2i)))*8i)

/(f*(d*1i - c)^(5/2)) - (((a^3*c^2 - a^3*d^2 + a^3*c*d*2i)*2i)/(3*d^2*f*(c - d*1i)) - (a^3*(c + d*tan(e + f*x)

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

\(\int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1131]

Optimal result