3.12.13 \(\int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx\) [1113]

Optimal. Leaf size=216 \[ -\frac {8 i a^3 (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {8 i a^3 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {8 a^3 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f} \]

[Out]

-8*I*a^3*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f+8*I*a^3*(c-I*d)^2*(c+d*tan(f*x+e))^(1/2
)/f+8/3*a^3*(I*c+d)*(c+d*tan(f*x+e))^(3/2)/f+8/5*I*a^3*(c+d*tan(f*x+e))^(5/2)/f+4/63*a^3*(I*c-10*d)*(c+d*tan(f
*x+e))^(7/2)/d^2/f-2/9*(a^3+I*a^3*tan(f*x+e))*(c+d*tan(f*x+e))^(7/2)/d/f

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Rubi [A]
time = 0.43, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3637, 3673, 3609, 3618, 65, 214} \begin {gather*} \frac {4 a^3 (-10 d+i c) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {8 a^3 (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {8 i a^3 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}-\frac {8 i a^3 (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*I)*a^3*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f + ((8*I)*a^3*(c - I*d)^2*Sqrt[c
+ d*Tan[e + f*x]])/f + (8*a^3*(I*c + d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (((8*I)/5)*a^3*(c + d*Tan[e + f*x
])^(5/2))/f + (4*a^3*(I*c - 10*d)*(c + d*Tan[e + f*x])^(7/2))/(63*d^2*f) - (2*(a^3 + I*a^3*Tan[e + f*x])*(c +
d*Tan[e + f*x])^(7/2))/(9*d*f)

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 214

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

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3618

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]

Rule 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3673

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x))^{5/2} \, dx &=-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {(2 a) \int (a+i a \tan (e+f x)) (a (i c+8 d)+a (c+10 i d) \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx}{9 d}\\ &=\frac {4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {(2 a) \int (c+d \tan (e+f x))^{5/2} \left (18 a^2 d+18 i a^2 d \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac {8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {(2 a) \int (c+d \tan (e+f x))^{3/2} \left (18 a^2 (c-i d) d+18 a^2 d (i c+d) \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac {8 a^3 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {(2 a) \int \sqrt {c+d \tan (e+f x)} \left (18 a^2 (c-i d)^2 d+18 i a^2 (c-i d)^2 d \tan (e+f x)\right ) \, dx}{9 d}\\ &=\frac {8 i a^3 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {8 a^3 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {(2 a) \int \frac {18 a^2 (c-i d)^3 d-18 a^2 d (i c+d)^3 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{9 d}\\ &=\frac {8 i a^3 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {8 a^3 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}+\frac {\left (72 i a^5 (c-i d)^6 d\right ) \text {Subst}\left (\int \frac {1}{\left (324 a^4 d^2 (i c+d)^6+18 a^2 (c-i d)^3 d x\right ) \sqrt {c-\frac {x}{18 a^2 (i c+d)^3}}} \, dx,x,-18 a^2 d (i c+d)^3 \tan (e+f x)\right )}{f}\\ &=\frac {8 i a^3 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {8 a^3 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\left (2592 a^7 (c-i d)^9 d\right ) \text {Subst}\left (\int \frac {1}{324 a^4 c (c-i d)^3 d (i c+d)^3+324 a^4 d^2 (i c+d)^6-324 a^4 (c-i d)^3 d (i c+d)^3 x^2} \, dx,x,\sqrt {c+d \tan (e+f x)}\right )}{f}\\ &=-\frac {8 i a^3 (c-i d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {8 i a^3 (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {8 a^3 (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {8 i a^3 (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {4 a^3 (i c-10 d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{7/2}}{9 d f}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(528\) vs. \(2(216)=432\).
time = 9.92, size = 528, normalized size = 2.44 \begin {gather*} -\frac {8 i (c-i d)^{5/2} e^{-3 i e} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{\sqrt {c-i d}}\right ) \cos ^3(e+f x) (a+i a \tan (e+f x))^3}{f (\cos (f x)+i \sin (f x))^3}+\frac {\cos ^3(e+f x) \left (\sec (e) \sec ^2(e+f x) \left (75 c^2 \cos (e)-405 i c d \cos (e)-322 d^2 \cos (e)+95 c d \sin (e)-135 i d^2 \sin (e)\right ) \left (-\frac {2}{315} i \cos (3 e)-\frac {2}{315} \sin (3 e)\right )+\sec (e) \left (10 i c^4 \cos (e)-135 c^3 d \cos (e)+2007 i c^2 d^2 \cos (e)+3345 c d^3 \cos (e)-1547 i d^4 \cos (e)-5 i c^3 d \sin (e)-405 c^2 d^2 \sin (e)+1019 i c d^3 \sin (e)+555 d^4 \sin (e)\right ) \left (\frac {2 \cos (3 e)}{315 d^2}-\frac {2 i \sin (3 e)}{315 d^2}\right )+\sec ^4(e+f x) \left (-\frac {2}{9} i d^2 \cos (3 e)-\frac {2}{9} d^2 \sin (3 e)\right )+\sec (e) \sec ^3(e+f x) \left (\frac {2}{63} \cos (3 e)-\frac {2}{63} i \sin (3 e)\right ) \left (-19 i c d \sin (f x)-27 d^2 \sin (f x)\right )+\sec (e) \sec (e+f x) \left (\frac {2 \cos (3 e)}{315 d}-\frac {2 i \sin (3 e)}{315 d}\right ) \left (-5 i c^3 \sin (f x)-405 c^2 d \sin (f x)+1019 i c d^2 \sin (f x)+555 d^3 \sin (f x)\right )\right ) \sqrt {\sec (e+f x) (c \cos (e+f x)+d \sin (e+f x))} (a+i a \tan (e+f x))^3}{f (\cos (f x)+i \sin (f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*I)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))]/Sqrt[c -
I*d]]*Cos[e + f*x]^3*(a + I*a*Tan[e + f*x])^3)/(E^((3*I)*e)*f*(Cos[f*x] + I*Sin[f*x])^3) + (Cos[e + f*x]^3*(Se
c[e]*Sec[e + f*x]^2*(75*c^2*Cos[e] - (405*I)*c*d*Cos[e] - 322*d^2*Cos[e] + 95*c*d*Sin[e] - (135*I)*d^2*Sin[e])
*(((-2*I)/315)*Cos[3*e] - (2*Sin[3*e])/315) + Sec[e]*((10*I)*c^4*Cos[e] - 135*c^3*d*Cos[e] + (2007*I)*c^2*d^2*
Cos[e] + 3345*c*d^3*Cos[e] - (1547*I)*d^4*Cos[e] - (5*I)*c^3*d*Sin[e] - 405*c^2*d^2*Sin[e] + (1019*I)*c*d^3*Si
n[e] + 555*d^4*Sin[e])*((2*Cos[3*e])/(315*d^2) - (((2*I)/315)*Sin[3*e])/d^2) + Sec[e + f*x]^4*(((-2*I)/9)*d^2*
Cos[3*e] - (2*d^2*Sin[3*e])/9) + Sec[e]*Sec[e + f*x]^3*((2*Cos[3*e])/63 - ((2*I)/63)*Sin[3*e])*((-19*I)*c*d*Si
n[f*x] - 27*d^2*Sin[f*x]) + Sec[e]*Sec[e + f*x]*((2*Cos[3*e])/(315*d) - (((2*I)/315)*Sin[3*e])/d)*((-5*I)*c^3*
Sin[f*x] - 405*c^2*d*Sin[f*x] + (1019*I)*c*d^2*Sin[f*x] + 555*d^3*Sin[f*x]))*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x]
+ d*Sin[e + f*x])]*(a + I*a*Tan[e + f*x])^3)/(f*(Cos[f*x] + I*Sin[f*x])^3)

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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. 1042 vs. \(2 (183 ) = 366\).
time = 0.32, size = 1043, normalized size = 4.83 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*a^3/d^2*(-1/9*I*(c+d*tan(f*x+e))^(9/2)+1/7*I*c*(c+d*tan(f*x+e))^(7/2)+4/5*I*d^2*(c+d*tan(f*x+e))^(5/2)-3/7
*d*(c+d*tan(f*x+e))^(7/2)+4/3*I*c*d^2*(c+d*tan(f*x+e))^(3/2)+4*I*c^2*d^2*(c+d*tan(f*x+e))^(1/2)-4*I*d^4*(c+d*t
an(f*x+e))^(1/2)+4/3*d^3*(c+d*tan(f*x+e))^(3/2)+8*c*d^3*(c+d*tan(f*x+e))^(1/2)-4*d^2*(1/2/(2*(c^2+d^2)^(1/2)+2
*c)^(1/2)/(c^2+d^2)^(1/2)*(1/2*(I*c^3*(c^2+d^2)^(1/2)-3*I*c*d^2*(c^2+d^2)^(1/2)+I*c^4-I*d^4+3*c^2*d*(c^2+d^2)^
(1/2)-d^3*(c^2+d^2)^(1/2)+2*c^3*d+2*c*d^3)*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))+2*(I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^4-I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^4+2*(2*(c^2+d^2)
^(1/2)+2*c)^(1/2)*c^3*d+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d^3-1/2*(I*c^3*(c^2+d^2)^(1/2)-3*I*c*d^2*(c^2+d^2)^(
1/2)+I*c^4-I*d^4+3*c^2*d*(c^2+d^2)^(1/2)-d^3*(c^2+d^2)^(1/2)+2*c^3*d+2*c*d^3)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(
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)))+1/2/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*(1/2*(-I*c^3*(c^2+d^2)^(1/2)+3*I*c*d^2*(c^2+d
^2)^(1/2)-I*c^4+I*d^4-3*c^2*d*(c^2+d^2)^(1/2)+d^3*(c^2+d^2)^(1/2)-2*c^3*d-2*c*d^3)*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))+2*(I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^4-I*(2*(c^2+
d^2)^(1/2)+2*c)^(1/2)*d^4+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3*d+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d^3+1/2*(-I*
c^3*(c^2+d^2)^(1/2)+3*I*c*d^2*(c^2+d^2)^(1/2)-I*c^4+I*d^4-3*c^2*d*(c^2+d^2)^(1/2)+d^3*(c^2+d^2)^(1/2)-2*c^3*d-
2*c*d^3)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(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)))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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. 1115 vs. \(2 (183) = 366\).
time = 2.54, size = 1115, normalized size = 5.16 \begin {gather*} -\frac {2 \, {\left (315 \, {\left (d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {a^{6} c^{5} - 5 i \, a^{6} c^{4} d - 10 \, a^{6} c^{3} d^{2} + 10 i \, a^{6} c^{2} d^{3} + 5 \, a^{6} c d^{4} - i \, a^{6} d^{5}}{f^{2}}} \log \left (\frac {2 \, {\left (a^{3} c^{3} - 2 i \, a^{3} c^{2} d - a^{3} c d^{2} - {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{6} c^{5} - 5 i \, a^{6} c^{4} d - 10 \, a^{6} c^{3} d^{2} + 10 i \, a^{6} c^{2} d^{3} + 5 \, a^{6} c d^{4} - i \, a^{6} d^{5}}{f^{2}}} + {\left (a^{3} c^{3} - 3 i \, a^{3} c^{2} d - 3 \, a^{3} c d^{2} + i \, a^{3} d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{3} c^{2} - 2 i \, a^{3} c d - a^{3} d^{2}}\right ) - 315 \, {\left (d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {a^{6} c^{5} - 5 i \, a^{6} c^{4} d - 10 \, a^{6} c^{3} d^{2} + 10 i \, a^{6} c^{2} d^{3} + 5 \, a^{6} c d^{4} - i \, a^{6} d^{5}}{f^{2}}} \log \left (\frac {2 \, {\left (a^{3} c^{3} - 2 i \, a^{3} c^{2} d - a^{3} c d^{2} - {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{6} c^{5} - 5 i \, a^{6} c^{4} d - 10 \, a^{6} c^{3} d^{2} + 10 i \, a^{6} c^{2} d^{3} + 5 \, a^{6} c d^{4} - i \, a^{6} d^{5}}{f^{2}}} + {\left (a^{3} c^{3} - 3 i \, a^{3} c^{2} d - 3 \, a^{3} c d^{2} + i \, a^{3} d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{3} c^{2} - 2 i \, a^{3} c d - a^{3} d^{2}}\right ) + 2 \, {\left (-5 i \, a^{3} c^{4} + 65 \, a^{3} c^{3} d - 801 i \, a^{3} c^{2} d^{2} - 1163 \, a^{3} c d^{3} + 496 i \, a^{3} d^{4} + {\left (-5 i \, a^{3} c^{4} + 70 \, a^{3} c^{3} d - 1206 i \, a^{3} c^{2} d^{2} - 2182 \, a^{3} c d^{3} + 1051 i \, a^{3} d^{4}\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-20 i \, a^{3} c^{4} + 275 \, a^{3} c^{3} d - 4269 i \, a^{3} c^{2} d^{2} - 6709 \, a^{3} c d^{3} + 2735 i \, a^{3} d^{4}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, {\left (-10 i \, a^{3} c^{4} + 135 \, a^{3} c^{3} d - 1907 i \, a^{3} c^{2} d^{2} - 2805 \, a^{3} c d^{3} + 1211 i \, a^{3} d^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-20 i \, a^{3} c^{4} + 265 \, a^{3} c^{3} d - 3459 i \, a^{3} c^{2} d^{2} - 5051 \, a^{3} c d^{3} + 2165 i \, a^{3} d^{4}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )}}{315 \, {\left (d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(315*(d^2*f*e^(8*I*f*x + 8*I*e) + 4*d^2*f*e^(6*I*f*x + 6*I*e) + 6*d^2*f*e^(4*I*f*x + 4*I*e) + 4*d^2*f*e
^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-(a^6*c^5 - 5*I*a^6*c^4*d - 10*a^6*c^3*d^2 + 10*I*a^6*c^2*d^3 + 5*a^6*c*d^4 -
I*a^6*d^5)/f^2)*log(2*(a^3*c^3 - 2*I*a^3*c^2*d - a^3*c*d^2 - (I*f*e^(2*I*f*x + 2*I*e) + I*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))*sqrt(-(a^6*c^5 - 5*I*a^6*c^4*d - 10*a^6*c^3*d^2 + 10
*I*a^6*c^2*d^3 + 5*a^6*c*d^4 - I*a^6*d^5)/f^2) + (a^3*c^3 - 3*I*a^3*c^2*d - 3*a^3*c*d^2 + I*a^3*d^3)*e^(2*I*f*
x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a^3*c^2 - 2*I*a^3*c*d - a^3*d^2)) - 315*(d^2*f*e^(8*I*f*x + 8*I*e) + 4*d^2*f
*e^(6*I*f*x + 6*I*e) + 6*d^2*f*e^(4*I*f*x + 4*I*e) + 4*d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)*sqrt(-(a^6*c^5 - 5*I
*a^6*c^4*d - 10*a^6*c^3*d^2 + 10*I*a^6*c^2*d^3 + 5*a^6*c*d^4 - I*a^6*d^5)/f^2)*log(2*(a^3*c^3 - 2*I*a^3*c^2*d
- a^3*c*d^2 - (-I*f*e^(2*I*f*x + 2*I*e) - I*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))*sqrt(-(a^6*c^5 - 5*I*a^6*c^4*d - 10*a^6*c^3*d^2 + 10*I*a^6*c^2*d^3 + 5*a^6*c*d^4 - I*a^6*d^5)/f^2)
+ (a^3*c^3 - 3*I*a^3*c^2*d - 3*a^3*c*d^2 + I*a^3*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a^3*c^2 - 2*I
*a^3*c*d - a^3*d^2)) + 2*(-5*I*a^3*c^4 + 65*a^3*c^3*d - 801*I*a^3*c^2*d^2 - 1163*a^3*c*d^3 + 496*I*a^3*d^4 + (
-5*I*a^3*c^4 + 70*a^3*c^3*d - 1206*I*a^3*c^2*d^2 - 2182*a^3*c*d^3 + 1051*I*a^3*d^4)*e^(8*I*f*x + 8*I*e) + (-20
*I*a^3*c^4 + 275*a^3*c^3*d - 4269*I*a^3*c^2*d^2 - 6709*a^3*c*d^3 + 2735*I*a^3*d^4)*e^(6*I*f*x + 6*I*e) + 3*(-1
0*I*a^3*c^4 + 135*a^3*c^3*d - 1907*I*a^3*c^2*d^2 - 2805*a^3*c*d^3 + 1211*I*a^3*d^4)*e^(4*I*f*x + 4*I*e) + (-20
*I*a^3*c^4 + 265*a^3*c^3*d - 3459*I*a^3*c^2*d^2 - 5051*a^3*c*d^3 + 2165*I*a^3*d^4)*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)))/(d^2*f*e^(8*I*f*x + 8*I*e) + 4*d^2*f*e^(6
*I*f*x + 6*I*e) + 6*d^2*f*e^(4*I*f*x + 4*I*e) + 4*d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i a^{3} \left (\int i c^{2} \sqrt {c + d \tan {\left (e + f x \right )}}\, dx + \int \left (- 3 c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx + \int c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- 3 d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\right )\, dx + \int d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{5}{\left (e + f x \right )}\, dx + \int \left (- 3 i c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int i d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- 3 i d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{4}{\left (e + f x \right )}\right )\, dx + \int \left (- 6 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{4}{\left (e + f x \right )}\, dx + \int 2 i c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\, dx + \int \left (- 6 i c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a**3*(Integral(I*c**2*sqrt(c + d*tan(e + f*x)), x) + Integral(-3*c**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)
, x) + Integral(c**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**3, x) + Integral(-3*d**2*sqrt(c + d*tan(e + f*x))*
tan(e + f*x)**3, x) + Integral(d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**5, x) + Integral(-3*I*c**2*sqrt(c +
d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(I*d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral
(-3*I*d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**4, x) + Integral(-6*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x
)**2, x) + Integral(2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**4, x) + Integral(2*I*c*d*sqrt(c + d*tan(e + f
*x))*tan(e + f*x), x) + Integral(-6*I*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**3, x))

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (183) = 366\).
time = 1.35, size = 423, normalized size = 1.96 \begin {gather*} -\frac {16 \, {\left (-i \, a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 i \, a^{3} c d^{2} + a^{3} d^{3}\right )} \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 )}{\sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 \, {\left (35 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} d^{16} f^{8} - 45 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c d^{16} f^{8} + 135 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} d^{17} f^{8} - 252 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} d^{18} f^{8} - 420 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c d^{18} f^{8} - 1260 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{3} c^{2} d^{18} f^{8} - 420 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} d^{19} f^{8} - 2520 \, \sqrt {d \tan \left (f x + e\right ) + c} a^{3} c d^{19} f^{8} + 1260 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{3} d^{20} f^{8}\right )}}{315 \, d^{18} f^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16*(-I*a^3*c^3 - 3*a^3*c^2*d + 3*I*a^3*c*d^2 + a^3*d^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))))/(sqrt(-2*c + 2*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) -
2/315*(35*I*(d*tan(f*x + e) + c)^(9/2)*a^3*d^16*f^8 - 45*I*(d*tan(f*x + e) + c)^(7/2)*a^3*c*d^16*f^8 + 135*(d*
tan(f*x + e) + c)^(7/2)*a^3*d^17*f^8 - 252*I*(d*tan(f*x + e) + c)^(5/2)*a^3*d^18*f^8 - 420*I*(d*tan(f*x + e) +
c)^(3/2)*a^3*c*d^18*f^8 - 1260*I*sqrt(d*tan(f*x + e) + c)*a^3*c^2*d^18*f^8 - 420*(d*tan(f*x + e) + c)^(3/2)*a
^3*d^19*f^8 - 2520*sqrt(d*tan(f*x + e) + c)*a^3*c*d^19*f^8 + 1260*I*sqrt(d*tan(f*x + e) + c)*a^3*d^20*f^8)/(d^
18*f^9)

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Mupad [B]
time = 27.63, size = 400, normalized size = 1.85 \begin {gather*} -\left (\frac {\left (c-d\,1{}\mathrm {i}\right )\,\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d^2\,f}\right )}{5}+\frac {a^3\,{\left (c+d\,1{}\mathrm {i}\right )}^2\,2{}\mathrm {i}}{5\,d^2\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}-\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{7\,d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{7\,d^2\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}-{\left (c-d\,1{}\mathrm {i}\right )}^2\,\left (\left (c-d\,1{}\mathrm {i}\right )\,\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d^2\,f}\right )+\frac {a^3\,{\left (c+d\,1{}\mathrm {i}\right )}^2\,2{}\mathrm {i}}{d^2\,f}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}-\frac {\left (c-d\,1{}\mathrm {i}\right )\,\left (\left (c-d\,1{}\mathrm {i}\right )\,\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d^2\,f}\right )+\frac {a^3\,{\left (c+d\,1{}\mathrm {i}\right )}^2\,2{}\mathrm {i}}{d^2\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3}-\frac {a^3\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{9/2}\,2{}\mathrm {i}}{9\,d^2\,f}-\frac {\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,{\left (d+c\,1{}\mathrm {i}\right )}^{5/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{4\,\left (-c^3\,1{}\mathrm {i}-3\,c^2\,d+c\,d^2\,3{}\mathrm {i}+d^3\right )}\right )\,{\left (d+c\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (((c - d*1i)*((a^3*(c - d*1i)*2i)/(d^2*f) - (a^3*(c + d*1i)*4i)/(d^2*f)))/5 + (a^3*(c + d*1i)^2*2i)/(5*d^2*f
))*(c + d*tan(e + f*x))^(5/2) - ((a^3*(c - d*1i)*2i)/(7*d^2*f) - (a^3*(c + d*1i)*4i)/(7*d^2*f))*(c + d*tan(e +
f*x))^(7/2) - (c - d*1i)^2*((c - d*1i)*((a^3*(c - d*1i)*2i)/(d^2*f) - (a^3*(c + d*1i)*4i)/(d^2*f)) + (a^3*(c
+ d*1i)^2*2i)/(d^2*f))*(c + d*tan(e + f*x))^(1/2) - ((c - d*1i)*((c - d*1i)*((a^3*(c - d*1i)*2i)/(d^2*f) - (a^
3*(c + d*1i)*4i)/(d^2*f)) + (a^3*(c + d*1i)^2*2i)/(d^2*f))*(c + d*tan(e + f*x))^(3/2))/3 - (a^3*(c + d*tan(e +
f*x))^(9/2)*2i)/(9*d^2*f) - (16i^(1/2)*a^3*atan((16i^(1/2)*(c*1i + d)^(5/2)*(c + d*tan(e + f*x))^(1/2)*1i)/(4
*(c*d^2*3i - 3*c^2*d - c^3*1i + d^3)))*(c*1i + d)^(5/2)*2i)/f

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