3.2.85 \(\int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^3} \, dx\) [185]

Optimal. Leaf size=368 \[ \frac {\left (\frac {15}{8}+\frac {7 i}{4}\right ) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {\left (\frac {15}{8}+\frac {7 i}{4}\right ) \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 d^{3/2} f}+\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {15}{4 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {1}{6 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3}+\frac {5}{12 a d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {7}{6 d f \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )} \]

[Out]

(15/16+7/8*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/d^(3/2)/f*2^(1/2)-(15/16+7/8*I)*arctan(1+2^(1
/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/d^(3/2)/f*2^(1/2)+(-15/32+7/16*I)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2
)+d^(1/2)*tan(f*x+e))/a^3/d^(3/2)/f*2^(1/2)+(15/32-7/16*I)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan
(f*x+e))/a^3/d^(3/2)/f*2^(1/2)-15/4/a^3/d/f/(d*tan(f*x+e))^(1/2)+1/6/d/f/(d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e
))^3+5/12/a/d/f/(d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^2+7/6/d/f/(d*tan(f*x+e))^(1/2)/(a^3+I*a^3*tan(f*x+e))

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Rubi [A]
time = 0.45, antiderivative size = 368, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3640, 3677, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\left (\frac {15}{8}+\frac {7 i}{4}\right ) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {\left (\frac {15}{8}+\frac {7 i}{4}\right ) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 d^{3/2} f}+\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {15}{4 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {7}{6 d f \left (a^3+i a^3 \tan (e+f x)\right ) \sqrt {d \tan (e+f x)}}+\frac {5}{12 a d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}+\frac {1}{6 d f (a+i a \tan (e+f x))^3 \sqrt {d \tan (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])^3),x]

[Out]

((15/8 + (7*I)/4)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*d^(3/2)*f) - ((15/8 + (7*I)
/4)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*d^(3/2)*f) - ((15/16 - (7*I)/8)*Log[Sqrt[
d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*d^(3/2)*f) + ((15/16 - (7*I)/8)*Log[Sq
rt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*d^(3/2)*f) - 15/(4*a^3*d*f*Sqrt[d*T
an[e + f*x]]) + 1/(6*d*f*Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])^3) + 5/(12*a*d*f*Sqrt[d*Tan[e + f*x]]*(a
+ I*a*Tan[e + f*x])^2) + 7/(6*d*f*Sqrt[d*Tan[e + f*x]]*(a^3 + I*a^3*Tan[e + f*x]))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3610

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

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3640

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*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] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3677

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*
f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

Rubi steps

\begin {align*} \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^3} \, dx &=\frac {1}{6 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3}+\frac {\int \frac {\frac {13 a d}{2}-\frac {7}{2} i a d \tan (e+f x)}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx}{6 a^2 d}\\ &=\frac {1}{6 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3}+\frac {5}{12 a d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {\int \frac {31 a^2 d^2-25 i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx}{24 a^4 d^2}\\ &=\frac {1}{6 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3}+\frac {5}{12 a d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {7}{6 d f \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\int \frac {90 a^3 d^3-84 i a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{48 a^6 d^3}\\ &=-\frac {15}{4 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {1}{6 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3}+\frac {5}{12 a d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {7}{6 d f \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\int \frac {-84 i a^3 d^4-90 a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{48 a^6 d^5}\\ &=-\frac {15}{4 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {1}{6 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3}+\frac {5}{12 a d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {7}{6 d f \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {-84 i a^3 d^5-90 a^3 d^4 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{24 a^6 d^5 f}\\ &=-\frac {15}{4 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {1}{6 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3}+\frac {5}{12 a d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {7}{6 d f \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}+-\frac {\left (\frac {15}{8}+\frac {7 i}{4}\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^3 d f}+\frac {\left (\frac {15}{8}-\frac {7 i}{4}\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^3 d f}\\ &=-\frac {15}{4 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {1}{6 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3}+\frac {5}{12 a d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {7}{6 d f \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}+-\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 d^{3/2} f}+-\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 d^{3/2} f}+-\frac {\left (\frac {15}{16}+\frac {7 i}{8}\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^3 d f}+-\frac {\left (\frac {15}{16}+\frac {7 i}{8}\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^3 d f}\\ &=-\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 d^{3/2} f}+\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {15}{4 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {1}{6 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3}+\frac {5}{12 a d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {7}{6 d f \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}+-\frac {\left (\frac {15}{8}+\frac {7 i}{4}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 d^{3/2} f}+\frac {\left (\frac {15}{8}+\frac {7 i}{4}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 d^{3/2} f}\\ &=\frac {\left (\frac {15}{8}+\frac {7 i}{4}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {\left (\frac {15}{8}+\frac {7 i}{4}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 d^{3/2} f}+\frac {\left (\frac {15}{16}-\frac {7 i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 d^{3/2} f}-\frac {15}{4 a^3 d f \sqrt {d \tan (e+f x)}}+\frac {1}{6 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3}+\frac {5}{12 a d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2}+\frac {7}{6 d f \sqrt {d \tan (e+f x)} \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 2.23, size = 234, normalized size = 0.64 \begin {gather*} \frac {e^{-6 i (e+f x)} \left (1+9 e^{2 i (e+f x)}+49 e^{4 i (e+f x)}-105 e^{6 i (e+f x)}-146 e^{8 i (e+f x)}-87 e^{6 i (e+f x)} \sqrt {-1+e^{4 i (e+f x)}} \text {ArcTan}\left (\sqrt {-1+e^{4 i (e+f x)}}\right )+6 e^{6 i (e+f x)} \sqrt {-1+e^{2 i (e+f x)}} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )\right )}{48 a^3 d \left (1+e^{2 i (e+f x)}\right ) f \sqrt {d \tan (e+f x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])^3),x]

[Out]

(1 + 9*E^((2*I)*(e + f*x)) + 49*E^((4*I)*(e + f*x)) - 105*E^((6*I)*(e + f*x)) - 146*E^((8*I)*(e + f*x)) - 87*E
^((6*I)*(e + f*x))*Sqrt[-1 + E^((4*I)*(e + f*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(e + f*x))]] + 6*E^((6*I)*(e + f*x
))*Sqrt[-1 + E^((2*I)*(e + f*x))]*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(e + f*x)))/(1 + E
^((2*I)*(e + f*x)))]])/(48*a^3*d*E^((6*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))*f*Sqrt[d*Tan[e + f*x]])

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Maple [A]
time = 0.20, size = 147, normalized size = 0.40

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (-\frac {1}{d^{5} \sqrt {d \tan \left (f x +e \right )}}-\frac {\frac {14 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {98 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-20 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (-i d +d \tan \left (f x +e \right )\right )^{3}}+\frac {29 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{5}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{5} \sqrt {i d}}\right )}{f \,a^{3}}\)	\(147\)
default	\(\frac {2 d^{4} \left (-\frac {1}{d^{5} \sqrt {d \tan \left (f x +e \right )}}-\frac {\frac {14 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {98 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-20 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (-i d +d \tan \left (f x +e \right )\right )^{3}}+\frac {29 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{5}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{5} \sqrt {i d}}\right )}{f \,a^{3}}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/a^3*d^4*(-1/d^5/(d*tan(f*x+e))^(1/2)-1/16/d^5*((14*(d*tan(f*x+e))^(5/2)-98/3*I*d*(d*tan(f*x+e))^(3/2)-20*d
^2*(d*tan(f*x+e))^(1/2))/(-I*d+d*tan(f*x+e))^3+29/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))-1/16
/d^5/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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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. 742 vs. \(2 (291) = 582\).
time = 0.39, size = 742, normalized size = 2.02 \begin {gather*} \frac {12 \, {\left (a^{3} d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} - a^{3} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt {\frac {i}{64 \, a^{6} d^{3} f^{2}}} \log \left (-2 \, {\left (8 \, {\left (i \, a^{3} d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{3} d^{2} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{64 \, a^{6} d^{3} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 12 \, {\left (a^{3} d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} - a^{3} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt {\frac {i}{64 \, a^{6} d^{3} f^{2}}} \log \left (-2 \, {\left (8 \, {\left (-i \, a^{3} d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} d^{2} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{64 \, a^{6} d^{3} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) + 12 \, {\left (a^{3} d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} - a^{3} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt {-\frac {841 i}{64 \, a^{6} d^{3} f^{2}}} \log \left (\frac {{\left (8 \, {\left (a^{3} d f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {841 i}{64 \, a^{6} d^{3} f^{2}}} + 29\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} d f}\right ) - 12 \, {\left (a^{3} d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} - a^{3} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt {-\frac {841 i}{64 \, a^{6} d^{3} f^{2}}} \log \left (-\frac {{\left (8 \, {\left (a^{3} d f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {841 i}{64 \, a^{6} d^{3} f^{2}}} - 29\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} d f}\right ) + \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-146 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 105 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 49 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 9 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}}{48 \, {\left (a^{3} d^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} - a^{3} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(12*(a^3*d^2*f*e^(8*I*f*x + 8*I*e) - a^3*d^2*f*e^(6*I*f*x + 6*I*e))*sqrt(1/64*I/(a^6*d^3*f^2))*log(-2*(8*
(I*a^3*d^2*f*e^(2*I*f*x + 2*I*e) + I*a^3*d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1
))*sqrt(1/64*I/(a^6*d^3*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 12*(a^3*d^2*f*e^(8*I*f*x + 8*
I*e) - a^3*d^2*f*e^(6*I*f*x + 6*I*e))*sqrt(1/64*I/(a^6*d^3*f^2))*log(-2*(8*(-I*a^3*d^2*f*e^(2*I*f*x + 2*I*e) -
I*a^3*d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/64*I/(a^6*d^3*f^2)) + I*
d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) + 12*(a^3*d^2*f*e^(8*I*f*x + 8*I*e) - a^3*d^2*f*e^(6*I*f*x + 6*I*
e))*sqrt(-841/64*I/(a^6*d^3*f^2))*log(1/8*(8*(a^3*d*f*e^(2*I*f*x + 2*I*e) + a^3*d*f)*sqrt((-I*d*e^(2*I*f*x + 2
*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-841/64*I/(a^6*d^3*f^2)) + 29)*e^(-2*I*f*x - 2*I*e)/(a^3*d*f)) -
12*(a^3*d^2*f*e^(8*I*f*x + 8*I*e) - a^3*d^2*f*e^(6*I*f*x + 6*I*e))*sqrt(-841/64*I/(a^6*d^3*f^2))*log(-1/8*(8*(
a^3*d*f*e^(2*I*f*x + 2*I*e) + a^3*d*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-
841/64*I/(a^6*d^3*f^2)) - 29)*e^(-2*I*f*x - 2*I*e)/(a^3*d*f)) + sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*
f*x + 2*I*e) + 1))*(-146*I*e^(8*I*f*x + 8*I*e) - 105*I*e^(6*I*f*x + 6*I*e) + 49*I*e^(4*I*f*x + 4*I*e) + 9*I*e^
(2*I*f*x + 2*I*e) + I))/(a^3*d^2*f*e^(8*I*f*x + 8*I*e) - a^3*d^2*f*e^(6*I*f*x + 6*I*e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {i \int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{3}{\left (e + f x \right )} - 3 i \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )} - 3 \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )} + i \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**3,x)

[Out]

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

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Giac [A]
time = 0.82, size = 252, normalized size = 0.68 \begin {gather*} -\frac {\frac {87 \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} \sqrt {d} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 i \, \sqrt {2} \arctan \left (\frac {8 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} \sqrt {d} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {48}{\sqrt {d \tan \left (f x + e\right )} a^{3} f} + \frac {2 \, {\left (21 i \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )^{2} + 49 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 30 i \, \sqrt {d \tan \left (f x + e\right )} d^{2}\right )}}{{\left (-i \, d \tan \left (f x + e\right ) - d\right )}^{3} a^{3} f}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(87*sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d))
)/(a^3*sqrt(d)*f*(-I*d/sqrt(d^2) + 1)) + 3*I*sqrt(2)*arctan(8*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-4*I*sqrt(2)*d
^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^3*sqrt(d)*f*(-I*d/sqrt(d^2) + 1)) + 48/(sqrt(d*tan(f*x + e))*a^3*f)
+ 2*(21*I*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e)^2 + 49*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e) - 30*I*sqrt(d*tan
(f*x + e))*d^2)/((-I*d*tan(f*x + e) - d)^3*a^3*f))/d

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Mupad [B]
time = 6.12, size = 227, normalized size = 0.62 \begin {gather*} \frac {\frac {15\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^3}{4\,a^3\,f}-\frac {17\,d^2\,\mathrm {tan}\left (e+f\,x\right )}{2\,a^3\,f}+\frac {d^2\,2{}\mathrm {i}}{a^3\,f}-\frac {d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2\,121{}\mathrm {i}}{12\,a^3\,f}}{3\,d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}-{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{7/2}+d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,3{}\mathrm {i}-d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}+2\,\mathrm {atanh}\left (16\,a^3\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {1{}\mathrm {i}}{256\,a^6\,d^3\,f^2}}\right )\,\sqrt {\frac {1{}\mathrm {i}}{256\,a^6\,d^3\,f^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {841{}\mathrm {i}}{256\,a^6\,d^3\,f^2}}}{29}\right )\,\sqrt {-\frac {841{}\mathrm {i}}{256\,a^6\,d^3\,f^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d^2*2i)/(a^3*f) - (17*d^2*tan(e + f*x))/(2*a^3*f) - (d^2*tan(e + f*x)^2*121i)/(12*a^3*f) + (15*d^2*tan(e + f
*x)^3)/(4*a^3*f))/(d*(d*tan(e + f*x))^(5/2)*3i - (d*tan(e + f*x))^(7/2) - d^3*(d*tan(e + f*x))^(1/2)*1i + 3*d^
2*(d*tan(e + f*x))^(3/2)) + 2*atanh(16*a^3*d*f*(d*tan(e + f*x))^(1/2)*(1i/(256*a^6*d^3*f^2))^(1/2))*(1i/(256*a
^6*d^3*f^2))^(1/2) + 2*atanh((16*a^3*d*f*(d*tan(e + f*x))^(1/2)*(-841i/(256*a^6*d^3*f^2))^(1/2))/29)*(-841i/(2
56*a^6*d^3*f^2))^(1/2)

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