3.2.84 \(\int \frac {1}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3} \, dx\) [184]

Optimal. Leaf size=343 \[ -\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 \sqrt {d} f}-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )} \]

[Out]

(-7/32+5/32*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f*2^(1/2)/d^(1/2)+(7/32-5/32*I)*arctan(1+2^(
1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f*2^(1/2)/d^(1/2)-(7/64+5/64*I)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)
+d^(1/2)*tan(f*x+e))/a^3/f*2^(1/2)/d^(1/2)+(7/64+5/64*I)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f
*x+e))/a^3/f*2^(1/2)/d^(1/2)+1/6*(d*tan(f*x+e))^(1/2)/d/f/(a+I*a*tan(f*x+e))^3+1/3*(d*tan(f*x+e))^(1/2)/a/d/f/
(a+I*a*tan(f*x+e))^2+5/8*(d*tan(f*x+e))^(1/2)/d/f/(a^3+I*a^3*tan(f*x+e))

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Rubi [A]
time = 0.38, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3640, 3677, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-7/16 + (5*I)/16)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*Sqrt[d]*f) + ((7/16 - (5*
I)/16)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*Sqrt[d]*f) - ((7/32 + (5*I)/32)*Log[Sq
rt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*Sqrt[d]*f) + ((7/32 + (5*I)/32)*Log
[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*Sqrt[d]*f) + Sqrt[d*Tan[e + f*x]
]/(6*d*f*(a + I*a*Tan[e + f*x])^3) + Sqrt[d*Tan[e + f*x]]/(3*a*d*f*(a + I*a*Tan[e + f*x])^2) + (5*Sqrt[d*Tan[e
+ f*x]])/(8*d*f*(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 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}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3} \, dx &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\int \frac {\frac {11 a d}{2}-\frac {5}{2} i a d \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2} \, dx}{6 a^2 d}\\ &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {\int \frac {18 a^2 d^2-12 i a^2 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{24 a^4 d^2}\\ &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\int \frac {21 a^3 d^3-15 i a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{48 a^6 d^3}\\ &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {21 a^3 d^4-15 i a^3 d^3 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{24 a^6 d^3 f}\\ &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^3 f}+\frac {\left (\frac {7}{16}+\frac {5 i}{16}\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^3 f}\\ &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\left (\frac {7}{32}-\frac {5 i}{32}\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 f}+\frac {\left (\frac {7}{32}-\frac {5 i}{32}\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 f}+-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\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 \sqrt {d} f}+-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\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 \sqrt {d} f}\\ &=-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+-\frac {\left (\frac {7}{16}-\frac {5 i}{16}\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 \sqrt {d} f}+\frac {\left (\frac {7}{16}-\frac {5 i}{16}\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 \sqrt {d} f}\\ &=-\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 \sqrt {d} f}-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 1.31, size = 234, normalized size = 0.68 \begin {gather*} \frac {\sec ^4(e+f x) \left (19 \cos (4 (e+f x))-(15+21 i) \text {ArcSin}(\cos (e+f x)-\sin (e+f x)) \sqrt {\sin (2 (e+f x))} (\cos (3 (e+f x))+i \sin (3 (e+f x)))+i \left (19 i+(21+15 i) \cos (3 (e+f x)) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sqrt {\sin (2 (e+f x))}+12 \sin (2 (e+f x))-(15-21 i) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sqrt {\sin (2 (e+f x))} \sin (3 (e+f x))+21 \sin (4 (e+f x))\right )\right )}{96 a^3 f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^4*(19*Cos[4*(e + f*x)] - (15 + 21*I)*ArcSin[Cos[e + f*x] - Sin[e + f*x]]*Sqrt[Sin[2*(e + f*x)]]*
(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) + I*(19*I + (21 + 15*I)*Cos[3*(e + f*x)]*Log[Cos[e + f*x] + Sin[e + f*
x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] + 12*Sin[2*(e + f*x)] - (15 - 21*I)*Log[Cos[e + f*x] + Sin
[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]]*Sin[3*(e + f*x)] + 21*Sin[4*(e + f*x)])))/(96*a^3*f
*Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x])^3)

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Maple [A]
time = 0.25, size = 133, normalized size = 0.39

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\frac {\frac {-5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\frac {38 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+9 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (i d \tan \left (f x +e \right )+d \right )^{3}}-\frac {6 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{4}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{4} \sqrt {i d}}\right )}{f \,a^{3}}\)	\(133\)
default	\(\frac {2 d^{4} \left (\frac {\frac {-5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\frac {38 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+9 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (i d \tan \left (f x +e \right )+d \right )^{3}}-\frac {6 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{4}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{4} \sqrt {i d}}\right )}{f \,a^{3}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/a^3*d^4*(1/16/d^4*((-5*(d*tan(f*x+e))^(5/2)+38/3*I*d*(d*tan(f*x+e))^(3/2)+9*d^2*(d*tan(f*x+e))^(1/2))/(I*d
*tan(f*x+e)+d)^3-6*I/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))+1/16*I/d^4/(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))^(1/2)/(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. 596 vs. \(2 (269) = 538\).
time = 0.39, size = 596, normalized size = 1.74 \begin {gather*} -\frac {{\left (12 \, a^{3} d f \sqrt {-\frac {i}{64 \, a^{6} d f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-2 \, {\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 {i}{64 \, a^{6} d 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 \, a^{3} d f \sqrt {-\frac {i}{64 \, a^{6} d f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (2 \, {\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 {i}{64 \, a^{6} d 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 \, a^{3} d f \sqrt {\frac {9 i}{16 \, a^{6} d f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (4 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} 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 {9 i}{16 \, a^{6} d f^{2}}} + 3 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} f}\right ) + 12 \, a^{3} d f \sqrt {\frac {9 i}{16 \, a^{6} d f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {{\left (4 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} 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 {9 i}{16 \, a^{6} d f^{2}}} - 3 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} 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 (20 \, e^{\left (6 i \, f x + 6 i \, e\right )} + 26 \, e^{\left (4 i \, f x + 4 i \, e\right )} + 7 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(12*a^3*d*f*sqrt(-1/64*I/(a^6*d*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(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(-1/64*I/(a^6*d*f^2)) + I*d*e^(2*I*f*
x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 12*a^3*d*f*sqrt(-1/64*I/(a^6*d*f^2))*e^(6*I*f*x + 6*I*e)*log(2*(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(-1/64*I/
(a^6*d*f^2)) - I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 12*a^3*d*f*sqrt(9/16*I/(a^6*d*f^2))*e^(6*I*f*x
+ 6*I*e)*log(1/4*(4*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2
*I*e) + 1))*sqrt(9/16*I/(a^6*d*f^2)) + 3*I)*e^(-2*I*f*x - 2*I*e)/(a^3*f)) + 12*a^3*d*f*sqrt(9/16*I/(a^6*d*f^2)
)*e^(6*I*f*x + 6*I*e)*log(-1/4*(4*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e
^(2*I*f*x + 2*I*e) + 1))*sqrt(9/16*I/(a^6*d*f^2)) - 3*I)*e^(-2*I*f*x - 2*I*e)/(a^3*f)) - sqrt((-I*d*e^(2*I*f*x
+ 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(20*e^(6*I*f*x + 6*I*e) + 26*e^(4*I*f*x + 4*I*e) + 7*e^(2*I*f*x +
2*I*e) + 1))*e^(-6*I*f*x - 6*I*e)/(a^3*d*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.73, size = 227, normalized size = 0.66 \begin {gather*} \frac {i \, \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 )}{8 \, a^{3} \sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {3 i \, \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 )}{4 \, a^{3} \sqrt {d} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {15 i \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )^{2} + 38 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 27 i \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{24 \, {\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*I*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)) - 3/4*I*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)) - 1/24*(15*I*sqrt(d*tan(f*x + e))*d^2*ta
n(f*x + e)^2 + 38*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e) - 27*I*sqrt(d*tan(f*x + e))*d^2)/((d*tan(f*x + e) - I*
d)^3*a^3*f)

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Mupad [B]
time = 5.68, size = 206, normalized size = 0.60 \begin {gather*} \frac {\frac {19\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{12\,a^3\,f}+\frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,5{}\mathrm {i}}{8\,a^3\,f}-\frac {d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,9{}\mathrm {i}}{8\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}}-\mathrm {atan}\left (\frac {8\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {9{}\mathrm {i}}{64\,a^6\,d\,f^2}}}{3}\right )\,\sqrt {\frac {9{}\mathrm {i}}{64\,a^6\,d\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (16\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {1{}\mathrm {i}}{256\,a^6\,d\,f^2}}\right )\,\sqrt {-\frac {1{}\mathrm {i}}{256\,a^6\,d\,f^2}}\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((d*tan(e + f*x))^(5/2)*5i)/(8*a^3*f) - (d^2*(d*tan(e + f*x))^(1/2)*9i)/(8*a^3*f) + (19*d*(d*tan(e + f*x))^(3
/2))/(12*a^3*f))/(3*d^3*tan(e + f*x) - d^3*1i + d^3*tan(e + f*x)^2*3i - d^3*tan(e + f*x)^3) - atan((8*a^3*f*(d
*tan(e + f*x))^(1/2)*(9i/(64*a^6*d*f^2))^(1/2))/3)*(9i/(64*a^6*d*f^2))^(1/2)*2i + atan(16*a^3*f*(d*tan(e + f*x
))^(1/2)*(-1i/(256*a^6*d*f^2))^(1/2))*(-1i/(256*a^6*d*f^2))^(1/2)*2i

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