3.2.81 \(\int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx\) [181]

Optimal. Leaf size=329 \[ \frac {d^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.40, antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {3639, 3676, 3677, 12, 16, 3557, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {d^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{16 \sqrt {2} a^3 f}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(8*Sqrt[2]*a^3*f) - (d^(5/2)*ArcTan[1 + (Sqrt[2]*
Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(8*Sqrt[2]*a^3*f) - (d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt
[d*Tan[e + f*x]]])/(16*Sqrt[2]*a^3*f) + (d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f
*x]]])/(16*Sqrt[2]*a^3*f) - (d*(d*Tan[e + f*x])^(3/2))/(6*f*(a + I*a*Tan[e + f*x])^3) + ((I/4)*d^2*Sqrt[d*Tan[
e + f*x]])/(a*f*(a + I*a*Tan[e + f*x])^2) - ((I/4)*d^2*Sqrt[d*Tan[e + f*x]])/(f*(a^3 + I*a^3*Tan[e + f*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3639

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

Rule 3676

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*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f
*m)), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d
*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]

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 {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {\sqrt {d \tan (e+f x)} \left (-\frac {3 a d^2}{2}+\frac {9}{2} i a d^2 \tan (e+f x)\right )}{(a+i a \tan (e+f x))^2} \, dx}{6 a^2}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}+\frac {\int \frac {-3 i a^2 d^3-9 a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{24 a^4}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\int -\frac {6 a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{48 a^6 d}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^3 \int \frac {\tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^3}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^2 \int \sqrt {d \tan (e+f x)} \, dx}{8 a^3}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^3 \text {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{8 a^3 f}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^3 \text {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^3 f}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {d^3 \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^3 f}-\frac {d^3 \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^3 f}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^{5/2} \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 )}{16 \sqrt {2} a^3 f}-\frac {d^{5/2} \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 )}{16 \sqrt {2} a^3 f}-\frac {d^3 \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{16 a^3 f}-\frac {d^3 \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{16 a^3 f}\\ &=-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^{5/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}+\frac {d^{5/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}\\ &=\frac {d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 1.15, size = 232, normalized size = 0.71 \begin {gather*} \frac {d^3 \sec ^4(e+f x) \left (i-i \cos (4 (e+f x))+6 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))}-6 \sin (2 (e+f x))+6 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)))-6 \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))+3 \sin (4 (e+f x))\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[(d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(d^3*Sec[e + f*x]^4*(I - I*Cos[4*(e + f*x)] + (6*I)*Cos[3*(e + f*x)]*Log[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Si
n[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] - 6*Sin[2*(e + f*x)] + (6*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)]) - 6*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)] + 3*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 = 0.19, size = 116, normalized size = 0.35

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (-\frac {\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}}{\left (-i d +d \tan \left (f x +e \right )\right )^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d \sqrt {i d}}\right )}{f \,a^{3}}\)	\(116\)
default	\(\frac {2 d^{4} \left (-\frac {\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}}{\left (-i d +d \tan \left (f x +e \right )\right )^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d \sqrt {i d}}\right )}{f \,a^{3}}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/a^3*d^4*(-1/16/d*((2*(d*tan(f*x+e))^(5/2)-2/3*I*d*(d*tan(f*x+e))^(3/2))/(-I*d+d*tan(f*x+e))^3+1/(-I*d)^(1/
2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))-1/16/d/(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((d*tan(f*x+e))^(5/2)/(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. 615 vs. \(2 (267) = 534\).
time = 0.38, size = 615, normalized size = 1.87 \begin {gather*} \frac {{\left (12 \, a^{3} f \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, 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 {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2}}\right ) - 12 \, a^{3} f \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, 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 {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2}}\right ) + 12 \, a^{3} f \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (d^{3} + 8 \, {\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 {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) - 12 \, a^{3} f \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (d^{3} - 8 \, {\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 {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) + {\left (-2 i \, d^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + i \, d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, d^{2}\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}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(12*a^3*f*sqrt(1/64*I*d^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(I*d^3*e^(2*I*f*x + 2*I*e) + 8*(I*a^3*f*e
^(2*I*f*x + 2*I*e) + I*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(1/64*I*d^5
/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d^2) - 12*a^3*f*sqrt(1/64*I*d^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(I*d^3
*e^(2*I*f*x + 2*I*e) + 8*(-I*a^3*f*e^(2*I*f*x + 2*I*e) - I*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(1/64*I*d^5/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d^2) + 12*a^3*f*sqrt(-1/64*I*d^5/(a^6*f^
2))*e^(6*I*f*x + 6*I*e)*log(1/8*(d^3 + 8*(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(-1/64*I*d^5/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/(a^3*f)) - 12*a^3*f*sqrt(-1/
64*I*d^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*(d^3 - 8*(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(-1/64*I*d^5/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/(a^3*f)) + (
-2*I*d^2*e^(6*I*f*x + 6*I*e) + I*d^2*e^(4*I*f*x + 4*I*e) + 2*I*d^2*e^(2*I*f*x + 2*I*e) - I*d^2)*sqrt((-I*d*e^(
2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.81, size = 216, normalized size = 0.66 \begin {gather*} -\frac {1}{24} \, d^{2} {\left (\frac {3 \, \sqrt {2} \sqrt {d} \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} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 \, \sqrt {2} \sqrt {d} \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} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {2 \, {\left (3 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{2} - i \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} f}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*d^2*(3*sqrt(2)*sqrt(d)*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*f*(I*d/sqrt(d^2) + 1)) + 3*sqrt(2)*sqrt(d)*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*f*(-I*d/sqrt(d^2) + 1)) + 2*(3*sqrt(d*tan(f*x + e))*d^3*tan(f
*x + e)^2 - I*sqrt(d*tan(f*x + e))*d^3*tan(f*x + e))/((d*tan(f*x + e) - I*d)^3*a^3*f))

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Mupad [B]
time = 4.16, size = 158, normalized size = 0.48 \begin {gather*} \frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {-\frac {d^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{4\,a^3\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,1{}\mathrm {i}}{12\,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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-1)^(1/4)*d^(5/2)*atanh(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/(8*a^3*f) - ((-1)^(1/4)*d^(5/2)*atan((
(-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/(8*a^3*f) - ((d^4*(d*tan(e + f*x))^(3/2)*1i)/(12*a^3*f) - (d^3*(d
*tan(e + f*x))^(5/2))/(4*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)

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