3.2.79 \(\int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx\) [179]

Optimal. Leaf size=370 \[ \frac {\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {7}{8}-\frac {15 i}{16}\right ) d^{9/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}+\frac {\left (\frac {7}{8}-\frac {15 i}{16}\right ) d^{9/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}+\frac {15 i d^4 \sqrt {d \tan (e+f x)}}{4 a^3 f}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac {7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )} \]

[Out]

(7/8+15/16*I)*d^(9/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f*2^(1/2)-(7/8+15/16*I)*d^(9/2)*arcta
n(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f*2^(1/2)+(-7/16+15/32*I)*d^(9/2)*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)+(7/16-15/32*I)*d^(9/2)*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)+15/4*I*d^4*(d*tan(f*x+e))^(1/2)/a^3/f-1/6*d*(d*tan(f*x+e))^(7/2)/f/(a+I*a*tan(f
*x+e))^3+5/12*I*d^2*(d*tan(f*x+e))^(5/2)/a/f/(a+I*a*tan(f*x+e))^2+7/6*d^3*(d*tan(f*x+e))^(3/2)/f/(a^3+I*a^3*ta
n(f*x+e))

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Rubi [A]
time = 0.43, antiderivative size = 370, 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 = {3639, 3676, 3609, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {7}{8}-\frac {15 i}{16}\right ) d^{9/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 f}+\frac {\left (\frac {7}{8}-\frac {15 i}{16}\right ) d^{9/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 f}+\frac {15 i d^4 \sqrt {d \tan (e+f x)}}{4 a^3 f}+\frac {7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(9/2)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

((7/4 + (15*I)/8)*d^(9/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*f) - ((7/4 + (15*I)
/8)*d^(9/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*f) - ((7/8 - (15*I)/16)*d^(9/2)*L
og[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*f) + ((7/8 - (15*I)/16)*d^(9/2
)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*f) + (((15*I)/4)*d^4*Sqrt[d
*Tan[e + f*x]])/(a^3*f) - (d*(d*Tan[e + f*x])^(7/2))/(6*f*(a + I*a*Tan[e + f*x])^3) + (((5*I)/12)*d^2*(d*Tan[e
+ f*x])^(5/2))/(a*f*(a + I*a*Tan[e + f*x])^2) + (7*d^3*(d*Tan[e + f*x])^(3/2))/(6*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 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 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 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]

Rubi steps

\begin {align*} \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx &=-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {(d \tan (e+f x))^{5/2} \left (-\frac {7 a d^2}{2}+\frac {13}{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))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac {\int \frac {(d \tan (e+f x))^{3/2} \left (-25 i a^2 d^3-31 a^2 d^3 \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{24 a^4}\\ &=-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac {7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\int \sqrt {d \tan (e+f x)} \left (84 a^3 d^4-90 i a^3 d^4 \tan (e+f x)\right ) \, dx}{48 a^6}\\ &=\frac {15 i d^4 \sqrt {d \tan (e+f x)}}{4 a^3 f}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac {7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\int \frac {90 i a^3 d^5+84 a^3 d^5 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{48 a^6}\\ &=\frac {15 i d^4 \sqrt {d \tan (e+f x)}}{4 a^3 f}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac {7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {90 i a^3 d^6+84 a^3 d^5 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{24 a^6 f}\\ &=\frac {15 i d^4 \sqrt {d \tan (e+f x)}}{4 a^3 f}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac {7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}--\frac {\left (\left (\frac {7}{4}-\frac {15 i}{8}\right ) d^5\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 (\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^5\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 {15 i d^4 \sqrt {d \tan (e+f x)}}{4 a^3 f}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac {7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\left (\left (\frac {7}{8}-\frac {15 i}{16}\right ) d^{9/2}\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 f}-\frac {\left (\left (\frac {7}{8}-\frac {15 i}{16}\right ) d^{9/2}\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 f}-\frac {\left (\left (\frac {7}{8}+\frac {15 i}{16}\right ) d^5\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 (\left (\frac {7}{8}+\frac {15 i}{16}\right ) d^5\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}{8}-\frac {15 i}{16}\right ) d^{9/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}+\frac {\left (\frac {7}{8}-\frac {15 i}{16}\right ) d^{9/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}+\frac {15 i d^4 \sqrt {d \tan (e+f x)}}{4 a^3 f}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac {7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}--\frac {\left (\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2}\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 f}-\frac {\left (\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2}\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 f}\\ &=\frac {\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {7}{8}-\frac {15 i}{16}\right ) d^{9/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}+\frac {\left (\frac {7}{8}-\frac {15 i}{16}\right ) d^{9/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}+\frac {15 i d^4 \sqrt {d \tan (e+f x)}}{4 a^3 f}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac {7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 3.69, size = 236, normalized size = 0.64 \begin {gather*} \frac {i d^4 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 ) \sqrt {d \tan (e+f x)}}{48 a^3 \left (-1+e^{2 i (e+f x)}\right ) f} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Tan[e + f*x])^(9/2)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

((I/48)*d^4*(-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)))]])*Sqrt[d*Tan[e + f*x]])/(a^3*E^((6*I)*(e + f*x))*(-1 + E^((2*I)*(e + f*x)))*
f)

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Maple [A]
time = 0.19, size = 141, normalized size = 0.38

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (i \sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {-20 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\frac {98 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+14 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (-i d +d \tan \left (f x +e \right )\right )^{3}}+\frac {29 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}\right )}{16}+\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}\right )}{f \,a^{3}}\)	\(141\)
default	\(\frac {2 d^{4} \left (i \sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {-20 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\frac {98 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+14 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (-i d +d \tan \left (f x +e \right )\right )^{3}}+\frac {29 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}\right )}{16}+\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}\right )}{f \,a^{3}}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2/f/a^3*d^4*(I*(d*tan(f*x+e))^(1/2)-1/16*d*((-20*(d*tan(f*x+e))^(5/2)+98/3*I*d*(d*tan(f*x+e))^(3/2)+14*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/(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))^(9/2)/(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. 620 vs. \(2 (289) = 578\).
time = 0.40, size = 620, normalized size = 1.68 \begin {gather*} -\frac {{\left (12 \, a^{3} \sqrt {\frac {i \, d^{9}}{64 \, a^{6} f^{2}}} f e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{5} 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^{9}}{64 \, a^{6} f^{2}}} \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 (-2 i \, f x - 2 i \, e\right )}}{d^{4}}\right ) - 12 \, a^{3} \sqrt {\frac {i \, d^{9}}{64 \, a^{6} f^{2}}} f e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{5} 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^{9}}{64 \, a^{6} f^{2}}} \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 (-2 i \, f x - 2 i \, e\right )}}{d^{4}}\right ) - 12 \, a^{3} \sqrt {-\frac {841 i \, d^{9}}{64 \, a^{6} f^{2}}} f e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (29 \, d^{5} + 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {-\frac {841 i \, d^{9}}{64 \, a^{6} f^{2}}} \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 (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) + 12 \, a^{3} \sqrt {-\frac {841 i \, d^{9}}{64 \, a^{6} f^{2}}} f e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (29 \, d^{5} - 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {-\frac {841 i \, d^{9}}{64 \, a^{6} f^{2}}} \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 (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) - {\left (146 i \, d^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 41 i \, d^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - 8 i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d^{4}\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))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/48*(12*a^3*sqrt(1/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6*I*e)*log(-2*(I*d^5*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(1/64*I*d^9/(a^6*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x +
2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/d^4) - 12*a^3*sqrt(1/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6*I*e)*log(-2*(I*d^
5*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(1/64*I*d^9/(a^6*f^2))*sqrt((-I*d*e^(2*
I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/d^4) - 12*a^3*sqrt(-841/64*I*d^9/(a^6*f
^2))*f*e^(6*I*f*x + 6*I*e)*log(1/8*(29*d^5 + 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-841/64*I*d^9/(a^6*f^2
))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a^3*f)) + 12*a^3*sq
rt(-841/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6*I*e)*log(1/8*(29*d^5 - 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt
(-841/64*I*d^9/(a^6*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*
e)/(a^3*f)) - (146*I*d^4*e^(6*I*f*x + 6*I*e) + 41*I*d^4*e^(4*I*f*x + 4*I*e) - 8*I*d^4*e^(2*I*f*x + 2*I*e) + I*
d^4)*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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(9/2)/(a+I*a*tan(f*x+e))**3,x)

[Out]

Timed out

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Giac [A]
time = 0.93, size = 251, normalized size = 0.68 \begin {gather*} -\frac {1}{24} \, d^{4} {\left (\frac {87 i \, \sqrt {2} \sqrt {d} \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} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 i \, \sqrt {2} \sqrt {d} \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} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {48 i \, \sqrt {d \tan \left (f x + e\right )}}{a^{3} f} - \frac {2 \, {\left (30 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{2} - 49 i \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right ) - 21 \, \sqrt {d \tan \left (f x + e\right )} d^{3}\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))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 5.48, size = 240, normalized size = 0.65 \begin {gather*} \mathrm {atan}\left (\frac {a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^9\,1{}\mathrm {i}}{256\,a^6\,f^2}}\,16{}\mathrm {i}}{d^5}\right )\,\sqrt {\frac {d^9\,1{}\mathrm {i}}{256\,a^6\,f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^9\,841{}\mathrm {i}}{256\,a^6\,f^2}}\,16{}\mathrm {i}}{29\,d^5}\right )\,\sqrt {-\frac {d^9\,841{}\mathrm {i}}{256\,a^6\,f^2}}\,2{}\mathrm {i}+\frac {\frac {7\,d^7\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,a^3\,f}-\frac {5\,d^5\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{2\,a^3\,f}+\frac {d^6\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,49{}\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}}+\frac {d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{a^3\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(9/2)/(a + a*tan(e + f*x)*1i)^3,x)

[Out]

atan((a^3*f*(d*tan(e + f*x))^(1/2)*((d^9*1i)/(256*a^6*f^2))^(1/2)*16i)/d^5)*((d^9*1i)/(256*a^6*f^2))^(1/2)*2i
- atan((a^3*f*(d*tan(e + f*x))^(1/2)*(-(d^9*841i)/(256*a^6*f^2))^(1/2)*16i)/(29*d^5))*(-(d^9*841i)/(256*a^6*f^
2))^(1/2)*2i + ((7*d^7*(d*tan(e + f*x))^(1/2))/(4*a^3*f) + (d^6*(d*tan(e + f*x))^(3/2)*49i)/(12*a^3*f) - (5*d^
5*(d*tan(e + f*x))^(5/2))/(2*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
) + (d^4*(d*tan(e + f*x))^(1/2)*2i)/(a^3*f)

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