3.164 \(\int \frac {(d \tan (e+f x))^{7/2}}{a+i a \tan (e+f x)} \, dx\)

Optimal. Leaf size=312 \[ \frac {\left (\frac {5}{4}-\frac {7 i}{4}\right ) d^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}-\frac {\left (\frac {5}{4}-\frac {7 i}{4}\right ) d^{7/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a f}+\frac {\left (\frac {5}{8}+\frac {7 i}{8}\right ) d^{7/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a f}-\frac {\left (\frac {5}{8}+\frac {7 i}{8}\right ) d^{7/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a f}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{2 a f}-\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 312, 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 = {3550, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {\left (\frac {5}{4}-\frac {7 i}{4}\right ) d^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}-\frac {\left (\frac {5}{4}-\frac {7 i}{4}\right ) d^{7/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a f}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{2 a f}-\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}+\frac {\left (\frac {5}{8}+\frac {7 i}{8}\right ) d^{7/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a f}-\frac {\left (\frac {5}{8}+\frac {7 i}{8}\right ) d^{7/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a f}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(7/2)/(a + I*a*Tan[e + f*x]),x]

[Out]

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

Rule 204

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

Rule 617

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 628

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 1162

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 1165

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 1168

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 3528

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 3534

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 3550

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

Rubi steps

\begin {align*} \int \frac {(d \tan (e+f x))^{7/2}}{a+i a \tan (e+f x)} \, dx &=-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}+\frac {\int (d \tan (e+f x))^{3/2} \left (\frac {5 a d^2}{2}-\frac {7}{2} i a d^2 \tan (e+f x)\right ) \, dx}{2 a^2}\\ &=-\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}+\frac {\int \sqrt {d \tan (e+f x)} \left (\frac {7}{2} i a d^3+\frac {5}{2} a d^3 \tan (e+f x)\right ) \, dx}{2 a^2}\\ &=\frac {5 d^3 \sqrt {d \tan (e+f x)}}{2 a f}-\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}+\frac {\int \frac {-\frac {5 a d^4}{2}+\frac {7}{2} i a d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{2 a^2}\\ &=\frac {5 d^3 \sqrt {d \tan (e+f x)}}{2 a f}-\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {5 a d^5}{2}+\frac {7}{2} i a d^4 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}\\ &=\frac {5 d^3 \sqrt {d \tan (e+f x)}}{2 a f}-\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}+-\frac {\left (\left (\frac {5}{4}+\frac {7 i}{4}\right ) d^4\right ) \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a f}+-\frac {\left (\left (\frac {5}{4}-\frac {7 i}{4}\right ) d^4\right ) \operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a f}\\ &=\frac {5 d^3 \sqrt {d \tan (e+f x)}}{2 a f}-\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}+\frac {\left (\left (\frac {5}{8}+\frac {7 i}{8}\right ) d^{7/2}\right ) \operatorname {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 f}+\frac {\left (\left (\frac {5}{8}+\frac {7 i}{8}\right ) d^{7/2}\right ) \operatorname {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 f}+-\frac {\left (\left (\frac {5}{8}-\frac {7 i}{8}\right ) d^4\right ) \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a f}+-\frac {\left (\left (\frac {5}{8}-\frac {7 i}{8}\right ) d^4\right ) \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a f}\\ &=\frac {\left (\frac {5}{8}+\frac {7 i}{8}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a f}-\frac {\left (\frac {5}{8}+\frac {7 i}{8}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a f}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{2 a f}-\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}+-\frac {\left (\left (\frac {5}{4}-\frac {7 i}{4}\right ) d^{7/2}\right ) \operatorname {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 f}+\frac {\left (\left (\frac {5}{4}-\frac {7 i}{4}\right ) d^{7/2}\right ) \operatorname {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 f}\\ &=\frac {\left (\frac {5}{4}-\frac {7 i}{4}\right ) d^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}-\frac {\left (\frac {5}{4}-\frac {7 i}{4}\right ) d^{7/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a f}+\frac {\left (\frac {5}{8}+\frac {7 i}{8}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a f}-\frac {\left (\frac {5}{8}+\frac {7 i}{8}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a f}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{2 a f}-\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{6 a f}-\frac {d (d \tan (e+f x))^{5/2}}{2 f (a+i a \tan (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 1.80, size = 275, normalized size = 0.88 \[ -\frac {d^4 \sec ^3(e+f x) \left (54 i \sin (e+f x)+22 i \sin (3 (e+f x))-16 \cos (e+f x)+16 \cos (3 (e+f x))+(42+30 i) \sqrt {\sin (2 (e+f x))} \cos (e+f x) \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (e+f x)+i \sin (e+f x))+(15+21 i) \sin ^{\frac {3}{2}}(2 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )+(21-15 i) \sqrt {\sin (2 (e+f x))} \cos (2 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )+(21-15 i) \sqrt {\sin (2 (e+f x))} \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{48 a f (\tan (e+f x)-i) \sqrt {d \tan (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Tan[e + f*x])^(7/2)/(a + I*a*Tan[e + f*x]),x]

[Out]

-1/48*(d^4*Sec[e + f*x]^3*(-16*Cos[e + f*x] + 16*Cos[3*(e + f*x)] + (54*I)*Sin[e + f*x] + (21 - 15*I)*Log[Cos[
e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] + (21 - 15*I)*Cos[2*(e + f*x)]*Log[Co
s[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] + (42 + 30*I)*ArcSin[Cos[e + f*x] -
 Sin[e + f*x]]*Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e + f*x])*Sqrt[Sin[2*(e + f*x)]] + (15 + 21*I)*Log[Cos[e + f
*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sin[2*(e + f*x)]^(3/2) + (22*I)*Sin[3*(e + f*x)]))/(a*f*Sqrt[d*Ta
n[e + f*x]]*(-I + Tan[e + f*x]))

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fricas [B]  time = 0.79, size = 612, normalized size = 1.96 \[ \frac {3 \, \sqrt {\frac {9 i \, d^{7}}{a^{2} f^{2}}} {\left (a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (-3 i \, d^{4} + \sqrt {\frac {9 i \, d^{7}}{a^{2} f^{2}}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) - 3 \, \sqrt {\frac {9 i \, d^{7}}{a^{2} f^{2}}} {\left (a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (-3 i \, d^{4} - \sqrt {\frac {9 i \, d^{7}}{a^{2} f^{2}}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) + 3 \, \sqrt {-\frac {i \, d^{7}}{4 \, a^{2} f^{2}}} {\left (a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (-2 i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, \sqrt {-\frac {i \, d^{7}}{4 \, a^{2} f^{2}}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3}}\right ) - 3 \, \sqrt {-\frac {i \, d^{7}}{4 \, a^{2} f^{2}}} {\left (a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (-2 i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, \sqrt {-\frac {i \, d^{7}}{4 \, a^{2} f^{2}}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a 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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3}}\right ) + {\left (19 \, d^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 38 \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 \, d^{3}\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}}}{12 \, {\left (a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/12*(3*sqrt(9*I*d^7/(a^2*f^2))*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*log((-3*I*d^4 + sqrt(9*I*d
^7/(a^2*f^2))*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*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*f)) - 3*sqrt(9*I*d^7/(a^2*f^2))*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*
log((-3*I*d^4 - sqrt(9*I*d^7/(a^2*f^2))*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*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*f)) + 3*sqrt(-1/4*I*d^7/(a^2*f^2))*(a*f*e^(4*I*f*x + 4*I*e
) + a*f*e^(2*I*f*x + 2*I*e))*log((-2*I*d^4*e^(2*I*f*x + 2*I*e) + 4*sqrt(-1/4*I*d^7/(a^2*f^2))*(a*f*e^(2*I*f*x
+ 2*I*e) + a*f)*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^3) -
3*sqrt(-1/4*I*d^7/(a^2*f^2))*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*log((-2*I*d^4*e^(2*I*f*x + 2*
I*e) - 4*sqrt(-1/4*I*d^7/(a^2*f^2))*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*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^3) + (19*d^3*e^(4*I*f*x + 4*I*e) + 38*d^3*e^(2*I*f*x + 2*I*e) +
 3*d^3)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*
I*f*x + 2*I*e))

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giac [A]  time = 1.59, size = 242, normalized size = 0.78 \[ -\frac {1}{6} \, d^{3} {\left (\frac {18 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 i \, \sqrt {d \tan \left (f x + e\right )} d}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a f} + \frac {4 \, {\left (i \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{3} f^{2} \tan \left (f x + e\right ) - 3 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{3} f^{2}\right )}}{a^{3} d^{3} f^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

-1/6*d^3*(18*sqrt(2)*sqrt(d)*arctan(16*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(
d^2)*sqrt(d)))/(a*f*(I*d/sqrt(d^2) + 1)) + 3*sqrt(2)*sqrt(d)*arctan(16*I*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-8*I*
sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a*f*(-I*d/sqrt(d^2) + 1)) + 3*I*sqrt(d*tan(f*x + e))*d/((d*ta
n(f*x + e) - I*d)*a*f) + 4*(I*sqrt(d*tan(f*x + e))*a^2*d^3*f^2*tan(f*x + e) - 3*sqrt(d*tan(f*x + e))*a^2*d^3*f
^2)/(a^3*d^3*f^3))

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maple [A]  time = 0.28, size = 154, normalized size = 0.49 \[ -\frac {2 i d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f a}+\frac {2 d^{3} \sqrt {d \tan \left (f x +e \right )}}{a f}-\frac {i d^{4} \sqrt {d \tan \left (f x +e \right )}}{2 f a \left (d \tan \left (f x +e \right )-i d \right )}+\frac {3 i d^{4} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{f a \sqrt {-i d}}+\frac {i d^{4} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{2 f a \sqrt {i d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e)),x)

[Out]

-2/3*I/f/a*d^2*(d*tan(f*x+e))^(3/2)+2*d^3*(d*tan(f*x+e))^(1/2)/a/f-1/2*I/f/a*d^4*(d*tan(f*x+e))^(1/2)/(d*tan(f
*x+e)-I*d)+3*I/f/a*d^4/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))+1/2*I/f/a*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 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e)),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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mupad [B]  time = 6.21, size = 180, normalized size = 0.58 \[ \mathrm {atan}\left (\frac {2\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{4\,a^2\,f^2}}}{3\,d^4}\right )\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{4\,a^2\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {4\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{16\,a^2\,f^2}}}{d^4}\right )\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{16\,a^2\,f^2}}\,2{}\mathrm {i}+\frac {2\,d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{a\,f}-\frac {d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,2{}\mathrm {i}}{3\,a\,f}+\frac {d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{2\,a\,f\,\left (-d\,\mathrm {tan}\left (e+f\,x\right )+d\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(7/2)/(a + a*tan(e + f*x)*1i),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}{\tan {\left (e + f x \right )} - i}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**(7/2)/(a+I*a*tan(f*x+e)),x)

[Out]

-I*Integral((d*tan(e + f*x))**(7/2)/(tan(e + f*x) - I), x)/a

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