3.180 \(\int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.61, 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 = {3558, 3595, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^3 f}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 f}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(7/2)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

((5/16 - (7*I)/16)*d^(7/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*f) - ((5/16 - (7*I
)/16)*d^(7/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^3*f) + ((5/32 + (7*I)/32)*d^(7/2)
*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*f) - ((5/32 + (7*I)/32)*d^(7
/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^3*f) - (d*(d*Tan[e + f*x])^
(5/2))/(6*f*(a + I*a*Tan[e + f*x])^3) + ((I/3)*d^2*(d*Tan[e + f*x])^(3/2))/(a*f*(a + I*a*Tan[e + f*x])^2) + (5
*d^3*Sqrt[d*Tan[e + f*x]])/(8*f*(a^3 + I*a^3*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 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 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((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 3595

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))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx &=-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {(d \tan (e+f x))^{3/2} \left (-\frac {5 a d^2}{2}+\frac {11}{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))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {\int \frac {\sqrt {d \tan (e+f x)} \left (-12 i a^2 d^3-18 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))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\int \frac {15 a^3 d^4-21 i a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{48 a^6}\\ &=-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {15 a^3 d^5-21 i a^3 d^4 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{24 a^6 f}\\ &=-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\left (\left (\frac {5}{16}-\frac {7 i}{16}\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^3 f}-\frac {\left (\left (\frac {5}{16}+\frac {7 i}{16}\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^3 f}\\ &=-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}--\frac {\left (\left (\frac {5}{32}+\frac {7 i}{32}\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^3 f}--\frac {\left (\left (\frac {5}{32}+\frac {7 i}{32}\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^3 f}-\frac {\left (\left (\frac {5}{32}-\frac {7 i}{32}\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^3 f}-\frac {\left (\left (\frac {5}{32}-\frac {7 i}{32}\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^3 f}\\ &=\frac {\left (\frac {5}{32}+\frac {7 i}{32}\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^3 f}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\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^3 f}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}--\frac {\left (\left (\frac {5}{16}-\frac {7 i}{16}\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^3 f}-\frac {\left (\left (\frac {5}{16}-\frac {7 i}{16}\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^3 f}\\ &=\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\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^3 f}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\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^3 f}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.39, size = 234, normalized size = 0.68 \[ \frac {d^4 \sec ^4(e+f x) \left (-12 i \sin (2 (e+f x))+21 i \sin (4 (e+f x))+19 \cos (4 (e+f x))+(21+15 i) \sqrt {\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+(15+21 i) \sqrt {\sin (2 (e+f x))} \sin (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )+(21-15 i) \sqrt {\sin (2 (e+f x))} \cos (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )-19\right )}{96 a^3 f (\tan (e+f x)-i)^3 \sqrt {d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(7/2)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(d^4*Sec[e + f*x]^4*(-19 + 19*Cos[4*(e + f*x)] + (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*I)*Sin[2*(e + f*x)] + (21 + 15*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)]) + (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*I)*Sin[4*(e + f*x)]))/(9
6*a^3*f*Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x])^3)

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fricas [B]  time = 0.62, size = 583, normalized size = 1.70 \[ \frac {{\left (12 \, a^{3} f \sqrt {-\frac {i \, d^{7}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (-2 i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 16 \, {\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^{7}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3}}\right ) - 12 \, a^{3} f \sqrt {-\frac {i \, d^{7}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (-2 i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 16 \, {\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^{7}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3}}\right ) + 12 \, a^{3} f \sqrt {\frac {9 i \, d^{7}}{16 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (-3 i \, d^{4} + 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 \, d^{7}}{16 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} f}\right ) - 12 \, a^{3} f \sqrt {\frac {9 i \, d^{7}}{16 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (-3 i \, d^{4} - 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 \, d^{7}}{16 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} f}\right ) + {\left (20 \, d^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 14 \, d^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 5 \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 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}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} f} \]

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))^3,x, algorithm="fricas")

[Out]

1/48*(12*a^3*f*sqrt(-1/64*I*d^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log((-2*I*d^4*e^(2*I*f*x + 2*I*e) + 16*(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^7/
(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d^3) - 12*a^3*f*sqrt(-1/64*I*d^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log((-2*I*d^4
*e^(2*I*f*x + 2*I*e) - 16*(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^7/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d^3) + 12*a^3*f*sqrt(9/16*I*d^7/(a^6*f^2))*
e^(6*I*f*x + 6*I*e)*log(1/4*(-3*I*d^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*d^7/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/(a^3*f)) - 12*a^3*f*sqrt(9/1
6*I*d^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/4*(-3*I*d^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*d^7/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/(a^3*f))
+ (20*d^3*e^(6*I*f*x + 6*I*e) + 14*d^3*e^(4*I*f*x + 4*I*e) - 5*d^3*e^(2*I*f*x + 2*I*e) + d^3)*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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giac [A]  time = 2.49, size = 231, normalized size = 0.67 \[ -\frac {1}{24} \, d^{3} {\left (-\frac {3 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {18 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {27 i \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{2} + 38 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right ) - 15 i \, \sqrt {d \tan \left (f x + e\right )} d^{3}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} f}\right )} \]

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))^3,x, algorithm="giac")

[Out]

-1/24*d^3*(-3*I*sqrt(2)*sqrt(d)*arctan(16*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^3*f*(I*d/sqrt(d^2) + 1)) - 18*I*sqrt(2)*sqrt(d)*arctan(16*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^3*f*(-I*d/sqrt(d^2) + 1)) + (27*I*sqrt(d*tan(f*x + e))*
d^3*tan(f*x + e)^2 + 38*sqrt(d*tan(f*x + e))*d^3*tan(f*x + e) - 15*I*sqrt(d*tan(f*x + e))*d^3)/((d*tan(f*x + e
) - I*d)^3*a^3*f))

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maple [A]  time = 0.35, size = 184, normalized size = 0.54 \[ -\frac {9 i d^{4} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{8 f \,a^{3} \left (d \tan \left (f x +e \right )-i d \right )^{3}}-\frac {19 d^{5} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{12 f \,a^{3} \left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {5 i d^{6} \sqrt {d \tan \left (f x +e \right )}}{8 f \,a^{3} \left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {3 i d^{4} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{4 f \,a^{3} \sqrt {-i d}}+\frac {i d^{4} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 f \,a^{3} \sqrt {i d}} \]

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))^3,x)

[Out]

-9/8*I/f/a^3*d^4/(d*tan(f*x+e)-I*d)^3*(d*tan(f*x+e))^(5/2)-19/12/f/a^3*d^5/(d*tan(f*x+e)-I*d)^3*(d*tan(f*x+e))
^(3/2)+5/8*I/f/a^3*d^6/(d*tan(f*x+e)-I*d)^3*(d*tan(f*x+e))^(1/2)+3/4*I/f/a^3*d^4/(-I*d)^(1/2)*arctan((d*tan(f*
x+e))^(1/2)/(-I*d)^(1/2))+1/8*I/f/a^3*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))^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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mupad [B]  time = 5.78, size = 217, normalized size = 0.63 \[ \mathrm {atan}\left (\frac {8\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{64\,a^6\,f^2}}}{3\,d^4}\right )\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{64\,a^6\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{256\,a^6\,f^2}}}{d^4}\right )\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{256\,a^6\,f^2}}\,2{}\mathrm {i}+\frac {\frac {19\,d^5\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{12\,a^3\,f}-\frac {d^6\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,5{}\mathrm {i}}{8\,a^3\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,9{}\mathrm {i}}{8\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}} \]

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)^3,x)

[Out]

atan((8*a^3*f*(d*tan(e + f*x))^(1/2)*((d^7*9i)/(64*a^6*f^2))^(1/2))/(3*d^4))*((d^7*9i)/(64*a^6*f^2))^(1/2)*2i
+ atan((16*a^3*f*(d*tan(e + f*x))^(1/2)*(-(d^7*1i)/(256*a^6*f^2))^(1/2))/d^4)*(-(d^7*1i)/(256*a^6*f^2))^(1/2)*
2i + ((19*d^5*(d*tan(e + f*x))^(3/2))/(12*a^3*f) - (d^6*(d*tan(e + f*x))^(1/2)*5i)/(8*a^3*f) + (d^4*(d*tan(e +
f*x))^(5/2)*9i)/(8*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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {i \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \]

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))**3,x)

[Out]

I*Integral((d*tan(e + f*x))**(7/2)/(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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