3.2.73 \(\int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx\) [173]

Optimal. Leaf size=301 \[ \frac {\left (\frac {9}{16}+\frac {5 i}{16}\right ) d^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {9}{16}+\frac {5 i}{16}\right ) d^{5/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {9}{32}-\frac {5 i}{32}\right ) d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {9}{32}-\frac {5 i}{32}\right ) d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2} \]

[Out]

(9/32+5/32*I)*d^(5/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/f*2^(1/2)-(9/32+5/32*I)*d^(5/2)*arcta
n(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^2/f*2^(1/2)+(-9/64+5/64*I)*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^2/f*2^(1/2)+(9/64-5/64*I)*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^2/f*2^(1/2)+5/8*I*d^2*(d*tan(f*x+e))^(1/2)/a^2/f/(1+I*tan(f*x+e))-1/4*d*(d*tan(f*x+e))^(3/2)/
f/(a+I*a*tan(f*x+e))^2

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Rubi [A]
time = 0.28, antiderivative size = 301, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3639, 3676, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\left (\frac {9}{16}+\frac {5 i}{16}\right ) d^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {9}{16}+\frac {5 i}{16}\right ) d^{5/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {9}{32}-\frac {5 i}{32}\right ) d^{5/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {9}{32}-\frac {5 i}{32}\right ) d^{5/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}+\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((9/16 + (5*I)/16)*d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*f) - ((9/16 + (5*I
)/16)*d^(5/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a^2*f) - ((9/32 - (5*I)/32)*d^(5/2)
*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*f) + ((9/32 - (5*I)/32)*d^(5
/2)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a^2*f) + (((5*I)/8)*d^2*Sqrt[
d*Tan[e + f*x]])/(a^2*f*(1 + I*Tan[e + f*x])) - (d*(d*Tan[e + f*x])^(3/2))/(4*f*(a + I*a*Tan[e + f*x])^2)

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 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))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx &=-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}-\frac {\int \frac {\sqrt {d \tan (e+f x)} \left (-\frac {3 a d^2}{2}+\frac {7}{2} i a d^2 \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{4 a^2}\\ &=\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\int \frac {-\frac {5}{2} i a^2 d^3-\frac {9}{2} a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^4}\\ &=\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\text {Subst}\left (\int \frac {-\frac {5}{2} i a^2 d^4-\frac {9}{2} a^2 d^3 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^4 f}\\ &=\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac {\left (\left (\frac {9}{16}+\frac {5 i}{16}\right ) d^3\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}+\frac {\left (\left (\frac {9}{16}-\frac {5 i}{16}\right ) d^3\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}\\ &=\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac {\left (\left (\frac {9}{32}-\frac {5 i}{32}\right ) d^{5/2}\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^2 f}+-\frac {\left (\left (\frac {9}{32}-\frac {5 i}{32}\right ) d^{5/2}\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^2 f}+-\frac {\left (\left (\frac {9}{32}+\frac {5 i}{32}\right ) d^3\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^2 f}+-\frac {\left (\left (\frac {9}{32}+\frac {5 i}{32}\right ) d^3\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^2 f}\\ &=-\frac {\left (\frac {9}{32}-\frac {5 i}{32}\right ) d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {9}{32}-\frac {5 i}{32}\right ) d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac {\left (\left (\frac {9}{16}+\frac {5 i}{16}\right ) d^{5/2}\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^2 f}+\frac {\left (\left (\frac {9}{16}+\frac {5 i}{16}\right ) d^{5/2}\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^2 f}\\ &=\frac {\left (\frac {9}{16}+\frac {5 i}{16}\right ) d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {9}{16}+\frac {5 i}{16}\right ) d^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {9}{32}-\frac {5 i}{32}\right ) d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {9}{32}-\frac {5 i}{32}\right ) d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]
time = 1.07, size = 231, normalized size = 0.77 \begin {gather*} -\frac {d^3 \sec ^3(e+f x) \left (-7 \cos (e+f x)+7 \cos (3 (e+f x))+5 i \sin (e+f x)+(9-5 i) \cos (2 (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))}+(9+5 i) \text {ArcSin}(\cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x))) \sqrt {\sin (2 (e+f x))}+(5+9 i) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sin ^{\frac {3}{2}}(2 (e+f x))+5 i \sin (3 (e+f x))\right )}{32 a^2 f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(d^3*Sec[e + f*x]^3*(-7*Cos[e + f*x] + 7*Cos[3*(e + f*x)] + (5*I)*Sin[e + f*x] + (9 - 5*I)*Cos[2*(e + f*
x)]*Log[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sqrt[Sin[2*(e + f*x)]] + (9 + 5*I)*ArcSin[Cos[e
+ f*x] - Sin[e + f*x]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*Sqrt[Sin[2*(e + f*x)]] + (5 + 9*I)*Log[Cos[e +
f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*Sin[2*(e + f*x)]^(3/2) + (5*I)*Sin[3*(e + f*x)]))/(a^2*f*Sqrt[d*
Tan[e + f*x]]*(-I + Tan[e + f*x])^2)

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Maple [A]
time = 0.17, size = 108, normalized size = 0.36

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\frac {\frac {7 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}-\frac {5 i d \sqrt {d \tan \left (f x +e \right )}}{2}}{8 \left (i d \tan \left (f x +e \right )+d \right )^{2}}-\frac {7 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{16 \sqrt {-i d}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\)	\(108\)
default	\(\frac {2 d^{3} \left (-\frac {\frac {7 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}-\frac {5 i d \sqrt {d \tan \left (f x +e \right )}}{2}}{8 \left (i d \tan \left (f x +e \right )+d \right )^{2}}-\frac {7 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{16 \sqrt {-i d}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\)	\(108\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

2/f/a^2*d^3*(-1/8*(7/2*(d*tan(f*x+e))^(3/2)-5/2*I*d*(d*tan(f*x+e))^(1/2))/(I*d*tan(f*x+e)+d)^2-7/16/(-I*d)^(1/
2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))-1/8/(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))^2,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. 604 vs. \(2 (229) = 458\).
time = 0.38, size = 604, normalized size = 2.01 \begin {gather*} \frac {{\left (4 \, a^{2} f \sqrt {\frac {i \, d^{5}}{16 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, {\left (i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{2} 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}}{16 \, a^{4} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2}}\right ) - 4 \, a^{2} f \sqrt {\frac {i \, d^{5}}{16 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, {\left (-i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} 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}}{16 \, a^{4} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2}}\right ) + 4 \, a^{2} f \sqrt {-\frac {49 i \, d^{5}}{64 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (7 \, d^{3} + 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} 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 {49 i \, d^{5}}{64 \, a^{4} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} f}\right ) - 4 \, a^{2} f \sqrt {-\frac {49 i \, d^{5}}{64 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (7 \, d^{3} - 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} 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 {49 i \, d^{5}}{64 \, a^{4} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} f}\right ) + {\left (6 i \, d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 5 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 (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

1/16*(4*a^2*f*sqrt(1/16*I*d^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-2*(I*d^3*e^(2*I*f*x + 2*I*e) + 4*(I*a^2*f*e^
(2*I*f*x + 2*I*e) + I*a^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/16*I*d^5/
(a^4*f^2)))*e^(-2*I*f*x - 2*I*e)/d^2) - 4*a^2*f*sqrt(1/16*I*d^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-2*(I*d^3*e
^(2*I*f*x + 2*I*e) + 4*(-I*a^2*f*e^(2*I*f*x + 2*I*e) - I*a^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*
f*x + 2*I*e) + 1))*sqrt(1/16*I*d^5/(a^4*f^2)))*e^(-2*I*f*x - 2*I*e)/d^2) + 4*a^2*f*sqrt(-49/64*I*d^5/(a^4*f^2)
)*e^(4*I*f*x + 4*I*e)*log(1/8*(7*d^3 + 8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) +
I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-49/64*I*d^5/(a^4*f^2)))*e^(-2*I*f*x - 2*I*e)/(a^2*f)) - 4*a^2*f*sqrt(-49
/64*I*d^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/8*(7*d^3 - 8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt((-I*d*e^(
2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-49/64*I*d^5/(a^4*f^2)))*e^(-2*I*f*x - 2*I*e)/(a^2*f))
+ (6*I*d^2*e^(4*I*f*x + 4*I*e) + 5*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^(-4*I*f*x - 4*I*e)/(a^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \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))**2,x)

[Out]

-Integral((d*tan(e + f*x))**(5/2)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x)/a**2

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Giac [A]
time = 0.75, size = 206, normalized size = 0.68 \begin {gather*} \frac {1}{8} \, d^{2} {\left (\frac {7 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^{2} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {2 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^{2} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {7 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 5 i \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} 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))^2,x, algorithm="giac")

[Out]

1/8*d^2*(7*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^2*f*(I*d/sqrt(d^2) + 1)) + 2*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^2*f*(-I*d/sqrt(d^2) + 1)) + (7*sqrt(d*tan(f*x + e))*d^2
*tan(f*x + e) - 5*I*sqrt(d*tan(f*x + e))*d^2)/((d*tan(f*x + e) - I*d)^2*a^2*f))

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Mupad [B]
time = 5.67, size = 177, normalized size = 0.59 \begin {gather*} \frac {-\frac {7\,d^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8\,a^2\,f}+\frac {d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,5{}\mathrm {i}}{8\,a^2\,f}}{-d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+d^2\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}+d^2}+2\,\mathrm {atanh}\left (\frac {8\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^5\,1{}\mathrm {i}}{64\,a^4\,f^2}}}{d^3}\right )\,\sqrt {\frac {d^5\,1{}\mathrm {i}}{64\,a^4\,f^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^5\,49{}\mathrm {i}}{256\,a^4\,f^2}}}{7\,d^3}\right )\,\sqrt {-\frac {d^5\,49{}\mathrm {i}}{256\,a^4\,f^2}} \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)^2,x)

[Out]

((d^4*(d*tan(e + f*x))^(1/2)*5i)/(8*a^2*f) - (7*d^3*(d*tan(e + f*x))^(3/2))/(8*a^2*f))/(d^2*tan(e + f*x)*2i +
d^2 - d^2*tan(e + f*x)^2) + 2*atanh((8*a^2*f*(d*tan(e + f*x))^(1/2)*((d^5*1i)/(64*a^4*f^2))^(1/2))/d^3)*((d^5*
1i)/(64*a^4*f^2))^(1/2) + 2*atanh((16*a^2*f*(d*tan(e + f*x))^(1/2)*(-(d^5*49i)/(256*a^4*f^2))^(1/2))/(7*d^3))*
(-(d^5*49i)/(256*a^4*f^2))^(1/2)

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