\(\int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx\) [370]

Optimal result

Integrand size = 25, antiderivative size = 189 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=-\frac {31 d^{9/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {d^{9/2} \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}+\frac {27 d^4 \sqrt {d \tan (e+f x)}}{8 a^3 f}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a+a \tan (e+f x))^2} \]

[Out]

-31/8*d^(9/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f+1/4*d^(9/2)*arctanh(1/2*(d^(1/2)+d^(1/2)*tan(f*x+e))*
2^(1/2)/(d*tan(f*x+e))^(1/2))/a^3/f*2^(1/2)+27/8*d^4*(d*tan(f*x+e))^(1/2)/a^3/f-9/8*d^3*(d*tan(f*x+e))^(3/2)/a
^3/f/(1+tan(f*x+e))-1/4*d^2*(d*tan(f*x+e))^(5/2)/a/f/(a+a*tan(f*x+e))^2

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3646, 3726, 3728, 3735, 3613, 214, 3715, 65, 211} \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=-\frac {31 d^{9/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {d^{9/2} \text {arctanh}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}+\frac {27 d^4 \sqrt {d \tan (e+f x)}}{8 a^3 f}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{8 a^3 f (\tan (e+f x)+1)}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2} \]

[In]

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

[Out]

(-31*d^(9/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*f) + (d^(9/2)*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x
])/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(2*Sqrt[2]*a^3*f) + (27*d^4*Sqrt[d*Tan[e + f*x]])/(8*a^3*f) - (9*d^3*(d*Ta
n[e + f*x])^(3/2))/(8*a^3*f*(1 + Tan[e + f*x])) - (d^2*(d*Tan[e + f*x])^(5/2))/(4*a*f*(a + a*Tan[e + f*x])^2)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 211

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

Rule 214

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

Rule 3613

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2*(d^2/f),
Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x
] && EqQ[c^2 - d^2, 0]

Rule 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3726

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Ta
n[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3728

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d
*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3735

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*ta
n[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*x])^n*Simp[a*(A - C) - (A*b - b*
C)*Tan[e + f*x], x], x], x] + Dist[(A*b^2 + a^2*C)/(a^2 + b^2), Int[(c + d*Tan[e + f*x])^n*((1 + Tan[e + f*x]^
2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^
2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a+a \tan (e+f x))^2}+\frac {\int \frac {(d \tan (e+f x))^{3/2} \left (\frac {5 a^2 d^3}{2}-2 a^2 d^3 \tan (e+f x)+\frac {9}{2} a^2 d^3 \tan ^2(e+f x)\right )}{(a+a \tan (e+f x))^2} \, dx}{4 a^3} \\ & = -\frac {9 d^3 (d \tan (e+f x))^{3/2}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a+a \tan (e+f x))^2}+\frac {\int \frac {\sqrt {d \tan (e+f x)} \left (\frac {27 a^4 d^4}{2}-4 a^4 d^4 \tan (e+f x)+\frac {27}{2} a^4 d^4 \tan ^2(e+f x)\right )}{a+a \tan (e+f x)} \, dx}{8 a^6} \\ & = \frac {27 d^4 \sqrt {d \tan (e+f x)}}{8 a^3 f}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a+a \tan (e+f x))^2}+\frac {\int \frac {-\frac {27}{4} a^5 d^5-\frac {35}{4} a^5 d^5 \tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{4 a^7} \\ & = \frac {27 d^4 \sqrt {d \tan (e+f x)}}{8 a^3 f}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a+a \tan (e+f x))^2}+\frac {\int \frac {2 a^6 d^5-2 a^6 d^5 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^9}-\frac {\left (31 d^5\right ) \int \frac {1+\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{16 a^2} \\ & = \frac {27 d^4 \sqrt {d \tan (e+f x)}}{8 a^3 f}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a+a \tan (e+f x))^2}-\frac {\left (31 d^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {d x} (a+a x)} \, dx,x,\tan (e+f x)\right )}{16 a^2 f}-\frac {\left (a^3 d^{10}\right ) \text {Subst}\left (\int \frac {1}{-8 a^{12} d^{10}+d x^2} \, dx,x,\frac {2 a^6 d^5+2 a^6 d^5 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f} \\ & = \frac {d^{9/2} \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}+\frac {27 d^4 \sqrt {d \tan (e+f x)}}{8 a^3 f}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a+a \tan (e+f x))^2}-\frac {\left (31 d^4\right ) \text {Subst}\left (\int \frac {1}{a+\frac {a x^2}{d}} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^2 f} \\ & = -\frac {31 d^{9/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {d^{9/2} \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}+\frac {27 d^4 \sqrt {d \tan (e+f x)}}{8 a^3 f}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a+a \tan (e+f x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.18 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.75 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=-\frac {d^4 \sec ^2(e+f x) \left (2 \sqrt {2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )-2 \sqrt {2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )+62 \sqrt {d} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (\cos (e+f x)+\sin (e+f x))^2-43 \sqrt {d \tan (e+f x)}-11 \cos (2 (e+f x)) \sqrt {d \tan (e+f x)}+\sin (2 (e+f x)) \left (2 \sqrt {2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )-2 \sqrt {2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )-45 \sqrt {d \tan (e+f x)}\right )\right )}{16 a^3 f (1+\tan (e+f x))^2} \]

[In]

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

[Out]

-1/16*(d^4*Sec[e + f*x]^2*(2*Sqrt[2]*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]
] - 2*Sqrt[2]*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]] + 62*Sqrt[d]*ArcTan[S
qrt[d*Tan[e + f*x]]/Sqrt[d]]*(Cos[e + f*x] + Sin[e + f*x])^2 - 43*Sqrt[d*Tan[e + f*x]] - 11*Cos[2*(e + f*x)]*S
qrt[d*Tan[e + f*x]] + Sin[2*(e + f*x)]*(2*Sqrt[2]*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*
Tan[e + f*x]]] - 2*Sqrt[2]*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]] - 45*Sqr
t[d*Tan[e + f*x]])))/(a^3*f*(1 + Tan[e + f*x])^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(354\) vs. \(2(156)=312\).

Time = 1.10 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.88

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {-\frac {13 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}-\frac {11 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {31 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}\right )}{4}+\frac {d \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{4}\right )}{f \,a^{3}}\)	\(355\)
default	\(\frac {2 d^{4} \left (\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {-\frac {13 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}-\frac {11 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {31 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}\right )}{4}+\frac {d \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{4}\right )}{f \,a^{3}}\)	\(355\)

[In]

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

[Out]

2/f/a^3*d^4*((d*tan(f*x+e))^(1/2)-1/4*d*((-13/4*(d*tan(f*x+e))^(3/2)-11/4*d*(d*tan(f*x+e))^(1/2))/(d*tan(f*x+e
)+d)^2+31/4/d^(1/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2)))+1/4*d*(1/8/d*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(
d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d
^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^
(1/2)+1))-1/8/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*t
an(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(
1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 472, normalized size of antiderivative = 2.50 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=\left [-\frac {4 \, {\left (\sqrt {2} d^{4} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} d^{4} \tan \left (f x + e\right ) + \sqrt {2} d^{4}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {-d}}{2 \, d \tan \left (f x + e\right )}\right ) - 31 \, {\left (d^{4} \tan \left (f x + e\right )^{2} + 2 \, d^{4} \tan \left (f x + e\right ) + d^{4}\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) - 2 \, {\left (16 \, d^{4} \tan \left (f x + e\right )^{2} + 45 \, d^{4} \tan \left (f x + e\right ) + 27 \, d^{4}\right )} \sqrt {d \tan \left (f x + e\right )}}{16 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}, -\frac {31 \, {\left (d^{4} \tan \left (f x + e\right )^{2} + 2 \, d^{4} \tan \left (f x + e\right ) + d^{4}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right ) - {\left (\sqrt {2} d^{4} \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} d^{4} \tan \left (f x + e\right ) + \sqrt {2} d^{4}\right )} \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (\sqrt {2} \tan \left (f x + e\right ) + \sqrt {2}\right )} \sqrt {d} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (16 \, d^{4} \tan \left (f x + e\right )^{2} + 45 \, d^{4} \tan \left (f x + e\right ) + 27 \, d^{4}\right )} \sqrt {d \tan \left (f x + e\right )}}{8 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}\right ] \]

[In]

integrate((d*tan(f*x+e))^(9/2)/(a+a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/16*(4*(sqrt(2)*d^4*tan(f*x + e)^2 + 2*sqrt(2)*d^4*tan(f*x + e) + sqrt(2)*d^4)*sqrt(-d)*arctan(1/2*sqrt(d*t
an(f*x + e))*(sqrt(2)*tan(f*x + e) + sqrt(2))*sqrt(-d)/(d*tan(f*x + e))) - 31*(d^4*tan(f*x + e)^2 + 2*d^4*tan(
f*x + e) + d^4)*sqrt(-d)*log((d*tan(f*x + e) - 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)) - 2*(1
6*d^4*tan(f*x + e)^2 + 45*d^4*tan(f*x + e) + 27*d^4)*sqrt(d*tan(f*x + e)))/(a^3*f*tan(f*x + e)^2 + 2*a^3*f*tan
(f*x + e) + a^3*f), -1/8*(31*(d^4*tan(f*x + e)^2 + 2*d^4*tan(f*x + e) + d^4)*sqrt(d)*arctan(sqrt(d*tan(f*x + e
))/sqrt(d)) - (sqrt(2)*d^4*tan(f*x + e)^2 + 2*sqrt(2)*d^4*tan(f*x + e) + sqrt(2)*d^4)*sqrt(d)*log((d*tan(f*x +
e)^2 + 2*sqrt(d*tan(f*x + e))*(sqrt(2)*tan(f*x + e) + sqrt(2))*sqrt(d) + 4*d*tan(f*x + e) + d)/(tan(f*x + e)^
2 + 1)) - (16*d^4*tan(f*x + e)^2 + 45*d^4*tan(f*x + e) + 27*d^4)*sqrt(d*tan(f*x + e)))/(a^3*f*tan(f*x + e)^2 +
2*a^3*f*tan(f*x + e) + a^3*f)]

Sympy [F]

\[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=\frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}{\tan ^{3}{\left (e + f x \right )} + 3 \tan ^{2}{\left (e + f x \right )} + 3 \tan {\left (e + f x \right )} + 1}\, dx}{a^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.06 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=\frac {\frac {d^{6} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a^{3}} - \frac {31 \, d^{\frac {11}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3}} + \frac {16 \, \sqrt {d \tan \left (f x + e\right )} d^{5}}{a^{3}} + \frac {13 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{6} + 11 \, \sqrt {d \tan \left (f x + e\right )} d^{7}}{a^{3} d^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}}}{8 \, d f} \]

[In]

integrate((d*tan(f*x+e))^(9/2)/(a+a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

1/8*(d^6*(sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) - sqrt(2)*log(d*tan(f
*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d))/a^3 - 31*d^(11/2)*arctan(sqrt(d*tan(f*x + e))/sqr
t(d))/a^3 + 16*sqrt(d*tan(f*x + e))*d^5/a^3 + (13*(d*tan(f*x + e))^(3/2)*d^6 + 11*sqrt(d*tan(f*x + e))*d^7)/(a
^3*d^2*tan(f*x + e)^2 + 2*a^3*d^2*tan(f*x + e) + a^3*d^2))/(d*f)

Giac [F(-2)]

Exception generated. \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((d*tan(f*x+e))^(9/2)/(a+a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Not invertible Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 6.08 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.93 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=\frac {\frac {11\,d^6\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8}+\frac {13\,d^5\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8}}{f\,a^3\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,f\,a^3\,d^2\,\mathrm {tan}\left (e+f\,x\right )+f\,a^3\,d^2}+\frac {2\,d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{a^3\,f}-\frac {31\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {\sqrt {2}\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,d^{49/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,969{}\mathrm {i}}{32\,\left (\frac {969\,d^{25}\,\mathrm {tan}\left (e+f\,x\right )}{32}+\frac {969\,d^{25}}{32}\right )}\right )\,1{}\mathrm {i}}{4\,a^3\,f} \]

[In]

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

[Out]

((11*d^6*(d*tan(e + f*x))^(1/2))/8 + (13*d^5*(d*tan(e + f*x))^(3/2))/8)/(a^3*d^2*f + a^3*d^2*f*tan(e + f*x)^2
+ 2*a^3*d^2*f*tan(e + f*x)) + (2*d^4*(d*tan(e + f*x))^(1/2))/(a^3*f) - (31*d^(9/2)*atan((d*tan(e + f*x))^(1/2)
/d^(1/2)))/(8*a^3*f) - (2^(1/2)*d^(9/2)*atan((2^(1/2)*d^(49/2)*(d*tan(e + f*x))^(1/2)*969i)/(32*((969*d^25*tan
(e + f*x))/32 + (969*d^25)/32)))*1i)/(4*a^3*f)