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

Optimal. Leaf size=189 \[ -\frac{31 d^{9/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}+\frac{27 d^4 \sqrt{d \tan (e+f x)}}{8 a^3 f}-\frac{9 d^3 (d \tan (e+f x))^{3/2}}{8 a^3 f (\tan (e+f x)+1)}+\frac{d^{9/2} \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 f}-\frac{d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2} \]

[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)

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Rubi [A]  time = 0.755399, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3565, 3645, 3647, 3654, 3532, 208, 3634, 63, 205} \[ -\frac{31 d^{9/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}+\frac{27 d^4 \sqrt{d \tan (e+f x)}}{8 a^3 f}-\frac{9 d^3 (d \tan (e+f x))^{3/2}}{8 a^3 f (\tan (e+f x)+1)}+\frac{d^{9/2} \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 f}-\frac{d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[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 3565

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 3645

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*T
an[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 3647

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 3654

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]

Rule 3532

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 208

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

Rule 3634

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 63

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 205

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

Rubi steps

\begin{align*} \int \frac{(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx &=-\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 ) \operatorname{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 ) \operatorname{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} \tanh ^{-1}\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 ) \operatorname{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} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}+\frac{d^{9/2} \tanh ^{-1}\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]  time = 6.26722, size = 346, normalized size = 1.83 \[ \frac{\cot (e+f x) \csc ^3(e+f x) (d \tan (e+f x))^{9/2} (\sin (e+f x)+\cos (e+f x))^3 \left (-\frac{11 \sin (e+f x)}{8 (\sin (e+f x)+\cos (e+f x))}-\frac{1}{8 (\sin (e+f x)+\cos (e+f x))^2}+\frac{7}{2}\right )}{f (a \tan (e+f x)+a)^3}+\frac{\sec ^3(e+f x) (d \tan (e+f x))^{9/2} (\sin (e+f x)+\cos (e+f x))^3 \left (\frac{2 \sqrt{2} \cos (2 (e+f x)) \csc (e+f x) \sec ^3(e+f x) \left (\log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )-\log \left (-\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}-1\right )\right )}{(1-\tan (e+f x)) \left (\tan ^2(e+f x)+1\right ) (\cot (e+f x)+1)}-\frac{62 \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right ) (\tan (e+f x)+1) \csc (e+f x) \sec ^3(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 (\cot (e+f x)+1)}\right )}{16 f \tan ^{\frac{9}{2}}(e+f x) (a \tan (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[e + f*x]*Csc[e + f*x]^3*(Cos[e + f*x] + Sin[e + f*x])^3*(7/2 - 1/(8*(Cos[e + f*x] + Sin[e + f*x])^2) - (1
1*Sin[e + f*x])/(8*(Cos[e + f*x] + Sin[e + f*x])))*(d*Tan[e + f*x])^(9/2))/(f*(a + a*Tan[e + f*x])^3) + (Sec[e
+ f*x]^3*(Cos[e + f*x] + Sin[e + f*x])^3*(d*Tan[e + f*x])^(9/2)*((-62*ArcTan[Sqrt[Tan[e + f*x]]]*Csc[e + f*x]
*Sec[e + f*x]^3*(1 + Tan[e + f*x]))/((1 + Cot[e + f*x])*(1 + Tan[e + f*x]^2)^2) + (2*Sqrt[2]*Cos[2*(e + f*x)]*
Csc[e + f*x]*(-Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[e + f*x]] + Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[
e + f*x]])*Sec[e + f*x]^3)/((1 + Cot[e + f*x])*(1 - Tan[e + f*x])*(1 + Tan[e + f*x]^2))))/(16*f*Tan[e + f*x]^(
9/2)*(a + a*Tan[e + f*x])^3)

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Maple [B]  time = 0.039, size = 461, normalized size = 2.4 \begin{align*} 2\,{\frac{{d}^{4}\sqrt{d\tan \left ( fx+e \right ) }}{{a}^{3}f}}+{\frac{{d}^{4}\sqrt{2}}{16\,{a}^{3}f}\sqrt [4]{{d}^{2}}\ln \left ({ \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ) }+{\frac{{d}^{4}\sqrt{2}}{8\,{a}^{3}f}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{{d}^{4}\sqrt{2}}{8\,{a}^{3}f}\sqrt [4]{{d}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{{d}^{5}\sqrt{2}}{16\,{a}^{3}f}\ln \left ({ \left ( d\tan \left ( fx+e \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( fx+e \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{{d}^{5}\sqrt{2}}{8\,{a}^{3}f}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{{d}^{5}\sqrt{2}}{8\,{a}^{3}f}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{13\,{d}^{5}}{8\,{a}^{3}f \left ( d\tan \left ( fx+e \right ) +d \right ) ^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{11\,{d}^{6}}{8\,{a}^{3}f \left ( d\tan \left ( fx+e \right ) +d \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{31}{8\,{a}^{3}f}{d}^{{\frac{9}{2}}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{d}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d^4*(d*tan(f*x+e))^(1/2)/a^3/f+1/16/f/a^3*d^4*(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)))+1/8/f/a^3*d^
4*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/8/f/a^3*d^4*(d^2)^(1/4)*2^(1/2)*arc
tan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-1/16/f/a^3*d^5/(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
)))-1/8/f/a^3*d^5/(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+1/8/f/a^3*d^5/(d^2)^(
1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)+13/8/f/a^3*d^5/(d*tan(f*x+e)+d)^2*(d*tan(f*x+
e))^(3/2)+11/8/f/a^3*d^6/(d*tan(f*x+e)+d)^2*(d*tan(f*x+e))^(1/2)-31/8*d^(9/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1
/2))/a^3/f

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.95026, size = 1235, normalized size = 6.53 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.39279, size = 466, normalized size = 2.47 \begin{align*} \frac{1}{16} \, d^{4}{\left (\frac{2 \, \sqrt{2}{\left (d \sqrt{{\left | d \right |}} -{\left | d \right |}^{\frac{3}{2}}\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | d \right |}} + 2 \, \sqrt{d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{a^{3} d f} + \frac{2 \, \sqrt{2}{\left (d \sqrt{{\left | d \right |}} -{\left | d \right |}^{\frac{3}{2}}\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | d \right |}} - 2 \, \sqrt{d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{a^{3} d f} - \frac{62 \, \sqrt{d} \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right )}{a^{3} f} + \frac{\sqrt{2}{\left (d \sqrt{{\left | d \right |}} +{\left | d \right |}^{\frac{3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{a^{3} d f} - \frac{\sqrt{2}{\left (d \sqrt{{\left | d \right |}} +{\left | d \right |}^{\frac{3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{a^{3} d f} + \frac{32 \, \sqrt{d \tan \left (f x + e\right )}}{a^{3} f} + \frac{2 \,{\left (13 \, \sqrt{d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) + 11 \, \sqrt{d \tan \left (f x + e\right )} d^{2}\right )}}{{\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*d^4*(2*sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(f*
x + e)))/sqrt(abs(d)))/(a^3*d*f) + 2*sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt
(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a^3*d*f) - 62*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(
a^3*f) + sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(abs(d)
) + abs(d))/(a^3*d*f) - sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x +
e))*sqrt(abs(d)) + abs(d))/(a^3*d*f) + 32*sqrt(d*tan(f*x + e))/(a^3*f) + 2*(13*sqrt(d*tan(f*x + e))*d^2*tan(f*
x + e) + 11*sqrt(d*tan(f*x + e))*d^2)/((d*tan(f*x + e) + d)^2*a^3*f))