3.374 \(\int \frac{\sqrt{d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=161 \[ \frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}-\frac{3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3 \tan (e+f x)+a^3\right )}+\frac{\sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 f}-\frac{\sqrt{d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2} \]

[Out]

(Sqrt[d]*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*f) + (Sqrt[d]*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(
Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(2*Sqrt[2]*a^3*f) - Sqrt[d*Tan[e + f*x]]/(4*a*f*(a + a*Tan[e + f*x])^2) - (3*S
qrt[d*Tan[e + f*x]])/(8*f*(a^3 + a^3*Tan[e + f*x]))

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Rubi [A]  time = 0.537122, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3568, 3649, 3654, 3532, 208, 3634, 63, 205} \[ \frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}-\frac{3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3 \tan (e+f x)+a^3\right )}+\frac{\sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 f}-\frac{\sqrt{d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d]*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*f) + (Sqrt[d]*ArcTanh[(Sqrt[d] + Sqrt[d]*Tan[e + f*x])/(
Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(2*Sqrt[2]*a^3*f) - Sqrt[d*Tan[e + f*x]]/(4*a*f*(a + a*Tan[e + f*x])^2) - (3*S
qrt[d*Tan[e + f*x]])/(8*f*(a^3 + a^3*Tan[e + f*x]))

Rule 3568

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n)/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(a^2
+ b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*(m + 1) - b*d*n - (b*c - a*d)*
(m + 1)*Tan[e + f*x] - b*d*(m + 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] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[2*m]

Rule 3649

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*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[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{\sqrt{d \tan (e+f x)}}{(a+a \tan (e+f x))^3} \, dx &=-\frac{\sqrt{d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac{\int \frac{-\frac{a d}{2}-2 a d \tan (e+f x)+\frac{3}{2} a d \tan ^2(e+f x)}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))^2} \, dx}{4 a^2}\\ &=-\frac{\sqrt{d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac{3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}-\frac{\int \frac{-\frac{5}{2} a^3 d^2+\frac{3}{2} a^3 d^2 \tan ^2(e+f x)}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{8 a^5 d}\\ &=-\frac{\sqrt{d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac{3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}-\frac{\int \frac{-4 a^4 d^2+4 a^4 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{16 a^7 d}+\frac{d \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{\sqrt{d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac{3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}+\frac{d \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 (2 a d^3\right ) \operatorname{Subst}\left (\int \frac{1}{-32 a^8 d^4+d x^2} \, dx,x,\frac{-4 a^4 d^2-4 a^4 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}}\right )}{f}\\ &=\frac{\sqrt{d} \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{\sqrt{d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac{3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}+\frac{\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{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}+\frac{\sqrt{d} \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{\sqrt{d \tan (e+f x)}}{4 a f (a+a \tan (e+f x))^2}-\frac{3 \sqrt{d \tan (e+f x)}}{8 f \left (a^3+a^3 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.797928, size = 253, normalized size = 1.57 \[ -\frac{\sqrt{d \tan (e+f x)} \left (5 \sqrt{\tan (e+f x)}+2 \sqrt{2} \log \left (-\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}-1\right )-2 \sqrt{2} \log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )-2 (\sin (2 (e+f x))+1) \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right )+5 \cos (2 (e+f x)) \sqrt{\tan (e+f x)}+\sin (2 (e+f x)) \left (3 \sqrt{\tan (e+f x)}+2 \sqrt{2} \left (\log \left (-\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}-1\right )-\log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )\right )\right )\right )}{16 a^3 f \sqrt{\tan (e+f x)} (\sin (e+f x)+\cos (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*Sqrt[2]*Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[e + f*x]] - 2*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]
] + Tan[e + f*x]] - 2*ArcTan[Sqrt[Tan[e + f*x]]]*(1 + Sin[2*(e + f*x)]) + Sin[2*(e + f*x)]*(2*Sqrt[2]*(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]]) + 3*Sqrt[Ta
n[e + f*x]]) + 5*Sqrt[Tan[e + f*x]] + 5*Cos[2*(e + f*x)]*Sqrt[Tan[e + f*x]])*Sqrt[d*Tan[e + f*x]])/(16*a^3*f*(
Cos[e + f*x] + Sin[e + f*x])^2*Sqrt[Tan[e + f*x]])

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Maple [B]  time = 0.042, size = 423, normalized size = 2.6 \begin{align*}{\frac{\sqrt{2}}{16\,f{a}^{3}}\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{\sqrt{2}}{8\,f{a}^{3}}\sqrt [4]{{d}^{2}}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ) }-{\frac{\sqrt{2}}{8\,f{a}^{3}}\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\sqrt{2}}{16\,f{a}^{3}}\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\sqrt{2}}{8\,f{a}^{3}}\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\sqrt{2}}{8\,f{a}^{3}}\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{3\,d}{8\,f{a}^{3} \left ( d\tan \left ( fx+e \right ) +d \right ) ^{2}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{d}^{2}}{8\,f{a}^{3} \left ( d\tan \left ( fx+e \right ) +d \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{1}{8\,f{a}^{3}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{d}}}} \right ) \sqrt{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e))^3,x)

[Out]

1/16/f/a^3*(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^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^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)
+1)-1/16/f/a^3*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)))-1/8/f/a^3*d/(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/(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x
+e))^(1/2)+1)-3/8/f/a^3*d/(d*tan(f*x+e)+d)^2*(d*tan(f*x+e))^(3/2)-5/8/f/a^3*d^2/(d*tan(f*x+e)+d)^2*(d*tan(f*x+
e))^(1/2)+1/8*arctan((d*tan(f*x+e))^(1/2)/d^(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))^(1/2)/(a+a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88879, size = 1075, normalized size = 6.68 \begin{align*} \left [-\frac{4 \,{\left (\sqrt{2} \tan \left (f x + e\right )^{2} + 2 \, \sqrt{2} \tan \left (f x + e\right ) + \sqrt{2}\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 ) -{\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\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 \, \sqrt{d \tan \left (f x + e\right )}{\left (3 \, \tan \left (f x + e\right ) + 5\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{{\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt{d} \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right ) +{\left (\sqrt{2} \tan \left (f x + e\right )^{2} + 2 \, \sqrt{2} \tan \left (f x + e\right ) + \sqrt{2}\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 ) - \sqrt{d \tan \left (f x + e\right )}{\left (3 \, \tan \left (f x + e\right ) + 5\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))^(1/2)/(a+a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/16*(4*(sqrt(2)*tan(f*x + e)^2 + 2*sqrt(2)*tan(f*x + e) + sqrt(2))*sqrt(-d)*arctan(1/2*sqrt(d*tan(f*x + e))
*(sqrt(2)*tan(f*x + e) + sqrt(2))*sqrt(-d)/(d*tan(f*x + e))) - (tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(-d)*
log((d*tan(f*x + e) + 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)) + 2*sqrt(d*tan(f*x + e))*(3*tan
(f*x + e) + 5))/(a^3*f*tan(f*x + e)^2 + 2*a^3*f*tan(f*x + e) + a^3*f), 1/8*((tan(f*x + e)^2 + 2*tan(f*x + e) +
1)*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d)) + (sqrt(2)*tan(f*x + e)^2 + 2*sqrt(2)*tan(f*x + e) + sqrt(2))
*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)) - sqrt(d*tan(f*x + e))*(3*tan(f*x + e) + 5))/(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sqrt{d \tan{\left (e + f x \right )}}}{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.30432, size = 439, normalized size = 2.73 \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^{5} 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^{5} f} + \frac{2 \, \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right )}{a^{3} d^{\frac{7}{2}} 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^{5} 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^{5} f} - \frac{2 \,{\left (3 \, \sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + 5 \, \sqrt{d \tan \left (f x + e\right )} d\right )}}{{\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} d^{3} f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(1/2)/(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^5*f) + 2*sqrt(2)*(d*sqrt(abs(d)) - abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*sq
rt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a^3*d^5*f) + 2*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3*d
^(7/2)*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^5*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^5*f) - 2*(3*sqrt(d*tan(f*x + e))*d*tan(f*x + e) + 5*sqrt(d*tan(f*x + e)
)*d)/((d*tan(f*x + e) + d)^2*a^3*d^3*f))