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

Optimal. Leaf size=164 \[ -\frac{5 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}-\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 f}-\frac{d \sqrt{d \tan (e+f x)}}{8 f \left (a^3 \tan (e+f x)+a^3\right )}+\frac{d \sqrt{d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2} \]

[Out]

(-5*d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*f) - (d^(3/2)*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])
/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(2*Sqrt[2]*a^3*f) + (d*Sqrt[d*Tan[e + f*x]])/(4*a*f*(a + a*Tan[e + f*x])^2)
- (d*Sqrt[d*Tan[e + f*x]])/(8*f*(a^3 + a^3*Tan[e + f*x]))

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Rubi [A]  time = 0.594045, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3567, 3649, 3653, 3532, 205, 3634, 63} \[ -\frac{5 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 f}-\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 f}-\frac{d \sqrt{d \tan (e+f x)}}{8 f \left (a^3 \tan (e+f x)+a^3\right )}+\frac{d \sqrt{d \tan (e+f x)}}{4 a f (a \tan (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*f) - (d^(3/2)*ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])
/(Sqrt[2]*Sqrt[d*Tan[e + f*x]])])/(2*Sqrt[2]*a^3*f) + (d*Sqrt[d*Tan[e + f*x]])/(4*a*f*(a + a*Tan[e + f*x])^2)
- (d*Sqrt[d*Tan[e + f*x]])/(8*f*(a^3 + a^3*Tan[e + f*x]))

Rule 3567

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 + 1)*(c + d*Tan[e + f*x])^(n - 1))/(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 - 2)*Simp[a*c^2*(m + 1) + a*
d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*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] && LtQ[1, n, 2] && 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 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + 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, B, 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 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]

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]

Rubi steps

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

Mathematica [A]  time = 2.23369, size = 127, normalized size = 0.77 \[ \frac{d \sqrt{d \tan (e+f x)} \left (\frac{-2 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right )+2 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right )-5 \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right )}{\sqrt{\tan (e+f x)}}+\frac{(\cot (e+f x)-1) \cot (e+f x)}{(\cot (e+f x)+1)^2}\right )}{8 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(((-1 + Cot[e + f*x])*Cot[e + f*x])/(1 + Cot[e + f*x])^2 + (-2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]
]] + 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] - 5*ArcTan[Sqrt[Tan[e + f*x]]])/Sqrt[Tan[e + f*x]])*Sqrt
[d*Tan[e + f*x]])/(8*a^3*f)

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Maple [B]  time = 0.037, size = 434, normalized size = 2.7 \begin{align*}{\frac{d\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{d\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}}{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}^{2}\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}^{2}\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}^{2}\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}^{2}}{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{{d}^{3}}{8\,f{a}^{3} \left ( d\tan \left ( fx+e \right ) +d \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{5}{8\,f{a}^{3}}{d}^{{\frac{3}{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))^(3/2)/(a+a*tan(f*x+e))^3,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82422, size = 1107, normalized size = 6.75 \begin{align*} \left [\frac{2 \,{\left (\sqrt{2} d \tan \left (f x + e\right )^{2} + 2 \, \sqrt{2} d \tan \left (f x + e\right ) + \sqrt{2} d\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 ) + 5 \,{\left (d \tan \left (f x + e\right )^{2} + 2 \, d \tan \left (f x + e\right ) + d\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 (d \tan \left (f x + e\right ) - d\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{5 \,{\left (d \tan \left (f x + e\right )^{2} + 2 \, d \tan \left (f x + e\right ) + d\right )} \sqrt{d} \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right ) - 2 \,{\left (\sqrt{2} d \tan \left (f x + e\right )^{2} + 2 \, \sqrt{2} d \tan \left (f x + e\right ) + \sqrt{2} d\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 )}}{2 \, \sqrt{d} \tan \left (f x + e\right )}\right ) +{\left (d \tan \left (f x + e\right ) - d\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))^(3/2)/(a+a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/16*(2*(sqrt(2)*d*tan(f*x + e)^2 + 2*sqrt(2)*d*tan(f*x + e) + sqrt(2)*d)*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)) +
5*(d*tan(f*x + e)^2 + 2*d*tan(f*x + e) + d)*sqrt(-d)*log((d*tan(f*x + e) - 2*sqrt(d*tan(f*x + e))*sqrt(-d) -
d)/(tan(f*x + e) + 1)) - 2*(d*tan(f*x + e) - d)*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*(5*(d*tan(f*x + e)^2 + 2*d*tan(f*x + e) + d)*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))
- 2*(sqrt(2)*d*tan(f*x + e)^2 + 2*sqrt(2)*d*tan(f*x + e) + sqrt(2)*d)*sqrt(d)*arctan(1/2*sqrt(d*tan(f*x + e))*
(sqrt(2)*tan(f*x + e) - sqrt(2))/(sqrt(d)*tan(f*x + e))) + (d*tan(f*x + e) - d)*sqrt(d*tan(f*x + e)))/(a^3*f*t
an(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{\left (d \tan{\left (e + f x \right )}\right )^{\frac{3}{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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*tan(e + f*x))**(3/2)/(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 = 1.35586, size = 437, normalized size = 2.66 \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^{4} 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^{4} f} - \frac{10 \, \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right )}{a^{3} d^{\frac{5}{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^{4} 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^{4} f} - \frac{2 \,{\left (\sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) - \sqrt{d \tan \left (f x + e\right )} d\right )}}{{\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} d^{2} f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)/(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^4*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^4*f) - 10*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3*
d^(5/2)*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(ab
s(d)) + abs(d))/(a^3*d^4*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^4*f) - 2*(sqrt(d*tan(f*x + e))*d*tan(f*x + e) - sqrt(d*tan(f*x + e))*d
)/((d*tan(f*x + e) + d)^2*a^3*d^2*f))