3.375 \(\int \frac{1}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=165 \[ \frac{11 \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 \sqrt{d} f}+\frac{\tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 \sqrt{d} f}+\frac{7 \sqrt{d \tan (e+f x)}}{8 a^3 d f (\tan (e+f x)+1)}+\frac{\sqrt{d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2} \]

[Out]

(11*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*Sqrt[d]*f) + ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sqrt[2]
*Sqrt[d*Tan[e + f*x]])]/(2*Sqrt[2]*a^3*Sqrt[d]*f) + (7*Sqrt[d*Tan[e + f*x]])/(8*a^3*d*f*(1 + Tan[e + f*x])) +
Sqrt[d*Tan[e + f*x]]/(4*a*d*f*(a + a*Tan[e + f*x])^2)

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Rubi [A]  time = 0.619819, antiderivative size = 165, 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 = {3569, 3649, 3653, 3532, 205, 3634, 63} \[ \frac{11 \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 \sqrt{d} f}+\frac{\tan ^{-1}\left (\frac{\sqrt{d}-\sqrt{d} \tan (e+f x)}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{2 \sqrt{2} a^3 \sqrt{d} f}+\frac{7 \sqrt{d \tan (e+f x)}}{8 a^3 d f (\tan (e+f x)+1)}+\frac{\sqrt{d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])^3),x]

[Out]

(11*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(8*a^3*Sqrt[d]*f) + ArcTan[(Sqrt[d] - Sqrt[d]*Tan[e + f*x])/(Sqrt[2]
*Sqrt[d*Tan[e + f*x]])]/(2*Sqrt[2]*a^3*Sqrt[d]*f) + (7*Sqrt[d*Tan[e + f*x]])/(8*a^3*d*f*(1 + Tan[e + f*x])) +
Sqrt[d*Tan[e + f*x]]/(4*a*d*f*(a + a*Tan[e + f*x])^2)

Rule 3569

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

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{1}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))^3} \, dx &=\frac{\sqrt{d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}+\frac{\int \frac{\frac{7 a^2 d}{2}-2 a^2 d \tan (e+f x)+\frac{3}{2} a^2 d \tan ^2(e+f x)}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))^2} \, dx}{4 a^3 d}\\ &=\frac{7 \sqrt{d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac{\sqrt{d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}+\frac{\int \frac{\frac{7 a^4 d^2}{2}-4 a^4 d^2 \tan (e+f x)+\frac{7}{2} a^4 d^2 \tan ^2(e+f x)}{\sqrt{d \tan (e+f x)} (a+a \tan (e+f x))} \, dx}{8 a^6 d^2}\\ &=\frac{7 \sqrt{d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac{\sqrt{d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}+\frac{11 \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{\int \frac{-4 a^5 d^2-4 a^5 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{16 a^8 d^2}\\ &=\frac{7 \sqrt{d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac{\sqrt{d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}+\frac{11 \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^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{32 a^{10} d^4+d x^2} \, dx,x,\frac{-4 a^5 d^2+4 a^5 d^2 \tan (e+f x)}{\sqrt{d \tan (e+f x)}}\right )}{f}\\ &=\frac{\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 \sqrt{d} f}+\frac{7 \sqrt{d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac{\sqrt{d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}+\frac{11 \operatorname{Subst}\left (\int \frac{1}{a+\frac{a x^2}{d}} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{8 a^2 d f}\\ &=\frac{11 \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{8 a^3 \sqrt{d} f}+\frac{\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 \sqrt{d} f}+\frac{7 \sqrt{d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac{\sqrt{d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 1.00607, size = 217, normalized size = 1.32 \[ \frac{\sqrt{\tan (e+f x)} \left (22 \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right )+9 \sqrt{\tan (e+f x)}+22 \sin (2 (e+f x)) \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right )+7 \sin (2 (e+f x)) \sqrt{\tan (e+f x)}+9 \cos (2 (e+f x)) \sqrt{\tan (e+f x)}+4 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right ) (\sin (e+f x)+\cos (e+f x))^2-4 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right ) (\sin (e+f x)+\cos (e+f x))^2\right )}{16 a^3 f \sqrt{d \tan (e+f x)} (\sin (e+f x)+\cos (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[d*Tan[e + f*x]]*(a + a*Tan[e + f*x])^3),x]

[Out]

((22*ArcTan[Sqrt[Tan[e + f*x]]] + 4*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]*(Cos[e + f*x] + Sin[e + f*x
])^2 - 4*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*(Cos[e + f*x] + Sin[e + f*x])^2 + 22*ArcTan[Sqrt[Tan[e
+ f*x]]]*Sin[2*(e + f*x)] + 9*Sqrt[Tan[e + f*x]] + 9*Cos[2*(e + f*x)]*Sqrt[Tan[e + f*x]] + 7*Sin[2*(e + f*x)]
*Sqrt[Tan[e + f*x]])*Sqrt[Tan[e + f*x]])/(16*a^3*f*(Cos[e + f*x] + Sin[e + f*x])^2*Sqrt[d*Tan[e + f*x]])

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Maple [B]  time = 0.051, size = 426, normalized size = 2.6 \begin{align*} -{\frac{\sqrt{2}}{16\,f{a}^{3}d}\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}d}\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}d}\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}}{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{\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{\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{7}{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{9\,d}{8\,f{a}^{3} \left ( d\tan \left ( fx+e \right ) +d \right ) ^{2}}\sqrt{d\tan \left ( fx+e \right ) }}+{\frac{11}{8\,f{a}^{3}}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{d}}}} \right ){\frac{1}{\sqrt{d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(f*x+e))^(1/2)/(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*ta
n(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)^(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)+7/8/f/a^3/(d*tan(f*x+e)+d)^2*(d*tan(f*x+e))^(3/2)+9/8/f/a^3*d/(d*tan(f*x+e)+d)^2*(d*tan(f*x+e))
^(1/2)+11/8*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f/d^(1/2)

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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(1/(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 = 1.86119, size = 1046, normalized size = 6.34 \begin{align*} \left [-\frac{2 \, \sqrt{2}{\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} + 2 \, \sqrt{2} \sqrt{d \tan \left (f x + e\right )} \sqrt{-d}{\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 11 \,{\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 (7 \, \tan \left (f x + e\right ) + 9\right )}}{16 \,{\left (a^{3} d f \tan \left (f x + e\right )^{2} + 2 \, a^{3} d f \tan \left (f x + e\right ) + a^{3} d f\right )}}, -\frac{2 \, \sqrt{2}{\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt{d} \arctan \left (\frac{\sqrt{2} \sqrt{d \tan \left (f x + e\right )}{\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt{d} \tan \left (f x + e\right )}\right ) - 11 \,{\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 ) - \sqrt{d \tan \left (f x + e\right )}{\left (7 \, \tan \left (f x + e\right ) + 9\right )}}{8 \,{\left (a^{3} d f \tan \left (f x + e\right )^{2} + 2 \, a^{3} d f \tan \left (f x + e\right ) + a^{3} d f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(1/2)/(a+a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/16*(2*sqrt(2)*(tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(-d)*log((d*tan(f*x + e)^2 + 2*sqrt(2)*sqrt(d*tan(
f*x + e))*sqrt(-d)*(tan(f*x + e) - 1) - 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) + 11*(tan(f*x + e)^2 + 2*t
an(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*s
qrt(d*tan(f*x + e))*(7*tan(f*x + e) + 9))/(a^3*d*f*tan(f*x + e)^2 + 2*a^3*d*f*tan(f*x + e) + a^3*d*f), -1/8*(2
*sqrt(2)*(tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) -
1)/(sqrt(d)*tan(f*x + e))) - 11*(tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqr
t(d)) - sqrt(d*tan(f*x + e))*(7*tan(f*x + e) + 9))/(a^3*d*f*tan(f*x + e)^2 + 2*a^3*d*f*tan(f*x + e) + a^3*d*f)
]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\sqrt{d \tan{\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )} + 3 \sqrt{d \tan{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )} + 3 \sqrt{d \tan{\left (e + f x \right )}} \tan{\left (e + f x \right )} + \sqrt{d \tan{\left (e + f x \right )}}}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.34466, size = 439, 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^{6} 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^{6} f} - \frac{22 \, \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right )}{a^{3} d^{\frac{9}{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^{6} 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^{6} f} - \frac{2 \,{\left (7 \, \sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + 9 \, \sqrt{d \tan \left (f x + e\right )} d\right )}}{{\left (d \tan \left (f x + e\right ) + d\right )}^{2} a^{3} d^{4} f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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^6*f) + 2*sqrt(2)*(d*sqrt(abs(d)) + abs(d)^(3/2))*arctan(-1/2*sqrt(2)*(sqrt(2)*s
qrt(abs(d)) - 2*sqrt(d*tan(f*x + e)))/sqrt(abs(d)))/(a^3*d^6*f) - 22*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a^3
*d^(9/2)*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(a
bs(d)) + abs(d))/(a^3*d^6*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^6*f) - 2*(7*sqrt(d*tan(f*x + e))*d*tan(f*x + e) + 9*sqrt(d*tan(f*x +
e))*d)/((d*tan(f*x + e) + d)^2*a^3*d^4*f))