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

Optimal. Leaf size=87 \[ \frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{a f}-\frac{d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{\sqrt{2} a f} \]

[Out]

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

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Rubi [A]  time = 0.223015, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3573, 3532, 208, 3634, 63, 205} \[ \frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{a f}-\frac{d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \tan (e+f x)+\sqrt{d}}{\sqrt{2} \sqrt{d \tan (e+f x)}}\right )}{\sqrt{2} a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3573

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1
/(c^2 + d^2), Int[Simp[a^2*c - b^2*c + 2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e +
f*x]], x], x] + Dist[(b*c - a*d)^2/(c^2 + d^2), Int[(1 + Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan
[e + f*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]

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

Mathematica [A]  time = 0.330334, size = 107, normalized size = 1.23 \[ \frac{(d \tan (e+f x))^{3/2} \left (4 \tan ^{-1}\left (\sqrt{\tan (e+f x)}\right )+\sqrt{2} \left (\log \left (-\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}-1\right )-\log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )\right )\right )}{4 a f \tan ^{\frac{3}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.028, size = 367, normalized size = 4.2 \begin{align*} -{\frac{d\sqrt{2}}{8\,fa}\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}}{4\,fa}\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}}{4\,fa}\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}}{8\,fa}\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}}{4\,fa}\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}}{4\,fa}\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{1}{fa}{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)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7204, size = 583, normalized size = 6.7 \begin{align*} \left [\frac{\sqrt{2} \sqrt{-d} 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 ) + \sqrt{-d} 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 \, a f}, \frac{\sqrt{2} d^{\frac{3}{2}} \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 ) + 4 \, d^{\frac{3}{2}} \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right )}{4 \, a 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)),x, algorithm="fricas")

[Out]

[1/2*(sqrt(2)*sqrt(-d)*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))) + sqrt(-d)*d*log((d*tan(f*x + e) + 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1)))/(a*f), 1/4
*(sqrt(2)*d^(3/2)*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)) + 4*d^(3/2)*arctan(sqrt(d*tan(f*x + e))/sqrt(d)))/(a*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{\left (e + f x \right )} + 1}\, dx}{a} \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)),x)

[Out]

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

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Giac [B]  time = 1.36942, size = 358, normalized size = 4.11 \begin{align*} -\frac{1}{8} \, d^{2}{\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 d^{2} 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 d^{2} f} - \frac{8 \, \arctan \left (\frac{\sqrt{d \tan \left (f x + e\right )}}{\sqrt{d}}\right )}{a \sqrt{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 d^{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 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)),x, algorithm="giac")

[Out]

-1/8*d^2*(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*d^2*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*d^2*f) - 8*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a*sqrt(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*d^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*d^2*f))