3.38 \(\int \frac{\tan (d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\)

Optimal. Leaf size=79 \[ -\frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt{a-b+c}} \]

[Out]

-ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]
)]/(2*Sqrt[a - b + c]*e)

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Rubi [A]  time = 0.113837, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {3700, 1247, 724, 206} \[ -\frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt{a-b+c}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[d + e*x]/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]

[Out]

-ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]
)]/(2*Sqrt[a - b + c]*e)

Rule 3700

Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Dist[f/e, Subst[Int[((x/f)^m*(a + b*x^n + c*x^(2*n))^p)/(f^2 + x^2
), x], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.111007, size = 79, normalized size = 1. \[ -\frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt{a-b+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[d + e*x]/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]

[Out]

-ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]
)]/(2*Sqrt[a - b + c]*e)

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Maple [A]  time = 0.19, size = 102, normalized size = 1.3 \begin{align*} -{\frac{1}{2\,e}\ln \left ({\frac{1}{1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2}} \left ( 2\,a-2\,b+2\,c+ \left ( b-2\,c \right ) \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ) +2\,\sqrt{a-b+c}\sqrt{ \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ) ^{2}c+ \left ( b-2\,c \right ) \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ) +a-b+c} \right ) } \right ){\frac{1}{\sqrt{a-b+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x)

[Out]

-1/2/e/(a-b+c)^(1/2)*ln((2*a-2*b+2*c+(b-2*c)*(1+tan(e*x+d)^2)+2*(a-b+c)^(1/2)*((1+tan(e*x+d)^2)^2*c+(b-2*c)*(1
+tan(e*x+d)^2)+a-b+c)^(1/2))/(1+tan(e*x+d)^2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (e x + d\right )}{\sqrt{c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(tan(e*x + d)/sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a), x)

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Fricas [A]  time = 3.86014, size = 733, normalized size = 9.28 \begin{align*} \left [\frac{\log \left (\frac{{\left (b^{2} + 4 \,{\left (a - 2 \, b\right )} c + 8 \, c^{2}\right )} \tan \left (e x + d\right )^{4} + 2 \,{\left (4 \, a b - 3 \, b^{2} - 4 \,{\left (a - b\right )} c\right )} \tan \left (e x + d\right )^{2} - 4 \, \sqrt{c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}{\left ({\left (b - 2 \, c\right )} \tan \left (e x + d\right )^{2} + 2 \, a - b\right )} \sqrt{a - b + c} + 8 \, a^{2} - 8 \, a b + b^{2} + 4 \, a c}{\tan \left (e x + d\right )^{4} + 2 \, \tan \left (e x + d\right )^{2} + 1}\right )}{4 \, \sqrt{a - b + c} e}, -\frac{\sqrt{-a + b - c} \arctan \left (-\frac{\sqrt{c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}{\left ({\left (b - 2 \, c\right )} \tan \left (e x + d\right )^{2} + 2 \, a - b\right )} \sqrt{-a + b - c}}{2 \,{\left ({\left ({\left (a - b\right )} c + c^{2}\right )} \tan \left (e x + d\right )^{4} +{\left (a b - b^{2} + b c\right )} \tan \left (e x + d\right )^{2} + a^{2} - a b + a c\right )}}\right )}{2 \,{\left (a - b + c\right )} e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="fricas")

[Out]

[1/4*log(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + d)^4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*tan(e*x + d)^2 - 4*sq
rt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(a - b + c) + 8*a^2 - 8*a
*b + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1))/(sqrt(a - b + c)*e), -1/2*sqrt(-a + b - c)*arctan(-
1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(-a + b - c)/(((a -
b)*c + c^2)*tan(e*x + d)^4 + (a*b - b^2 + b*c)*tan(e*x + d)^2 + a^2 - a*b + a*c))/((a - b + c)*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (d + e x \right )}}{\sqrt{a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(e*x+d)/(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(1/2),x)

[Out]

Integral(tan(d + e*x)/sqrt(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (e x + d\right )}{\sqrt{c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(tan(e*x + d)/sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a), x)