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]

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

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Rubi [A]  time = 0.11, antiderivative size = 79, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.11, size = 79, normalized size = 1.00 \[ -\frac {\tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt {a-b+c}} \]

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]

-1/2*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])]/(Sqrt[a - b + c]*e)

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fricas [A]  time = 0.82, size = 299, normalized size = 3.78 \[ \left [\frac {\log \left (\frac {{\left (b^{2} + 4 \, {\left (a - 2 \, b\right )} c + 8 \, c^{2}\right )} \tan \left (e x + d\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2} - 4 \, {\left (a - b\right )} c\right )} \tan \left (e x + d\right )^{2} - 4 \, \sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} {\left ({\left (b - 2 \, c\right )} \tan \left (e x + d\right )^{2} + 2 \, a - b\right )} \sqrt {a - b + c} + 8 \, a^{2} - 8 \, a b + b^{2} + 4 \, a c}{\tan \left (e x + d\right )^{4} + 2 \, \tan \left (e x + d\right )^{2} + 1}\right )}{4 \, \sqrt {a - b + c} e}, -\frac {\sqrt {-a + b - c} \arctan \left (-\frac {\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} {\left ({\left (b - 2 \, c\right )} \tan \left (e x + d\right )^{2} + 2 \, a - b\right )} \sqrt {-a + b - c}}{2 \, {\left ({\left ({\left (a - b\right )} c + c^{2}\right )} \tan \left (e x + d\right )^{4} + {\left (a b - b^{2} + b c\right )} \tan \left (e x + d\right )^{2} + a^{2} - a b + a c\right )}}\right )}{2 \, {\left (a - b + c\right )} e}\right ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (e x + d\right )}{\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \]

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)

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maple [A]  time = 0.39, size = 102, normalized size = 1.29 \[ -\frac {\ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+2 \sqrt {a -b +c}\, \sqrt {\left (1+\tan ^{2}\left (e x +d \right )\right )^{2} c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+a -b +c}}{1+\tan ^{2}\left (e x +d \right )}\right )}{2 e \sqrt {a -b +c}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (e x + d\right )}{\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {tan}\left (d+e\,x\right )}{\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan {\left (d + e x \right )}}{\sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}}}\, dx \]

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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