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

Optimal. Leaf size=294 \[ \frac {\sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+a-c} \tanh ^{-1}\left (\frac {-\sqrt {a^2-2 a c+b^2+c^2}+a+b \tan (d+e x)-c}{\sqrt {2} \sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+a-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt {a^2-2 a c+b^2+c^2}}-\frac {\sqrt {\sqrt {a^2-2 a c+b^2+c^2}+a-c} \tanh ^{-1}\left (\frac {\sqrt {a^2-2 a c+b^2+c^2}+a+b \tan (d+e x)-c}{\sqrt {2} \sqrt {\sqrt {a^2-2 a c+b^2+c^2}+a-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt {a^2-2 a c+b^2+c^2}} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3700, 1036, 1030, 208} \[ \frac {\sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+a-c} \tanh ^{-1}\left (\frac {-\sqrt {a^2-2 a c+b^2+c^2}+a+b \tan (d+e x)-c}{\sqrt {2} \sqrt {-\sqrt {a^2-2 a c+b^2+c^2}+a-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt {a^2-2 a c+b^2+c^2}}-\frac {\sqrt {\sqrt {a^2-2 a c+b^2+c^2}+a-c} \tanh ^{-1}\left (\frac {\sqrt {a^2-2 a c+b^2+c^2}+a+b \tan (d+e x)-c}{\sqrt {2} \sqrt {\sqrt {a^2-2 a c+b^2+c^2}+a-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt {a^2-2 a c+b^2+c^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1030

Int[((g_) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*
a*g*h, Subst[Int[1/Simp[2*a^2*g*h*c + a*e*x^2, x], x], x, Simp[a*h - g*c*x, x]/Sqrt[d + e*x + f*x^2]], x] /; F
reeQ[{a, c, d, e, f, g, h}, x] && EqQ[a*h^2*e + 2*g*h*(c*d - a*f) - g^2*c*e, 0]

Rule 1036

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[(c*d - a*f)^2 + a*c*e^2, 2]}, Dist[1/(2*q), Int[Simp[-(a*h*e) - g*(c*d - a*f - q) + (h*(c*d - a*f + q) -
g*c*e)*x, x]/((a + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] - Dist[1/(2*q), Int[Simp[-(a*h*e) - g*(c*d - a*f + q
) + (h*(c*d - a*f - q) - g*c*e)*x, x]/((a + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g,
h}, x] && NeQ[e^2 - 4*d*f, 0] && NegQ[-(a*c)]

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

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Mathematica [C]  time = 0.11, size = 173, normalized size = 0.59 \[ \frac {-\frac {\tanh ^{-1}\left (\frac {2 a+(b-2 i c) \tan (d+e x)-i b}{2 \sqrt {a-i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 \sqrt {a-i b-c}}-\frac {\tanh ^{-1}\left (\frac {2 a+(b+2 i c) \tan (d+e x)+i b}{2 \sqrt {a+i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 \sqrt {a+i b-c}}}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1/2*ArcTanh[(2*a - I*b + (b - (2*I)*c)*Tan[d + e*x])/(2*Sqrt[a - I*b - c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d
+ e*x]^2])]/Sqrt[a - I*b - c] - ArcTanh[(2*a + I*b + (b + (2*I)*c)*Tan[d + e*x])/(2*Sqrt[a + I*b - c]*Sqrt[a +
b*Tan[d + e*x] + c*Tan[d + e*x]^2])]/(2*Sqrt[a + I*b - c]))/e

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^2-1)]Discontinuities at zeroes of t_nostep^2-1 were not checkedWarning, integration of abs or sign
assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep^2-1)]Evaluation time:
2.95Not invertible Error: Bad Argument Value

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maple [B]  time = 0.83, size = 9338543, normalized size = 31763.75 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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