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

Optimal. Leaf size=574 \[ -\frac {\sqrt {-a \left (2 c-\sqrt {a^2-2 a c+b^2+c^2}\right )+c \left (c-\sqrt {a^2-2 a c+b^2+c^2}\right )+a^2+b^2} \tan ^{-1}\left (\frac {b \sqrt {a^2-2 a c+b^2+c^2}-\left ((a-c) \left (\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2\right ) \tan (d+e x)}{\sqrt {2} \sqrt [4]{a^2-2 a c+b^2+c^2} \sqrt {-a \left (2 c-\sqrt {a^2-2 a c+b^2+c^2}\right )+c \left (c-\sqrt {a^2-2 a c+b^2+c^2}\right )+a^2+b^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt [4]{a^2-2 a c+b^2+c^2}}-\frac {\sqrt {-a \left (\sqrt {a^2-2 a c+b^2+c^2}+2 c\right )+c \left (\sqrt {a^2-2 a c+b^2+c^2}+c\right )+a^2+b^2} \tanh ^{-1}\left (\frac {\left ((a-c) \left (-\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2\right ) \tan (d+e x)+b \sqrt {a^2-2 a c+b^2+c^2}}{\sqrt {2} \sqrt [4]{a^2-2 a c+b^2+c^2} \sqrt {-a \left (\sqrt {a^2-2 a c+b^2+c^2}+2 c\right )+c \left (\sqrt {a^2-2 a c+b^2+c^2}+c\right )+a^2+b^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt [4]{a^2-2 a c+b^2+c^2}}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{e} \]

[Out]

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

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Rubi [A]  time = 23.27, antiderivative size = 574, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {990, 621, 206, 1036, 1030, 208, 205} \[ -\frac {\sqrt {-a \left (2 c-\sqrt {a^2-2 a c+b^2+c^2}\right )+c \left (c-\sqrt {a^2-2 a c+b^2+c^2}\right )+a^2+b^2} \tan ^{-1}\left (\frac {b \sqrt {a^2-2 a c+b^2+c^2}-\left ((a-c) \left (\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2\right ) \tan (d+e x)}{\sqrt {2} \sqrt [4]{a^2-2 a c+b^2+c^2} \sqrt {-a \left (2 c-\sqrt {a^2-2 a c+b^2+c^2}\right )+c \left (c-\sqrt {a^2-2 a c+b^2+c^2}\right )+a^2+b^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt [4]{a^2-2 a c+b^2+c^2}}-\frac {\sqrt {-a \left (\sqrt {a^2-2 a c+b^2+c^2}+2 c\right )+c \left (\sqrt {a^2-2 a c+b^2+c^2}+c\right )+a^2+b^2} \tanh ^{-1}\left (\frac {\left ((a-c) \left (-\sqrt {a^2-2 a c+b^2+c^2}+a-c\right )+b^2\right ) \tan (d+e x)+b \sqrt {a^2-2 a c+b^2+c^2}}{\sqrt {2} \sqrt [4]{a^2-2 a c+b^2+c^2} \sqrt {-a \left (\sqrt {a^2-2 a c+b^2+c^2}+2 c\right )+c \left (\sqrt {a^2-2 a c+b^2+c^2}+c\right )+a^2+b^2} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{\sqrt {2} e \sqrt [4]{a^2-2 a c+b^2+c^2}}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a^2 + b^2 + c*(c - Sqrt[a^2 + b^2 - 2*a*c + c^2]) - a*(2*c - Sqrt[a^2 + b^2 - 2*a*c + c^2])]*ArcTan[(b
*Sqrt[a^2 + b^2 - 2*a*c + c^2] - (b^2 + (a - c)*(a - c + Sqrt[a^2 + b^2 - 2*a*c + c^2]))*Tan[d + e*x])/(Sqrt[2
]*(a^2 + b^2 - 2*a*c + c^2)^(1/4)*Sqrt[a^2 + b^2 + c*(c - Sqrt[a^2 + b^2 - 2*a*c + c^2]) - a*(2*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]*(a^2 + b^2 - 2*a*c + c^2)^(1/4)*
e)) + (Sqrt[c]*ArcTanh[(b + 2*c*Tan[d + e*x])/(2*Sqrt[c]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/e - (S
qrt[a^2 + b^2 + c*(c + Sqrt[a^2 + b^2 - 2*a*c + c^2]) - a*(2*c + Sqrt[a^2 + b^2 - 2*a*c + c^2])]*ArcTanh[(b*Sq
rt[a^2 + b^2 - 2*a*c + c^2] + (b^2 + (a - c)*(a - c - Sqrt[a^2 + b^2 - 2*a*c + c^2]))*Tan[d + e*x])/(Sqrt[2]*(
a^2 + b^2 - 2*a*c + c^2)^(1/4)*Sqrt[a^2 + b^2 + c*(c + Sqrt[a^2 + b^2 - 2*a*c + c^2]) - a*(2*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]*(a^2 + b^2 - 2*a*c + c^2)^(1/4)*e)

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 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 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 621

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

Rule 990

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

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)]

Rubi steps

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

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Mathematica [C]  time = 0.20, size = 228, normalized size = 0.40 \[ \frac {-i \sqrt {a-i b-c} \tanh ^{-1}\left (\frac {2 a+(b-2 i c) \tan (d+e x)-i b}{2 \sqrt {a-i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )+i \sqrt {a+i b-c} \tanh ^{-1}\left (\frac {2 a+(b+2 i c) \tan (d+e x)+i b}{2 \sqrt {a+i b-c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )+2 \sqrt {c} \tanh ^{-1}\left (\frac {b+2 c \tan (d+e x)}{2 \sqrt {c} \sqrt {a+b \tan (d+e x)+c \tan ^2(d+e x)}}\right )}{2 e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*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])] + I*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[c]*ArcTanh[(b + 2*c*Tan[d + e*x])/(2*Sqrt[c
]*Sqrt[a + b*Tan[d + e*x] + c*Tan[d + e*x]^2])])/(2*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((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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c \tan \left (e x + d\right )^{2} + b \tan \left (e x + d\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.83, size = 17246812, normalized size = 30046.71 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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((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((c-b-a)*(c+b-a)>0)', see `assu
me?` for more details)Is (c-b-a)*(c+b-a)    positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \[ \int \sqrt {a + b \tan {\left (d + e x \right )} + c \tan ^{2}{\left (d + e x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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