3.1615 \(\int \frac{b+2 c x}{\sqrt{d+e x} (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=175 \[ -\frac{2 \sqrt{2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{2 \sqrt{2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} \]

[Out]

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

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Rubi [A]  time = 0.176051, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {826, 1166, 208} \[ -\frac{2 \sqrt{2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{2 \sqrt{2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 826

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

Rule 1166

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

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]

Rubi steps

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

Mathematica [A]  time = 0.662598, size = 165, normalized size = 0.94 \[ 2 \sqrt{2} \sqrt{c} \left (-\frac{\tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 158, normalized size = 0.9 \begin{align*} -2\,{\frac{c\sqrt{2}}{\sqrt{ \left ( -be+2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}{\it Artanh} \left ({\frac{\sqrt{ex+d}c\sqrt{2}}{\sqrt{ \left ( -be+2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}} \right ) }+2\,{\frac{c\sqrt{2}}{\sqrt{ \left ( be-2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c\sqrt{2}}{\sqrt{ \left ( be-2\,cd+\sqrt{-{e}^{2} \left ( 4\,ac-{b}^{2} \right ) } \right ) c}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.54873, size = 2784, normalized size = 15.91 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*sqrt((2*c*d - b*e + (c*d^2 - b*d*e + a*e^2)*sqrt((b^2 - 4*a*c)*e^2/(c^2*d^4 - 2*b*c*d^3*e - 2*a*b
*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2)))/(c*d^2 - b*d*e + a*e^2))*log(sqrt(2)*(2*c*d - b*e - (c*d^2 - b*d*e
+ a*e^2)*sqrt((b^2 - 4*a*c)*e^2/(c^2*d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2)))*sqr
t((2*c*d - b*e + (c*d^2 - b*d*e + a*e^2)*sqrt((b^2 - 4*a*c)*e^2/(c^2*d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3 + a^2*e^4
+ (b^2 + 2*a*c)*d^2*e^2)))/(c*d^2 - b*d*e + a*e^2)) + 4*sqrt(e*x + d)*c) + 1/2*sqrt(2)*sqrt((2*c*d - b*e + (c
*d^2 - b*d*e + a*e^2)*sqrt((b^2 - 4*a*c)*e^2/(c^2*d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^
2*e^2)))/(c*d^2 - b*d*e + a*e^2))*log(-sqrt(2)*(2*c*d - b*e - (c*d^2 - b*d*e + a*e^2)*sqrt((b^2 - 4*a*c)*e^2/(
c^2*d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2)))*sqrt((2*c*d - b*e + (c*d^2 - b*d*e +
a*e^2)*sqrt((b^2 - 4*a*c)*e^2/(c^2*d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2)))/(c*d^2
- b*d*e + a*e^2)) + 4*sqrt(e*x + d)*c) - 1/2*sqrt(2)*sqrt((2*c*d - b*e - (c*d^2 - b*d*e + a*e^2)*sqrt((b^2 -
4*a*c)*e^2/(c^2*d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2)))/(c*d^2 - b*d*e + a*e^2))*
log(sqrt(2)*(2*c*d - b*e + (c*d^2 - b*d*e + a*e^2)*sqrt((b^2 - 4*a*c)*e^2/(c^2*d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3
+ a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2)))*sqrt((2*c*d - b*e - (c*d^2 - b*d*e + a*e^2)*sqrt((b^2 - 4*a*c)*e^2/(c^2*
d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2)))/(c*d^2 - b*d*e + a*e^2)) + 4*sqrt(e*x + d
)*c) + 1/2*sqrt(2)*sqrt((2*c*d - b*e - (c*d^2 - b*d*e + a*e^2)*sqrt((b^2 - 4*a*c)*e^2/(c^2*d^4 - 2*b*c*d^3*e -
2*a*b*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2)))/(c*d^2 - b*d*e + a*e^2))*log(-sqrt(2)*(2*c*d - b*e + (c*d^2
- b*d*e + a*e^2)*sqrt((b^2 - 4*a*c)*e^2/(c^2*d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2
)))*sqrt((2*c*d - b*e - (c*d^2 - b*d*e + a*e^2)*sqrt((b^2 - 4*a*c)*e^2/(c^2*d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3 +
a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2)))/(c*d^2 - b*d*e + a*e^2)) + 4*sqrt(e*x + d)*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out