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

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

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Rubi [A]  time = 0.18, antiderivative size = 175, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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 )}} \end {gather*}

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 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 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]

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*}

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Mathematica [A]  time = 0.63, size = 165, normalized size = 0.94 \begin {gather*} 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 ) \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.64, size = 173, normalized size = 0.99 \begin {gather*} \frac {2 \sqrt {2} \sqrt {c} \tan ^{-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 \sqrt {b^2-4 a c}+b e-2 c d}}+\frac {2 \sqrt {2} \sqrt {c} \tan ^{-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 \sqrt {b^2-4 a c}+b e-2 c d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*Sqrt[c]*ArcTan[(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*e - Sqrt[b^2 - 4*a*c]*e] + (2*Sqrt[2]*Sqrt[c]*ArcTan[(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*e + Sqrt[b^2 - 4*a*c]*e]

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fricas [B]  time = 0.44, size = 1317, normalized size = 7.53

result too large to display

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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giac [A]  time = 0.30, size = 254, normalized size = 1.45 \begin {gather*} -\frac {2 \, \sqrt {-4 \, c^{2} d + 2 \, {\left (b c - \sqrt {b^{2} - 4 \, a c} c\right )} e} c \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {x e + d}}{\sqrt {-\frac {2 \, c d - b e + \sqrt {{\left (2 \, c d - b e\right )}^{2} - 4 \, {\left (c d^{2} - b d e + a e^{2}\right )} c}}{c}}}\right )}{{\left (2 \, c d - {\left (b - \sqrt {b^{2} - 4 \, a c}\right )} e\right )} {\left | c \right |}} - \frac {2 \, \sqrt {-4 \, c^{2} d + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} e} c \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {x e + d}}{\sqrt {-\frac {2 \, c d - b e - \sqrt {{\left (2 \, c d - b e\right )}^{2} - 4 \, {\left (c d^{2} - b d e + a e^{2}\right )} c}}{c}}}\right )}{{\left (2 \, c d - {\left (b + \sqrt {b^{2} - 4 \, a c}\right )} e\right )} {\left | c \right |}} \end {gather*}

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]

-2*sqrt(-4*c^2*d + 2*(b*c - sqrt(b^2 - 4*a*c)*c)*e)*c*arctan(2*sqrt(1/2)*sqrt(x*e + d)/sqrt(-(2*c*d - b*e + sq
rt((2*c*d - b*e)^2 - 4*(c*d^2 - b*d*e + a*e^2)*c))/c))/((2*c*d - (b - sqrt(b^2 - 4*a*c))*e)*abs(c)) - 2*sqrt(-
4*c^2*d + 2*(b*c + sqrt(b^2 - 4*a*c)*c)*e)*c*arctan(2*sqrt(1/2)*sqrt(x*e + d)/sqrt(-(2*c*d - b*e - sqrt((2*c*d
- b*e)^2 - 4*(c*d^2 - b*d*e + a*e^2)*c))/c))/((2*c*d - (b + sqrt(b^2 - 4*a*c))*e)*abs(c))

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maple [A]  time = 0.07, size = 158, normalized size = 0.90 \begin {gather*} -\frac {2 \sqrt {2}\, c \arctanh \left (\frac {\sqrt {e x +d}\, \sqrt {2}\, c}{\sqrt {\left (-b e +2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}\right )}{\sqrt {\left (-b e +2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}+\frac {2 \sqrt {2}\, c \arctan \left (\frac {\sqrt {e x +d}\, \sqrt {2}\, c}{\sqrt {\left (b e -2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}}\right )}{\sqrt {\left (b e -2 c d +\sqrt {-\left (4 a c -b^{2}\right ) e^{2}}\right ) c}} \end {gather*}

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, c x + b}{{\left (c x^{2} + b x + a\right )} \sqrt {e x + d}}\,{d x} \end {gather*}

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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mupad [B]  time = 2.37, size = 205, normalized size = 1.17 \begin {gather*} \mathrm {atan}\left (\sqrt {\frac {2\,c\,d-b\,e+e\,\sqrt {b^2-4\,a\,c}}{2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2}}\,\sqrt {d+e\,x}\,1{}\mathrm {i}\right )\,\sqrt {\frac {2\,c\,d-b\,e+e\,\sqrt {b^2-4\,a\,c}}{2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\sqrt {-\frac {b\,e-2\,c\,d+e\,\sqrt {b^2-4\,a\,c}}{2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2}}\,\sqrt {d+e\,x}\,1{}\mathrm {i}\right )\,\sqrt {-\frac {b\,e-2\,c\,d+e\,\sqrt {b^2-4\,a\,c}}{2\,c\,d^2-2\,b\,d\,e+2\,a\,e^2}}\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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