3.2467 \(\int \frac{(A+B x) (d+e x)}{\sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=116 \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (4 c (2 A c d-a B e)-4 b c (A e+B d)+3 b^2 B e\right )}{8 c^{5/2}}-\frac{\sqrt{a+b x+c x^2} (-4 c (A e+B d)+3 b B e-2 B c e x)}{4 c^2} \]

[Out]

-((3*b*B*e - 4*c*(B*d + A*e) - 2*B*c*e*x)*Sqrt[a + b*x + c*x^2])/(4*c^2) + ((3*b^2*B*e - 4*b*c*(B*d + A*e) + 4
*c*(2*A*c*d - a*B*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(5/2))

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Rubi [A]  time = 0.0876261, antiderivative size = 115, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {779, 621, 206} \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (-4 a B c e-4 b c (A e+B d)+8 A c^2 d+3 b^2 B e\right )}{8 c^{5/2}}-\frac{\sqrt{a+b x+c x^2} (-4 c (A e+B d)+3 b B e-2 B c e x)}{4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*b*B*e - 4*c*(B*d + A*e) - 2*B*c*e*x)*Sqrt[a + b*x + c*x^2])/(4*c^2) + ((8*A*c^2*d + 3*b^2*B*e - 4*a*B*c*e
- 4*b*c*(B*d + A*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(5/2))

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

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 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])

Rubi steps

\begin{align*} \int \frac{(A+B x) (d+e x)}{\sqrt{a+b x+c x^2}} \, dx &=-\frac{(3 b B e-4 c (B d+A e)-2 B c e x) \sqrt{a+b x+c x^2}}{4 c^2}+\frac{\left (8 A c^2 d+3 b^2 B e-4 a B c e-4 b c (B d+A e)\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{8 c^2}\\ &=-\frac{(3 b B e-4 c (B d+A e)-2 B c e x) \sqrt{a+b x+c x^2}}{4 c^2}+\frac{\left (8 A c^2 d+3 b^2 B e-4 a B c e-4 b c (B d+A e)\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{4 c^2}\\ &=-\frac{(3 b B e-4 c (B d+A e)-2 B c e x) \sqrt{a+b x+c x^2}}{4 c^2}+\frac{\left (8 A c^2 d+3 b^2 B e-4 a B c e-4 b c (B d+A e)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0827641, size = 115, normalized size = 0.99 \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) \left (-2 a B c e-2 b c (A e+B d)+4 A c^2 d+\frac{3}{2} b^2 B e\right )+\sqrt{c} \sqrt{a+x (b+c x)} (4 A c e+B (-3 b e+4 c d+2 c e x))}{4 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*Sqrt[a + x*(b + c*x)]*(4*A*c*e + B*(4*c*d - 3*b*e + 2*c*e*x)) + (4*A*c^2*d + (3*b^2*B*e)/2 - 2*a*B*c*
e - 2*b*c*(B*d + A*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(4*c^(5/2))

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Maple [B]  time = 0.006, size = 243, normalized size = 2.1 \begin{align*}{\frac{Bex}{2\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,bBe}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{b}^{2}Be}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{aBe}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{Ae}{c}\sqrt{c{x}^{2}+bx+a}}+{\frac{Bd}{c}\sqrt{c{x}^{2}+bx+a}}-{\frac{Abe}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}-{\frac{Bbd}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{Ad\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*e*x/c*(c*x^2+b*x+a)^(1/2)-3/4*B*e*b/c^2*(c*x^2+b*x+a)^(1/2)+3/8*B*e*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(
c*x^2+b*x+a)^(1/2))-1/2*B*e*a/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/c*(c*x^2+b*x+a)^(1/2)*A*e+
1/c*(c*x^2+b*x+a)^(1/2)*B*d-1/2*b/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*A*e-1/2*b/c^(3/2)*ln((1/
2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*B*d+A*d*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60711, size = 639, normalized size = 5.51 \begin{align*} \left [\frac{{\left (4 \,{\left (B b c - 2 \, A c^{2}\right )} d -{\left (3 \, B b^{2} - 4 \,{\left (B a + A b\right )} c\right )} e\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right ) + 4 \,{\left (2 \, B c^{2} e x + 4 \, B c^{2} d -{\left (3 \, B b c - 4 \, A c^{2}\right )} e\right )} \sqrt{c x^{2} + b x + a}}{16 \, c^{3}}, \frac{{\left (4 \,{\left (B b c - 2 \, A c^{2}\right )} d -{\left (3 \, B b^{2} - 4 \,{\left (B a + A b\right )} c\right )} e\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \,{\left (2 \, B c^{2} e x + 4 \, B c^{2} d -{\left (3 \, B b c - 4 \, A c^{2}\right )} e\right )} \sqrt{c x^{2} + b x + a}}{8 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((4*(B*b*c - 2*A*c^2)*d - (3*B*b^2 - 4*(B*a + A*b)*c)*e)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt
(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(2*B*c^2*e*x + 4*B*c^2*d - (3*B*b*c - 4*A*c^2)*e)*sqrt(c*x^
2 + b*x + a))/c^3, 1/8*((4*(B*b*c - 2*A*c^2)*d - (3*B*b^2 - 4*(B*a + A*b)*c)*e)*sqrt(-c)*arctan(1/2*sqrt(c*x^2
+ b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(2*B*c^2*e*x + 4*B*c^2*d - (3*B*b*c - 4*A*c^2)*e
)*sqrt(c*x^2 + b*x + a))/c^3]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19437, size = 161, normalized size = 1.39 \begin{align*} \frac{1}{4} \, \sqrt{c x^{2} + b x + a}{\left (\frac{2 \, B x e}{c} + \frac{4 \, B c d - 3 \, B b e + 4 \, A c e}{c^{2}}\right )} + \frac{{\left (4 \, B b c d - 8 \, A c^{2} d - 3 \, B b^{2} e + 4 \, B a c e + 4 \, A b c e\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{8 \, c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(c*x^2 + b*x + a)*(2*B*x*e/c + (4*B*c*d - 3*B*b*e + 4*A*c*e)/c^2) + 1/8*(4*B*b*c*d - 8*A*c^2*d - 3*B*b
^2*e + 4*B*a*c*e + 4*A*b*c*e)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)