3.39 \(\int \frac{(g+h x) \sqrt{a+b x+c x^2}}{(a d+b d x+c d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=136 \[ \frac{h \sqrt{a+b x+c x^2} \log \left (a+b x+c x^2\right )}{2 c d \sqrt{a d+b d x+c d x^2}}-\frac{\sqrt{a+b x+c x^2} (2 c g-b h) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c d \sqrt{b^2-4 a c} \sqrt{a d+b d x+c d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.11423, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {999, 634, 618, 206, 628} \[ \frac{h \sqrt{a+b x+c x^2} \log \left (a+b x+c x^2\right )}{2 c d \sqrt{a d+b d x+c d x^2}}-\frac{\sqrt{a+b x+c x^2} (2 c g-b h) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c d \sqrt{b^2-4 a c} \sqrt{a d+b d x+c d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((g + h*x)*Sqrt[a + b*x + c*x^2])/(a*d + b*d*x + c*d*x^2)^(3/2),x]

[Out]

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

Rule 999

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_)
, x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x + c*x^2)^FracPart[p])/(d^IntPart[p]*(d + e*x + f*x^2)^FracPart[p]),
Int[(g + h*x)^m*(d + e*x + f*x^2)^(p + q), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p, q}, x] && EqQ[c*d - a*
f, 0] && EqQ[b*d - a*e, 0] &&  !IntegerQ[p] &&  !IntegerQ[q] &&  !GtQ[c/f, 0]

Rule 634

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], 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])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.095123, size = 108, normalized size = 0.79 \[ \frac{(a+x (b+c x))^{3/2} \left ((4 c g-2 b h) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )+h \sqrt{4 a c-b^2} \log (a+x (b+c x))\right )}{2 c \sqrt{4 a c-b^2} (d (a+x (b+c x)))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((g + h*x)*Sqrt[a + b*x + c*x^2])/(a*d + b*d*x + c*d*x^2)^(3/2),x]

[Out]

((a + x*(b + c*x))^(3/2)*((4*c*g - 2*b*h)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]] + Sqrt[-b^2 + 4*a*c]*h*Log[a
+ x*(b + c*x)]))/(2*c*Sqrt[-b^2 + 4*a*c]*(d*(a + x*(b + c*x)))^(3/2))

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Maple [A]  time = 0.234, size = 122, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,c{d}^{2}}\sqrt{d \left ( c{x}^{2}+bx+a \right ) } \left ( 2\,\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) bh-4\,\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) cg-h\ln \left ( c{x}^{2}+bx+a \right ) \sqrt{4\,ac-{b}^{2}} \right ){\frac{1}{\sqrt{c{x}^{2}+bx+a}}}{\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x+g)*(c*x^2+b*x+a)^(1/2)/(c*d*x^2+b*d*x+a*d)^(3/2),x)

[Out]

-1/2/(c*x^2+b*x+a)^(1/2)*(d*(c*x^2+b*x+a))^(1/2)*(2*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b*h-4*arctan((2*c*x+b)
/(4*a*c-b^2)^(1/2))*c*g-h*ln(c*x^2+b*x+a)*(4*a*c-b^2)^(1/2))/d^2/c/(4*a*c-b^2)^(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((h*x+g)*(c*x^2+b*x+a)^(1/2)/(c*d*x^2+b*d*x+a*d)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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((h*x+g)*(c*x^2+b*x+a)^(1/2)/(c*d*x^2+b*d*x+a*d)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g + h x\right ) \sqrt{a + b x + c x^{2}}}{\left (d \left (a + b x + c x^{2}\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)*(c*x**2+b*x+a)**(1/2)/(c*d*x**2+b*d*x+a*d)**(3/2),x)

[Out]

Integral((g + h*x)*sqrt(a + b*x + c*x**2)/(d*(a + b*x + c*x**2))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{2} + b x + a}{\left (h x + g\right )}}{{\left (c d x^{2} + b d x + a d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)*(c*x^2+b*x+a)^(1/2)/(c*d*x^2+b*d*x+a*d)^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^2 + b*x + a)*(h*x + g)/(c*d*x^2 + b*d*x + a*d)^(3/2), x)