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

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

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Rubi [A]  time = 0.11, antiderivative size = 136, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.09, size = 108, normalized size = 0.79 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.28, size = 127, normalized size = 0.93 \begin {gather*} \frac {d^{3/2} (a+x (b+c x))^{3/2} \left (\frac {(2 c g-b h) \tan ^{-1}\left (\frac {2 c x}{\sqrt {4 a c-b^2}}+\frac {b}{\sqrt {4 a c-b^2}}\right )}{c d^{3/2} \sqrt {4 a c-b^2}}+\frac {h \log \left (a+b x+c x^2\right )}{2 c d^{3/2}}\right )}{(d (a+x (b+c x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 50.45, size = 0, normalized size = 0.00 \begin {gather*} {\rm integral}\left (\frac {\sqrt {c d x^{2} + b d x + a d} \sqrt {c x^{2} + b x + a} {\left (h x + g\right )}}{c^{2} d^{2} x^{4} + 2 \, b c d^{2} x^{3} + 2 \, a b d^{2} x + {\left (b^{2} + 2 \, a c\right )} d^{2} x^{2} + a^{2} d^{2}}, x\right ) \end {gather*}

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]

integral(sqrt(c*d*x^2 + b*d*x + a*d)*sqrt(c*x^2 + b*x + a)*(h*x + g)/(c^2*d^2*x^4 + 2*b*c*d^2*x^3 + 2*a*b*d^2*
x + (b^2 + 2*a*c)*d^2*x^2 + a^2*d^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.03, size = 121, normalized size = 0.89 \begin {gather*} \frac {\sqrt {\left (c \,x^{2}+b x +a \right ) d}\, \left (-2 b h \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )+4 c g \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )+\sqrt {4 a c -b^{2}}\, h \ln \left (c \,x^{2}+b x +a \right )\right )}{2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {4 a c -b^{2}}\, c \,d^{2}} \end {gather*}

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)*(h*ln(c*x^2+b*x+a)*(4*a*c-b^2)^(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)/d^2/c/(4*a*c-b^2)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (g+h\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{{\left (c\,d\,x^2+b\,d\,x+a\,d\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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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