3.1.39 \(\int \frac {(g+h x) \sqrt {a+b x+c x^2}}{(a d+b d x+c d x^2)^{3/2}} \, dx\) [39]

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

[Out]

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

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Rubi [A]
time = 0.07, 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 = {1013, 648, 632, 212, 642} \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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

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 642

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 648

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 1013

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 ) \text {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 = 0.07, 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 {-b^2+4 a c}}\right )+\sqrt {-b^2+4 a c} h \log (a+x (b+c x))\right )}{2 c \sqrt {-b^2+4 a c} (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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Maple [A]
time = 0.18, size = 121, normalized size = 0.89

method	result	size
default	\(\frac {\sqrt {d \left (c \,x^{2}+b x +a \right )}\, \left (h \ln \left (c \,x^{2}+b x +a \right ) \sqrt {4 a c -b^{2}}-2 \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) b h +4 \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) c g \right )}{2 \sqrt {c \,x^{2}+b x +a}\, d^{2} c \sqrt {4 a c -b^{2}}}\)	\(121\)
risch	\(\frac {\sqrt {c \,x^{2}+b x +a}\, \left (4 a c h -b^{2} h +\sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\right ) \ln \left (-4 a b c h +8 a \,c^{2} g +b^{3} h -2 b^{2} c g -2 \sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\, c x -\sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\, b \right )}{2 d \sqrt {d \left (c \,x^{2}+b x +a \right )}\, c \left (4 a c -b^{2}\right )}-\frac {\sqrt {c \,x^{2}+b x +a}\, \left (-4 a c h +b^{2} h +\sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\right ) \ln \left (-4 a b c h +8 a \,c^{2} g +b^{3} h -2 b^{2} c g +2 \sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\, c x +\sqrt {-\left (b h -2 c g \right )^{2} \left (4 a c -b^{2}\right )}\, b \right )}{2 d \sqrt {d \left (c \,x^{2}+b x +a \right )}\, c \left (4 a c -b^{2}\right )}\)	\(328\)

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,method=_RETURNVERBOSE)

[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 deta

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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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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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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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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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