3.21.53 \(\int \frac {(d+e x)^2}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=129 \[ \frac {2 e \sqrt {a+b x+c x^2} (2 c d-b e)}{c \left (b^2-4 a c\right )}-\frac {2 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {e^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {738, 640, 621, 206} \begin {gather*} \frac {2 e \sqrt {a+b x+c x^2} (2 c d-b e)}{c \left (b^2-4 a c\right )}-\frac {2 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {e^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 640

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

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 127, normalized size = 0.98 \begin {gather*} \frac {\frac {2 \sqrt {c} \left (a b e^2-2 a c e (2 d+e x)+b^2 e^2 x+b c d (d-2 e x)+2 c^2 d^2 x\right )}{\sqrt {a+x (b+c x)}}-e^2 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{c^{3/2} \left (4 a c-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.56, size = 125, normalized size = 0.97 \begin {gather*} \frac {2 \left (a b e^2-4 a c d e-2 a c e^2 x+b^2 e^2 x+b c d^2-2 b c d e x+2 c^2 d^2 x\right )}{c \left (4 a c-b^2\right ) \sqrt {a+b x+c x^2}}-\frac {e^2 \log \left (-2 c^{3/2} \sqrt {a+b x+c x^2}+b c+2 c^2 x\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*c*d^2 - 4*a*c*d*e + a*b*e^2 + 2*c^2*d^2*x - 2*b*c*d*e*x + b^2*e^2*x - 2*a*c*e^2*x))/(c*(-b^2 + 4*a*c)*Sq
rt[a + b*x + c*x^2]) - (e^2*Log[b*c + 2*c^2*x - 2*c^(3/2)*Sqrt[a + b*x + c*x^2]])/c^(3/2)

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fricas [A]  time = 0.56, size = 461, normalized size = 3.57 \begin {gather*} \left [\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} x^{2} + {\left (b^{3} - 4 \, a b c\right )} e^{2} x + {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\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 (b c^{2} d^{2} - 4 \, a c^{2} d e + a b c e^{2} + {\left (2 \, c^{3} d^{2} - 2 \, b c^{2} d e + {\left (b^{2} c - 2 \, a c^{2}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )}}, -\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} x^{2} + {\left (b^{3} - 4 \, a b c\right )} e^{2} x + {\left (a b^{2} - 4 \, a^{2} c\right )} e^{2}\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 (b c^{2} d^{2} - 4 \, a c^{2} d e + a b c e^{2} + {\left (2 \, c^{3} d^{2} - 2 \, b c^{2} d e + {\left (b^{2} c - 2 \, a c^{2}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[1/2*(((b^2*c - 4*a*c^2)*e^2*x^2 + (b^3 - 4*a*b*c)*e^2*x + (a*b^2 - 4*a^2*c)*e^2)*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*(b*c^2*d^2 - 4*a*c^2*d*e + a*b*c*e^2 + (
2*c^3*d^2 - 2*b*c^2*d*e + (b^2*c - 2*a*c^2)*e^2)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 -
4*a*c^4)*x^2 + (b^3*c^2 - 4*a*b*c^3)*x), -(((b^2*c - 4*a*c^2)*e^2*x^2 + (b^3 - 4*a*b*c)*e^2*x + (a*b^2 - 4*a^
2*c)*e^2)*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*(b*c^2*d
^2 - 4*a*c^2*d*e + a*b*c*e^2 + (2*c^3*d^2 - 2*b*c^2*d*e + (b^2*c - 2*a*c^2)*e^2)*x)*sqrt(c*x^2 + b*x + a))/(a*
b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^2 + (b^3*c^2 - 4*a*b*c^3)*x)]

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giac [A]  time = 0.26, size = 132, normalized size = 1.02 \begin {gather*} -\frac {2 \, {\left (\frac {{\left (2 \, c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2} - 2 \, a c e^{2}\right )} x}{b^{2} c - 4 \, a c^{2}} + \frac {b c d^{2} - 4 \, a c d e + a b e^{2}}{b^{2} c - 4 \, a c^{2}}\right )}}{\sqrt {c x^{2} + b x + a}} - \frac {e^{2} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((2*c^2*d^2 - 2*b*c*d*e + b^2*e^2 - 2*a*c*e^2)*x/(b^2*c - 4*a*c^2) + (b*c*d^2 - 4*a*c*d*e + a*b*e^2)/(b^2*c
- 4*a*c^2))/sqrt(c*x^2 + b*x + a) - e^2*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(3/2)

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maple [B]  time = 0.06, size = 264, normalized size = 2.05 \begin {gather*} \frac {b^{2} e^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {4 b d e x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {b^{3} e^{2}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {2 b^{2} d e}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {e^{2} x}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {2 \left (2 c x +b \right ) d^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}+\frac {b \,e^{2}}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {2 d e}{\sqrt {c \,x^{2}+b x +a}\, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/(c*x^2+b*x+a)^(3/2),x)

[Out]

-e^2*x/c/(c*x^2+b*x+a)^(1/2)+1/2*e^2*b/c^2/(c*x^2+b*x+a)^(1/2)+e^2*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+1/2
*e^2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+e^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-2*d*e/c/(
c*x^2+b*x+a)^(1/2)-4*d*e*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-2*d*e*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+2*d^2
*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(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((e*x+d)^2/(c*x^2+b*x+a)^(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 zero or nonzero?

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mupad [B]  time = 1.47, size = 150, normalized size = 1.16 \begin {gather*} \frac {e^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{c^{3/2}}+\frac {d^2\,\left (\frac {b}{2}+c\,x\right )}{\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^2+b\,x+a}}-\frac {2\,d\,e\,\left (4\,a+2\,b\,x\right )}{\left (4\,a\,c-b^2\right )\,\sqrt {c\,x^2+b\,x+a}}+\frac {e^2\,\left (\frac {a\,b}{2}-x\,\left (a\,c-\frac {b^2}{2}\right )\right )}{c\,\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^2+b\,x+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(a + b*x + c*x^2)^(3/2),x)

[Out]

(e^2*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2)))/c^(3/2) + (d^2*(b/2 + c*x))/((a*c - b^2/4)*(a + b*x +
c*x^2)^(1/2)) - (2*d*e*(4*a + 2*b*x))/((4*a*c - b^2)*(a + b*x + c*x^2)^(1/2)) + (e^2*((a*b)/2 - x*(a*c - b^2/
2)))/(c*(a*c - b^2/4)*(a + b*x + c*x^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{2}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((d + e*x)**2/(a + b*x + c*x**2)**(3/2), x)

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