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

Optimal. Leaf size=93 \[ \frac {2 e (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}}-\frac {2 (d+e x)^2}{\sqrt {a+b x+c x^2}}+\frac {4 e^2 \sqrt {a+b x+c x^2}}{c} \]

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Rubi [A]  time = 0.05, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {768, 640, 621, 206} \begin {gather*} \frac {2 e (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}}-\frac {2 (d+e x)^2}{\sqrt {a+b x+c x^2}}+\frac {4 e^2 \sqrt {a+b x+c x^2}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^2)/Sqrt[a + b*x + c*x^2] + (4*e^2*Sqrt[a + b*x + c*x^2])/c + (2*e*(2*c*d - b*e)*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 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(b+2 c x) (d+e x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 (d+e x)^2}{\sqrt {a+b x+c x^2}}+(4 e) \int \frac {d+e x}{\sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 (d+e x)^2}{\sqrt {a+b x+c x^2}}+\frac {4 e^2 \sqrt {a+b x+c x^2}}{c}+\frac {(2 e (2 c d-b e)) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{c}\\ &=-\frac {2 (d+e x)^2}{\sqrt {a+b x+c x^2}}+\frac {4 e^2 \sqrt {a+b x+c x^2}}{c}+\frac {(4 e (2 c d-b e)) \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)^2}{\sqrt {a+b x+c x^2}}+\frac {4 e^2 \sqrt {a+b x+c x^2}}{c}+\frac {2 e (2 c d-b e) \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 = 0.17, size = 95, normalized size = 1.02 \begin {gather*} \frac {2 e (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{c^{3/2}}+\frac {4 e^2 (a+b x)-2 c \left (d^2+2 d e x-e^2 x^2\right )}{c \sqrt {a+x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.80, size = 105, normalized size = 1.13 \begin {gather*} -\frac {2 \left (2 c d e-b e^2\right ) \log \left (-2 c^{3/2} \sqrt {a+b x+c x^2}+b c+2 c^2 x\right )}{c^{3/2}}-\frac {2 \left (-2 a e^2-2 b e^2 x+c d^2+2 c d e x-c e^2 x^2\right )}{c \sqrt {a+b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(c*d^2 - 2*a*e^2 + 2*c*d*e*x - 2*b*e^2*x - c*e^2*x^2))/(c*Sqrt[a + b*x + c*x^2]) - (2*(2*c*d*e - b*e^2)*Lo
g[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 [B]  time = 0.66, size = 363, normalized size = 3.90 \begin {gather*} \left [-\frac {{\left (2 \, a c d e - a b e^{2} + {\left (2 \, c^{2} d e - b c e^{2}\right )} x^{2} + {\left (2 \, b c d e - b^{2} e^{2}\right )} x\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 ) - 2 \, {\left (c^{2} e^{2} x^{2} - c^{2} d^{2} + 2 \, a c e^{2} - 2 \, {\left (c^{2} d e - b c e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{c^{3} x^{2} + b c^{2} x + a c^{2}}, -\frac {2 \, {\left ({\left (2 \, a c d e - a b e^{2} + {\left (2 \, c^{2} d e - b c e^{2}\right )} x^{2} + {\left (2 \, b c d e - b^{2} e^{2}\right )} x\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 ) - {\left (c^{2} e^{2} x^{2} - c^{2} d^{2} + 2 \, a c e^{2} - 2 \, {\left (c^{2} d e - b c e^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}\right )}}{c^{3} x^{2} + b c^{2} x + a c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.33, size = 197, normalized size = 2.12 \begin {gather*} \frac {2 \, {\left ({\left (\frac {{\left (b^{2} c e^{2} - 4 \, a c^{2} e^{2}\right )} x}{b^{2} c - 4 \, a c^{2}} - \frac {2 \, {\left (b^{2} c d e - 4 \, a c^{2} d e - b^{3} e^{2} + 4 \, a b c e^{2}\right )}}{b^{2} c - 4 \, a c^{2}}\right )} x - \frac {b^{2} c d^{2} - 4 \, a c^{2} d^{2} - 2 \, a b^{2} e^{2} + 8 \, a^{2} c e^{2}}{b^{2} c - 4 \, a c^{2}}\right )}}{\sqrt {c x^{2} + b x + a}} - \frac {2 \, {\left (2 \, c d e - b e^{2}\right )} \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((2*c*x+b)*(e*x+d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

2*(((b^2*c*e^2 - 4*a*c^2*e^2)*x/(b^2*c - 4*a*c^2) - 2*(b^2*c*d*e - 4*a*c^2*d*e - b^3*e^2 + 4*a*b*c*e^2)/(b^2*c
- 4*a*c^2))*x - (b^2*c*d^2 - 4*a*c^2*d^2 - 2*a*b^2*e^2 + 8*a^2*c*e^2)/(b^2*c - 4*a*c^2))/sqrt(c*x^2 + b*x + a
) - 2*(2*c*d*e - b*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 = 427, normalized size = 4.59 \begin {gather*} \frac {8 a b \,e^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {2 b^{3} e^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {4 b c \,d^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {4 a \,b^{2} e^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {b^{4} e^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {2 b^{2} d^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {2 e^{2} x^{2}}{\sqrt {c \,x^{2}+b x +a}}+\frac {2 b \,e^{2} x}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {2 \left (2 c x +b \right ) b \,d^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {4 d e x}{\sqrt {c \,x^{2}+b x +a}}-\frac {2 b \,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 {4 d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+\frac {4 a \,e^{2}}{\sqrt {c \,x^{2}+b x +a}\, c}-\frac {b^{2} e^{2}}{\sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {2 d^{2}}{\sqrt {c \,x^{2}+b x +a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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