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

Optimal. Leaf size=60 \[ \frac {2 e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}-\frac {2 (d+e x)}{\sqrt {a+b x+c x^2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {768, 621, 206} \begin {gather*} \frac {2 e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}-\frac {2 (d+e x)}{\sqrt {a+b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.27, size = 56, normalized size = 0.93 \begin {gather*} \frac {2 e \log \left (2 \sqrt {c} \sqrt {a+x (b+c x)}+b+2 c x\right )}{\sqrt {c}}-\frac {2 (d+e x)}{\sqrt {a+x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.47, size = 58, normalized size = 0.97 \begin {gather*} -\frac {2 (d+e x)}{\sqrt {a+b x+c x^2}}-\frac {2 e \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{\sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.57, size = 211, normalized size = 3.52 \begin {gather*} \left [\frac {{\left (c e x^{2} + b e x + a e\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 e x + c d\right )} \sqrt {c x^{2} + b x + a}}{c^{2} x^{2} + b c x + a c}, -\frac {2 \, {\left ({\left (c e x^{2} + b e x + a e\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 e x + c d\right )} \sqrt {c x^{2} + b x + a}\right )}}{c^{2} x^{2} + b c x + a c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.30, size = 101, normalized size = 1.68 \begin {gather*} -\frac {2 \, e \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{\sqrt {c}} - \frac {2 \, {\left (\frac {{\left (b^{2} e - 4 \, a c e\right )} x}{b^{2} - 4 \, a c} + \frac {b^{2} d - 4 \, a c d}{b^{2} - 4 \, a c}\right )}}{\sqrt {c x^{2} + b x + a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 158, normalized size = 2.63 \begin {gather*} -\frac {4 b c d x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {2 b^{2} d}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {2 \left (2 c x +b \right ) b d}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {2 e x}{\sqrt {c \,x^{2}+b x +a}}+\frac {2 e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}-\frac {2 d}{\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)/(c*x^2+b*x+a)^(3/2),x)

[Out]

-2*e*x/(c*x^2+b*x+a)^(1/2)+2/c^(1/2)*e*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-2/(c*x^2+b*x+a)^(1/2)*d-4*b
/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*c*d-2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d+2*b*d*(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((2*c*x+b)*(e*x+d)/(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 = 2.62, size = 161, normalized size = 2.68 \begin {gather*} \frac {2\,b^2\,d-4\,a\,b\,e-2\,b^2\,e\,x+4\,b\,c\,d\,x}{\left (4\,a\,c-b^2\right )\,\sqrt {c\,x^2+b\,x+a}}+\frac {2\,e\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{\sqrt {c}}+\frac {2\,e\,\left (\frac {a\,b}{2}-x\,\left (a\,c-\frac {b^2}{2}\right )\right )}{\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^2+b\,x+a}}-\frac {2\,c\,d\,\left (4\,a+2\,b\,x\right )}{\left (4\,a\,c-b^2\right )\,\sqrt {c\,x^2+b\,x+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^2*d - 4*a*b*e - 2*b^2*e*x + 4*b*c*d*x)/((4*a*c - b^2)*(a + b*x + c*x^2)^(1/2)) + (2*e*log((b/2 + c*x)/c^(
1/2) + (a + b*x + c*x^2)^(1/2)))/c^(1/2) + (2*e*((a*b)/2 - x*(a*c - b^2/2)))/((a*c - b^2/4)*(a + b*x + c*x^2)^
(1/2)) - (2*c*d*(4*a + 2*b*x))/((4*a*c - b^2)*(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 (b + 2 c x\right ) \left (d + e x\right )}{\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)/(c*x**2+b*x+a)**(3/2),x)

[Out]

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

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