3.1244 \(\int \frac {(b d+2 c d x)^2}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=66 \[ 4 \sqrt {c} d^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )-\frac {2 d^2 (b+2 c x)}{\sqrt {a+b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 66, 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 = {686, 621, 206} \[ 4 \sqrt {c} d^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )-\frac {2 d^2 (b+2 c x)}{\sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^2*(b + 2*c*x))/Sqrt[a + b*x + c*x^2] + 4*Sqrt[c]*d^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*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 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 686

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

Rubi steps

\begin {align*} \int \frac {(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 d^2 (b+2 c x)}{\sqrt {a+b x+c x^2}}+\left (4 c d^2\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {2 d^2 (b+2 c x)}{\sqrt {a+b x+c x^2}}+\left (8 c d^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 )\\ &=-\frac {2 d^2 (b+2 c x)}{\sqrt {a+b x+c x^2}}+4 \sqrt {c} d^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 118, normalized size = 1.79 \[ d^2 \left (\frac {4 \sqrt {c} \sqrt {a+x (b+c x)} \sinh ^{-1}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {4 a-\frac {b^2}{c}}}\right )}{\sqrt {4 a-\frac {b^2}{c}} \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}}}-\frac {2 (b+2 c x)}{\sqrt {a+x (b+c x)}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.37, size = 225, normalized size = 3.41 \[ \left [\frac {2 \, {\left ({\left (c d^{2} x^{2} + b d^{2} x + a d^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {c x^{2} + b x + a}\right )}}{c x^{2} + b x + a}, -\frac {2 \, {\left (2 \, {\left (c d^{2} x^{2} + b d^{2} x + a d^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {c x^{2} + b x + a}\right )}}{c x^{2} + b x + a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.29, size = 113, normalized size = 1.71 \[ -4 \, \sqrt {c} d^{2} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right ) - \frac {2 \, {\left (\frac {2 \, {\left (b^{2} c d^{2} - 4 \, a c^{2} d^{2}\right )} x}{b^{2} - 4 \, a c} + \frac {b^{3} d^{2} - 4 \, a b c d^{2}}{b^{2} - 4 \, a c}\right )}}{\sqrt {c x^{2} + b x + a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 72, normalized size = 1.09 \[ -\frac {4 c \,d^{2} x}{\sqrt {c \,x^{2}+b x +a}}+4 \sqrt {c}\, d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )-\frac {2 b \,d^{2}}{\sqrt {c \,x^{2}+b x +a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*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 = 0.79, size = 156, normalized size = 2.36 \[ 4\,\sqrt {c}\,d^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )+\frac {b^2\,d^2\,\left (\frac {b}{2}+c\,x\right )}{\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^2+b\,x+a}}+\frac {4\,c\,d^2\,\left (\frac {a\,b}{2}-x\,\left (a\,c-\frac {b^2}{2}\right )\right )}{\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^2+b\,x+a}}-\frac {4\,b\,c\,d^2\,\left (4\,a+2\,b\,x\right )}{\left (4\,a\,c-b^2\right )\,\sqrt {c\,x^2+b\,x+a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*c^(1/2)*d^2*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2)) + (b^2*d^2*(b/2 + c*x))/((a*c - b^2/4)*(a + b
*x + c*x^2)^(1/2)) + (4*c*d^2*((a*b)/2 - x*(a*c - b^2/2)))/((a*c - b^2/4)*(a + b*x + c*x^2)^(1/2)) - (4*b*c*d^
2*(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 \[ d^{2} \left (\int \frac {b^{2}}{a \sqrt {a + b x + c x^{2}} + b x \sqrt {a + b x + c x^{2}} + c x^{2} \sqrt {a + b x + c x^{2}}}\, dx + \int \frac {4 c^{2} x^{2}}{a \sqrt {a + b x + c x^{2}} + b x \sqrt {a + b x + c x^{2}} + c x^{2} \sqrt {a + b x + c x^{2}}}\, dx + \int \frac {4 b c x}{a \sqrt {a + b x + c x^{2}} + b x \sqrt {a + b x + c x^{2}} + c x^{2} \sqrt {a + b x + c x^{2}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*(Integral(b**2/(a*sqrt(a + b*x + c*x**2) + b*x*sqrt(a + b*x + c*x**2) + c*x**2*sqrt(a + b*x + c*x**2)), x
) + Integral(4*c**2*x**2/(a*sqrt(a + b*x + c*x**2) + b*x*sqrt(a + b*x + c*x**2) + c*x**2*sqrt(a + b*x + c*x**2
)), x) + Integral(4*b*c*x/(a*sqrt(a + b*x + c*x**2) + b*x*sqrt(a + b*x + c*x**2) + c*x**2*sqrt(a + b*x + c*x**
2)), x))

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