3.13.46 \(\int \frac {1}{(b d+2 c d x) (a+b x+c x^2)^{3/2}} \, dx\) [1246]

Optimal. Leaf size=86 \[ -\frac {2}{\left (b^2-4 a c\right ) d \sqrt {a+b x+c x^2}}-\frac {4 \sqrt {c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2} d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 86, 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 = {701, 702, 211} \begin {gather*} -\frac {4 \sqrt {c} \text {ArcTan}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )^{3/2}}-\frac {2}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^(3/2)),x]

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 701

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

Rule 702

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[4*c, Subst[Int[1/(b^2*e
- 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2}{\left (b^2-4 a c\right ) d \sqrt {a+b x+c x^2}}-\frac {(4 c) \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=-\frac {2}{\left (b^2-4 a c\right ) d \sqrt {a+b x+c x^2}}-\frac {\left (16 c^2\right ) \text {Subst}\left (\int \frac {1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt {a+b x+c x^2}\right )}{b^2-4 a c}\\ &=-\frac {2}{\left (b^2-4 a c\right ) d \sqrt {a+b x+c x^2}}-\frac {4 \sqrt {c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2} d}\\ \end {align*}

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Mathematica [A]
time = 0.54, size = 89, normalized size = 1.03 \begin {gather*} \frac {-\frac {2}{\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)}}+\frac {8 \sqrt {c} \tan ^{-1}\left (\frac {b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^(3/2)),x]

[Out]

(-2/((b^2 - 4*a*c)*Sqrt[a + x*(b + c*x)]) + (8*Sqrt[c]*ArcTan[(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])/Sq
rt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2))/d

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(161\) vs. \(2(74)=148\).
time = 0.73, size = 162, normalized size = 1.88

method	result	size
default	\(\frac {\frac {4 c}{\left (4 a c -b^{2}\right ) \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}-\frac {8 c \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{\left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}}}{2 d c}\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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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(1/(2*c*d*x+b*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 deta

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Fricas [A]
time = 2.96, size = 311, normalized size = 3.62 \begin {gather*} \left [-\frac {2 \, {\left ({\left (c x^{2} + b x + a\right )} \sqrt {-\frac {c}{b^{2} - 4 \, a c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c + 4 \, \sqrt {c x^{2} + b x + a} {\left (b^{2} - 4 \, a c\right )} \sqrt {-\frac {c}{b^{2} - 4 \, a c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + \sqrt {c x^{2} + b x + a}\right )}}{{\left (b^{2} c - 4 \, a c^{2}\right )} d x^{2} + {\left (b^{3} - 4 \, a b c\right )} d x + {\left (a b^{2} - 4 \, a^{2} c\right )} d}, -\frac {2 \, {\left (2 \, {\left (c x^{2} + b x + a\right )} \sqrt {\frac {c}{b^{2} - 4 \, a c}} \arctan \left (-\frac {\sqrt {c x^{2} + b x + a} {\left (b^{2} - 4 \, a c\right )} \sqrt {\frac {c}{b^{2} - 4 \, a c}}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + \sqrt {c x^{2} + b x + a}\right )}}{{\left (b^{2} c - 4 \, a c^{2}\right )} d x^{2} + {\left (b^{3} - 4 \, a b c\right )} d x + {\left (a b^{2} - 4 \, a^{2} c\right )} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-2*((c*x^2 + b*x + a)*sqrt(-c/(b^2 - 4*a*c))*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c + 4*sqrt(c*x^2 + b*x + a
)*(b^2 - 4*a*c)*sqrt(-c/(b^2 - 4*a*c)))/(4*c^2*x^2 + 4*b*c*x + b^2)) + sqrt(c*x^2 + b*x + a))/((b^2*c - 4*a*c^
2)*d*x^2 + (b^3 - 4*a*b*c)*d*x + (a*b^2 - 4*a^2*c)*d), -2*(2*(c*x^2 + b*x + a)*sqrt(c/(b^2 - 4*a*c))*arctan(-1
/2*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt(c/(b^2 - 4*a*c))/(c^2*x^2 + b*c*x + a*c)) + sqrt(c*x^2 + b*x + a))
/((b^2*c - 4*a*c^2)*d*x^2 + (b^3 - 4*a*b*c)*d*x + (a*b^2 - 4*a^2*c)*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{a b \sqrt {a + b x + c x^{2}} + 2 a c x \sqrt {a + b x + c x^{2}} + b^{2} x \sqrt {a + b x + c x^{2}} + 3 b c x^{2} \sqrt {a + b x + c x^{2}} + 2 c^{2} x^{3} \sqrt {a + b x + c x^{2}}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral(1/(a*b*sqrt(a + b*x + c*x**2) + 2*a*c*x*sqrt(a + b*x + c*x**2) + b**2*x*sqrt(a + b*x + c*x**2) + 3*b*
c*x**2*sqrt(a + b*x + c*x**2) + 2*c**2*x**3*sqrt(a + b*x + c*x**2)), x)/d

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (74) = 148\).
time = 4.34, size = 158, normalized size = 1.84 \begin {gather*} \frac {8 \, c \arctan \left (\frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} c + b \sqrt {c}}{\sqrt {b^{2} c - 4 \, a c^{2}}}\right )}{\sqrt {b^{2} c - 4 \, a c^{2}} {\left (b^{2} d - 4 \, a c d\right )}} - \frac {2 \, {\left (b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d\right )}}{{\left (b^{6} d^{2} - 12 \, a b^{4} c d^{2} + 48 \, a^{2} b^{2} c^{2} d^{2} - 64 \, a^{3} c^{3} d^{2}\right )} \sqrt {c x^{2} + b x + a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

8*c*arctan((2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c + b*sqrt(c))/sqrt(b^2*c - 4*a*c^2))/(sqrt(b^2*c - 4*a*c^2)
*(b^2*d - 4*a*c*d)) - 2*(b^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d)/((b^6*d^2 - 12*a*b^4*c*d^2 + 48*a^2*b^2*c^2*d^2 -
64*a^3*c^3*d^2)*sqrt(c*x^2 + b*x + a))

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

[Out]

int(1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^(3/2)), x)

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