3.13.10 \(\int \frac {(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^4} \, dx\) [1210]

Optimal. Leaf size=107 \[ -\frac {\sqrt {a+b x+c x^2}}{8 c^2 d^4 (b+2 c x)}-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}+\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} d^4} \]

[Out]

-1/6*(c*x^2+b*x+a)^(3/2)/c/d^4/(2*c*x+b)^3+1/16*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)/d^4
-1/8*(c*x^2+b*x+a)^(1/2)/c^2/d^4/(2*c*x+b)

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Rubi [A]
time = 0.03, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {698, 635, 212} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} d^4}-\frac {\sqrt {a+b x+c x^2}}{8 c^2 d^4 (b+2 c x)}-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^4,x]

[Out]

-1/8*Sqrt[a + b*x + c*x^2]/(c^2*d^4*(b + 2*c*x)) - (a + b*x + c*x^2)^(3/2)/(6*c*d^4*(b + 2*c*x)^3) + ArcTanh[(
b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])]/(16*c^(5/2)*d^4)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 698

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^4} \, dx &=-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}+\frac {\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^2} \, dx}{4 c d^2}\\ &=-\frac {\sqrt {a+b x+c x^2}}{8 c^2 d^4 (b+2 c x)}-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}+\frac {\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c^2 d^4}\\ &=-\frac {\sqrt {a+b x+c x^2}}{8 c^2 d^4 (b+2 c x)}-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}+\frac {\text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{8 c^2 d^4}\\ &=-\frac {\sqrt {a+b x+c x^2}}{8 c^2 d^4 (b+2 c x)}-\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}+\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} d^4}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 95, normalized size = 0.89 \begin {gather*} \frac {-\frac {\sqrt {a+x (b+c x)} \left (3 b^2+16 b c x+4 c \left (a+4 c x^2\right )\right )}{24 c^2 (b+2 c x)^3}-\frac {\log \left (c^2 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{16 c^{5/2}}}{d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^4,x]

[Out]

(-1/24*(Sqrt[a + x*(b + c*x)]*(3*b^2 + 16*b*c*x + 4*c*(a + 4*c*x^2)))/(c^2*(b + 2*c*x)^3) - Log[c^2*(b + 2*c*x
- 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]/(16*c^(5/2)))/d^4

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(310\) vs. \(2(89)=178\).
time = 0.70, size = 311, normalized size = 2.91

method	result	size
default	\(\frac {-\frac {4 c \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{3 \left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{3}}+\frac {8 c^{2} \left (-\frac {4 c \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )}+\frac {16 c^{2} \left (\frac {\left (x +\frac {b}{2 c}\right ) \left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{4}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (x +\frac {b}{2 c}\right ) \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}}{2}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\sqrt {c}\, \left (x +\frac {b}{2 c}\right )+\sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{4 a c -b^{2}}\right )}{3 \left (4 a c -b^{2}\right )}}{16 d^{4} c^{4}}\)	\(311\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,x,method=_RETURNVERBOSE)

[Out]

1/16/d^4/c^4*(-4/3/(4*a*c-b^2)*c/(x+1/2*b/c)^3*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+8/3*c^2/(4*a*c-b^2)*(
-4/(4*a*c-b^2)*c/(x+1/2*b/c)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+16*c^2/(4*a*c-b^2)*(1/4*(x+1/2*b/c)*((x
+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+3/16*(4*a*c-b^2)/c*(1/2*(x+1/2*b/c)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)
^(1/2)+1/8*(4*a*c-b^2)/c^(3/2)*ln(c^(1/2)*(x+1/2*b/c)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(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((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,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 = 5.21, size = 345, normalized size = 3.22 \begin {gather*} \left [\frac {3 \, {\left (8 \, c^{3} x^{3} + 12 \, b c^{2} x^{2} + 6 \, b^{2} c x + b^{3}\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 (16 \, c^{3} x^{2} + 16 \, b c^{2} x + 3 \, b^{2} c + 4 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{96 \, {\left (8 \, c^{6} d^{4} x^{3} + 12 \, b c^{5} d^{4} x^{2} + 6 \, b^{2} c^{4} d^{4} x + b^{3} c^{3} d^{4}\right )}}, -\frac {3 \, {\left (8 \, c^{3} x^{3} + 12 \, b c^{2} x^{2} + 6 \, b^{2} c x + b^{3}\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 (16 \, c^{3} x^{2} + 16 \, b c^{2} x + 3 \, b^{2} c + 4 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{48 \, {\left (8 \, c^{6} d^{4} x^{3} + 12 \, b c^{5} d^{4} x^{2} + 6 \, b^{2} c^{4} d^{4} x + b^{3} c^{3} d^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,x, algorithm="fricas")

[Out]

[1/96*(3*(8*c^3*x^3 + 12*b*c^2*x^2 + 6*b^2*c*x + b^3)*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*(16*c^3*x^2 + 16*b*c^2*x + 3*b^2*c + 4*a*c^2)*sqrt(c*x^2 + b*x + a))
/(8*c^6*d^4*x^3 + 12*b*c^5*d^4*x^2 + 6*b^2*c^4*d^4*x + b^3*c^3*d^4), -1/48*(3*(8*c^3*x^3 + 12*b*c^2*x^2 + 6*b^
2*c*x + b^3)*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*(16*c
^3*x^2 + 16*b*c^2*x + 3*b^2*c + 4*a*c^2)*sqrt(c*x^2 + b*x + a))/(8*c^6*d^4*x^3 + 12*b*c^5*d^4*x^2 + 6*b^2*c^4*
d^4*x + b^3*c^3*d^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a \sqrt {a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx + \int \frac {b x \sqrt {a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx + \int \frac {c x^{2} \sqrt {a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx}{d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(3/2)/(2*c*d*x+b*d)**4,x)

[Out]

(Integral(a*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b*c**3*x**3 + 16*c**4*x**4), x)
+ Integral(b*x*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b*c**3*x**3 + 16*c**4*x**4)
, x) + Integral(c*x**2*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b*c**3*x**3 + 16*c**
4*x**4), x))/d**4

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%%{16,[4]%%%},[8,4,8,0]%%%}+%%%{%%%{-256,[5]%%%},[8,4,6
,1]%%%}+%%%

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (b\,d+2\,c\,d\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^4,x)

[Out]

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^4, x)

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