3.1207 \(\int \frac {(a+b x+c x^2)^{3/2}}{b d+2 c d x} \, dx\)

Optimal. Leaf size=115 \[ \frac {\left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{16 c^{5/2} d}-\frac {\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}{8 c^2 d}+\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 115, 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 = {685, 688, 205} \[ -\frac {\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}{8 c^2 d}+\frac {\left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{16 c^{5/2} d}+\frac {\left (a+b x+c x^2\right )^{3/2}}{6 c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 685

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 + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), 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] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

Rule 688

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

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Mathematica [A]  time = 0.10, size = 103, normalized size = 0.90 \[ \frac {2 \sqrt {c} \sqrt {a+x (b+c x)} \left (4 c \left (4 a+c x^2\right )-3 b^2+4 b c x\right )+3 \left (b^2-4 a c\right )^{3/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+x (b+c x)}}{\sqrt {b^2-4 a c}}\right )}{48 c^{5/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-3*b^2 + 4*b*c*x + 4*c*(4*a + c*x^2)) + 3*(b^2 - 4*a*c)^(3/2)*ArcTan[(2*Sqrt
[c]*Sqrt[a + x*(b + c*x)])/Sqrt[b^2 - 4*a*c]])/(48*c^(5/2)*d)

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fricas [A]  time = 1.14, size = 246, normalized size = 2.14 \[ \left [-\frac {3 \, {\left (b^{2} - 4 \, a c\right )} \sqrt {-\frac {b^{2} - 4 \, a c}{c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {c x^{2} + b x + a} c \sqrt {-\frac {b^{2} - 4 \, a c}{c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) - 4 \, {\left (4 \, c^{2} x^{2} + 4 \, b c x - 3 \, b^{2} + 16 \, a c\right )} \sqrt {c x^{2} + b x + a}}{96 \, c^{2} d}, -\frac {3 \, {\left (b^{2} - 4 \, a c\right )} \sqrt {\frac {b^{2} - 4 \, a c}{c}} \arctan \left (\frac {\sqrt {\frac {b^{2} - 4 \, a c}{c}}}{2 \, \sqrt {c x^{2} + b x + a}}\right ) - 2 \, {\left (4 \, c^{2} x^{2} + 4 \, b c x - 3 \, b^{2} + 16 \, a c\right )} \sqrt {c x^{2} + b x + a}}{48 \, c^{2} d}\right ] \]

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),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.28, size = 149, normalized size = 1.30 \[ \frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (4 \, x {\left (\frac {x}{d} + \frac {b}{c d}\right )} - \frac {3 \, b^{2} c^{3} d^{3} - 16 \, a c^{4} d^{3}}{c^{5} d^{4}}\right )} + \frac {{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \arctan \left (-\frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} c + b \sqrt {c}}{\sqrt {b^{2} c - 4 \, a c^{2}}}\right )}{8 \, \sqrt {b^{2} c - 4 \, a c^{2}} c^{2} d} \]

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),x, algorithm="giac")

[Out]

1/24*sqrt(c*x^2 + b*x + a)*(4*x*(x/d + b/(c*d)) - (3*b^2*c^3*d^3 - 16*a*c^4*d^3)/(c^5*d^4)) + 1/8*(b^4 - 8*a*b
^2*c + 16*a^2*c^2)*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)*c^2*d)

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maple [B]  time = 0.05, size = 430, normalized size = 3.74 \[ -\frac {a^{2} \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{\sqrt {\frac {4 a c -b^{2}}{c}}\, c d}+\frac {a \,b^{2} \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{2 \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{2} d}-\frac {b^{4} \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{16 \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{3} d}+\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, a}{4 c d}-\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, b^{2}}{16 c^{2} d}+\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{6 c d} \]

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),x)

[Out]

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

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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((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d),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 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{b\,d+2\,c\,d\,x} \,d x \]

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),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {b x \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {c x^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx}{d} \]

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),x)

[Out]

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

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