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

Optimal. Leaf size=113 \[ -\frac {3 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32 c^{5/2} d^2}+\frac {3 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)} \]

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Rubi [A]  time = 0.05, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {684, 612, 621, 206} \begin {gather*} -\frac {3 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32 c^{5/2} d^2}+\frac {3 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(16*c^2*d^2) - (a + b*x + c*x^2)^(3/2)/(2*c*d^2*(b + 2*c*x)) - (3*(b^2 -
4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(32*c^(5/2)*d^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 612

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

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 684

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)^2} \, dx &=-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}+\frac {3 \int \sqrt {a+b x+c x^2} \, dx}{4 c d^2}\\ &=\frac {3 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{32 c^2 d^2}\\ &=\frac {3 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}-\frac {\left (3 \left (b^2-4 a c\right )\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 )}{16 c^2 d^2}\\ &=\frac {3 (b+2 c x) \sqrt {a+b x+c x^2}}{16 c^2 d^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{2 c d^2 (b+2 c x)}-\frac {3 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{32 c^{5/2} d^2}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 95, normalized size = 0.84 \begin {gather*} \frac {\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{16 c^2 d^2 (b+2 c x) \sqrt {\frac {c (a+x (b+c x))}{4 a c-b^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 4*a*c)*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-3/2, -1/2, 1/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(16*c^2*
d^2*(b + 2*c*x)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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IntegrateAlgebraic [A]  time = 0.54, size = 104, normalized size = 0.92 \begin {gather*} \frac {3 \left (b^2-4 a c\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{32 c^{5/2} d^2}+\frac {\sqrt {a+b x+c x^2} \left (-8 a c+3 b^2+4 b c x+4 c^2 x^2\right )}{16 c^2 d^2 (b+2 c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(3*b^2 - 8*a*c + 4*b*c*x + 4*c^2*x^2))/(16*c^2*d^2*(b + 2*c*x)) + (3*(b^2 - 4*a*c)*Log[
b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(32*c^(5/2)*d^2)

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fricas [A]  time = 0.52, size = 281, normalized size = 2.49 \begin {gather*} \left [-\frac {3 \, {\left (b^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x\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 (4 \, c^{3} x^{2} + 4 \, b c^{2} x + 3 \, b^{2} c - 8 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{64 \, {\left (2 \, c^{4} d^{2} x + b c^{3} d^{2}\right )}}, \frac {3 \, {\left (b^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x\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 (4 \, c^{3} x^{2} + 4 \, b c^{2} x + 3 \, b^{2} c - 8 \, a c^{2}\right )} \sqrt {c x^{2} + b x + a}}{32 \, {\left (2 \, c^{4} d^{2} x + b c^{3} d^{2}\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)^2,x, algorithm="fricas")

[Out]

[-1/64*(3*(b^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*x)*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*(4*c^3*x^2 + 4*b*c^2*x + 3*b^2*c - 8*a*c^2)*sqrt(c*x^2 + b*x + a))/(2*c^
4*d^2*x + b*c^3*d^2), 1/32*(3*(b^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*x)*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*(4*c^3*x^2 + 4*b*c^2*x + 3*b^2*c - 8*a*c^2)*sqrt(c*x^2 + b
*x + a))/(2*c^4*d^2*x + b*c^3*d^2)]

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giac [B]  time = 0.51, size = 440, normalized size = 3.89 \begin {gather*} \frac {1}{32} \, d^{2} {\left (\frac {3 \, {\left (b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d) - 4 \, a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d)\right )} \arctan \left (\frac {\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2} d^{4} {\left | c \right |}} + \frac {2 \, {\left (\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d) - 4 \, \sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d)\right )}}{c^{3} d^{4} {\left | c \right |}} + \frac {\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d) - 4 \, \sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\relax (c) \mathrm {sgn}\relax (d)}{{\left (\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}}\right )} c^{2} d^{4} {\left | c \right |}}\right )} {\left | c \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)^2,x, algorithm="giac")

[Out]

1/32*d^2*(3*(b^2*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 4*a*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d))*arctan(sqr
t(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)/sqrt(-c))/(sqrt(-c)*c^2*d^4*abs(c)) + 2*(s
qrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*b^2*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d)
- 4*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*a*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*
sgn(d))/(c^3*d^4*abs(c)) + (sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*b^2*sgn(1/(
2*c*d*x + b*d))*sgn(c)*sgn(d) - 4*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)*a*c*s
gn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d))/((b^2*c*d^2/(2*c*d*x + b*d)^2 - 4*a*c^2*d^2/(2*c*d*x + b*d)^2)*c^2*d^4*ab
s(c)))*abs(c)

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maple [B]  time = 0.05, size = 570, normalized size = 5.04 \begin {gather*} \frac {3 a^{2} \ln \left (\left (x +\frac {b}{2 c}\right ) \sqrt {c}+\sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\right )}{2 \left (4 a c -b^{2}\right ) \sqrt {c}\, d^{2}}-\frac {3 a \,b^{2} \ln \left (\left (x +\frac {b}{2 c}\right ) \sqrt {c}+\sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\right )}{4 \left (4 a c -b^{2}\right ) c^{\frac {3}{2}} d^{2}}+\frac {3 b^{4} \ln \left (\left (x +\frac {b}{2 c}\right ) \sqrt {c}+\sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\right )}{32 \left (4 a c -b^{2}\right ) c^{\frac {5}{2}} d^{2}}+\frac {3 \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\, a x}{2 \left (4 a c -b^{2}\right ) d^{2}}-\frac {3 \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\, b^{2} x}{8 \left (4 a c -b^{2}\right ) c \,d^{2}}+\frac {3 \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\, a b}{4 \left (4 a c -b^{2}\right ) c \,d^{2}}-\frac {3 \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\, b^{3}}{16 \left (4 a c -b^{2}\right ) c^{2} d^{2}}+\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}} x}{\left (4 a c -b^{2}\right ) d^{2}}+\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}} b}{2 \left (4 a c -b^{2}\right ) c \,d^{2}}-\frac {\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 ) c \,d^{2}} \end {gather*}

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)^2,x)

[Out]

-1/c/d^2/(4*a*c-b^2)/(x+1/2*b/c)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+1/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+
1/4*(4*a*c-b^2)/c)^(3/2)*x+1/2/c/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b+3/2/d^2/(4*a*c-b^
2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*x*a-3/8/c/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/
2)*x*b^2+3/4/c/d^2/(4*a*c-b^2)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*b*a-3/16/c^2/d^2/(4*a*c-b^2)*((x+1/2*
b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*b^3+3/2/c^(1/2)/d^2/(4*a*c-b^2)*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*
(4*a*c-b^2)/c)^(1/2))*a^2-3/4/c^(3/2)/d^2/(4*a*c-b^2)*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/
c)^(1/2))*b^2*a+3/32/c^(5/2)/d^2/(4*a*c-b^2)*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2))
*b^4

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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)^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 [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 )}^2} \,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)^2,x)

[Out]

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

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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^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx + \int \frac {b x \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx + \int \frac {c x^{2} \sqrt {a + b x + c x^{2}}}{b^{2} + 4 b c x + 4 c^{2} x^{2}}\, dx}{d^{2}} \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)**2,x)

[Out]

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

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