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

Optimal. Leaf size=115 \[ -\frac {3 \sqrt {b^2-4 a c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{32 c^{5/2} d^3}+\frac {3 \sqrt {a+b x+c x^2}}{16 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2} \]

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Rubi [A]  time = 0.07, antiderivative size = 115, 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, 685, 688, 205} \begin {gather*} -\frac {3 \sqrt {b^2-4 a c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{32 c^{5/2} d^3}+\frac {3 \sqrt {a+b x+c x^2}}{16 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

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)^3} \, dx &=-\frac {\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2}+\frac {3 \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx}{8 c d^2}\\ &=\frac {3 \sqrt {a+b x+c x^2}}{16 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{32 c^2 d^2}\\ &=\frac {3 \sqrt {a+b x+c x^2}}{16 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \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 )}{8 c d^2}\\ &=\frac {3 \sqrt {a+b x+c x^2}}{16 c^2 d^3}-\frac {\left (a+b x+c x^2\right )^{3/2}}{4 c d^3 (b+2 c x)^2}-\frac {3 \sqrt {b^2-4 a c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{32 c^{5/2} d^3}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 62, normalized size = 0.54 \begin {gather*} \frac {2 (a+x (b+c x))^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {4 c (a+x (b+c x))}{4 a c-b^2}\right )}{5 d^3 \left (b^2-4 a c\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + x*(b + c*x))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/(5*(b^2 - 4
*a*c)^2*d^3)

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IntegrateAlgebraic [A]  time = 1.09, size = 145, normalized size = 1.26 \begin {gather*} \frac {3 \sqrt {b^2-4 a c} \tan ^{-1}\left (-\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+\frac {2 c x}{\sqrt {b^2-4 a c}}+\frac {b}{\sqrt {b^2-4 a c}}\right )}{16 c^{5/2} d^3}+\frac {\sqrt {a+b x+c x^2} \left (-4 a c+3 b^2+8 b c x+8 c^2 x^2\right )}{16 c^2 d^3 (b+2 c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 320, normalized size = 2.78 \begin {gather*} \left [\frac {3 \, {\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\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 (8 \, c^{2} x^{2} + 8 \, b c x + 3 \, b^{2} - 4 \, a c\right )} \sqrt {c x^{2} + b x + a}}{64 \, {\left (4 \, c^{4} d^{3} x^{2} + 4 \, b c^{3} d^{3} x + b^{2} c^{2} d^{3}\right )}}, \frac {3 \, {\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\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 (8 \, c^{2} x^{2} + 8 \, b c x + 3 \, b^{2} - 4 \, a c\right )} \sqrt {c x^{2} + b x + a}}{32 \, {\left (4 \, c^{4} d^{3} x^{2} + 4 \, b c^{3} d^{3} x + b^{2} c^{2} d^{3}\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)^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.36, size = 387, normalized size = 3.37 \begin {gather*} -\frac {3 \, {\left (b^{2} - 4 \, a c\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 )}{16 \, \sqrt {b^{2} c - 4 \, a c^{2}} c^{2} d^{3}} + \frac {\sqrt {c x^{2} + b x + a}}{8 \, c^{2} d^{3}} - \frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} b^{2} c^{\frac {3}{2}} - 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} a c^{\frac {5}{2}} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} b^{3} c - 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} a b c^{2} + {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b^{4} \sqrt {c} - 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} a b^{2} c^{\frac {3}{2}} - 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} a^{2} c^{\frac {5}{2}} + a b^{3} c - 4 \, a^{2} b c^{2}}{16 \, {\left (2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} c^{\frac {3}{2}} + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b c + b^{2} \sqrt {c} - 2 \, a c^{\frac {3}{2}}\right )}^{2} c^{\frac {3}{2}} d^{3}} \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)^3,x, algorithm="giac")

[Out]

-3/16*(b^2 - 4*a*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)*c^2*d^3) + 1/8*sqrt(c*x^2 + b*x + a)/(c^2*d^3) - 1/16*(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))
^3*b^2*c^(3/2) - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*c^(5/2) + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b
^3*c - 12*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b*c^2 + (sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^4*sqrt(c) - 2*
(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b^2*c^(3/2) - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^2*c^(5/2) + a*b^3*
c - 4*a^2*b*c^2)/((2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c^(3/2) + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b*c
+ b^2*sqrt(c) - 2*a*c^(3/2))^2*c^(3/2)*d^3)

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maple [B]  time = 0.06, size = 562, normalized size = 4.89 \begin {gather*} -\frac {3 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 )}{2 \left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}\, c \,d^{3}}+\frac {3 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 )}{4 \left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{2} d^{3}}-\frac {3 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 )}{32 \left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}\, c^{3} d^{3}}+\frac {3 \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, a}{8 \left (4 a c -b^{2}\right ) c \,d^{3}}-\frac {3 \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}\, b^{2}}{32 \left (4 a c -b^{2}\right ) c^{2} d^{3}}+\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{4 \left (4 a c -b^{2}\right ) c \,d^{3}}-\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{4 \left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2} c^{2} d^{3}} \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)^3,x)

[Out]

-1/4/d^3/c^2/(4*a*c-b^2)/(x+1/2*b/c)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+1/4/d^3/c/(4*a*c-b^2)*((x+1/2
*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)+3/8/d^3/c/(4*a*c-b^2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*a-3/32/d^3/c^
2/(4*a*c-b^2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2)*b^2-3/2/d^3/c/(4*a*c-b^2)/((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+3/4/d^3/c^2/
(4*a*c-b^2)/((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^2-3/32/d^3/c^3/(4*a*c-b^2)/((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 \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)^3,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 )}^3} \,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)^3,x)

[Out]

int((a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^3, 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^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {b x \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx + \int \frac {c x^{2} \sqrt {a + b x + c x^{2}}}{b^{3} + 6 b^{2} c x + 12 b c^{2} x^{2} + 8 c^{3} x^{3}}\, dx}{d^{3}} \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)**3,x)

[Out]

(Integral(a*sqrt(a + b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(b*x*sqrt(
a + b*x + c*x**2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x) + Integral(c*x**2*sqrt(a + b*x + c*x*
*2)/(b**3 + 6*b**2*c*x + 12*b*c**2*x**2 + 8*c**3*x**3), x))/d**3

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