3.1200 \(\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^6} \, dx\)

Optimal. Leaf size=79 \[ \frac {4 \left (a+b x+c x^2\right )^{3/2}}{15 d^6 \left (b^2-4 a c\right )^2 (b+2 c x)^3}+\frac {2 \left (a+b x+c x^2\right )^{3/2}}{5 d^6 \left (b^2-4 a c\right ) (b+2 c x)^5} \]

[Out]

2/5*(c*x^2+b*x+a)^(3/2)/(-4*a*c+b^2)/d^6/(2*c*x+b)^5+4/15*(c*x^2+b*x+a)^(3/2)/(-4*a*c+b^2)^2/d^6/(2*c*x+b)^3

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Rubi [A]  time = 0.03, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {693, 682} \[ \frac {4 \left (a+b x+c x^2\right )^{3/2}}{15 d^6 \left (b^2-4 a c\right )^2 (b+2 c x)^3}+\frac {2 \left (a+b x+c x^2\right )^{3/2}}{5 d^6 \left (b^2-4 a c\right ) (b+2 c x)^5} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x + c*x^2]/(b*d + 2*c*d*x)^6,x]

[Out]

(2*(a + b*x + c*x^2)^(3/2))/(5*(b^2 - 4*a*c)*d^6*(b + 2*c*x)^5) + (4*(a + b*x + c*x^2)^(3/2))/(15*(b^2 - 4*a*c
)^2*d^6*(b + 2*c*x)^3)

Rule 682

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] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*
a*c, 0] && EqQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0] && NeQ[p, -1]

Rule 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*
c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^6} \, dx &=\frac {2 \left (a+b x+c x^2\right )^{3/2}}{5 \left (b^2-4 a c\right ) d^6 (b+2 c x)^5}+\frac {2 \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^4} \, dx}{5 \left (b^2-4 a c\right ) d^2}\\ &=\frac {2 \left (a+b x+c x^2\right )^{3/2}}{5 \left (b^2-4 a c\right ) d^6 (b+2 c x)^5}+\frac {4 \left (a+b x+c x^2\right )^{3/2}}{15 \left (b^2-4 a c\right )^2 d^6 (b+2 c x)^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 62, normalized size = 0.78 \[ \frac {2 (a+x (b+c x))^{3/2} \left (4 c \left (2 c x^2-3 a\right )+5 b^2+8 b c x\right )}{15 d^6 \left (b^2-4 a c\right )^2 (b+2 c x)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x + c*x^2]/(b*d + 2*c*d*x)^6,x]

[Out]

(2*(a + x*(b + c*x))^(3/2)*(5*b^2 + 8*b*c*x + 4*c*(-3*a + 2*c*x^2)))/(15*(b^2 - 4*a*c)^2*d^6*(b + 2*c*x)^5)

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fricas [B]  time = 4.37, size = 274, normalized size = 3.47 \[ \frac {2 \, {\left (8 \, c^{3} x^{4} + 16 \, b c^{2} x^{3} + 5 \, a b^{2} - 12 \, a^{2} c + {\left (13 \, b^{2} c - 4 \, a c^{2}\right )} x^{2} + {\left (5 \, b^{3} - 4 \, a b c\right )} x\right )} \sqrt {c x^{2} + b x + a}}{15 \, {\left (32 \, {\left (b^{4} c^{5} - 8 \, a b^{2} c^{6} + 16 \, a^{2} c^{7}\right )} d^{6} x^{5} + 80 \, {\left (b^{5} c^{4} - 8 \, a b^{3} c^{5} + 16 \, a^{2} b c^{6}\right )} d^{6} x^{4} + 80 \, {\left (b^{6} c^{3} - 8 \, a b^{4} c^{4} + 16 \, a^{2} b^{2} c^{5}\right )} d^{6} x^{3} + 40 \, {\left (b^{7} c^{2} - 8 \, a b^{5} c^{3} + 16 \, a^{2} b^{3} c^{4}\right )} d^{6} x^{2} + 10 \, {\left (b^{8} c - 8 \, a b^{6} c^{2} + 16 \, a^{2} b^{4} c^{3}\right )} d^{6} x + {\left (b^{9} - 8 \, a b^{7} c + 16 \, a^{2} b^{5} c^{2}\right )} d^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^6,x, algorithm="fricas")

[Out]

2/15*(8*c^3*x^4 + 16*b*c^2*x^3 + 5*a*b^2 - 12*a^2*c + (13*b^2*c - 4*a*c^2)*x^2 + (5*b^3 - 4*a*b*c)*x)*sqrt(c*x
^2 + b*x + a)/(32*(b^4*c^5 - 8*a*b^2*c^6 + 16*a^2*c^7)*d^6*x^5 + 80*(b^5*c^4 - 8*a*b^3*c^5 + 16*a^2*b*c^6)*d^6
*x^4 + 80*(b^6*c^3 - 8*a*b^4*c^4 + 16*a^2*b^2*c^5)*d^6*x^3 + 40*(b^7*c^2 - 8*a*b^5*c^3 + 16*a^2*b^3*c^4)*d^6*x
^2 + 10*(b^8*c - 8*a*b^6*c^2 + 16*a^2*b^4*c^3)*d^6*x + (b^9 - 8*a*b^7*c + 16*a^2*b^5*c^2)*d^6)

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giac [B]  time = 0.45, size = 423, normalized size = 5.35 \[ \frac {60 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{6} c^{\frac {7}{2}} + 180 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} b c^{3} + 220 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} b^{2} c^{\frac {5}{2}} + 20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} a c^{\frac {7}{2}} + 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} b^{3} c^{2} + 40 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} a b c^{3} + 50 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} b^{4} c^{\frac {3}{2}} + 20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} a b^{2} c^{\frac {5}{2}} + 20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} a^{2} c^{\frac {7}{2}} + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b^{5} c + 20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} a^{2} b c^{3} + b^{6} \sqrt {c} - 2 \, a b^{4} c^{\frac {3}{2}} + 8 \, a^{2} b^{2} c^{\frac {5}{2}} - 4 \, a^{3} c^{\frac {7}{2}}}{30 \, {\left (2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} c + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b \sqrt {c} + b^{2} - 2 \, a c\right )}^{5} c^{2} d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^6,x, algorithm="giac")

[Out]

1/30*(60*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*c^(7/2) + 180*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*b*c^3 + 220
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*b^2*c^(5/2) + 20*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a*c^(7/2) + 140*
(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^3*c^2 + 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b*c^3 + 50*(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))^2*b^4*c^(3/2) + 20*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b^2*c^(5/2) + 20*(sqrt(
c)*x - sqrt(c*x^2 + b*x + a))^2*a^2*c^(7/2) + 10*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^5*c + 20*(sqrt(c)*x - s
qrt(c*x^2 + b*x + a))*a^2*b*c^3 + b^6*sqrt(c) - 2*a*b^4*c^(3/2) + 8*a^2*b^2*c^(5/2) - 4*a^3*c^(7/2))/((2*(sqrt
(c)*x - sqrt(c*x^2 + b*x + a))^2*c + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b*sqrt(c) + b^2 - 2*a*c)^5*c^2*d^6)

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maple [A]  time = 0.05, size = 70, normalized size = 0.89 \[ -\frac {2 \left (-8 c^{2} x^{2}-8 b c x +12 a c -5 b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{15 \left (2 c x +b \right )^{5} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(1/2)/(2*c*d*x+b*d)^6,x)

[Out]

-2/15*(-8*c^2*x^2-8*b*c*x+12*a*c-5*b^2)*(c*x^2+b*x+a)^(3/2)/(2*c*x+b)^5/d^6/(16*a^2*c^2-8*a*b^2*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)^(1/2)/(2*c*d*x+b*d)^6,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 [B]  time = 1.00, size = 589, normalized size = 7.46 \[ \frac {\left (\frac {b\,\left (\frac {4\,c^2\,\left (2\,b^2+4\,a\,c\right )}{15\,d^6\,{\left (4\,a\,c-b^2\right )}^2\,\left (32\,a\,c^3-8\,b^2\,c^2\right )}-\frac {8\,b^2\,c^2}{15\,d^6\,{\left (4\,a\,c-b^2\right )}^2\,\left (32\,a\,c^3-8\,b^2\,c^2\right )}\right )}{2\,c}-\frac {8\,a\,b\,c^2}{15\,d^6\,{\left (4\,a\,c-b^2\right )}^2\,\left (32\,a\,c^3-8\,b^2\,c^2\right )}\right )\,\sqrt {c\,x^2+b\,x+a}}{{\left (b+2\,c\,x\right )}^2}-\frac {\left (\frac {2\,a}{15\,d^6\,{\left (4\,a\,c-b^2\right )}^3}-\frac {b^2}{30\,c\,d^6\,{\left (4\,a\,c-b^2\right )}^3}\right )\,\sqrt {c\,x^2+b\,x+a}}{b+2\,c\,x}+\frac {\left (\frac {b\,\left (\frac {4\,c^2\,\left (2\,b^2+4\,a\,c\right )}{5\,d^6\,\left (4\,a\,c-b^2\right )\,\left (64\,a\,c^3-16\,b^2\,c^2\right )}-\frac {8\,b^2\,c^2}{5\,d^6\,\left (4\,a\,c-b^2\right )\,\left (64\,a\,c^3-16\,b^2\,c^2\right )}\right )}{2\,c}-\frac {8\,a\,b\,c^2}{5\,d^6\,\left (4\,a\,c-b^2\right )\,\left (64\,a\,c^3-16\,b^2\,c^2\right )}\right )\,\sqrt {c\,x^2+b\,x+a}}{{\left (b+2\,c\,x\right )}^4}-\frac {\left (\frac {8\,a\,c^2}{d^6\,\left (80\,a\,c^3-20\,b^2\,c^2\right )}-\frac {2\,b^2\,c}{d^6\,\left (80\,a\,c^3-20\,b^2\,c^2\right )}\right )\,\sqrt {c\,x^2+b\,x+a}}{{\left (b+2\,c\,x\right )}^5}-\frac {\left (\frac {8\,a\,c^2}{5\,d^6\,\left (4\,a\,c-b^2\right )\,\left (48\,a\,c^3-12\,b^2\,c^2\right )}-\frac {2\,b^2\,c}{5\,d^6\,\left (4\,a\,c-b^2\right )\,\left (48\,a\,c^3-12\,b^2\,c^2\right )}\right )\,\sqrt {c\,x^2+b\,x+a}}{{\left (b+2\,c\,x\right )}^3}+\frac {\sqrt {c\,x^2+b\,x+a}}{10\,c\,d^6\,{\left (4\,a\,c-b^2\right )}^2\,\left (b+2\,c\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(1/2)/(b*d + 2*c*d*x)^6,x)

[Out]

(((b*((4*c^2*(4*a*c + 2*b^2))/(15*d^6*(4*a*c - b^2)^2*(32*a*c^3 - 8*b^2*c^2)) - (8*b^2*c^2)/(15*d^6*(4*a*c - b
^2)^2*(32*a*c^3 - 8*b^2*c^2))))/(2*c) - (8*a*b*c^2)/(15*d^6*(4*a*c - b^2)^2*(32*a*c^3 - 8*b^2*c^2)))*(a + b*x
+ c*x^2)^(1/2))/(b + 2*c*x)^2 - (((2*a)/(15*d^6*(4*a*c - b^2)^3) - b^2/(30*c*d^6*(4*a*c - b^2)^3))*(a + b*x +
c*x^2)^(1/2))/(b + 2*c*x) + (((b*((4*c^2*(4*a*c + 2*b^2))/(5*d^6*(4*a*c - b^2)*(64*a*c^3 - 16*b^2*c^2)) - (8*b
^2*c^2)/(5*d^6*(4*a*c - b^2)*(64*a*c^3 - 16*b^2*c^2))))/(2*c) - (8*a*b*c^2)/(5*d^6*(4*a*c - b^2)*(64*a*c^3 - 1
6*b^2*c^2)))*(a + b*x + c*x^2)^(1/2))/(b + 2*c*x)^4 - (((8*a*c^2)/(d^6*(80*a*c^3 - 20*b^2*c^2)) - (2*b^2*c)/(d
^6*(80*a*c^3 - 20*b^2*c^2)))*(a + b*x + c*x^2)^(1/2))/(b + 2*c*x)^5 - (((8*a*c^2)/(5*d^6*(4*a*c - b^2)*(48*a*c
^3 - 12*b^2*c^2)) - (2*b^2*c)/(5*d^6*(4*a*c - b^2)*(48*a*c^3 - 12*b^2*c^2)))*(a + b*x + c*x^2)^(1/2))/(b + 2*c
*x)^3 + (a + b*x + c*x^2)^(1/2)/(10*c*d^6*(4*a*c - b^2)^2*(b + 2*c*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {a + b x + c x^{2}}}{b^{6} + 12 b^{5} c x + 60 b^{4} c^{2} x^{2} + 160 b^{3} c^{3} x^{3} + 240 b^{2} c^{4} x^{4} + 192 b c^{5} x^{5} + 64 c^{6} x^{6}}\, dx}{d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(1/2)/(2*c*d*x+b*d)**6,x)

[Out]

Integral(sqrt(a + b*x + c*x**2)/(b**6 + 12*b**5*c*x + 60*b**4*c**2*x**2 + 160*b**3*c**3*x**3 + 240*b**2*c**4*x
**4 + 192*b*c**5*x**5 + 64*c**6*x**6), x)/d**6

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