3.2.17 \(\int \frac {d+e x+f x^2}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=131 \[ \frac {2 \left (c \left (2 a e-b \left (\frac {a f}{c}+d\right )\right )-x \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 (b+2 c x) \left (4 a f+\frac {b^2 f}{c}-4 b e+8 c d\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1660, 12, 613} \begin {gather*} \frac {2 \left (c \left (2 a e-b \left (\frac {a f}{c}+d\right )\right )-x \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {2 (b+2 c x) \left (4 a f+\frac {b^2 f}{c}-4 b e+8 c d\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(2*(c*(2*a*e - b*(d + (a*f)/c)) - (2*c^2*d - b*c*e + b^2*f - 2*a*c*f)*x))/(3*c*(b^2 - 4*a*c)*(a + b*x + c*x^2)
^(3/2)) + (2*(8*c*d - 4*b*e + 4*a*f + (b^2*f)/c)*(b + 2*c*x))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps

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

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Mathematica [A]  time = 0.38, size = 147, normalized size = 1.12 \begin {gather*} \frac {8 b \left (2 a^2 f+3 a c \left (d-e x+f x^2\right )-2 c^2 x^2 (e x-3 d)\right )+16 c \left (-a^2 e+a c x \left (3 d+f x^2\right )+2 c^2 d x^3\right )-4 b^2 \left (a (e-6 f x)-c x \left (3 d-6 e x+f x^2\right )\right )-2 b^3 (d+3 x (e-f x))}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*x^2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*b^3*(d + 3*x*(e - f*x)) + 16*c*(-(a^2*e) + 2*c^2*d*x^3 + a*c*x*(3*d + f*x^2)) - 4*b^2*(a*(e - 6*f*x) - c*x
*(3*d - 6*e*x + f*x^2)) + 8*b*(2*a^2*f - 2*c^2*x^2*(-3*d + e*x) + 3*a*c*(d - e*x + f*x^2)))/(3*(b^2 - 4*a*c)^2
*(a + x*(b + c*x))^(3/2))

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IntegrateAlgebraic [A]  time = 1.17, size = 176, normalized size = 1.34 \begin {gather*} -\frac {2 \left (-8 a^2 b f+8 a^2 c e+2 a b^2 e-12 a b^2 f x-12 a b c d+12 a b c e x-12 a b c f x^2-24 a c^2 d x-8 a c^2 f x^3+b^3 d+3 b^3 e x-3 b^3 f x^2-6 b^2 c d x+12 b^2 c e x^2-2 b^2 c f x^3-24 b c^2 d x^2+8 b c^2 e x^3-16 c^3 d x^3\right )}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x + f*x^2)/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*(b^3*d - 12*a*b*c*d + 2*a*b^2*e + 8*a^2*c*e - 8*a^2*b*f - 6*b^2*c*d*x - 24*a*c^2*d*x + 3*b^3*e*x + 12*a*b*
c*e*x - 12*a*b^2*f*x - 24*b*c^2*d*x^2 + 12*b^2*c*e*x^2 - 3*b^3*f*x^2 - 12*a*b*c*f*x^2 - 16*c^3*d*x^3 + 8*b*c^2
*e*x^3 - 2*b^2*c*f*x^3 - 8*a*c^2*f*x^3))/(3*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)^(3/2))

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fricas [B]  time = 2.60, size = 286, normalized size = 2.18 \begin {gather*} \frac {2 \, {\left (8 \, a^{2} b f + 2 \, {\left (8 \, c^{3} d - 4 \, b c^{2} e + {\left (b^{2} c + 4 \, a c^{2}\right )} f\right )} x^{3} + 3 \, {\left (8 \, b c^{2} d - 4 \, b^{2} c e + {\left (b^{3} + 4 \, a b c\right )} f\right )} x^{2} - {\left (b^{3} - 12 \, a b c\right )} d - 2 \, {\left (a b^{2} + 4 \, a^{2} c\right )} e + 3 \, {\left (4 \, a b^{2} f + 2 \, {\left (b^{2} c + 4 \, a c^{2}\right )} d - {\left (b^{3} + 4 \, a b c\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

2/3*(8*a^2*b*f + 2*(8*c^3*d - 4*b*c^2*e + (b^2*c + 4*a*c^2)*f)*x^3 + 3*(8*b*c^2*d - 4*b^2*c*e + (b^3 + 4*a*b*c
)*f)*x^2 - (b^3 - 12*a*b*c)*d - 2*(a*b^2 + 4*a^2*c)*e + 3*(4*a*b^2*f + 2*(b^2*c + 4*a*c^2)*d - (b^3 + 4*a*b*c)
*e)*x)*sqrt(c*x^2 + b*x + a)/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 +
2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*
a^3*b*c^2)*x)

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giac [A]  time = 0.32, size = 240, normalized size = 1.83 \begin {gather*} \frac {2 \, {\left ({\left ({\left (\frac {2 \, {\left (8 \, c^{3} d + b^{2} c f + 4 \, a c^{2} f - 4 \, b c^{2} e\right )} x}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}} + \frac {3 \, {\left (8 \, b c^{2} d + b^{3} f + 4 \, a b c f - 4 \, b^{2} c e\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x + \frac {3 \, {\left (2 \, b^{2} c d + 8 \, a c^{2} d + 4 \, a b^{2} f - b^{3} e - 4 \, a b c e\right )}}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )} x - \frac {b^{3} d - 12 \, a b c d - 8 \, a^{2} b f + 2 \, a b^{2} e + 8 \, a^{2} c e}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e*x+d)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

2/3*(((2*(8*c^3*d + b^2*c*f + 4*a*c^2*f - 4*b*c^2*e)*x/(b^4 - 8*a*b^2*c + 16*a^2*c^2) + 3*(8*b*c^2*d + b^3*f +
4*a*b*c*f - 4*b^2*c*e)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x + 3*(2*b^2*c*d + 8*a*c^2*d + 4*a*b^2*f - b^3*e - 4*a
*b*c*e)/(b^4 - 8*a*b^2*c + 16*a^2*c^2))*x - (b^3*d - 12*a*b*c*d - 8*a^2*b*f + 2*a*b^2*e + 8*a^2*c*e)/(b^4 - 8*
a*b^2*c + 16*a^2*c^2))/(c*x^2 + b*x + a)^(3/2)

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maple [A]  time = 0.01, size = 185, normalized size = 1.41 \begin {gather*} \frac {\frac {16}{3} a \,c^{2} f \,x^{3}+\frac {4}{3} b^{2} c f \,x^{3}-\frac {16}{3} b \,c^{2} e \,x^{3}+\frac {32}{3} c^{3} d \,x^{3}+8 a b c f \,x^{2}+2 b^{3} f \,x^{2}-8 b^{2} c e \,x^{2}+16 b \,c^{2} d \,x^{2}+8 a \,b^{2} f x -8 a b c e x +16 a \,c^{2} d x -2 b^{3} e x +4 b^{2} c d x +\frac {16}{3} a^{2} b f -\frac {16}{3} a^{2} c e -\frac {4}{3} a \,b^{2} e +8 a b c d -\frac {2}{3} b^{3} d}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^2+e*x+d)/(c*x^2+b*x+a)^(5/2),x)

[Out]

2/3/(c*x^2+b*x+a)^(3/2)*(8*a*c^2*f*x^3+2*b^2*c*f*x^3-8*b*c^2*e*x^3+16*c^3*d*x^3+12*a*b*c*f*x^2+3*b^3*f*x^2-12*
b^2*c*e*x^2+24*b*c^2*d*x^2+12*a*b^2*f*x-12*a*b*c*e*x+24*a*c^2*d*x-3*b^3*e*x+6*b^2*c*d*x+8*a^2*b*f-8*a^2*c*e-2*
a*b^2*e+12*a*b*c*d-b^3*d)/(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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e*x+d)/(c*x^2+b*x+a)^(5/2),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 [B]  time = 3.71, size = 175, normalized size = 1.34 \begin {gather*} \frac {2\,\left (8\,f\,a^2\,b-8\,e\,a^2\,c+12\,f\,a\,b^2\,x-2\,e\,a\,b^2+12\,f\,a\,b\,c\,x^2-12\,e\,a\,b\,c\,x+12\,d\,a\,b\,c+8\,f\,a\,c^2\,x^3+24\,d\,a\,c^2\,x+3\,f\,b^3\,x^2-3\,e\,b^3\,x-d\,b^3+2\,f\,b^2\,c\,x^3-12\,e\,b^2\,c\,x^2+6\,d\,b^2\,c\,x-8\,e\,b\,c^2\,x^3+24\,d\,b\,c^2\,x^2+16\,d\,c^3\,x^3\right )}{3\,{\left (4\,a\,c-b^2\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x + f*x^2)/(a + b*x + c*x^2)^(5/2),x)

[Out]

(2*(16*c^3*d*x^3 - b^3*d + 3*b^3*f*x^2 - 2*a*b^2*e + 8*a^2*b*f - 8*a^2*c*e - 3*b^3*e*x + 24*a*c^2*d*x + 12*a*b
^2*f*x + 6*b^2*c*d*x + 24*b*c^2*d*x^2 - 12*b^2*c*e*x^2 + 8*a*c^2*f*x^3 - 8*b*c^2*e*x^3 + 2*b^2*c*f*x^3 + 12*a*
b*c*d - 12*a*b*c*e*x + 12*a*b*c*f*x^2))/(3*(4*a*c - b^2)^2*(a + b*x + c*x^2)^(3/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**2+e*x+d)/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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