### 3.115 $$\int \frac{A+B x+C x^2}{(a+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=67 $\frac{x (a C+2 A c)}{3 a^2 c \sqrt{a+c x^2}}-\frac{a B-x (A c-a C)}{3 a c \left (a+c x^2\right )^{3/2}}$

[Out]

-(a*B - (A*c - a*C)*x)/(3*a*c*(a + c*x^2)^(3/2)) + ((2*A*c + a*C)*x)/(3*a^2*c*Sqrt[a + c*x^2])

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Rubi [A]  time = 0.0422141, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.136, Rules used = {1814, 12, 191} $\frac{x (a C+2 A c)}{3 a^2 c \sqrt{a+c x^2}}-\frac{a B-x (A c-a C)}{3 a c \left (a+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(A + B*x + C*x^2)/(a + c*x^2)^(5/2),x]

[Out]

-(a*B - (A*c - a*C)*x)/(3*a*c*(a + c*x^2)^(3/2)) + ((2*A*c + a*C)*x)/(3*a^2*c*Sqrt[a + c*x^2])

Rule 1814

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

Rule 12

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{A+B x+C x^2}{\left (a+c x^2\right )^{5/2}} \, dx &=-\frac{a B-(A c-a C) x}{3 a c \left (a+c x^2\right )^{3/2}}-\frac{\int \frac{-2 A-\frac{a C}{c}}{\left (a+c x^2\right )^{3/2}} \, dx}{3 a}\\ &=-\frac{a B-(A c-a C) x}{3 a c \left (a+c x^2\right )^{3/2}}+\frac{(2 A c+a C) \int \frac{1}{\left (a+c x^2\right )^{3/2}} \, dx}{3 a c}\\ &=-\frac{a B-(A c-a C) x}{3 a c \left (a+c x^2\right )^{3/2}}+\frac{(2 A c+a C) x}{3 a^2 c \sqrt{a+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.033789, size = 50, normalized size = 0.75 $\frac{-a^2 B+a c x \left (3 A+C x^2\right )+2 A c^2 x^3}{3 a^2 c \left (a+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(A + B*x + C*x^2)/(a + c*x^2)^(5/2),x]

[Out]

(-(a^2*B) + 2*A*c^2*x^3 + a*c*x*(3*A + C*x^2))/(3*a^2*c*(a + c*x^2)^(3/2))

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Maple [A]  time = 0.052, size = 47, normalized size = 0.7 \begin{align*}{\frac{2\,A{x}^{3}{c}^{2}+Cac{x}^{3}+3\,Axac-B{a}^{2}}{3\,{a}^{2}c} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)/(c*x^2+a)^(5/2),x)

[Out]

1/3*(2*A*c^2*x^3+C*a*c*x^3+3*A*a*c*x-B*a^2)/(c*x^2+a)^(3/2)/a^2/c

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Maxima [A]  time = 0.979341, size = 112, normalized size = 1.67 \begin{align*} \frac{2 \, A x}{3 \, \sqrt{c x^{2} + a} a^{2}} + \frac{A x}{3 \,{\left (c x^{2} + a\right )}^{\frac{3}{2}} a} - \frac{C x}{3 \,{\left (c x^{2} + a\right )}^{\frac{3}{2}} c} + \frac{C x}{3 \, \sqrt{c x^{2} + a} a c} - \frac{B}{3 \,{\left (c x^{2} + a\right )}^{\frac{3}{2}} c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(c*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

2/3*A*x/(sqrt(c*x^2 + a)*a^2) + 1/3*A*x/((c*x^2 + a)^(3/2)*a) - 1/3*C*x/((c*x^2 + a)^(3/2)*c) + 1/3*C*x/(sqrt(
c*x^2 + a)*a*c) - 1/3*B/((c*x^2 + a)^(3/2)*c)

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Fricas [A]  time = 1.59628, size = 139, normalized size = 2.07 \begin{align*} \frac{{\left (3 \, A a c x +{\left (C a c + 2 \, A c^{2}\right )} x^{3} - B a^{2}\right )} \sqrt{c x^{2} + a}}{3 \,{\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(c*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

1/3*(3*A*a*c*x + (C*a*c + 2*A*c^2)*x^3 - B*a^2)*sqrt(c*x^2 + a)/(a^2*c^3*x^4 + 2*a^3*c^2*x^2 + a^4*c)

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Sympy [A]  time = 13.8247, size = 194, normalized size = 2.9 \begin{align*} A \left (\frac{3 a x}{3 a^{\frac{7}{2}} \sqrt{1 + \frac{c x^{2}}{a}} + 3 a^{\frac{5}{2}} c x^{2} \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{2 c x^{3}}{3 a^{\frac{7}{2}} \sqrt{1 + \frac{c x^{2}}{a}} + 3 a^{\frac{5}{2}} c x^{2} \sqrt{1 + \frac{c x^{2}}{a}}}\right ) + B \left (\begin{cases} - \frac{1}{3 a c \sqrt{a + c x^{2}} + 3 c^{2} x^{2} \sqrt{a + c x^{2}}} & \text{for}\: c \neq 0 \\\frac{x^{2}}{2 a^{\frac{5}{2}}} & \text{otherwise} \end{cases}\right ) + \frac{C x^{3}}{3 a^{\frac{5}{2}} \sqrt{1 + \frac{c x^{2}}{a}} + 3 a^{\frac{3}{2}} c x^{2} \sqrt{1 + \frac{c x^{2}}{a}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)/(c*x**2+a)**(5/2),x)

[Out]

A*(3*a*x/(3*a**(7/2)*sqrt(1 + c*x**2/a) + 3*a**(5/2)*c*x**2*sqrt(1 + c*x**2/a)) + 2*c*x**3/(3*a**(7/2)*sqrt(1
+ c*x**2/a) + 3*a**(5/2)*c*x**2*sqrt(1 + c*x**2/a))) + B*Piecewise((-1/(3*a*c*sqrt(a + c*x**2) + 3*c**2*x**2*s
qrt(a + c*x**2)), Ne(c, 0)), (x**2/(2*a**(5/2)), True)) + C*x**3/(3*a**(5/2)*sqrt(1 + c*x**2/a) + 3*a**(3/2)*c
*x**2*sqrt(1 + c*x**2/a))

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Giac [A]  time = 1.16968, size = 65, normalized size = 0.97 \begin{align*} \frac{x{\left (\frac{3 \, A}{a} + \frac{{\left (C a c + 2 \, A c^{2}\right )} x^{2}}{a^{2} c}\right )} - \frac{B}{c}}{3 \,{\left (c x^{2} + a\right )}^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)/(c*x^2+a)^(5/2),x, algorithm="giac")

[Out]

1/3*(x*(3*A/a + (C*a*c + 2*A*c^2)*x^2/(a^2*c)) - B/c)/(c*x^2 + a)^(3/2)