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

Optimal. Leaf size=99 \[ -\frac{a^2 \sqrt{c+d x^2}}{5 c x^5}-\frac{\sqrt{c+d x^2} \left (15 b^2 c^2-4 a d (5 b c-2 a d)\right )}{15 c^3 x}-\frac{2 a \sqrt{c+d x^2} (5 b c-2 a d)}{15 c^2 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0717884, antiderivative size = 100, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {462, 453, 264} \[ -\frac{\sqrt{c+d x^2} \left (8 a^2 d^2-20 a b c d+15 b^2 c^2\right )}{15 c^3 x}-\frac{a^2 \sqrt{c+d x^2}}{5 c x^5}-\frac{2 a \sqrt{c+d x^2} (5 b c-2 a d)}{15 c^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Sqrt[c + d*x^2])/(5*c*x^5) - (2*a*(5*b*c - 2*a*d)*Sqrt[c + d*x^2])/(15*c^2*x^3) - ((15*b^2*c^2 - 20*a*b*
c*d + 8*a^2*d^2)*Sqrt[c + d*x^2])/(15*c^3*x)

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 264

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2}{x^6 \sqrt{c+d x^2}} \, dx &=-\frac{a^2 \sqrt{c+d x^2}}{5 c x^5}+\frac{\int \frac{2 a (5 b c-2 a d)+5 b^2 c x^2}{x^4 \sqrt{c+d x^2}} \, dx}{5 c}\\ &=-\frac{a^2 \sqrt{c+d x^2}}{5 c x^5}-\frac{2 a (5 b c-2 a d) \sqrt{c+d x^2}}{15 c^2 x^3}-\frac{1}{15} \left (-15 b^2+\frac{4 a d (5 b c-2 a d)}{c^2}\right ) \int \frac{1}{x^2 \sqrt{c+d x^2}} \, dx\\ &=-\frac{a^2 \sqrt{c+d x^2}}{5 c x^5}-\frac{2 a (5 b c-2 a d) \sqrt{c+d x^2}}{15 c^2 x^3}-\frac{\left (15 b^2-\frac{4 a d (5 b c-2 a d)}{c^2}\right ) \sqrt{c+d x^2}}{15 c x}\\ \end{align*}

Mathematica [A]  time = 0.0225581, size = 74, normalized size = 0.75 \[ -\frac{\sqrt{c+d x^2} \left (a^2 \left (3 c^2-4 c d x^2+8 d^2 x^4\right )+10 a b c x^2 \left (c-2 d x^2\right )+15 b^2 c^2 x^4\right )}{15 c^3 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d*x^2]*(15*b^2*c^2*x^4 + 10*a*b*c*x^2*(c - 2*d*x^2) + a^2*(3*c^2 - 4*c*d*x^2 + 8*d^2*x^4)))/(15*c^3
*x^5)

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Maple [A]  time = 0.004, size = 78, normalized size = 0.8 \begin{align*} -{\frac{8\,{a}^{2}{d}^{2}{x}^{4}-20\,abcd{x}^{4}+15\,{b}^{2}{c}^{2}{x}^{4}-4\,{a}^{2}cd{x}^{2}+10\,a{c}^{2}b{x}^{2}+3\,{a}^{2}{c}^{2}}{15\,{x}^{5}{c}^{3}}\sqrt{d{x}^{2}+c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(d*x^2+c)^(1/2)*(8*a^2*d^2*x^4-20*a*b*c*d*x^4+15*b^2*c^2*x^4-4*a^2*c*d*x^2+10*a*b*c^2*x^2+3*a^2*c^2)/x^5
/c^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36814, size = 163, normalized size = 1.65 \begin{align*} -\frac{{\left ({\left (15 \, b^{2} c^{2} - 20 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + 3 \, a^{2} c^{2} + 2 \,{\left (5 \, a b c^{2} - 2 \, a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{15 \, c^{3} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*((15*b^2*c^2 - 20*a*b*c*d + 8*a^2*d^2)*x^4 + 3*a^2*c^2 + 2*(5*a*b*c^2 - 2*a^2*c*d)*x^2)*sqrt(d*x^2 + c)/
(c^3*x^5)

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Sympy [B]  time = 3.16681, size = 391, normalized size = 3.95 \begin{align*} - \frac{3 a^{2} c^{4} d^{\frac{9}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{15 c^{5} d^{4} x^{4} + 30 c^{4} d^{5} x^{6} + 15 c^{3} d^{6} x^{8}} - \frac{2 a^{2} c^{3} d^{\frac{11}{2}} x^{2} \sqrt{\frac{c}{d x^{2}} + 1}}{15 c^{5} d^{4} x^{4} + 30 c^{4} d^{5} x^{6} + 15 c^{3} d^{6} x^{8}} - \frac{3 a^{2} c^{2} d^{\frac{13}{2}} x^{4} \sqrt{\frac{c}{d x^{2}} + 1}}{15 c^{5} d^{4} x^{4} + 30 c^{4} d^{5} x^{6} + 15 c^{3} d^{6} x^{8}} - \frac{12 a^{2} c d^{\frac{15}{2}} x^{6} \sqrt{\frac{c}{d x^{2}} + 1}}{15 c^{5} d^{4} x^{4} + 30 c^{4} d^{5} x^{6} + 15 c^{3} d^{6} x^{8}} - \frac{8 a^{2} d^{\frac{17}{2}} x^{8} \sqrt{\frac{c}{d x^{2}} + 1}}{15 c^{5} d^{4} x^{4} + 30 c^{4} d^{5} x^{6} + 15 c^{3} d^{6} x^{8}} - \frac{2 a b \sqrt{d} \sqrt{\frac{c}{d x^{2}} + 1}}{3 c x^{2}} + \frac{4 a b d^{\frac{3}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{3 c^{2}} - \frac{b^{2} \sqrt{d} \sqrt{\frac{c}{d x^{2}} + 1}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*a**2*c**4*d**(9/2)*sqrt(c/(d*x**2) + 1)/(15*c**5*d**4*x**4 + 30*c**4*d**5*x**6 + 15*c**3*d**6*x**8) - 2*a**
2*c**3*d**(11/2)*x**2*sqrt(c/(d*x**2) + 1)/(15*c**5*d**4*x**4 + 30*c**4*d**5*x**6 + 15*c**3*d**6*x**8) - 3*a**
2*c**2*d**(13/2)*x**4*sqrt(c/(d*x**2) + 1)/(15*c**5*d**4*x**4 + 30*c**4*d**5*x**6 + 15*c**3*d**6*x**8) - 12*a*
*2*c*d**(15/2)*x**6*sqrt(c/(d*x**2) + 1)/(15*c**5*d**4*x**4 + 30*c**4*d**5*x**6 + 15*c**3*d**6*x**8) - 8*a**2*
d**(17/2)*x**8*sqrt(c/(d*x**2) + 1)/(15*c**5*d**4*x**4 + 30*c**4*d**5*x**6 + 15*c**3*d**6*x**8) - 2*a*b*sqrt(d
)*sqrt(c/(d*x**2) + 1)/(3*c*x**2) + 4*a*b*d**(3/2)*sqrt(c/(d*x**2) + 1)/(3*c**2) - b**2*sqrt(d)*sqrt(c/(d*x**2
) + 1)/c

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Giac [B]  time = 1.15187, size = 421, normalized size = 4.25 \begin{align*} \frac{2 \,{\left (15 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{8} b^{2} \sqrt{d} - 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} b^{2} c \sqrt{d} + 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} a b d^{\frac{3}{2}} + 90 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b^{2} c^{2} \sqrt{d} - 140 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a b c d^{\frac{3}{2}} + 80 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a^{2} d^{\frac{5}{2}} - 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{2} c^{3} \sqrt{d} + 100 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b c^{2} d^{\frac{3}{2}} - 40 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} c d^{\frac{5}{2}} + 15 \, b^{2} c^{4} \sqrt{d} - 20 \, a b c^{3} d^{\frac{3}{2}} + 8 \, a^{2} c^{2} d^{\frac{5}{2}}\right )}}{15 \,{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*sqrt(d) - 60*(sqrt(d)*x - sqrt(d*x^2 + c))^6*b^2*c*sqrt(d) + 60*(
sqrt(d)*x - sqrt(d*x^2 + c))^6*a*b*d^(3/2) + 90*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^2*c^2*sqrt(d) - 140*(sqrt(d)
*x - sqrt(d*x^2 + c))^4*a*b*c*d^(3/2) + 80*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*d^(5/2) - 60*(sqrt(d)*x - sqrt(
d*x^2 + c))^2*b^2*c^3*sqrt(d) + 100*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^2*d^(3/2) - 40*(sqrt(d)*x - sqrt(d*x
^2 + c))^2*a^2*c*d^(5/2) + 15*b^2*c^4*sqrt(d) - 20*a*b*c^3*d^(3/2) + 8*a^2*c^2*d^(5/2))/((sqrt(d)*x - sqrt(d*x
^2 + c))^2 - c)^5