3.668 \(\int \frac {(a+b x^2)^2}{x^4 (c+d x^2)^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 130, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {462, 453, 192, 191} \[ -\frac {a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}+\frac {2 x \left (b^2 c^2-8 a d (b c-a d)\right )}{3 c^4 \sqrt {c+d x^2}}+\frac {x \left (b^2-\frac {8 a d (b c-a d)}{c^2}\right )}{3 c \left (c+d x^2\right )^{3/2}}-\frac {2 a (b c-a d)}{c^2 x \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^4*(c + d*x^2)^(5/2)),x]

[Out]

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

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]

Rule 192

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

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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{5/2}} \, dx &=-\frac {a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}+\frac {\int \frac {6 a (b c-a d)+3 b^2 c x^2}{x^2 \left (c+d x^2\right )^{5/2}} \, dx}{3 c}\\ &=-\frac {a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}-\frac {2 a (b c-a d)}{c^2 x \left (c+d x^2\right )^{3/2}}-\left (-b^2+\frac {8 a d (b c-a d)}{c^2}\right ) \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx\\ &=-\frac {a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}-\frac {2 a (b c-a d)}{c^2 x \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2-\frac {8 a d (b c-a d)}{c^2}\right ) x}{3 c \left (c+d x^2\right )^{3/2}}+\frac {\left (2 \left (b^2-\frac {8 a d (b c-a d)}{c^2}\right )\right ) \int \frac {1}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c}\\ &=-\frac {a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}-\frac {2 a (b c-a d)}{c^2 x \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2-\frac {8 a d (b c-a d)}{c^2}\right ) x}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 \left (b^2-\frac {8 a d (b c-a d)}{c^2}\right ) x}{3 c^2 \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 107, normalized size = 0.82 \[ \frac {a^2 \left (-c^3+6 c^2 d x^2+24 c d^2 x^4+16 d^3 x^6\right )-2 a b c x^2 \left (3 c^2+12 c d x^2+8 d^2 x^4\right )+b^2 c^2 x^4 \left (3 c+2 d x^2\right )}{3 c^4 x^3 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^4*(c + d*x^2)^(5/2)),x]

[Out]

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

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fricas [A]  time = 0.86, size = 130, normalized size = 0.99 \[ \frac {{\left (2 \, {\left (b^{2} c^{2} d - 8 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{6} - a^{2} c^{3} + 3 \, {\left (b^{2} c^{3} - 8 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} x^{4} - 6 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3 \, {\left (c^{4} d^{2} x^{7} + 2 \, c^{5} d x^{5} + c^{6} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^4/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/3*(2*(b^2*c^2*d - 8*a*b*c*d^2 + 8*a^2*d^3)*x^6 - a^2*c^3 + 3*(b^2*c^3 - 8*a*b*c^2*d + 8*a^2*c*d^2)*x^4 - 6*(
a*b*c^3 - a^2*c^2*d)*x^2)*sqrt(d*x^2 + c)/(c^4*d^2*x^7 + 2*c^5*d*x^5 + c^6*x^3)

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giac [B]  time = 0.45, size = 258, normalized size = 1.97 \[ \frac {x {\left (\frac {2 \, {\left (b^{2} c^{5} d^{2} - 5 \, a b c^{4} d^{3} + 4 \, a^{2} c^{3} d^{4}\right )} x^{2}}{c^{7} d} + \frac {3 \, {\left (b^{2} c^{6} d - 4 \, a b c^{5} d^{2} + 3 \, a^{2} c^{4} d^{3}\right )}}{c^{7} d}\right )}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {4 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c \sqrt {d} - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} d^{\frac {3}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{2} \sqrt {d} + 9 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c d^{\frac {3}{2}} + 3 \, a b c^{3} \sqrt {d} - 4 \, a^{2} c^{2} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^4/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/3*x*(2*(b^2*c^5*d^2 - 5*a*b*c^4*d^3 + 4*a^2*c^3*d^4)*x^2/(c^7*d) + 3*(b^2*c^6*d - 4*a*b*c^5*d^2 + 3*a^2*c^4*
d^3)/(c^7*d))/(d*x^2 + c)^(3/2) + 4/3*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c*sqrt(d) - 3*(sqrt(d)*x - sqrt(d
*x^2 + c))^4*a^2*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^2*sqrt(d) + 9*(sqrt(d)*x - sqrt(d*x^2 + c))
^2*a^2*c*d^(3/2) + 3*a*b*c^3*sqrt(d) - 4*a^2*c^2*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^3*c^3)

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maple [A]  time = 0.01, size = 116, normalized size = 0.89 \[ -\frac {-16 a^{2} d^{3} x^{6}+16 a b c \,d^{2} x^{6}-2 b^{2} c^{2} d \,x^{6}-24 a^{2} c \,d^{2} x^{4}+24 a b \,c^{2} d \,x^{4}-3 b^{2} c^{3} x^{4}-6 a^{2} c^{2} d \,x^{2}+6 a b \,c^{3} x^{2}+a^{2} c^{3}}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^4/(d*x^2+c)^(5/2),x)

[Out]

-1/3*(-16*a^2*d^3*x^6+16*a*b*c*d^2*x^6-2*b^2*c^2*d*x^6-24*a^2*c*d^2*x^4+24*a*b*c^2*d*x^4-3*b^2*c^3*x^4-6*a^2*c
^2*d*x^2+6*a*b*c^3*x^2+a^2*c^3)/(d*x^2+c)^(3/2)/x^3/c^4

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maxima [A]  time = 0.93, size = 175, normalized size = 1.34 \[ \frac {2 \, b^{2} x}{3 \, \sqrt {d x^{2} + c} c^{2}} + \frac {b^{2} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} - \frac {16 \, a b d x}{3 \, \sqrt {d x^{2} + c} c^{3}} - \frac {8 \, a b d x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} + \frac {16 \, a^{2} d^{2} x}{3 \, \sqrt {d x^{2} + c} c^{4}} + \frac {8 \, a^{2} d^{2} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{3}} - \frac {2 \, a b}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c x} + \frac {2 \, a^{2} d}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2} x} - \frac {a^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^4/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.73, size = 187, normalized size = 1.43 \[ \frac {b^2\,c^4\,x^2-a^2\,c^3\,d-16\,a^2\,d\,{\left (d\,x^2+c\right )}^3+2\,a\,b\,c^4+b^2\,c^3\,x^2\,\left (d\,x^2+c\right )+16\,a\,b\,c\,{\left (d\,x^2+c\right )}^3+6\,a\,b\,c^3\,\left (d\,x^2+c\right )-2\,b^2\,c^2\,x^2\,{\left (d\,x^2+c\right )}^2-24\,a\,b\,c^2\,{\left (d\,x^2+c\right )}^2+24\,a^2\,c\,d\,{\left (d\,x^2+c\right )}^2-6\,a^2\,c^2\,d\,\left (d\,x^2+c\right )}{{\left (d\,x^2+c\right )}^{3/2}\,\left (3\,c^5\,x-3\,c^4\,x\,\left (d\,x^2+c\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/(x^4*(c + d*x^2)^(5/2)),x)

[Out]

(b^2*c^4*x^2 - a^2*c^3*d - 16*a^2*d*(c + d*x^2)^3 + 2*a*b*c^4 + b^2*c^3*x^2*(c + d*x^2) + 16*a*b*c*(c + d*x^2)
^3 + 6*a*b*c^3*(c + d*x^2) - 2*b^2*c^2*x^2*(c + d*x^2)^2 - 24*a*b*c^2*(c + d*x^2)^2 + 24*a^2*c*d*(c + d*x^2)^2
- 6*a^2*c^2*d*(c + d*x^2))/((c + d*x^2)^(3/2)*(3*c^5*x - 3*c^4*x*(c + d*x^2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right )^{2}}{x^{4} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**4/(d*x**2+c)**(5/2),x)

[Out]

Integral((a + b*x**2)**2/(x**4*(c + d*x**2)**(5/2)), x)

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