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

Optimal. Leaf size=72 \[ \frac {2 b (b c-a d)}{d^3 \sqrt {c+d x^2}}-\frac {(b c-a d)^2}{3 d^3 \left (c+d x^2\right )^{3/2}}+\frac {b^2 \sqrt {c+d x^2}}{d^3} \]

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Rubi [A]  time = 0.06, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {444, 43} \begin {gather*} \frac {2 b (b c-a d)}{d^3 \sqrt {c+d x^2}}-\frac {(b c-a d)^2}{3 d^3 \left (c+d x^2\right )^{3/2}}+\frac {b^2 \sqrt {c+d x^2}}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 444

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 67, normalized size = 0.93 \begin {gather*} \frac {-a^2 d^2-2 a b d \left (2 c+3 d x^2\right )+b^2 \left (8 c^2+12 c d x^2+3 d^2 x^4\right )}{3 d^3 \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.05, size = 72, normalized size = 1.00 \begin {gather*} \frac {-a^2 d^2-4 a b c d-6 a b d^2 x^2+8 b^2 c^2+12 b^2 c d x^2+3 b^2 d^2 x^4}{3 d^3 \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.12, size = 91, normalized size = 1.26 \begin {gather*} \frac {{\left (3 \, b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 4 \, a b c d - a^{2} d^{2} + 6 \, {\left (2 \, b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3 \, {\left (d^{5} x^{4} + 2 \, c d^{4} x^{2} + c^{2} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.46, size = 79, normalized size = 1.10 \begin {gather*} \frac {\sqrt {d x^{2} + c} b^{2}}{d^{3}} + \frac {6 \, {\left (d x^{2} + c\right )} b^{2} c - b^{2} c^{2} - 6 \, {\left (d x^{2} + c\right )} a b d + 2 \, a b c d - a^{2} d^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 68, normalized size = 0.94 \begin {gather*} -\frac {-3 b^{2} x^{4} d^{2}+6 a b \,d^{2} x^{2}-12 b^{2} c d \,x^{2}+a^{2} d^{2}+4 a b c d -8 b^{2} c^{2}}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-3*b^2*d^2*x^4+6*a*b*d^2*x^2-12*b^2*c*d*x^2+a^2*d^2+4*a*b*c*d-8*b^2*c^2)/(d*x^2+c)^(3/2)/d^3

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maxima [A]  time = 0.92, size = 114, normalized size = 1.58 \begin {gather*} \frac {b^{2} x^{4}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {4 \, b^{2} c x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}} - \frac {2 \, a b x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {8 \, b^{2} c^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{3}} - \frac {4 \, a b c}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}} - \frac {a^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.63, size = 76, normalized size = 1.06 \begin {gather*} \frac {3\,b^2\,{\left (d\,x^2+c\right )}^2-a^2\,d^2-b^2\,c^2+6\,b^2\,c\,\left (d\,x^2+c\right )-6\,a\,b\,d\,\left (d\,x^2+c\right )+2\,a\,b\,c\,d}{3\,d^3\,{\left (d\,x^2+c\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.38, size = 303, normalized size = 4.21 \begin {gather*} \begin {cases} - \frac {a^{2} d^{2}}{3 c d^{3} \sqrt {c + d x^{2}} + 3 d^{4} x^{2} \sqrt {c + d x^{2}}} - \frac {4 a b c d}{3 c d^{3} \sqrt {c + d x^{2}} + 3 d^{4} x^{2} \sqrt {c + d x^{2}}} - \frac {6 a b d^{2} x^{2}}{3 c d^{3} \sqrt {c + d x^{2}} + 3 d^{4} x^{2} \sqrt {c + d x^{2}}} + \frac {8 b^{2} c^{2}}{3 c d^{3} \sqrt {c + d x^{2}} + 3 d^{4} x^{2} \sqrt {c + d x^{2}}} + \frac {12 b^{2} c d x^{2}}{3 c d^{3} \sqrt {c + d x^{2}} + 3 d^{4} x^{2} \sqrt {c + d x^{2}}} + \frac {3 b^{2} d^{2} x^{4}}{3 c d^{3} \sqrt {c + d x^{2}} + 3 d^{4} x^{2} \sqrt {c + d x^{2}}} & \text {for}\: d \neq 0 \\\frac {\frac {a^{2} x^{2}}{2} + \frac {a b x^{4}}{2} + \frac {b^{2} x^{6}}{6}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*d**2/(3*c*d**3*sqrt(c + d*x**2) + 3*d**4*x**2*sqrt(c + d*x**2)) - 4*a*b*c*d/(3*c*d**3*sqrt(c
+ d*x**2) + 3*d**4*x**2*sqrt(c + d*x**2)) - 6*a*b*d**2*x**2/(3*c*d**3*sqrt(c + d*x**2) + 3*d**4*x**2*sqrt(c +
d*x**2)) + 8*b**2*c**2/(3*c*d**3*sqrt(c + d*x**2) + 3*d**4*x**2*sqrt(c + d*x**2)) + 12*b**2*c*d*x**2/(3*c*d**3
*sqrt(c + d*x**2) + 3*d**4*x**2*sqrt(c + d*x**2)) + 3*b**2*d**2*x**4/(3*c*d**3*sqrt(c + d*x**2) + 3*d**4*x**2*
sqrt(c + d*x**2)), Ne(d, 0)), ((a**2*x**2/2 + a*b*x**4/2 + b**2*x**6/6)/c**(5/2), True))

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