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

Optimal. Leaf size=61 \[ -\frac {2 a^2 c}{3 x^{3/2}}+\frac {2}{5} b x^{5/2} (2 a d+b c)+2 a \sqrt {x} (a d+2 b c)+\frac {2}{9} b^2 d x^{9/2} \]

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Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {448} \begin {gather*} -\frac {2 a^2 c}{3 x^{3/2}}+\frac {2}{5} b x^{5/2} (2 a d+b c)+2 a \sqrt {x} (a d+2 b c)+\frac {2}{9} b^2 d x^{9/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*c)/(3*x^(3/2)) + 2*a*(2*b*c + a*d)*Sqrt[x] + (2*b*(b*c + 2*a*d)*x^(5/2))/5 + (2*b^2*d*x^(9/2))/9

Rule 448

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )}{x^{5/2}} \, dx &=\int \left (\frac {a^2 c}{x^{5/2}}+\frac {a (2 b c+a d)}{\sqrt {x}}+b (b c+2 a d) x^{3/2}+b^2 d x^{7/2}\right ) \, dx\\ &=-\frac {2 a^2 c}{3 x^{3/2}}+2 a (2 b c+a d) \sqrt {x}+\frac {2}{5} b (b c+2 a d) x^{5/2}+\frac {2}{9} b^2 d x^{9/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 57, normalized size = 0.93 \begin {gather*} \frac {-30 a^2 \left (c-3 d x^2\right )+36 a b x^2 \left (5 c+d x^2\right )+2 b^2 x^4 \left (9 c+5 d x^2\right )}{45 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-30*a^2*(c - 3*d*x^2) + 36*a*b*x^2*(5*c + d*x^2) + 2*b^2*x^4*(9*c + 5*d*x^2))/(45*x^(3/2))

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IntegrateAlgebraic [A]  time = 0.04, size = 59, normalized size = 0.97 \begin {gather*} \frac {2 \left (-15 a^2 c+45 a^2 d x^2+90 a b c x^2+18 a b d x^4+9 b^2 c x^4+5 b^2 d x^6\right )}{45 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-15*a^2*c + 90*a*b*c*x^2 + 45*a^2*d*x^2 + 9*b^2*c*x^4 + 18*a*b*d*x^4 + 5*b^2*d*x^6))/(45*x^(3/2))

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fricas [A]  time = 0.82, size = 53, normalized size = 0.87 \begin {gather*} \frac {2 \, {\left (5 \, b^{2} d x^{6} + 9 \, {\left (b^{2} c + 2 \, a b d\right )} x^{4} - 15 \, a^{2} c + 45 \, {\left (2 \, a b c + a^{2} d\right )} x^{2}\right )}}{45 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5*b^2*d*x^6 + 9*(b^2*c + 2*a*b*d)*x^4 - 15*a^2*c + 45*(2*a*b*c + a^2*d)*x^2)/x^(3/2)

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giac [A]  time = 0.33, size = 53, normalized size = 0.87 \begin {gather*} \frac {2}{9} \, b^{2} d x^{\frac {9}{2}} + \frac {2}{5} \, b^{2} c x^{\frac {5}{2}} + \frac {4}{5} \, a b d x^{\frac {5}{2}} + 4 \, a b c \sqrt {x} + 2 \, a^{2} d \sqrt {x} - \frac {2 \, a^{2} c}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 56, normalized size = 0.92 \begin {gather*} -\frac {2 \left (-5 b^{2} d \,x^{6}-18 a b d \,x^{4}-9 b^{2} c \,x^{4}-45 a^{2} d \,x^{2}-90 a b c \,x^{2}+15 a^{2} c \right )}{45 x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45*(-5*b^2*d*x^6-18*a*b*d*x^4-9*b^2*c*x^4-45*a^2*d*x^2-90*a*b*c*x^2+15*a^2*c)/x^(3/2)

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maxima [A]  time = 1.10, size = 51, normalized size = 0.84 \begin {gather*} \frac {2}{9} \, b^{2} d x^{\frac {9}{2}} + \frac {2}{5} \, {\left (b^{2} c + 2 \, a b d\right )} x^{\frac {5}{2}} - \frac {2 \, a^{2} c}{3 \, x^{\frac {3}{2}}} + 2 \, {\left (2 \, a b c + a^{2} d\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 51, normalized size = 0.84 \begin {gather*} \sqrt {x}\,\left (2\,d\,a^2+4\,b\,c\,a\right )+x^{5/2}\,\left (\frac {2\,c\,b^2}{5}+\frac {4\,a\,d\,b}{5}\right )-\frac {2\,a^2\,c}{3\,x^{3/2}}+\frac {2\,b^2\,d\,x^{9/2}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.76, size = 76, normalized size = 1.25 \begin {gather*} - \frac {2 a^{2} c}{3 x^{\frac {3}{2}}} + 2 a^{2} d \sqrt {x} + 4 a b c \sqrt {x} + \frac {4 a b d x^{\frac {5}{2}}}{5} + \frac {2 b^{2} c x^{\frac {5}{2}}}{5} + \frac {2 b^{2} d x^{\frac {9}{2}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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