3.7.66 \(\int \frac {(a+b x^2)^2}{x^2 (c+d x^2)^{5/2}} \, dx\) [666]

Optimal. Leaf size=90 \[ -\frac {a^2}{c x \left (c+d x^2\right )^{3/2}}+\frac {x \left (2 a (b c-2 a d)+b^2 c x^2\right )}{3 c^2 \left (c+d x^2\right )^{3/2}}+\frac {4 a (b c-2 a d) x}{3 c^3 \sqrt {c+d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 90, normalized size of antiderivative = 1.00, 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 = {473, 386, 197} \begin {gather*} -\frac {a^2}{c x \left (c+d x^2\right )^{3/2}}+\frac {x \left (2 a (b c-2 a d)+b^2 c x^2\right )}{3 c^2 \left (c+d x^2\right )^{3/2}}+\frac {4 a x (b c-2 a d)}{3 c^3 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 197

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 386

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

Rule 473

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

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Mathematica [A]
time = 0.13, size = 76, normalized size = 0.84 \begin {gather*} \frac {b^2 c^2 x^4+2 a b c x^2 \left (3 c+2 d x^2\right )-a^2 \left (3 c^2+12 c d x^2+8 d^2 x^4\right )}{3 c^3 x \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 153, normalized size = 1.70

method	result	size
gosper	\(-\frac {8 a^{2} d^{2} x^{4}-4 a b c d \,x^{4}-b^{2} c^{2} x^{4}+12 a^{2} c d \,x^{2}-6 a b \,c^{2} x^{2}+3 a^{2} c^{2}}{3 x \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{3}}\)	\(78\)
trager	\(-\frac {8 a^{2} d^{2} x^{4}-4 a b c d \,x^{4}-b^{2} c^{2} x^{4}+12 a^{2} c d \,x^{2}-6 a b \,c^{2} x^{2}+3 a^{2} c^{2}}{3 x \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{3}}\)	\(78\)
risch	\(-\frac {a^{2} \sqrt {d \,x^{2}+c}}{c^{3} x}-\frac {\left (a d -b c \right ) \left (5 a d \,x^{2}+c \,x^{2} b +6 a c \right ) x \sqrt {d \,x^{2}+c}}{3 \left (d^{2} x^{4}+2 x^{2} d c +c^{2}\right ) c^{3}}\)	\(83\)
default	\(b^{2} \left (-\frac {x}{2 d \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {c \left (\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\right )}{2 d}\right )+2 a b \left (\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\right )+a^{2} \left (-\frac {1}{c x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {4 d \left (\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\right )}{c}\right )\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.35, size = 132, normalized size = 1.47 \begin {gather*} \frac {4 \, a b x}{3 \, \sqrt {d x^{2} + c} c^{2}} + \frac {2 \, a b x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} - \frac {b^{2} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {b^{2} x}{3 \, \sqrt {d x^{2} + c} c d} - \frac {8 \, a^{2} d x}{3 \, \sqrt {d x^{2} + c} c^{3}} - \frac {4 \, a^{2} d x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} - \frac {a^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.02, size = 92, normalized size = 1.02 \begin {gather*} \frac {{\left ({\left (b^{2} c^{2} + 4 \, a b c d - 8 \, a^{2} d^{2}\right )} x^{4} - 3 \, a^{2} c^{2} + 6 \, {\left (a b c^{2} - 2 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3 \, {\left (c^{3} d^{2} x^{5} + 2 \, c^{4} d x^{3} + c^{5} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.79, size = 117, normalized size = 1.30 \begin {gather*} \frac {x {\left (\frac {{\left (b^{2} c^{4} d + 4 \, a b c^{3} d^{2} - 5 \, a^{2} c^{2} d^{3}\right )} x^{2}}{c^{5} d} + \frac {6 \, {\left (a b c^{4} d - a^{2} c^{3} d^{2}\right )}}{c^{5} d}\right )}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {2 \, a^{2} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.35, size = 77, normalized size = 0.86 \begin {gather*} -\frac {3\,a^2\,c^2+12\,a^2\,c\,d\,x^2+8\,a^2\,d^2\,x^4-6\,a\,b\,c^2\,x^2-4\,a\,b\,c\,d\,x^4-b^2\,c^2\,x^4}{3\,c^3\,x\,{\left (d\,x^2+c\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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