3.7.46 \(\int \frac {(a+b x^2)^2}{x^6 \sqrt {c+d x^2}} \, dx\) [646]

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

[Out]

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

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Rubi [A]
time = 0.05, 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 = {473, 464, 270} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(a^2*Sqrt[c + d*x^2])/(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 270

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]

Rule 464

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 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^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*}

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Mathematica [A]
time = 0.12, size = 74, normalized size = 0.75 \begin {gather*} -\frac {\sqrt {c+d x^2} \left (15 b^2 c^2 x^4+10 a b c x^2 \left (c-2 d x^2\right )+a^2 \left (3 c^2-4 c d x^2+8 d^2 x^4\right )\right )}{15 c^3 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(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)))/(c
^3*x^5)

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Maple [A]
time = 0.09, size = 126, normalized size = 1.27

method	result	size
gosper	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (8 a^{2} d^{2} x^{4}-20 a b c d \,x^{4}+15 b^{2} c^{2} x^{4}-4 a^{2} c d \,x^{2}+10 a b \,c^{2} x^{2}+3 a^{2} c^{2}\right )}{15 x^{5} c^{3}}\)	\(78\)
trager	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (8 a^{2} d^{2} x^{4}-20 a b c d \,x^{4}+15 b^{2} c^{2} x^{4}-4 a^{2} c d \,x^{2}+10 a b \,c^{2} x^{2}+3 a^{2} c^{2}\right )}{15 x^{5} c^{3}}\)	\(78\)
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (8 a^{2} d^{2} x^{4}-20 a b c d \,x^{4}+15 b^{2} c^{2} x^{4}-4 a^{2} c d \,x^{2}+10 a b \,c^{2} x^{2}+3 a^{2} c^{2}\right )}{15 x^{5} c^{3}}\)	\(78\)
default	\(a^{2} \left (-\frac {\sqrt {d \,x^{2}+c}}{5 c \,x^{5}}-\frac {4 d \left (-\frac {\sqrt {d \,x^{2}+c}}{3 c \,x^{3}}+\frac {2 d \sqrt {d \,x^{2}+c}}{3 c^{2} x}\right )}{5 c}\right )+2 a b \left (-\frac {\sqrt {d \,x^{2}+c}}{3 c \,x^{3}}+\frac {2 d \sqrt {d \,x^{2}+c}}{3 c^{2} x}\right )-\frac {b^{2} \sqrt {d \,x^{2}+c}}{c x}\)	\(126\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.33, size = 124, normalized size = 1.25 \begin {gather*} -\frac {\sqrt {d x^{2} + c} b^{2}}{c x} + \frac {4 \, \sqrt {d x^{2} + c} a b d}{3 \, c^{2} x} - \frac {8 \, \sqrt {d x^{2} + c} a^{2} d^{2}}{15 \, c^{3} x} - \frac {2 \, \sqrt {d x^{2} + c} a b}{3 \, c x^{3}} + \frac {4 \, \sqrt {d x^{2} + c} a^{2} d}{15 \, c^{2} x^{3}} - \frac {\sqrt {d x^{2} + c} a^{2}}{5 \, c x^{5}} \end {gather*}

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]

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

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Fricas [A]
time = 1.35, size = 73, normalized size = 0.74 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (92) = 184\).
time = 1.94, size = 391, normalized size = 3.95 \begin {gather*} - \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (87) = 174\).
time = 1.03, size = 312, normalized size = 3.15 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.39, size = 77, normalized size = 0.78 \begin {gather*} -\frac {\sqrt {d\,x^2+c}\,\left (3\,a^2\,c^2-4\,a^2\,c\,d\,x^2+8\,a^2\,d^2\,x^4+10\,a\,b\,c^2\,x^2-20\,a\,b\,c\,d\,x^4+15\,b^2\,c^2\,x^4\right )}{15\,c^3\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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