3.7.58 \(\int \frac {(a+b x^2)^2}{x^6 (c+d x^2)^{3/2}} \, dx\) [658]

Optimal. Leaf size=141 \[ -\frac {a^2}{5 c x^5 \sqrt {c+d x^2}}-\frac {2 a (5 b c-3 a d)}{15 c^2 x^3 \sqrt {c+d x^2}}-\frac {15 b^2 c^2-8 a d (5 b c-3 a d)}{15 c^3 x \sqrt {c+d x^2}}-\frac {2 d \left (15 b^2 c^2-8 a d (5 b c-3 a d)\right ) x}{15 c^4 \sqrt {c+d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 141, normalized size of antiderivative = 1.00, 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 = {473, 464, 277, 197} \begin {gather*} -\frac {a^2}{5 c x^5 \sqrt {c+d x^2}}-\frac {15 b^2-\frac {8 a d (5 b c-3 a d)}{c^2}}{15 c x \sqrt {c+d x^2}}-\frac {2 d x \left (15 b^2 c^2-8 a d (5 b c-3 a d)\right )}{15 c^4 \sqrt {c+d x^2}}-\frac {2 a (5 b c-3 a d)}{15 c^2 x^3 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^6*(c + d*x^2)^(3/2)),x]

[Out]

-1/5*a^2/(c*x^5*Sqrt[c + d*x^2]) - (2*a*(5*b*c - 3*a*d))/(15*c^2*x^3*Sqrt[c + d*x^2]) - (15*b^2 - (8*a*d*(5*b*
c - 3*a*d))/c^2)/(15*c*x*Sqrt[c + d*x^2]) - (2*d*(15*b^2*c^2 - 8*a*d*(5*b*c - 3*a*d))*x)/(15*c^4*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 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(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 \left (c+d x^2\right )^{3/2}} \, dx &=-\frac {a^2}{5 c x^5 \sqrt {c+d x^2}}+\frac {\int \frac {2 a (5 b c-3 a d)+5 b^2 c x^2}{x^4 \left (c+d x^2\right )^{3/2}} \, dx}{5 c}\\ &=-\frac {a^2}{5 c x^5 \sqrt {c+d x^2}}-\frac {2 a (5 b c-3 a d)}{15 c^2 x^3 \sqrt {c+d x^2}}-\frac {1}{15} \left (-15 b^2+\frac {8 a d (5 b c-3 a d)}{c^2}\right ) \int \frac {1}{x^2 \left (c+d x^2\right )^{3/2}} \, dx\\ &=-\frac {a^2}{5 c x^5 \sqrt {c+d x^2}}-\frac {2 a (5 b c-3 a d)}{15 c^2 x^3 \sqrt {c+d x^2}}-\frac {15 b^2-\frac {8 a d (5 b c-3 a d)}{c^2}}{15 c x \sqrt {c+d x^2}}-\frac {\left (2 d \left (15 b^2-\frac {8 a d (5 b c-3 a d)}{c^2}\right )\right ) \int \frac {1}{\left (c+d x^2\right )^{3/2}} \, dx}{15 c}\\ &=-\frac {a^2}{5 c x^5 \sqrt {c+d x^2}}-\frac {2 a (5 b c-3 a d)}{15 c^2 x^3 \sqrt {c+d x^2}}-\frac {15 b^2-\frac {8 a d (5 b c-3 a d)}{c^2}}{15 c x \sqrt {c+d x^2}}-\frac {2 d \left (15 b^2-\frac {8 a d (5 b c-3 a d)}{c^2}\right ) x}{15 c^2 \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 103, normalized size = 0.73 \begin {gather*} \frac {-15 b^2 c^2 x^4 \left (c+2 d x^2\right )-10 a b c x^2 \left (c^2-4 c d x^2-8 d^2 x^4\right )-3 a^2 \left (c^3-2 c^2 d x^2+8 c d^2 x^4+16 d^3 x^6\right )}{15 c^4 x^5 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^6*(c + d*x^2)^(3/2)),x]

[Out]

(-15*b^2*c^2*x^4*(c + 2*d*x^2) - 10*a*b*c*x^2*(c^2 - 4*c*d*x^2 - 8*d^2*x^4) - 3*a^2*(c^3 - 2*c^2*d*x^2 + 8*c*d
^2*x^4 + 16*d^3*x^6))/(15*c^4*x^5*Sqrt[c + d*x^2])

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Maple [A]
time = 0.11, size = 188, normalized size = 1.33

method	result	size
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (33 a^{2} d^{2} x^{4}-50 a b c d \,x^{4}+15 b^{2} c^{2} x^{4}-9 a^{2} c d \,x^{2}+10 a b \,c^{2} x^{2}+3 a^{2} c^{2}\right )}{15 c^{4} x^{5}}-\frac {x \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) d}{\sqrt {d \,x^{2}+c}\, c^{4}}\)	\(116\)
gosper	\(-\frac {48 a^{2} d^{3} x^{6}-80 a b c \,d^{2} x^{6}+30 b^{2} c^{2} d \,x^{6}+24 a^{2} c \,d^{2} x^{4}-40 a b \,c^{2} d \,x^{4}+15 b^{2} c^{3} x^{4}-6 a^{2} c^{2} d \,x^{2}+10 a b \,c^{3} x^{2}+3 a^{2} c^{3}}{15 x^{5} \sqrt {d \,x^{2}+c}\, c^{4}}\)	\(117\)
trager	\(-\frac {48 a^{2} d^{3} x^{6}-80 a b c \,d^{2} x^{6}+30 b^{2} c^{2} d \,x^{6}+24 a^{2} c \,d^{2} x^{4}-40 a b \,c^{2} d \,x^{4}+15 b^{2} c^{3} x^{4}-6 a^{2} c^{2} d \,x^{2}+10 a b \,c^{3} x^{2}+3 a^{2} c^{3}}{15 x^{5} \sqrt {d \,x^{2}+c}\, c^{4}}\)	\(117\)
default	\(a^{2} \left (-\frac {1}{5 c \,x^{5} \sqrt {d \,x^{2}+c}}-\frac {6 d \left (-\frac {1}{3 c \,x^{3} \sqrt {d \,x^{2}+c}}-\frac {4 d \left (-\frac {1}{c x \sqrt {d \,x^{2}+c}}-\frac {2 d x}{c^{2} \sqrt {d \,x^{2}+c}}\right )}{3 c}\right )}{5 c}\right )+2 a b \left (-\frac {1}{3 c \,x^{3} \sqrt {d \,x^{2}+c}}-\frac {4 d \left (-\frac {1}{c x \sqrt {d \,x^{2}+c}}-\frac {2 d x}{c^{2} \sqrt {d \,x^{2}+c}}\right )}{3 c}\right )+b^{2} \left (-\frac {1}{c x \sqrt {d \,x^{2}+c}}-\frac {2 d x}{c^{2} \sqrt {d \,x^{2}+c}}\right )\)	\(188\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^6/(d*x^2+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 184, normalized size = 1.30 \begin {gather*} -\frac {2 \, b^{2} d x}{\sqrt {d x^{2} + c} c^{2}} + \frac {16 \, a b d^{2} x}{3 \, \sqrt {d x^{2} + c} c^{3}} - \frac {16 \, a^{2} d^{3} x}{5 \, \sqrt {d x^{2} + c} c^{4}} - \frac {b^{2}}{\sqrt {d x^{2} + c} c x} + \frac {8 \, a b d}{3 \, \sqrt {d x^{2} + c} c^{2} x} - \frac {8 \, a^{2} d^{2}}{5 \, \sqrt {d x^{2} + c} c^{3} x} - \frac {2 \, a b}{3 \, \sqrt {d x^{2} + c} c x^{3}} + \frac {2 \, a^{2} d}{5 \, \sqrt {d x^{2} + c} c^{2} x^{3}} - \frac {a^{2}}{5 \, \sqrt {d x^{2} + c} 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)^(3/2),x, algorithm="maxima")

[Out]

-2*b^2*d*x/(sqrt(d*x^2 + c)*c^2) + 16/3*a*b*d^2*x/(sqrt(d*x^2 + c)*c^3) - 16/5*a^2*d^3*x/(sqrt(d*x^2 + c)*c^4)
- b^2/(sqrt(d*x^2 + c)*c*x) + 8/3*a*b*d/(sqrt(d*x^2 + c)*c^2*x) - 8/5*a^2*d^2/(sqrt(d*x^2 + c)*c^3*x) - 2/3*a
*b/(sqrt(d*x^2 + c)*c*x^3) + 2/5*a^2*d/(sqrt(d*x^2 + c)*c^2*x^3) - 1/5*a^2/(sqrt(d*x^2 + c)*c*x^5)

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Fricas [A]
time = 1.38, size = 121, normalized size = 0.86 \begin {gather*} -\frac {{\left (2 \, {\left (15 \, b^{2} c^{2} d - 40 \, a b c d^{2} + 24 \, a^{2} d^{3}\right )} x^{6} + 3 \, a^{2} c^{3} + {\left (15 \, b^{2} c^{3} - 40 \, a b c^{2} d + 24 \, a^{2} c d^{2}\right )} x^{4} + 2 \, {\left (5 \, a b c^{3} - 3 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{15 \, {\left (c^{4} d x^{7} + c^{5} x^{5}\right )}} \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)^(3/2),x, algorithm="fricas")

[Out]

-1/15*(2*(15*b^2*c^2*d - 40*a*b*c*d^2 + 24*a^2*d^3)*x^6 + 3*a^2*c^3 + (15*b^2*c^3 - 40*a*b*c^2*d + 24*a^2*c*d^
2)*x^4 + 2*(5*a*b*c^3 - 3*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c)/(c^4*d*x^7 + c^5*x^5)

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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^{6} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \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)**(3/2),x)

[Out]

Integral((a + b*x**2)**2/(x**6*(c + d*x**2)**(3/2)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (125) = 250\).
time = 0.90, size = 452, normalized size = 3.21 \begin {gather*} -\frac {{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x}{\sqrt {d x^{2} + c} c^{4}} + \frac {2 \, {\left (15 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} b^{2} c^{2} \sqrt {d} - 30 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a b c d^{\frac {3}{2}} + 15 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a^{2} d^{\frac {5}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} b^{2} c^{3} \sqrt {d} + 180 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a b c^{2} d^{\frac {3}{2}} - 90 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a^{2} c d^{\frac {5}{2}} + 90 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b^{2} c^{4} \sqrt {d} - 320 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{3} d^{\frac {3}{2}} + 240 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c^{2} d^{\frac {5}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c^{5} \sqrt {d} + 220 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{4} d^{\frac {3}{2}} - 150 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c^{3} d^{\frac {5}{2}} + 15 \, b^{2} c^{6} \sqrt {d} - 50 \, a b c^{5} d^{\frac {3}{2}} + 33 \, a^{2} c^{4} d^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{5} c^{3}} \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)^(3/2),x, algorithm="giac")

[Out]

-(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x/(sqrt(d*x^2 + c)*c^4) + 2/15*(15*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*c^
2*sqrt(d) - 30*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a*b*c*d^(3/2) + 15*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a^2*d^(5/2)
- 60*(sqrt(d)*x - sqrt(d*x^2 + c))^6*b^2*c^3*sqrt(d) + 180*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a*b*c^2*d^(3/2) - 9
0*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a^2*c*d^(5/2) + 90*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^2*c^4*sqrt(d) - 320*(sq
rt(d)*x - sqrt(d*x^2 + c))^4*a*b*c^3*d^(3/2) + 240*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*c^2*d^(5/2) - 60*(sqrt(
d)*x - sqrt(d*x^2 + c))^2*b^2*c^5*sqrt(d) + 220*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^4*d^(3/2) - 150*(sqrt(d)
*x - sqrt(d*x^2 + c))^2*a^2*c^3*d^(5/2) + 15*b^2*c^6*sqrt(d) - 50*a*b*c^5*d^(3/2) + 33*a^2*c^4*d^(5/2))/(((sqr
t(d)*x - sqrt(d*x^2 + c))^2 - c)^5*c^3)

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Mupad [B]
time = 0.48, size = 116, normalized size = 0.82 \begin {gather*} -\frac {3\,a^2\,c^3-6\,a^2\,c^2\,d\,x^2+24\,a^2\,c\,d^2\,x^4+48\,a^2\,d^3\,x^6+10\,a\,b\,c^3\,x^2-40\,a\,b\,c^2\,d\,x^4-80\,a\,b\,c\,d^2\,x^6+15\,b^2\,c^3\,x^4+30\,b^2\,c^2\,d\,x^6}{15\,c^4\,x^5\,\sqrt {d\,x^2+c}} \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)^(3/2)),x)

[Out]

-(3*a^2*c^3 + 15*b^2*c^3*x^4 + 48*a^2*d^3*x^6 - 6*a^2*c^2*d*x^2 + 24*a^2*c*d^2*x^4 + 30*b^2*c^2*d*x^6 + 10*a*b
*c^3*x^2 - 40*a*b*c^2*d*x^4 - 80*a*b*c*d^2*x^6)/(15*c^4*x^5*(c + d*x^2)^(1/2))

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