3.647 \(\int \frac{(a+b x^2)^2}{x^7 \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=151 \[ -\frac{\sqrt{c+d x^2} \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right )}{16 c^3 x^2}+\frac{d \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{16 c^{7/2}}-\frac{a^2 \sqrt{c+d x^2}}{6 c x^6}-\frac{a \sqrt{c+d x^2} (12 b c-5 a d)}{24 c^2 x^4} \]

[Out]

-(a^2*Sqrt[c + d*x^2])/(6*c*x^6) - (a*(12*b*c - 5*a*d)*Sqrt[c + d*x^2])/(24*c^2*x^4) - ((8*b^2*c^2 - 12*a*b*c*
d + 5*a^2*d^2)*Sqrt[c + d*x^2])/(16*c^3*x^2) + (d*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]
/Sqrt[c]])/(16*c^(7/2))

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Rubi [A]  time = 0.161291, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 89, 78, 51, 63, 208} \[ -\frac{\sqrt{c+d x^2} \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right )}{16 c^3 x^2}+\frac{d \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{16 c^{7/2}}-\frac{a^2 \sqrt{c+d x^2}}{6 c x^6}-\frac{a \sqrt{c+d x^2} (12 b c-5 a d)}{24 c^2 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^7*Sqrt[c + d*x^2]),x]

[Out]

-(a^2*Sqrt[c + d*x^2])/(6*c*x^6) - (a*(12*b*c - 5*a*d)*Sqrt[c + d*x^2])/(24*c^2*x^4) - ((8*b^2*c^2 - 12*a*b*c*
d + 5*a^2*d^2)*Sqrt[c + d*x^2])/(16*c^3*x^2) + (d*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]
/Sqrt[c]])/(16*c^(7/2))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2}{x^7 \sqrt{c+d x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2}{x^4 \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=-\frac{a^2 \sqrt{c+d x^2}}{6 c x^6}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} a (12 b c-5 a d)+3 b^2 c x}{x^3 \sqrt{c+d x}} \, dx,x,x^2\right )}{6 c}\\ &=-\frac{a^2 \sqrt{c+d x^2}}{6 c x^6}-\frac{a (12 b c-5 a d) \sqrt{c+d x^2}}{24 c^2 x^4}+\frac{1}{16} \left (8 b^2-\frac{a d (12 b c-5 a d)}{c^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=-\frac{a^2 \sqrt{c+d x^2}}{6 c x^6}-\frac{a (12 b c-5 a d) \sqrt{c+d x^2}}{24 c^2 x^4}-\frac{\left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \sqrt{c+d x^2}}{16 c^3 x^2}+\frac{\left (d \left (-8 b^2+\frac{a d (12 b c-5 a d)}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )}{32 c}\\ &=-\frac{a^2 \sqrt{c+d x^2}}{6 c x^6}-\frac{a (12 b c-5 a d) \sqrt{c+d x^2}}{24 c^2 x^4}-\frac{\left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \sqrt{c+d x^2}}{16 c^3 x^2}+\frac{\left (-8 b^2+\frac{a d (12 b c-5 a d)}{c^2}\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )}{16 c}\\ &=-\frac{a^2 \sqrt{c+d x^2}}{6 c x^6}-\frac{a (12 b c-5 a d) \sqrt{c+d x^2}}{24 c^2 x^4}-\frac{\left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \sqrt{c+d x^2}}{16 c^3 x^2}+\frac{d \left (8 b^2-\frac{a d (12 b c-5 a d)}{c^2}\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{16 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.23156, size = 135, normalized size = 0.89 \[ \frac{\sqrt{c+d x^2} \left (\frac{3 d \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\sqrt{\frac{d x^2}{c}+1}\right )}{\sqrt{\frac{d x^2}{c}+1}}-\frac{c \left (a^2 \left (8 c^2-10 c d x^2+15 d^2 x^4\right )+12 a b c x^2 \left (2 c-3 d x^2\right )+24 b^2 c^2 x^4\right )}{x^6}\right )}{48 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^7*Sqrt[c + d*x^2]),x]

[Out]

(Sqrt[c + d*x^2]*(-((c*(24*b^2*c^2*x^4 + 12*a*b*c*x^2*(2*c - 3*d*x^2) + a^2*(8*c^2 - 10*c*d*x^2 + 15*d^2*x^4))
)/x^6) + (3*d*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*ArcTanh[Sqrt[1 + (d*x^2)/c]])/Sqrt[1 + (d*x^2)/c]))/(48*c^4
)

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Maple [A]  time = 0.012, size = 224, normalized size = 1.5 \begin{align*} -{\frac{{a}^{2}}{6\,c{x}^{6}}\sqrt{d{x}^{2}+c}}+{\frac{5\,{a}^{2}d}{24\,{c}^{2}{x}^{4}}\sqrt{d{x}^{2}+c}}-{\frac{5\,{a}^{2}{d}^{2}}{16\,{c}^{3}{x}^{2}}\sqrt{d{x}^{2}+c}}+{\frac{5\,{a}^{2}{d}^{3}}{16}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){c}^{-{\frac{7}{2}}}}-{\frac{ab}{2\,c{x}^{4}}\sqrt{d{x}^{2}+c}}+{\frac{3\,abd}{4\,{c}^{2}{x}^{2}}\sqrt{d{x}^{2}+c}}-{\frac{3\,ab{d}^{2}}{4}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){c}^{-{\frac{5}{2}}}}-{\frac{{b}^{2}}{2\,c{x}^{2}}\sqrt{d{x}^{2}+c}}+{\frac{{b}^{2}d}{2}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^7/(d*x^2+c)^(1/2),x)

[Out]

-1/6*a^2*(d*x^2+c)^(1/2)/c/x^6+5/24*a^2*d/c^2/x^4*(d*x^2+c)^(1/2)-5/16*a^2*d^2/c^3/x^2*(d*x^2+c)^(1/2)+5/16*a^
2*d^3/c^(7/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)-1/2*a*b/c/x^4*(d*x^2+c)^(1/2)+3/4*a*b*d/c^2/x^2*(d*x^2+c)^
(1/2)-3/4*a*b*d^2/c^(5/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)-1/2*b^2/c/x^2*(d*x^2+c)^(1/2)+1/2*b^2*d/c^(3/2
)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^7/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5465, size = 630, normalized size = 4.17 \begin{align*} \left [\frac{3 \,{\left (8 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} \sqrt{c} x^{6} \log \left (-\frac{d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) - 2 \,{\left (8 \, a^{2} c^{3} + 3 \,{\left (8 \, b^{2} c^{3} - 12 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} x^{4} + 2 \,{\left (12 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{96 \, c^{4} x^{6}}, -\frac{3 \,{\left (8 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} \sqrt{-c} x^{6} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) +{\left (8 \, a^{2} c^{3} + 3 \,{\left (8 \, b^{2} c^{3} - 12 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} x^{4} + 2 \,{\left (12 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{48 \, c^{4} x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^7/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(8*b^2*c^2*d - 12*a*b*c*d^2 + 5*a^2*d^3)*sqrt(c)*x^6*log(-(d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x
^2) - 2*(8*a^2*c^3 + 3*(8*b^2*c^3 - 12*a*b*c^2*d + 5*a^2*c*d^2)*x^4 + 2*(12*a*b*c^3 - 5*a^2*c^2*d)*x^2)*sqrt(d
*x^2 + c))/(c^4*x^6), -1/48*(3*(8*b^2*c^2*d - 12*a*b*c*d^2 + 5*a^2*d^3)*sqrt(-c)*x^6*arctan(sqrt(-c)/sqrt(d*x^
2 + c)) + (8*a^2*c^3 + 3*(8*b^2*c^3 - 12*a*b*c^2*d + 5*a^2*c*d^2)*x^4 + 2*(12*a*b*c^3 - 5*a^2*c^2*d)*x^2)*sqrt
(d*x^2 + c))/(c^4*x^6)]

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Sympy [B]  time = 140.095, size = 301, normalized size = 1.99 \begin{align*} - \frac{a^{2}}{6 \sqrt{d} x^{7} \sqrt{\frac{c}{d x^{2}} + 1}} + \frac{a^{2} \sqrt{d}}{24 c x^{5} \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{5 a^{2} d^{\frac{3}{2}}}{48 c^{2} x^{3} \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{5 a^{2} d^{\frac{5}{2}}}{16 c^{3} x \sqrt{\frac{c}{d x^{2}} + 1}} + \frac{5 a^{2} d^{3} \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )}}{16 c^{\frac{7}{2}}} - \frac{a b}{2 \sqrt{d} x^{5} \sqrt{\frac{c}{d x^{2}} + 1}} + \frac{a b \sqrt{d}}{4 c x^{3} \sqrt{\frac{c}{d x^{2}} + 1}} + \frac{3 a b d^{\frac{3}{2}}}{4 c^{2} x \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{3 a b d^{2} \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )}}{4 c^{\frac{5}{2}}} - \frac{b^{2} \sqrt{d} \sqrt{\frac{c}{d x^{2}} + 1}}{2 c x} + \frac{b^{2} d \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )}}{2 c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**7/(d*x**2+c)**(1/2),x)

[Out]

-a**2/(6*sqrt(d)*x**7*sqrt(c/(d*x**2) + 1)) + a**2*sqrt(d)/(24*c*x**5*sqrt(c/(d*x**2) + 1)) - 5*a**2*d**(3/2)/
(48*c**2*x**3*sqrt(c/(d*x**2) + 1)) - 5*a**2*d**(5/2)/(16*c**3*x*sqrt(c/(d*x**2) + 1)) + 5*a**2*d**3*asinh(sqr
t(c)/(sqrt(d)*x))/(16*c**(7/2)) - a*b/(2*sqrt(d)*x**5*sqrt(c/(d*x**2) + 1)) + a*b*sqrt(d)/(4*c*x**3*sqrt(c/(d*
x**2) + 1)) + 3*a*b*d**(3/2)/(4*c**2*x*sqrt(c/(d*x**2) + 1)) - 3*a*b*d**2*asinh(sqrt(c)/(sqrt(d)*x))/(4*c**(5/
2)) - b**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(2*c*x) + b**2*d*asinh(sqrt(c)/(sqrt(d)*x))/(2*c**(3/2))

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Giac [A]  time = 1.17442, size = 325, normalized size = 2.15 \begin{align*} -\frac{\frac{3 \,{\left (8 \, b^{2} c^{2} d^{2} - 12 \, a b c d^{3} + 5 \, a^{2} d^{4}\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c} c^{3}} + \frac{24 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} b^{2} c^{2} d^{2} - 48 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} b^{2} c^{3} d^{2} + 24 \, \sqrt{d x^{2} + c} b^{2} c^{4} d^{2} - 36 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} a b c d^{3} + 96 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a b c^{2} d^{3} - 60 \, \sqrt{d x^{2} + c} a b c^{3} d^{3} + 15 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} a^{2} d^{4} - 40 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a^{2} c d^{4} + 33 \, \sqrt{d x^{2} + c} a^{2} c^{2} d^{4}}{c^{3} d^{3} x^{6}}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^7/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-1/48*(3*(8*b^2*c^2*d^2 - 12*a*b*c*d^3 + 5*a^2*d^4)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^3) + (24*(d*x
^2 + c)^(5/2)*b^2*c^2*d^2 - 48*(d*x^2 + c)^(3/2)*b^2*c^3*d^2 + 24*sqrt(d*x^2 + c)*b^2*c^4*d^2 - 36*(d*x^2 + c)
^(5/2)*a*b*c*d^3 + 96*(d*x^2 + c)^(3/2)*a*b*c^2*d^3 - 60*sqrt(d*x^2 + c)*a*b*c^3*d^3 + 15*(d*x^2 + c)^(5/2)*a^
2*d^4 - 40*(d*x^2 + c)^(3/2)*a^2*c*d^4 + 33*sqrt(d*x^2 + c)*a^2*c^2*d^4)/(c^3*d^3*x^6))/d