3.582 \(\int \frac{A+B x^2}{x^7 (a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=153 \[ -\frac{5 b^2 (7 A b-6 a B)}{16 a^4 \sqrt{a+b x^2}}+\frac{5 b^2 (7 A b-6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{16 a^{9/2}}-\frac{5 b (7 A b-6 a B)}{48 a^3 x^2 \sqrt{a+b x^2}}+\frac{7 A b-6 a B}{24 a^2 x^4 \sqrt{a+b x^2}}-\frac{A}{6 a x^6 \sqrt{a+b x^2}} \]

[Out]

(-5*b^2*(7*A*b - 6*a*B))/(16*a^4*Sqrt[a + b*x^2]) - A/(6*a*x^6*Sqrt[a + b*x^2]) + (7*A*b - 6*a*B)/(24*a^2*x^4*
Sqrt[a + b*x^2]) - (5*b*(7*A*b - 6*a*B))/(48*a^3*x^2*Sqrt[a + b*x^2]) + (5*b^2*(7*A*b - 6*a*B)*ArcTanh[Sqrt[a
+ b*x^2]/Sqrt[a]])/(16*a^(9/2))

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Rubi [A]  time = 0.116951, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {446, 78, 51, 63, 208} \[ \frac{5 b^2 (7 A b-6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{16 a^{9/2}}-\frac{5 b \sqrt{a+b x^2} (7 A b-6 a B)}{16 a^4 x^2}+\frac{5 \sqrt{a+b x^2} (7 A b-6 a B)}{24 a^3 x^4}-\frac{7 A b-6 a B}{6 a^2 x^4 \sqrt{a+b x^2}}-\frac{A}{6 a x^6 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^7*(a + b*x^2)^(3/2)),x]

[Out]

-A/(6*a*x^6*Sqrt[a + b*x^2]) - (7*A*b - 6*a*B)/(6*a^2*x^4*Sqrt[a + b*x^2]) + (5*(7*A*b - 6*a*B)*Sqrt[a + b*x^2
])/(24*a^3*x^4) - (5*b*(7*A*b - 6*a*B)*Sqrt[a + b*x^2])/(16*a^4*x^2) + (5*b^2*(7*A*b - 6*a*B)*ArcTanh[Sqrt[a +
b*x^2]/Sqrt[a]])/(16*a^(9/2))

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 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{A+B x^2}{x^7 \left (a+b x^2\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{x^4 (a+b x)^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{A}{6 a x^6 \sqrt{a+b x^2}}+\frac{\left (-\frac{7 A b}{2}+3 a B\right ) \operatorname{Subst}\left (\int \frac{1}{x^3 (a+b x)^{3/2}} \, dx,x,x^2\right )}{6 a}\\ &=-\frac{A}{6 a x^6 \sqrt{a+b x^2}}-\frac{7 A b-6 a B}{6 a^2 x^4 \sqrt{a+b x^2}}-\frac{(5 (7 A b-6 a B)) \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a+b x}} \, dx,x,x^2\right )}{12 a^2}\\ &=-\frac{A}{6 a x^6 \sqrt{a+b x^2}}-\frac{7 A b-6 a B}{6 a^2 x^4 \sqrt{a+b x^2}}+\frac{5 (7 A b-6 a B) \sqrt{a+b x^2}}{24 a^3 x^4}+\frac{(5 b (7 A b-6 a B)) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^2\right )}{16 a^3}\\ &=-\frac{A}{6 a x^6 \sqrt{a+b x^2}}-\frac{7 A b-6 a B}{6 a^2 x^4 \sqrt{a+b x^2}}+\frac{5 (7 A b-6 a B) \sqrt{a+b x^2}}{24 a^3 x^4}-\frac{5 b (7 A b-6 a B) \sqrt{a+b x^2}}{16 a^4 x^2}-\frac{\left (5 b^2 (7 A b-6 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{32 a^4}\\ &=-\frac{A}{6 a x^6 \sqrt{a+b x^2}}-\frac{7 A b-6 a B}{6 a^2 x^4 \sqrt{a+b x^2}}+\frac{5 (7 A b-6 a B) \sqrt{a+b x^2}}{24 a^3 x^4}-\frac{5 b (7 A b-6 a B) \sqrt{a+b x^2}}{16 a^4 x^2}-\frac{(5 b (7 A b-6 a B)) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{16 a^4}\\ &=-\frac{A}{6 a x^6 \sqrt{a+b x^2}}-\frac{7 A b-6 a B}{6 a^2 x^4 \sqrt{a+b x^2}}+\frac{5 (7 A b-6 a B) \sqrt{a+b x^2}}{24 a^3 x^4}-\frac{5 b (7 A b-6 a B) \sqrt{a+b x^2}}{16 a^4 x^2}+\frac{5 b^2 (7 A b-6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{16 a^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0207722, size = 62, normalized size = 0.41 \[ \frac{b^2 x^6 (6 a B-7 A b) \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{b x^2}{a}+1\right )-a^3 A}{6 a^4 x^6 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^7*(a + b*x^2)^(3/2)),x]

[Out]

(-(a^3*A) + b^2*(-7*A*b + 6*a*B)*x^6*Hypergeometric2F1[-1/2, 3, 1/2, 1 + (b*x^2)/a])/(6*a^4*x^6*Sqrt[a + b*x^2
])

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Maple [A]  time = 0.01, size = 197, normalized size = 1.3 \begin{align*} -{\frac{A}{6\,a{x}^{6}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{7\,Ab}{24\,{a}^{2}{x}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{35\,A{b}^{2}}{48\,{a}^{3}{x}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{35\,A{b}^{3}}{16\,{a}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{35\,A{b}^{3}}{16}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{9}{2}}}}-{\frac{B}{4\,a{x}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{5\,Bb}{8\,{a}^{2}{x}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{15\,B{b}^{2}}{8\,{a}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{15\,B{b}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^7/(b*x^2+a)^(3/2),x)

[Out]

-1/6*A/a/x^6/(b*x^2+a)^(1/2)+7/24*A*b/a^2/x^4/(b*x^2+a)^(1/2)-35/48*A*b^2/a^3/x^2/(b*x^2+a)^(1/2)-35/16*A*b^3/
a^4/(b*x^2+a)^(1/2)+35/16*A*b^3/a^(9/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)-1/4*B/a/x^4/(b*x^2+a)^(1/2)+5/8*
B*b/a^2/x^2/(b*x^2+a)^(1/2)+15/8*B*b^2/a^3/(b*x^2+a)^(1/2)-15/8*B*b^2/a^(7/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(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)/x^7/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65938, size = 741, normalized size = 4.84 \begin{align*} \left [-\frac{15 \,{\left ({\left (6 \, B a b^{3} - 7 \, A b^{4}\right )} x^{8} +{\left (6 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6}\right )} \sqrt{a} \log \left (-\frac{b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (15 \,{\left (6 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6} - 8 \, A a^{4} + 5 \,{\left (6 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{4} - 2 \,{\left (6 \, B a^{4} - 7 \, A a^{3} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{96 \,{\left (a^{5} b x^{8} + a^{6} x^{6}\right )}}, \frac{15 \,{\left ({\left (6 \, B a b^{3} - 7 \, A b^{4}\right )} x^{8} +{\left (6 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (15 \,{\left (6 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6} - 8 \, A a^{4} + 5 \,{\left (6 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{4} - 2 \,{\left (6 \, B a^{4} - 7 \, A a^{3} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{48 \,{\left (a^{5} b x^{8} + a^{6} x^{6}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^7/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(15*((6*B*a*b^3 - 7*A*b^4)*x^8 + (6*B*a^2*b^2 - 7*A*a*b^3)*x^6)*sqrt(a)*log(-(b*x^2 + 2*sqrt(b*x^2 + a)
*sqrt(a) + 2*a)/x^2) - 2*(15*(6*B*a^2*b^2 - 7*A*a*b^3)*x^6 - 8*A*a^4 + 5*(6*B*a^3*b - 7*A*a^2*b^2)*x^4 - 2*(6*
B*a^4 - 7*A*a^3*b)*x^2)*sqrt(b*x^2 + a))/(a^5*b*x^8 + a^6*x^6), 1/48*(15*((6*B*a*b^3 - 7*A*b^4)*x^8 + (6*B*a^2
*b^2 - 7*A*a*b^3)*x^6)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (15*(6*B*a^2*b^2 - 7*A*a*b^3)*x^6 - 8*A*a^4
+ 5*(6*B*a^3*b - 7*A*a^2*b^2)*x^4 - 2*(6*B*a^4 - 7*A*a^3*b)*x^2)*sqrt(b*x^2 + a))/(a^5*b*x^8 + a^6*x^6)]

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Sympy [A]  time = 72.196, size = 236, normalized size = 1.54 \begin{align*} A \left (- \frac{1}{6 a \sqrt{b} x^{7} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{7 \sqrt{b}}{24 a^{2} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{35 b^{\frac{3}{2}}}{48 a^{3} x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{35 b^{\frac{5}{2}}}{16 a^{4} x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{35 b^{3} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{16 a^{\frac{9}{2}}}\right ) + B \left (- \frac{1}{4 a \sqrt{b} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{5 \sqrt{b}}{8 a^{2} x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{15 b^{\frac{3}{2}}}{8 a^{3} x \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{15 b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{8 a^{\frac{7}{2}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**7/(b*x**2+a)**(3/2),x)

[Out]

A*(-1/(6*a*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) + 7*sqrt(b)/(24*a**2*x**5*sqrt(a/(b*x**2) + 1)) - 35*b**(3/2)/(4
8*a**3*x**3*sqrt(a/(b*x**2) + 1)) - 35*b**(5/2)/(16*a**4*x*sqrt(a/(b*x**2) + 1)) + 35*b**3*asinh(sqrt(a)/(sqrt
(b)*x))/(16*a**(9/2))) + B*(-1/(4*a*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) + 5*sqrt(b)/(8*a**2*x**3*sqrt(a/(b*x**2
) + 1)) + 15*b**(3/2)/(8*a**3*x*sqrt(a/(b*x**2) + 1)) - 15*b**2*asinh(sqrt(a)/(sqrt(b)*x))/(8*a**(7/2)))

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Giac [A]  time = 1.1665, size = 243, normalized size = 1.59 \begin{align*} \frac{5 \,{\left (6 \, B a b^{2} - 7 \, A b^{3}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{16 \, \sqrt{-a} a^{4}} + \frac{B a b^{2} - A b^{3}}{\sqrt{b x^{2} + a} a^{4}} + \frac{42 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} B a b^{2} - 96 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a^{2} b^{2} + 54 \, \sqrt{b x^{2} + a} B a^{3} b^{2} - 57 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} A b^{3} + 136 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A a b^{3} - 87 \, \sqrt{b x^{2} + a} A a^{2} b^{3}}{48 \, a^{4} b^{3} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^7/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

5/16*(6*B*a*b^2 - 7*A*b^3)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^4) + (B*a*b^2 - A*b^3)/(sqrt(b*x^2 + a
)*a^4) + 1/48*(42*(b*x^2 + a)^(5/2)*B*a*b^2 - 96*(b*x^2 + a)^(3/2)*B*a^2*b^2 + 54*sqrt(b*x^2 + a)*B*a^3*b^2 -
57*(b*x^2 + a)^(5/2)*A*b^3 + 136*(b*x^2 + a)^(3/2)*A*a*b^3 - 87*sqrt(b*x^2 + a)*A*a^2*b^3)/(a^4*b^3*x^6)