3.596 \(\int \frac{A+B x^2}{x^5 (a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=146 \[ \frac{5 b (7 A b-4 a B)}{8 a^4 \sqrt{a+b x^2}}+\frac{5 b (7 A b-4 a B)}{24 a^3 \left (a+b x^2\right )^{3/2}}+\frac{7 A b-4 a B}{8 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac{5 b (7 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 a^{9/2}}-\frac{A}{4 a x^4 \left (a+b x^2\right )^{3/2}} \]

[Out]

(5*b*(7*A*b - 4*a*B))/(24*a^3*(a + b*x^2)^(3/2)) - A/(4*a*x^4*(a + b*x^2)^(3/2)) + (7*A*b - 4*a*B)/(8*a^2*x^2*
(a + b*x^2)^(3/2)) + (5*b*(7*A*b - 4*a*B))/(8*a^4*Sqrt[a + b*x^2]) - (5*b*(7*A*b - 4*a*B)*ArcTanh[Sqrt[a + b*x
^2]/Sqrt[a]])/(8*a^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.110352, antiderivative size = 150, normalized size of antiderivative = 1.03, 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} \[ \frac{5 \sqrt{a+b x^2} (7 A b-4 a B)}{8 a^4 x^2}-\frac{5 (7 A b-4 a B)}{12 a^3 x^2 \sqrt{a+b x^2}}-\frac{7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac{5 b (7 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 a^{9/2}}-\frac{A}{4 a x^4 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^5*(a + b*x^2)^(5/2)),x]

[Out]

-A/(4*a*x^4*(a + b*x^2)^(3/2)) - (7*A*b - 4*a*B)/(12*a^2*x^2*(a + b*x^2)^(3/2)) - (5*(7*A*b - 4*a*B))/(12*a^3*
x^2*Sqrt[a + b*x^2]) + (5*(7*A*b - 4*a*B)*Sqrt[a + b*x^2])/(8*a^4*x^2) - (5*b*(7*A*b - 4*a*B)*ArcTanh[Sqrt[a +
b*x^2]/Sqrt[a]])/(8*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^5 \left (a+b x^2\right )^{5/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{x^3 (a+b x)^{5/2}} \, dx,x,x^2\right )\\ &=-\frac{A}{4 a x^4 \left (a+b x^2\right )^{3/2}}+\frac{\left (-\frac{7 A b}{2}+2 a B\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{5/2}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac{A}{4 a x^4 \left (a+b x^2\right )^{3/2}}-\frac{7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac{(5 (7 A b-4 a B)) \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{3/2}} \, dx,x,x^2\right )}{24 a^2}\\ &=-\frac{A}{4 a x^4 \left (a+b x^2\right )^{3/2}}-\frac{7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac{5 (7 A b-4 a B)}{12 a^3 x^2 \sqrt{a+b x^2}}-\frac{(5 (7 A b-4 a B)) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^2\right )}{8 a^3}\\ &=-\frac{A}{4 a x^4 \left (a+b x^2\right )^{3/2}}-\frac{7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac{5 (7 A b-4 a B)}{12 a^3 x^2 \sqrt{a+b x^2}}+\frac{5 (7 A b-4 a B) \sqrt{a+b x^2}}{8 a^4 x^2}+\frac{(5 b (7 A b-4 a B)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{16 a^4}\\ &=-\frac{A}{4 a x^4 \left (a+b x^2\right )^{3/2}}-\frac{7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac{5 (7 A b-4 a B)}{12 a^3 x^2 \sqrt{a+b x^2}}+\frac{5 (7 A b-4 a B) \sqrt{a+b x^2}}{8 a^4 x^2}+\frac{(5 (7 A b-4 a B)) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{8 a^4}\\ &=-\frac{A}{4 a x^4 \left (a+b x^2\right )^{3/2}}-\frac{7 A b-4 a B}{12 a^2 x^2 \left (a+b x^2\right )^{3/2}}-\frac{5 (7 A b-4 a B)}{12 a^3 x^2 \sqrt{a+b x^2}}+\frac{5 (7 A b-4 a B) \sqrt{a+b x^2}}{8 a^4 x^2}-\frac{5 b (7 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 a^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0214161, size = 60, normalized size = 0.41 \[ \frac{b x^4 (7 A b-4 a B) \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};\frac{b x^2}{a}+1\right )-3 a^2 A}{12 a^3 x^4 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^5*(a + b*x^2)^(5/2)),x]

[Out]

(-3*a^2*A + b*(7*A*b - 4*a*B)*x^4*Hypergeometric2F1[-3/2, 2, -1/2, 1 + (b*x^2)/a])/(12*a^3*x^4*(a + b*x^2)^(3/
2))

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Maple [A]  time = 0.011, size = 187, normalized size = 1.3 \begin{align*} -{\frac{A}{4\,a{x}^{4}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{7\,Ab}{8\,{a}^{2}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{35\,A{b}^{2}}{24\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{35\,A{b}^{2}}{8\,{a}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{35\,A{b}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{9}{2}}}}-{\frac{B}{2\,a{x}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,Bb}{6\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,Bb}{2\,{a}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{5\,Bb}{2}\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^5/(b*x^2+a)^(5/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.73774, size = 880, normalized size = 6.03 \begin{align*} \left [-\frac{15 \,{\left ({\left (4 \, B a b^{3} - 7 \, A b^{4}\right )} x^{8} + 2 \,{\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6} +{\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{4}\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 (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6} + 6 \, A a^{4} + 20 \,{\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{4} + 3 \,{\left (4 \, B a^{4} - 7 \, A a^{3} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{48 \,{\left (a^{5} b^{2} x^{8} + 2 \, a^{6} b x^{6} + a^{7} x^{4}\right )}}, -\frac{15 \,{\left ({\left (4 \, B a b^{3} - 7 \, A b^{4}\right )} x^{8} + 2 \,{\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6} +{\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{4}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (15 \,{\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{6} + 6 \, A a^{4} + 20 \,{\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{4} + 3 \,{\left (4 \, B a^{4} - 7 \, A a^{3} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{24 \,{\left (a^{5} b^{2} x^{8} + 2 \, a^{6} b x^{6} + a^{7} x^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(15*((4*B*a*b^3 - 7*A*b^4)*x^8 + 2*(4*B*a^2*b^2 - 7*A*a*b^3)*x^6 + (4*B*a^3*b - 7*A*a^2*b^2)*x^4)*sqrt(
a)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(15*(4*B*a^2*b^2 - 7*A*a*b^3)*x^6 + 6*A*a^4 + 20*(4
*B*a^3*b - 7*A*a^2*b^2)*x^4 + 3*(4*B*a^4 - 7*A*a^3*b)*x^2)*sqrt(b*x^2 + a))/(a^5*b^2*x^8 + 2*a^6*b*x^6 + a^7*x
^4), -1/24*(15*((4*B*a*b^3 - 7*A*b^4)*x^8 + 2*(4*B*a^2*b^2 - 7*A*a*b^3)*x^6 + (4*B*a^3*b - 7*A*a^2*b^2)*x^4)*s
qrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (15*(4*B*a^2*b^2 - 7*A*a*b^3)*x^6 + 6*A*a^4 + 20*(4*B*a^3*b - 7*A*a
^2*b^2)*x^4 + 3*(4*B*a^4 - 7*A*a^3*b)*x^2)*sqrt(b*x^2 + a))/(a^5*b^2*x^8 + 2*a^6*b*x^6 + a^7*x^4)]

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Sympy [B]  time = 106.168, size = 1323, normalized size = 9.06 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**5/(b*x**2+a)**(5/2),x)

[Out]

A*(-6*a**(89/2)*b**75/(24*a**(93/2)*b**(151/2)*x**5*sqrt(a/(b*x**2) + 1) + 24*a**(91/2)*b**(153/2)*x**7*sqrt(a
/(b*x**2) + 1)) + 21*a**(87/2)*b**76*x**2/(24*a**(93/2)*b**(151/2)*x**5*sqrt(a/(b*x**2) + 1) + 24*a**(91/2)*b*
*(153/2)*x**7*sqrt(a/(b*x**2) + 1)) + 140*a**(85/2)*b**77*x**4/(24*a**(93/2)*b**(151/2)*x**5*sqrt(a/(b*x**2) +
1) + 24*a**(91/2)*b**(153/2)*x**7*sqrt(a/(b*x**2) + 1)) + 105*a**(83/2)*b**78*x**6/(24*a**(93/2)*b**(151/2)*x
**5*sqrt(a/(b*x**2) + 1) + 24*a**(91/2)*b**(153/2)*x**7*sqrt(a/(b*x**2) + 1)) - 105*a**42*b**(155/2)*x**5*sqrt
(a/(b*x**2) + 1)*asinh(sqrt(a)/(sqrt(b)*x))/(24*a**(93/2)*b**(151/2)*x**5*sqrt(a/(b*x**2) + 1) + 24*a**(91/2)*
b**(153/2)*x**7*sqrt(a/(b*x**2) + 1)) - 105*a**41*b**(157/2)*x**7*sqrt(a/(b*x**2) + 1)*asinh(sqrt(a)/(sqrt(b)*
x))/(24*a**(93/2)*b**(151/2)*x**5*sqrt(a/(b*x**2) + 1) + 24*a**(91/2)*b**(153/2)*x**7*sqrt(a/(b*x**2) + 1))) +
B*(-6*a**17*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/
2)*b**3*x**8) - 46*a**16*b*x**2*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**
2*x**6 + 12*a**(33/2)*b**3*x**8) - 15*a**16*b*x**2*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36
*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) + 30*a**16*b*x**2*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**
2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 70*a**15*b**2*x**4*sqrt(1 + b*x**
2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 45*a**15*b*
*2*x**4*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x*
*8) + 90*a**15*b**2*x**4*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b
**2*x**6 + 12*a**(33/2)*b**3*x**8) - 30*a**14*b**3*x**6*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b
*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 45*a**14*b**3*x**6*log(b*x**2/a)/(12*a**(39/2)*x**2
+ 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) + 90*a**14*b**3*x**6*log(sqrt(1 + b*
x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 15*
a**13*b**4*x**8*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)
*b**3*x**8) + 30*a**13*b**4*x**8*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**
(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8))

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Giac [A]  time = 1.19665, size = 223, normalized size = 1.53 \begin{align*} -\frac{5 \,{\left (4 \, B a b - 7 \, A b^{2}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a^{4}} - \frac{6 \,{\left (b x^{2} + a\right )} B a b + B a^{2} b - 9 \,{\left (b x^{2} + a\right )} A b^{2} - A a b^{2}}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{4}} - \frac{4 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a b - 4 \, \sqrt{b x^{2} + a} B a^{2} b - 11 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A b^{2} + 13 \, \sqrt{b x^{2} + a} A a b^{2}}{8 \, a^{4} b^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8*(4*B*a*b - 7*A*b^2)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^4) - 1/3*(6*(b*x^2 + a)*B*a*b + B*a^2*b
- 9*(b*x^2 + a)*A*b^2 - A*a*b^2)/((b*x^2 + a)^(3/2)*a^4) - 1/8*(4*(b*x^2 + a)^(3/2)*B*a*b - 4*sqrt(b*x^2 + a)*
B*a^2*b - 11*(b*x^2 + a)^(3/2)*A*b^2 + 13*sqrt(b*x^2 + a)*A*a*b^2)/(a^4*b^2*x^4)