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

Optimal. Leaf size=143 \[ -\frac{\left (a+b x^2\right )^{5/2} (4 a B+3 A b)}{8 a x^2}+\frac{5 b \left (a+b x^2\right )^{3/2} (4 a B+3 A b)}{24 a}+\frac{5}{8} b \sqrt{a+b x^2} (4 a B+3 A b)-\frac{5}{8} \sqrt{a} b (4 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4} \]

[Out]

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

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Rubi [A]  time = 0.10115, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {446, 78, 47, 50, 63, 208} \[ -\frac{\left (a+b x^2\right )^{5/2} (4 a B+3 A b)}{8 a x^2}+\frac{5 b \left (a+b x^2\right )^{3/2} (4 a B+3 A b)}{24 a}+\frac{5}{8} b \sqrt{a+b x^2} (4 a B+3 A b)-\frac{5}{8} \sqrt{a} b (4 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 47

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 )^{5/2} \left (A+B x^2\right )}{x^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2} (A+B x)}{x^3} \, dx,x,x^2\right )\\ &=-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac{\left (\frac{3 A b}{2}+2 a B\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^2} \, dx,x,x^2\right )}{4 a}\\ &=-\frac{(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac{(5 b (3 A b+4 a B)) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^2\right )}{16 a}\\ &=\frac{5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac{(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac{1}{16} (5 b (3 A b+4 a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac{5}{8} b (3 A b+4 a B) \sqrt{a+b x^2}+\frac{5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac{(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac{1}{16} (5 a b (3 A b+4 a B)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{5}{8} b (3 A b+4 a B) \sqrt{a+b x^2}+\frac{5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac{(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}+\frac{1}{8} (5 a (3 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 )\\ &=\frac{5}{8} b (3 A b+4 a B) \sqrt{a+b x^2}+\frac{5 b (3 A b+4 a B) \left (a+b x^2\right )^{3/2}}{24 a}-\frac{(3 A b+4 a B) \left (a+b x^2\right )^{5/2}}{8 a x^2}-\frac{A \left (a+b x^2\right )^{7/2}}{4 a x^4}-\frac{5}{8} \sqrt{a} b (3 A b+4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0307477, size = 60, normalized size = 0.42 \[ \frac{\left (a+b x^2\right )^{7/2} \left (b x^4 (4 a B+3 A b) \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{b x^2}{a}+1\right )-7 a^2 A\right )}{28 a^3 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(7/2)*(-7*a^2*A + b*(3*A*b + 4*a*B)*x^4*Hypergeometric2F1[2, 7/2, 9/2, 1 + (b*x^2)/a]))/(28*a^3*x
^4)

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Maple [A]  time = 0.009, size = 213, normalized size = 1.5 \begin{align*} -{\frac{A}{4\,a{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{3\,Ab}{8\,{a}^{2}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{3\,A{b}^{2}}{8\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,A{b}^{2}}{8\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{15\,A{b}^{2}}{8}\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{15\,A{b}^{2}}{8}\sqrt{b{x}^{2}+a}}-{\frac{B}{2\,a{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{Bb}{2\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,Bb}{6} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,Bb}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{5\,Bba}{2}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*A*(b*x^2+a)^(7/2)/a/x^4-3/8*A*b/a^2/x^2*(b*x^2+a)^(7/2)+3/8*A*b^2/a^2*(b*x^2+a)^(5/2)+5/8*A*b^2/a*(b*x^2+
a)^(3/2)-15/8*A*b^2*a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+15/8*A*b^2*(b*x^2+a)^(1/2)-1/2*B/a/x^2*(b*x^
2+a)^(7/2)+1/2*B*b/a*(b*x^2+a)^(5/2)+5/6*B*b*(b*x^2+a)^(3/2)-5/2*B*b*a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2)
)/x)+5/2*B*b*a*(b*x^2+a)^(1/2)

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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)^(5/2)*(B*x^2+A)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72509, size = 516, normalized size = 3.61 \begin{align*} \left [\frac{15 \,{\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt{a} x^{4} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (8 \, B b^{2} x^{6} + 8 \,{\left (7 \, B a b + 3 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 3 \,{\left (4 \, B a^{2} + 9 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{48 \, x^{4}}, \frac{15 \,{\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt{-a} x^{4} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (8 \, B b^{2} x^{6} + 8 \,{\left (7 \, B a b + 3 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 3 \,{\left (4 \, B a^{2} + 9 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{24 \, x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*(4*B*a*b + 3*A*b^2)*sqrt(a)*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(8*B*b^2*x^6
+ 8*(7*B*a*b + 3*A*b^2)*x^4 - 6*A*a^2 - 3*(4*B*a^2 + 9*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^4, 1/24*(15*(4*B*a*b +
3*A*b^2)*sqrt(-a)*x^4*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (8*B*b^2*x^6 + 8*(7*B*a*b + 3*A*b^2)*x^4 - 6*A*a^2 -
3*(4*B*a^2 + 9*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^4]

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Sympy [A]  time = 61.9685, size = 279, normalized size = 1.95 \begin{align*} - \frac{15 A \sqrt{a} b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{8} - \frac{A a^{3}}{4 \sqrt{b} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{3 A a^{2} \sqrt{b}}{8 x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{A a b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{x} + \frac{7 A a b^{\frac{3}{2}}}{8 x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{A b^{\frac{5}{2}} x}{\sqrt{\frac{a}{b x^{2}} + 1}} - \frac{5 B a^{\frac{3}{2}} b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2} - \frac{B a^{2} \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{2 x} + \frac{2 B a^{2} \sqrt{b}}{x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{2 B a b^{\frac{3}{2}} x}{\sqrt{\frac{a}{b x^{2}} + 1}} + B b^{2} \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\left (a + b x^{2}\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*A*sqrt(a)*b**2*asinh(sqrt(a)/(sqrt(b)*x))/8 - A*a**3/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) - 3*A*a**2*sqrt
(b)/(8*x**3*sqrt(a/(b*x**2) + 1)) - A*a*b**(3/2)*sqrt(a/(b*x**2) + 1)/x + 7*A*a*b**(3/2)/(8*x*sqrt(a/(b*x**2)
+ 1)) + A*b**(5/2)*x/sqrt(a/(b*x**2) + 1) - 5*B*a**(3/2)*b*asinh(sqrt(a)/(sqrt(b)*x))/2 - B*a**2*sqrt(b)*sqrt(
a/(b*x**2) + 1)/(2*x) + 2*B*a**2*sqrt(b)/(x*sqrt(a/(b*x**2) + 1)) + 2*B*a*b**(3/2)*x/sqrt(a/(b*x**2) + 1) + B*
b**2*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3/2)/(3*b), True))

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Giac [A]  time = 1.15734, size = 231, normalized size = 1.62 \begin{align*} \frac{8 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B b^{2} + 48 \, \sqrt{b x^{2} + a} B a b^{2} + 24 \, \sqrt{b x^{2} + a} A b^{3} + \frac{15 \,{\left (4 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{3 \,{\left (4 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a^{2} b^{2} - 4 \, \sqrt{b x^{2} + a} B a^{3} b^{2} + 9 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A a b^{3} - 7 \, \sqrt{b x^{2} + a} A a^{2} b^{3}\right )}}{b^{2} x^{4}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*(b*x^2 + a)^(3/2)*B*b^2 + 48*sqrt(b*x^2 + a)*B*a*b^2 + 24*sqrt(b*x^2 + a)*A*b^3 + 15*(4*B*a^2*b^2 + 3*
A*a*b^3)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) - 3*(4*(b*x^2 + a)^(3/2)*B*a^2*b^2 - 4*sqrt(b*x^2 + a)*B*a^
3*b^2 + 9*(b*x^2 + a)^(3/2)*A*a*b^3 - 7*sqrt(b*x^2 + a)*A*a^2*b^3)/(b^2*x^4))/b