3.547 \(\int \frac{(a+b x^2)^{5/2} (A+B x^2)}{x^4} \, dx\)

Optimal. Leaf size=146 \[ -\frac{\left (a+b x^2\right )^{5/2} (3 a B+4 A b)}{3 a x}+\frac{5 b x \left (a+b x^2\right )^{3/2} (3 a B+4 A b)}{12 a}+\frac{5}{8} b x \sqrt{a+b x^2} (3 a B+4 A b)+\frac{5}{8} a \sqrt{b} (3 a B+4 A b) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )-\frac{A \left (a+b x^2\right )^{7/2}}{3 a x^3} \]

[Out]

(5*b*(4*A*b + 3*a*B)*x*Sqrt[a + b*x^2])/8 + (5*b*(4*A*b + 3*a*B)*x*(a + b*x^2)^(3/2))/(12*a) - ((4*A*b + 3*a*B
)*(a + b*x^2)^(5/2))/(3*a*x) - (A*(a + b*x^2)^(7/2))/(3*a*x^3) + (5*a*Sqrt[b]*(4*A*b + 3*a*B)*ArcTanh[(Sqrt[b]
*x)/Sqrt[a + b*x^2]])/8

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Rubi [A]  time = 0.0600457, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {453, 277, 195, 217, 206} \[ -\frac{\left (a+b x^2\right )^{5/2} (3 a B+4 A b)}{3 a x}+\frac{5 b x \left (a+b x^2\right )^{3/2} (3 a B+4 A b)}{12 a}+\frac{5}{8} b x \sqrt{a+b x^2} (3 a B+4 A b)+\frac{5}{8} a \sqrt{b} (3 a B+4 A b) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )-\frac{A \left (a+b x^2\right )^{7/2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^4,x]

[Out]

(5*b*(4*A*b + 3*a*B)*x*Sqrt[a + b*x^2])/8 + (5*b*(4*A*b + 3*a*B)*x*(a + b*x^2)^(3/2))/(12*a) - ((4*A*b + 3*a*B
)*(a + b*x^2)^(5/2))/(3*a*x) - (A*(a + b*x^2)^(7/2))/(3*a*x^3) + (5*a*Sqrt[b]*(4*A*b + 3*a*B)*ArcTanh[(Sqrt[b]
*x)/Sqrt[a + b*x^2]])/8

Rule 453

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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.0366043, size = 84, normalized size = 0.58 \[ \frac{a \sqrt{a+b x^2} (-3 a B-4 A b) \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x^2}{a}\right )}{3 x \sqrt{\frac{b x^2}{a}+1}}-\frac{A \left (a+b x^2\right )^{7/2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^4,x]

[Out]

-(A*(a + b*x^2)^(7/2))/(3*a*x^3) + (a*(-4*A*b - 3*a*B)*Sqrt[a + b*x^2]*Hypergeometric2F1[-5/2, -1/2, 1/2, -((b
*x^2)/a)])/(3*x*Sqrt[1 + (b*x^2)/a])

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Maple [A]  time = 0.01, size = 204, normalized size = 1.4 \begin{align*} -{\frac{A}{3\,a{x}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{4\,Ab}{3\,{a}^{2}x} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{4\,A{b}^{2}x}{3\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,A{b}^{2}x}{3\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,A{b}^{2}x}{2}\sqrt{b{x}^{2}+a}}+{\frac{5\,Aa}{2}{b}^{{\frac{3}{2}}}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) }-{\frac{B}{ax} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{bBx}{a} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,bBx}{4} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{15\,Bbax}{8}\sqrt{b{x}^{2}+a}}+{\frac{15\,{a}^{2}B}{8}\sqrt{b}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) } \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^4,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66959, size = 514, normalized size = 3.52 \begin{align*} \left [\frac{15 \,{\left (3 \, B a^{2} + 4 \, A a b\right )} \sqrt{b} x^{3} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (6 \, B b^{2} x^{6} + 3 \,{\left (9 \, B a b + 4 \, A b^{2}\right )} x^{4} - 8 \, A a^{2} - 8 \,{\left (3 \, B a^{2} + 7 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{48 \, x^{3}}, -\frac{15 \,{\left (3 \, B a^{2} + 4 \, A a b\right )} \sqrt{-b} x^{3} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (6 \, B b^{2} x^{6} + 3 \,{\left (9 \, B a b + 4 \, A b^{2}\right )} x^{4} - 8 \, A a^{2} - 8 \,{\left (3 \, B a^{2} + 7 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{24 \, x^{3}}\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^4,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 10.4497, size = 299, normalized size = 2.05 \begin{align*} - \frac{2 A a^{\frac{3}{2}} b}{x \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{A \sqrt{a} b^{2} x \sqrt{1 + \frac{b x^{2}}{a}}}{2} - \frac{2 A \sqrt{a} b^{2} x}{\sqrt{1 + \frac{b x^{2}}{a}}} - \frac{A a^{2} \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{3 x^{2}} - \frac{A a b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{3} + \frac{5 A a b^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2} - \frac{B a^{\frac{5}{2}}}{x \sqrt{1 + \frac{b x^{2}}{a}}} + B a^{\frac{3}{2}} b x \sqrt{1 + \frac{b x^{2}}{a}} - \frac{7 B a^{\frac{3}{2}} b x}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 B \sqrt{a} b^{2} x^{3}}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{15 B a^{2} \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8} + \frac{B b^{3} x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \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**4,x)

[Out]

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

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Giac [A]  time = 1.14474, size = 321, normalized size = 2.2 \begin{align*} \frac{1}{8} \,{\left (2 \, B b^{2} x^{2} + \frac{9 \, B a b^{3} + 4 \, A b^{4}}{b^{2}}\right )} \sqrt{b x^{2} + a} x - \frac{5}{16} \,{\left (3 \, B a^{2} \sqrt{b} + 4 \, A a b^{\frac{3}{2}}\right )} \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right ) + \frac{2 \,{\left (3 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} B a^{3} \sqrt{b} + 9 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} A a^{2} b^{\frac{3}{2}} - 6 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} B a^{4} \sqrt{b} - 12 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} A a^{3} b^{\frac{3}{2}} + 3 \, B a^{5} \sqrt{b} + 7 \, A a^{4} b^{\frac{3}{2}}\right )}}{3 \,{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}^{3}} \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^4,x, algorithm="giac")

[Out]

1/8*(2*B*b^2*x^2 + (9*B*a*b^3 + 4*A*b^4)/b^2)*sqrt(b*x^2 + a)*x - 5/16*(3*B*a^2*sqrt(b) + 4*A*a*b^(3/2))*log((
sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2/3*(3*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a^3*sqrt(b) + 9*(sqrt(b)*x - sqrt(b
*x^2 + a))^4*A*a^2*b^(3/2) - 6*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^4*sqrt(b) - 12*(sqrt(b)*x - sqrt(b*x^2 + a)
)^2*A*a^3*b^(3/2) + 3*B*a^5*sqrt(b) + 7*A*a^4*b^(3/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^3