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]

5/12*b*(4*A*b+3*B*a)*x*(b*x^2+a)^(3/2)/a-1/3*(4*A*b+3*B*a)*(b*x^2+a)^(5/2)/a/x-1/3*A*(b*x^2+a)^(7/2)/a/x^3+5/8
*a*(4*A*b+3*B*a)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))*b^(1/2)+5/8*b*(4*A*b+3*B*a)*x*(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 146, normalized size of antiderivative = 1.00, 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 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 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])

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 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 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]

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*}

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Mathematica [C]  time = 0.04, 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]

-1/3*(A*(a + b*x^2)^(7/2))/(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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fricas [A]  time = 0.59, size = 220, normalized size = 1.51 \[ \left [\frac {15 \, {\left (3 \, B a^{2} + 4 \, A a b\right )} \sqrt {b} x^{3} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (6 \, B b^{2} x^{6} + 3 \, {\left (9 \, B a b + 4 \, A b^{2}\right )} x^{4} - 8 \, A a^{2} - 8 \, {\left (3 \, B a^{2} + 7 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, x^{3}}, -\frac {15 \, {\left (3 \, B a^{2} + 4 \, A a b\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (6 \, B b^{2} x^{6} + 3 \, {\left (9 \, B a b + 4 \, A b^{2}\right )} x^{4} - 8 \, A a^{2} - 8 \, {\left (3 \, B a^{2} + 7 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{24 \, x^{3}}\right ] \]

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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giac [A]  time = 0.47, size = 238, normalized size = 1.63 \[ \frac {1}{8} \, {\left (2 \, B b^{2} x^{2} + \frac {9 \, B a b^{3} + 4 \, A b^{4}}{b^{2}}\right )} \sqrt {b x^{2} + a} x - \frac {5}{16} \, {\left (3 \, B a^{2} \sqrt {b} + 4 \, A a b^{\frac {3}{2}}\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, {\left (3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a^{3} \sqrt {b} + 9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a^{2} b^{\frac {3}{2}} - 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{4} \sqrt {b} - 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{3} b^{\frac {3}{2}} + 3 \, B a^{5} \sqrt {b} + 7 \, A a^{4} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{3}} \]

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

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maple [A]  time = 0.01, size = 204, normalized size = 1.40 \[ \frac {5 A a \,b^{\frac {3}{2}} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2}+\frac {15 B \,a^{2} \sqrt {b}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8}+\frac {5 \sqrt {b \,x^{2}+a}\, A \,b^{2} x}{2}+\frac {15 \sqrt {b \,x^{2}+a}\, B a b x}{8}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{2} x}{3 a}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B b x}{4}+\frac {4 \left (b \,x^{2}+a \right )^{\frac {5}{2}} A \,b^{2} x}{3 a^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B b x}{a}-\frac {4 \left (b \,x^{2}+a \right )^{\frac {7}{2}} A b}{3 a^{2} x}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} B}{a x}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A}{3 a \,x^{3}} \]

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]

-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(b^(1/2)*x+(b*x^2+a)^(1/2))-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(b^(1/2)*x+(b
*x^2+a)^(1/2))

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maxima [A]  time = 1.05, size = 151, normalized size = 1.03 \[ \frac {5}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b x + \frac {15}{8} \, \sqrt {b x^{2} + a} B a b x + \frac {5}{2} \, \sqrt {b x^{2} + a} A b^{2} x + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2} x}{3 \, a} + \frac {15}{8} \, B a^{2} \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) + \frac {5}{2} \, A a b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{x} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{3 \, a x} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{3 \, a x^{3}} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{5/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^4,x)

[Out]

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^4, x)

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sympy [B]  time = 15.71, size = 299, normalized size = 2.05 \[ - \frac {2 A a^{\frac {3}{2}} b}{x \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {A \sqrt {a} b^{2} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} - \frac {2 A \sqrt {a} b^{2} x}{\sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3} + \frac {5 A a b^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2} - \frac {B a^{\frac {5}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + B a^{\frac {3}{2}} b x \sqrt {1 + \frac {b x^{2}}{a}} - \frac {7 B a^{\frac {3}{2}} b x}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B \sqrt {a} b^{2} x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {15 B a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8} + \frac {B b^{3} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

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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