3.6.12 \(\int \frac {(a+b x^2)^{3/2} (A+B x^2)}{x^4} \, dx\)

Optimal. Leaf size=119 \[ -\frac {\left (a+b x^2\right )^{3/2} (3 a B+2 A b)}{3 a x}+\frac {b x \sqrt {a+b x^2} (3 a B+2 A b)}{2 a}+\frac {1}{2} \sqrt {b} (3 a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {A \left (a+b x^2\right )^{5/2}}{3 a x^3} \]

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Rubi [A]  time = 0.05, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \begin {gather*} -\frac {\left (a+b x^2\right )^{3/2} (3 a B+2 A b)}{3 a x}+\frac {b x \sqrt {a+b x^2} (3 a B+2 A b)}{2 a}+\frac {1}{2} \sqrt {b} (3 a B+2 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {A \left (a+b x^2\right )^{5/2}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/x^4,x]

[Out]

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

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 )^{3/2} \left (A+B x^2\right )}{x^4} \, dx &=-\frac {A \left (a+b x^2\right )^{5/2}}{3 a x^3}-\frac {(-2 A b-3 a B) \int \frac {\left (a+b x^2\right )^{3/2}}{x^2} \, dx}{3 a}\\ &=-\frac {(2 A b+3 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {A \left (a+b x^2\right )^{5/2}}{3 a x^3}+\frac {(b (2 A b+3 a B)) \int \sqrt {a+b x^2} \, dx}{a}\\ &=\frac {b (2 A b+3 a B) x \sqrt {a+b x^2}}{2 a}-\frac {(2 A b+3 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {A \left (a+b x^2\right )^{5/2}}{3 a x^3}+\frac {1}{2} (b (2 A b+3 a B)) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=\frac {b (2 A b+3 a B) x \sqrt {a+b x^2}}{2 a}-\frac {(2 A b+3 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {A \left (a+b x^2\right )^{5/2}}{3 a x^3}+\frac {1}{2} (b (2 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 {b (2 A b+3 a B) x \sqrt {a+b x^2}}{2 a}-\frac {(2 A b+3 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {A \left (a+b x^2\right )^{5/2}}{3 a x^3}+\frac {1}{2} \sqrt {b} (2 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 = 0.08, size = 83, normalized size = 0.70 \begin {gather*} \frac {\sqrt {a+b x^2} (-3 a B-2 A b) \, _2F_1\left (-\frac {3}{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 )^{5/2}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/x^4,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.23, size = 88, normalized size = 0.74 \begin {gather*} \frac {1}{2} \left (-3 a \sqrt {b} B-2 A b^{3/2}\right ) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )+\frac {\sqrt {a+b x^2} \left (-2 a A-6 a B x^2-8 A b x^2+3 b B x^4\right )}{6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^2)^(3/2)*(A + B*x^2))/x^4,x]

[Out]

(Sqrt[a + b*x^2]*(-2*a*A - 8*A*b*x^2 - 6*a*B*x^2 + 3*b*B*x^4))/(6*x^3) + ((-2*A*b^(3/2) - 3*a*Sqrt[b]*B)*Log[-
(Sqrt[b]*x) + Sqrt[a + b*x^2]])/2

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fricas [A]  time = 0.98, size = 166, normalized size = 1.39 \begin {gather*} \left [\frac {3 \, {\left (3 \, B a + 2 \, 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 (3 \, B b x^{4} - 2 \, {\left (3 \, B a + 4 \, A b\right )} x^{2} - 2 \, A a\right )} \sqrt {b x^{2} + a}}{12 \, x^{3}}, -\frac {3 \, {\left (3 \, B a + 2 \, A b\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (3 \, B b x^{4} - 2 \, {\left (3 \, B a + 4 \, A b\right )} x^{2} - 2 \, A a\right )} \sqrt {b x^{2} + a}}{6 \, x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^4,x, algorithm="fricas")

[Out]

[1/12*(3*(3*B*a + 2*A*b)*sqrt(b)*x^3*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(3*B*b*x^4 - 2*(3*B*a
+ 4*A*b)*x^2 - 2*A*a)*sqrt(b*x^2 + a))/x^3, -1/6*(3*(3*B*a + 2*A*b)*sqrt(-b)*x^3*arctan(sqrt(-b)*x/sqrt(b*x^2
+ a)) - (3*B*b*x^4 - 2*(3*B*a + 4*A*b)*x^2 - 2*A*a)*sqrt(b*x^2 + a))/x^3]

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giac [B]  time = 0.52, size = 207, normalized size = 1.74 \begin {gather*} \frac {1}{2} \, \sqrt {b x^{2} + a} B b x - \frac {1}{4} \, {\left (3 \, B a \sqrt {b} + 2 \, 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^{2} \sqrt {b} + 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a b^{\frac {3}{2}} - 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{3} \sqrt {b} - 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{2} b^{\frac {3}{2}} + 3 \, B a^{4} \sqrt {b} + 4 \, A a^{3} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^4,x, algorithm="giac")

[Out]

1/2*sqrt(b*x^2 + a)*B*b*x - 1/4*(3*B*a*sqrt(b) + 2*A*b^(3/2))*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2/3*(3*(s
qrt(b)*x - sqrt(b*x^2 + a))^4*B*a^2*sqrt(b) + 6*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*a*b^(3/2) - 6*(sqrt(b)*x - s
qrt(b*x^2 + a))^2*B*a^3*sqrt(b) - 6*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^2*b^(3/2) + 3*B*a^4*sqrt(b) + 4*A*a^3*
b^(3/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^3

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maple [A]  time = 0.01, size = 168, normalized size = 1.41 \begin {gather*} A \,b^{\frac {3}{2}} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )+\frac {3 B a \sqrt {b}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2}+\frac {\sqrt {b \,x^{2}+a}\, A \,b^{2} x}{a}+\frac {3 \sqrt {b \,x^{2}+a}\, B b x}{2}+\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{2} x}{3 a^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B b x}{a}-\frac {2 \left (b \,x^{2}+a \right )^{\frac {5}{2}} A b}{3 a^{2} x}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B}{a x}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A}{3 a \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)*(B*x^2+A)/x^4,x)

[Out]

-B/a/x*(b*x^2+a)^(5/2)+B*b/a*x*(b*x^2+a)^(3/2)+3/2*B*b*x*(b*x^2+a)^(1/2)+3/2*B*b^(1/2)*a*ln(b^(1/2)*x+(b*x^2+a
)^(1/2))-1/3*A*(b*x^2+a)^(5/2)/a/x^3-2/3*A*b/a^2/x*(b*x^2+a)^(5/2)+2/3*A*b^2/a^2*x*(b*x^2+a)^(3/2)+A*b^2/a*x*(
b*x^2+a)^(1/2)+A*b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.01, size = 115, normalized size = 0.97 \begin {gather*} \frac {3}{2} \, \sqrt {b x^{2} + a} B b x + \frac {\sqrt {b x^{2} + a} A b^{2} x}{a} + \frac {3}{2} \, B a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) + A b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{x} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{3 \, a x} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{3 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^4,x, algorithm="maxima")

[Out]

3/2*sqrt(b*x^2 + a)*B*b*x + sqrt(b*x^2 + a)*A*b^2*x/a + 3/2*B*a*sqrt(b)*arcsinh(b*x/sqrt(a*b)) + A*b^(3/2)*arc
sinh(b*x/sqrt(a*b)) - (b*x^2 + a)^(3/2)*B/x - 2/3*(b*x^2 + a)^(3/2)*A*b/(a*x) - 1/3*(b*x^2 + a)^(5/2)*A/(a*x^3
)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{3/2}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^(3/2))/x^4,x)

[Out]

int(((A + B*x^2)*(a + b*x^2)^(3/2))/x^4, x)

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sympy [A]  time = 7.75, size = 202, normalized size = 1.70 \begin {gather*} - \frac {A \sqrt {a} b}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A a \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {A b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3} + A b^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {A b^{2} x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{\frac {3}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {B \sqrt {a} b x \sqrt {1 + \frac {b x^{2}}{a}}}{2} - \frac {B \sqrt {a} b x}{\sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/x**4,x)

[Out]

-A*sqrt(a)*b/(x*sqrt(1 + b*x**2/a)) - A*a*sqrt(b)*sqrt(a/(b*x**2) + 1)/(3*x**2) - A*b**(3/2)*sqrt(a/(b*x**2) +
1)/3 + A*b**(3/2)*asinh(sqrt(b)*x/sqrt(a)) - A*b**2*x/(sqrt(a)*sqrt(1 + b*x**2/a)) - B*a**(3/2)/(x*sqrt(1 + b
*x**2/a)) + B*sqrt(a)*b*x*sqrt(1 + b*x**2/a)/2 - B*sqrt(a)*b*x/sqrt(1 + b*x**2/a) + 3*B*a*sqrt(b)*asinh(sqrt(b
)*x/sqrt(a))/2

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