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

Optimal. Leaf size=136 \[ \frac {5 a^2 (a B+6 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}+\frac {x \left (a+b x^2\right )^{5/2} (a B+6 A b)}{6 a}+\frac {5}{24} x \left (a+b x^2\right )^{3/2} (a B+6 A b)+\frac {5}{16} a x \sqrt {a+b x^2} (a B+6 A b)-\frac {A \left (a+b x^2\right )^{7/2}}{a x} \]

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Rubi [A]  time = 0.05, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {453, 195, 217, 206} \begin {gather*} \frac {5 a^2 (a B+6 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}+\frac {x \left (a+b x^2\right )^{5/2} (a B+6 A b)}{6 a}+\frac {5}{24} x \left (a+b x^2\right )^{3/2} (a B+6 A b)+\frac {5}{16} a x \sqrt {a+b x^2} (a B+6 A b)-\frac {A \left (a+b x^2\right )^{7/2}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*(6*A*b + a*B)*x*Sqrt[a + b*x^2])/16 + (5*(6*A*b + a*B)*x*(a + b*x^2)^(3/2))/24 + ((6*A*b + a*B)*x*(a + b*
x^2)^(5/2))/(6*a) - (A*(a + b*x^2)^(7/2))/(a*x) + (5*a^2*(6*A*b + a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(
16*Sqrt[b])

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 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^2} \, dx &=-\frac {A \left (a+b x^2\right )^{7/2}}{a x}-\frac {(-6 A b-a B) \int \left (a+b x^2\right )^{5/2} \, dx}{a}\\ &=\frac {(6 A b+a B) x \left (a+b x^2\right )^{5/2}}{6 a}-\frac {A \left (a+b x^2\right )^{7/2}}{a x}+\frac {1}{6} (5 (6 A b+a B)) \int \left (a+b x^2\right )^{3/2} \, dx\\ &=\frac {5}{24} (6 A b+a B) x \left (a+b x^2\right )^{3/2}+\frac {(6 A b+a B) x \left (a+b x^2\right )^{5/2}}{6 a}-\frac {A \left (a+b x^2\right )^{7/2}}{a x}+\frac {1}{8} (5 a (6 A b+a B)) \int \sqrt {a+b x^2} \, dx\\ &=\frac {5}{16} a (6 A b+a B) x \sqrt {a+b x^2}+\frac {5}{24} (6 A b+a B) x \left (a+b x^2\right )^{3/2}+\frac {(6 A b+a B) x \left (a+b x^2\right )^{5/2}}{6 a}-\frac {A \left (a+b x^2\right )^{7/2}}{a x}+\frac {1}{16} \left (5 a^2 (6 A b+a B)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=\frac {5}{16} a (6 A b+a B) x \sqrt {a+b x^2}+\frac {5}{24} (6 A b+a B) x \left (a+b x^2\right )^{3/2}+\frac {(6 A b+a B) x \left (a+b x^2\right )^{5/2}}{6 a}-\frac {A \left (a+b x^2\right )^{7/2}}{a x}+\frac {1}{16} \left (5 a^2 (6 A b+a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=\frac {5}{16} a (6 A b+a B) x \sqrt {a+b x^2}+\frac {5}{24} (6 A b+a B) x \left (a+b x^2\right )^{3/2}+\frac {(6 A b+a B) x \left (a+b x^2\right )^{5/2}}{6 a}-\frac {A \left (a+b x^2\right )^{7/2}}{a x}+\frac {5 a^2 (6 A b+a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 125, normalized size = 0.92 \begin {gather*} \frac {\sqrt {a+b x^2} \left (\frac {(a B+6 A b) \left (15 a^{5/2} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+\sqrt {b} x \sqrt {\frac {b x^2}{a}+1} \left (33 a^2+26 a b x^2+8 b^2 x^4\right )\right )}{\sqrt {b} \sqrt {\frac {b x^2}{a}+1}}-\frac {48 A \left (a+b x^2\right )^3}{x}\right )}{48 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*((-48*A*(a + b*x^2)^3)/x + ((6*A*b + a*B)*(Sqrt[b]*x*Sqrt[1 + (b*x^2)/a]*(33*a^2 + 26*a*b*x^2
+ 8*b^2*x^4) + 15*a^(5/2)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]]))/(Sqrt[b]*Sqrt[1 + (b*x^2)/a])))/(48*a)

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IntegrateAlgebraic [A]  time = 0.25, size = 112, normalized size = 0.82 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-48 a^2 A+33 a^2 B x^2+54 a A b x^2+26 a b B x^4+12 A b^2 x^4+8 b^2 B x^6\right )}{48 x}-\frac {5 \left (a^3 B+6 a^2 A b\right ) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{16 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-48*a^2*A + 54*a*A*b*x^2 + 33*a^2*B*x^2 + 12*A*b^2*x^4 + 26*a*b*B*x^4 + 8*b^2*B*x^6))/(48*x)
- (5*(6*a^2*A*b + a^3*B)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16*Sqrt[b])

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fricas [A]  time = 1.13, size = 236, normalized size = 1.74 \begin {gather*} \left [\frac {15 \, {\left (B a^{3} + 6 \, A a^{2} b\right )} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, B b^{3} x^{6} + 2 \, {\left (13 \, B a b^{2} + 6 \, A b^{3}\right )} x^{4} - 48 \, A a^{2} b + 3 \, {\left (11 \, B a^{2} b + 18 \, A a b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{96 \, b x}, -\frac {15 \, {\left (B a^{3} + 6 \, A a^{2} b\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, B b^{3} x^{6} + 2 \, {\left (13 \, B a b^{2} + 6 \, A b^{3}\right )} x^{4} - 48 \, A a^{2} b + 3 \, {\left (11 \, B a^{2} b + 18 \, A a b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, b x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(15*(B*a^3 + 6*A*a^2*b)*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(8*B*b^3*x^6 + 2*(
13*B*a*b^2 + 6*A*b^3)*x^4 - 48*A*a^2*b + 3*(11*B*a^2*b + 18*A*a*b^2)*x^2)*sqrt(b*x^2 + a))/(b*x), -1/48*(15*(B
*a^3 + 6*A*a^2*b)*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*B*b^3*x^6 + 2*(13*B*a*b^2 + 6*A*b^3)*x^4
- 48*A*a^2*b + 3*(11*B*a^2*b + 18*A*a*b^2)*x^2)*sqrt(b*x^2 + a))/(b*x)]

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giac [A]  time = 0.54, size = 146, normalized size = 1.07 \begin {gather*} \frac {2 \, A a^{3} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{48} \, {\left (2 \, {\left (4 \, B b^{2} x^{2} + \frac {13 \, B a b^{5} + 6 \, A b^{6}}{b^{4}}\right )} x^{2} + \frac {3 \, {\left (11 \, B a^{2} b^{4} + 18 \, A a b^{5}\right )}}{b^{4}}\right )} \sqrt {b x^{2} + a} x - \frac {5 \, {\left (B a^{3} \sqrt {b} + 6 \, A a^{2} b^{\frac {3}{2}}\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right )}{32 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a^3*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a) + 1/48*(2*(4*B*b^2*x^2 + (13*B*a*b^5 + 6*A*b^6)/b^4)*x^2
+ 3*(11*B*a^2*b^4 + 18*A*a*b^5)/b^4)*sqrt(b*x^2 + a)*x - 5/32*(B*a^3*sqrt(b) + 6*A*a^2*b^(3/2))*log((sqrt(b)*
x - sqrt(b*x^2 + a))^2)/b

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maple [A]  time = 0.01, size = 158, normalized size = 1.16 \begin {gather*} \frac {15 A \,a^{2} \sqrt {b}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8}+\frac {5 B \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 \sqrt {b}}+\frac {15 \sqrt {b \,x^{2}+a}\, A a b x}{8}+\frac {5 \sqrt {b \,x^{2}+a}\, B \,a^{2} x}{16}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A b x}{4}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B a x}{24}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A b x}{a}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B x}{6}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x*B*(b*x^2+a)^(5/2)+5/24*B*a*x*(b*x^2+a)^(3/2)+5/16*B*a^2*x*(b*x^2+a)^(1/2)+5/16*B*a^3/b^(1/2)*ln(b^(1/2)*
x+(b*x^2+a)^(1/2))-A*(b*x^2+a)^(7/2)/a/x+A*b/a*x*(b*x^2+a)^(5/2)+5/4*A*b*x*(b*x^2+a)^(3/2)+15/8*A*b*a*x*(b*x^2
+a)^(1/2)+15/8*A*b^(1/2)*a^2*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 0.99, size = 124, normalized size = 0.91 \begin {gather*} \frac {1}{6} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B x + \frac {5}{24} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x + \frac {5}{16} \, \sqrt {b x^{2} + a} B a^{2} x + \frac {5}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b x + \frac {15}{8} \, \sqrt {b x^{2} + a} A a b x + \frac {5 \, B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {b}} + \frac {15}{8} \, A a^{2} \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b*x^2 + a)^(5/2)*B*x + 5/24*(b*x^2 + a)^(3/2)*B*a*x + 5/16*sqrt(b*x^2 + a)*B*a^2*x + 5/4*(b*x^2 + a)^(3/2
)*A*b*x + 15/8*sqrt(b*x^2 + a)*A*a*b*x + 5/16*B*a^3*arcsinh(b*x/sqrt(a*b))/sqrt(b) + 15/8*A*a^2*sqrt(b)*arcsin
h(b*x/sqrt(a*b)) - (b*x^2 + a)^(5/2)*A/x

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mupad [B]  time = 1.92, size = 80, normalized size = 0.59 \begin {gather*} \frac {B\,x\,{\left (b\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{5/2}}-\frac {A\,{\left (b\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*x*(a + b*x^2)^(5/2)*hypergeom([-5/2, 1/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^(5/2) - (A*(a + b*x^2)^(5/2)*h
ypergeom([-5/2, -1/2], 1/2, -(b*x^2)/a))/(x*((b*x^2)/a + 1)^(5/2))

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sympy [B]  time = 24.29, size = 306, normalized size = 2.25 \begin {gather*} - \frac {A a^{\frac {5}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + A a^{\frac {3}{2}} b x \sqrt {1 + \frac {b x^{2}}{a}} - \frac {7 A a^{\frac {3}{2}} b x}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 A \sqrt {a} b^{2} x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {15 A a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8} + \frac {A b^{3} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {B a^{\frac {5}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {3 B a^{\frac {5}{2}} x}{16 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {35 B a^{\frac {3}{2}} b x^{3}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {17 B \sqrt {a} b^{2} x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 \sqrt {b}} + \frac {B b^{3} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*a**(5/2)/(x*sqrt(1 + b*x**2/a)) + A*a**(3/2)*b*x*sqrt(1 + b*x**2/a) - 7*A*a**(3/2)*b*x/(8*sqrt(1 + b*x**2/a
)) + 3*A*sqrt(a)*b**2*x**3/(8*sqrt(1 + b*x**2/a)) + 15*A*a**2*sqrt(b)*asinh(sqrt(b)*x/sqrt(a))/8 + A*b**3*x**5
/(4*sqrt(a)*sqrt(1 + b*x**2/a)) + B*a**(5/2)*x*sqrt(1 + b*x**2/a)/2 + 3*B*a**(5/2)*x/(16*sqrt(1 + b*x**2/a)) +
35*B*a**(3/2)*b*x**3/(48*sqrt(1 + b*x**2/a)) + 17*B*sqrt(a)*b**2*x**5/(24*sqrt(1 + b*x**2/a)) + 5*B*a**3*asin
h(sqrt(b)*x/sqrt(a))/(16*sqrt(b)) + B*b**3*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))

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