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

Optimal. Leaf size=108 \[ a^{3/2} (-B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac {1}{2} \sqrt {a+b x^2} (2 a B+3 A b x)+\frac {3}{2} a A \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {813, 815, 844, 217, 206, 266, 63, 208} \begin {gather*} a^{3/2} (-B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac {1}{2} \sqrt {a+b x^2} (2 a B+3 A b x)+\frac {3}{2} a A \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx &=-\frac {(3 A-B x) \left (a+b x^2\right )^{3/2}}{3 x}-\frac {1}{2} \int \frac {(-2 a B-6 A b x) \sqrt {a+b x^2}}{x} \, dx\\ &=\frac {1}{2} (2 a B+3 A b x) \sqrt {a+b x^2}-\frac {(3 A-B x) \left (a+b x^2\right )^{3/2}}{3 x}-\frac {\int \frac {-4 a^2 b B-6 a A b^2 x}{x \sqrt {a+b x^2}} \, dx}{4 b}\\ &=\frac {1}{2} (2 a B+3 A b x) \sqrt {a+b x^2}-\frac {(3 A-B x) \left (a+b x^2\right )^{3/2}}{3 x}+\frac {1}{2} (3 a A b) \int \frac {1}{\sqrt {a+b x^2}} \, dx+\left (a^2 B\right ) \int \frac {1}{x \sqrt {a+b x^2}} \, dx\\ &=\frac {1}{2} (2 a B+3 A b x) \sqrt {a+b x^2}-\frac {(3 A-B x) \left (a+b x^2\right )^{3/2}}{3 x}+\frac {1}{2} (3 a A b) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )+\frac {1}{2} \left (a^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=\frac {1}{2} (2 a B+3 A b x) \sqrt {a+b x^2}-\frac {(3 A-B x) \left (a+b x^2\right )^{3/2}}{3 x}+\frac {3}{2} a A \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\frac {\left (a^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{b}\\ &=\frac {1}{2} (2 a B+3 A b x) \sqrt {a+b x^2}-\frac {(3 A-B x) \left (a+b x^2\right )^{3/2}}{3 x}+\frac {3}{2} a A \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-a^{3/2} B \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.18, size = 105, normalized size = 0.97 \begin {gather*} -a^{3/2} B \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {a^2 A \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {b x^2}{a}\right )}{x \sqrt {a+b x^2}}+\frac {1}{3} B \sqrt {a+b x^2} \left (4 a+b x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.40, size = 115, normalized size = 1.06 \begin {gather*} 2 a^{3/2} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}-\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {\sqrt {a+b x^2} \left (-6 a A+8 a B x+3 A b x^2+2 b B x^3\right )}{6 x}-\frac {3}{2} a A \sqrt {b} \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.85, size = 411, normalized size = 3.81 \begin {gather*} \left [\frac {9 \, A a \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 6 \, B a^{\frac {3}{2}} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{12 \, x}, -\frac {9 \, A a \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 3 \, B a^{\frac {3}{2}} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{6 \, x}, \frac {12 \, B \sqrt {-a} a x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 9 \, A a \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{12 \, x}, -\frac {9 \, A a \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 6 \, B \sqrt {-a} a x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{6 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(9*A*a*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 6*B*a^(3/2)*x*log(-(b*x^2 - 2*sqrt(b*
x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sqrt(b*x^2 + a))/x, -1/6*(9*A*a*sqr
t(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 3*B*a^(3/2)*x*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2)
- (2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sqrt(b*x^2 + a))/x, 1/12*(12*B*sqrt(-a)*a*x*arctan(sqrt(-a)/sqrt(
b*x^2 + a)) + 9*A*a*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*B*b*x^3 + 3*A*b*x^2 + 8*B
*a*x - 6*A*a)*sqrt(b*x^2 + a))/x, -1/6*(9*A*a*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 6*B*sqrt(-a)*a*x
*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sqrt(b*x^2 + a))/x]

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giac [A]  time = 0.60, size = 124, normalized size = 1.15 \begin {gather*} \frac {2 \, B a^{2} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3}{2} \, A a \sqrt {b} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \frac {2 \, A a^{2} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{6} \, \sqrt {b x^{2} + a} {\left (8 \, B a + {\left (2 \, B b x + 3 \, A b\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B*a^2*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 3/2*A*a*sqrt(b)*log(abs(-sqrt(b)*x + sqrt(b
*x^2 + a))) + 2*A*a^2*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a) + 1/6*sqrt(b*x^2 + a)*(8*B*a + (2*B*b*x +
3*A*b)*x)

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maple [A]  time = 0.01, size = 126, normalized size = 1.17 \begin {gather*} \frac {3 A a \sqrt {b}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2}-B \,a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )+\frac {3 \sqrt {b \,x^{2}+a}\, A b x}{2}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A b x}{a}+\sqrt {b \,x^{2}+a}\, B a +\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B}{3}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.39, size = 88, normalized size = 0.81 \begin {gather*} \frac {3}{2} \, \sqrt {b x^{2} + a} A b x + \frac {3}{2} \, A a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - B a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B + \sqrt {b x^{2} + a} B a - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.88, size = 86, normalized size = 0.80 \begin {gather*} \frac {B\,{\left (b\,x^2+a\right )}^{3/2}}{3}-B\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )+B\,a\,\sqrt {b\,x^2+a}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 13.29, size = 184, normalized size = 1.70 \begin {gather*} - \frac {A a^{\frac {3}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {A \sqrt {a} b x \sqrt {1 + \frac {b x^{2}}{a}}}{2} - \frac {A \sqrt {a} b x}{\sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 A a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2} - B a^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {B a^{2}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B a \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + B b \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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