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

Optimal. Leaf size=106 \[ -a^{3/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {3 a^2 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}+\frac {1}{8} a \sqrt {a+b x^2} (8 A+3 B x)+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 A+3 B x) \]

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Rubi [A]  time = 0.09, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {815, 844, 217, 206, 266, 63, 208} \begin {gather*} -a^{3/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {3 a^2 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}+\frac {1}{8} a \sqrt {a+b x^2} (8 A+3 B x)+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 A+3 B x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.31, size = 118, normalized size = 1.11 \begin {gather*} \frac {1}{24} \left (-24 a^{3/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {9 a^{5/2} B \sqrt {\frac {b x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {a+b x^2}}+\sqrt {a+b x^2} \left (32 a A+15 a B x+8 A b x^2+6 b B x^3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(32*a*A + 15*a*B*x + 8*A*b*x^2 + 6*b*B*x^3) + (9*a^(5/2)*B*Sqrt[1 + (b*x^2)/a]*ArcSinh[(Sqrt[
b]*x)/Sqrt[a]])/(Sqrt[b]*Sqrt[a + b*x^2]) - 24*a^(3/2)*A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/24

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IntegrateAlgebraic [A]  time = 0.40, size = 114, normalized size = 1.08 \begin {gather*} 2 a^{3/2} A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}-\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {3 a^2 B \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{8 \sqrt {b}}+\frac {1}{24} \sqrt {a+b x^2} \left (32 a A+15 a B x+8 A b x^2+6 b B x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(32*a*A + 15*a*B*x + 8*A*b*x^2 + 6*b*B*x^3))/24 + 2*a^(3/2)*A*ArcTanh[(Sqrt[b]*x)/Sqrt[a] - S
qrt[a + b*x^2]/Sqrt[a]] - (3*a^2*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*Sqrt[b])

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fricas [A]  time = 0.89, size = 439, normalized size = 4.14 \begin {gather*} \left [\frac {9 \, B a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 24 \, A a^{\frac {3}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{48 \, b}, -\frac {9 \, B a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 12 \, A a^{\frac {3}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{24 \, b}, \frac {48 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 9 \, B a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{48 \, b}, -\frac {9 \, B a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 24 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{24 \, b}\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,x, algorithm="fricas")

[Out]

[1/48*(9*B*a^2*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 24*A*a^(3/2)*b*log(-(b*x^2 - 2*sqrt(b
*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(6*B*b^2*x^3 + 8*A*b^2*x^2 + 15*B*a*b*x + 32*A*a*b)*sqrt(b*x^2 + a))/b, -1/2
4*(9*B*a^2*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 12*A*a^(3/2)*b*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a
) + 2*a)/x^2) - (6*B*b^2*x^3 + 8*A*b^2*x^2 + 15*B*a*b*x + 32*A*a*b)*sqrt(b*x^2 + a))/b, 1/48*(48*A*sqrt(-a)*a*
b*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + 9*B*a^2*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(6*B*
b^2*x^3 + 8*A*b^2*x^2 + 15*B*a*b*x + 32*A*a*b)*sqrt(b*x^2 + a))/b, -1/24*(9*B*a^2*sqrt(-b)*arctan(sqrt(-b)*x/s
qrt(b*x^2 + a)) - 24*A*sqrt(-a)*a*b*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (6*B*b^2*x^3 + 8*A*b^2*x^2 + 15*B*a*b*x
+ 32*A*a*b)*sqrt(b*x^2 + a))/b]

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giac [A]  time = 0.54, size = 100, normalized size = 0.94 \begin {gather*} \frac {2 \, A a^{2} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3 \, B a^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, \sqrt {b}} + \frac {1}{24} \, \sqrt {b x^{2} + a} {\left (32 \, A a + {\left (15 \, B a + 2 \, {\left (3 \, B b x + 4 \, A b\right )} x\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,x, algorithm="giac")

[Out]

2*A*a^2*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 3/8*B*a^2*log(abs(-sqrt(b)*x + sqrt(b*x^2 +
a)))/sqrt(b) + 1/24*sqrt(b*x^2 + a)*(32*A*a + (15*B*a + 2*(3*B*b*x + 4*A*b)*x)*x)

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maple [A]  time = 0.01, size = 107, normalized size = 1.01 \begin {gather*} -A \,a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )+\frac {3 B \,a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 \sqrt {b}}+\frac {3 \sqrt {b \,x^{2}+a}\, B a x}{8}+\sqrt {b \,x^{2}+a}\, A a +\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B x}{4}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x^2+a)^(3/2)*B*x+3/8*(b*x^2+a)^(1/2)*B*a*x+3/8*B*a^2/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+1/3*A*(b*x^2
+a)^(3/2)-A*a^(3/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)+A*(b*x^2+a)^(1/2)*a

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maxima [A]  time = 1.32, size = 88, normalized size = 0.83 \begin {gather*} \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B x + \frac {3}{8} \, \sqrt {b x^{2} + a} B a x + \frac {3 \, B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} - A 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}} A + \sqrt {b x^{2} + a} A a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x^2 + a)^(3/2)*B*x + 3/8*sqrt(b*x^2 + a)*B*a*x + 3/8*B*a^2*arcsinh(b*x/sqrt(a*b))/sqrt(b) - A*a^(3/2)*a
rcsinh(a/(sqrt(a*b)*abs(x))) + 1/3*(b*x^2 + a)^(3/2)*A + sqrt(b*x^2 + a)*A*a

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mupad [B]  time = 1.31, size = 83, normalized size = 0.78 \begin {gather*} \frac {A\,{\left (b\,x^2+a\right )}^{3/2}}{3}-A\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )+A\,a\,\sqrt {b\,x^2+a}+\frac {B\,x\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\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,x)

[Out]

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

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sympy [A]  time = 35.49, size = 218, normalized size = 2.06 \begin {gather*} - A a^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {A a^{2}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A a \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + A 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 ) + \frac {B a^{\frac {3}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {B a^{\frac {3}{2}} x}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B \sqrt {a} b x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 \sqrt {b}} + \frac {B b^{2} x^{5}}{4 \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+A)*(b*x**2+a)**(3/2)/x,x)

[Out]

-A*a**(3/2)*asinh(sqrt(a)/(sqrt(b)*x)) + A*a**2/(sqrt(b)*x*sqrt(a/(b*x**2) + 1)) + A*a*sqrt(b)*x/sqrt(a/(b*x**
2) + 1) + A*b*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3/2)/(3*b), True)) + B*a**(3/2)*x*sqrt(1 +
b*x**2/a)/2 + B*a**(3/2)*x/(8*sqrt(1 + b*x**2/a)) + 3*B*sqrt(a)*b*x**3/(8*sqrt(1 + b*x**2/a)) + 3*B*a**2*asin
h(sqrt(b)*x/sqrt(a))/(8*sqrt(b)) + B*b**2*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a))

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