\(\int \frac {(A+B x) (a+b x^2)^{3/2}}{x} \, dx\) [12]

Optimal result

Integrand size = 20, antiderivative size = 106 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=\frac {1}{8} a (8 A+3 B x) \sqrt {a+b x^2}+\frac {1}{12} (4 A+3 B x) \left (a+b x^2\right )^{3/2}+\frac {3 a^2 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}-a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {829, 858, 223, 212, 272, 65, 214} \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=-a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {3 a^2 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}+\frac {1}{8} a \sqrt {a+b x^2} (8 A+3 B x)+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 A+3 B x) \]

[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 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 223

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 272

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 829

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 858

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*} \text {integral}& = \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 ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )+\frac {1}{8} \left (3 a^2 B\right ) \text {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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {1}{24} \left (\sqrt {a+b x^2} \left (32 a A+15 a B x+8 A b x^2+6 b B x^3\right )-\frac {9 a^2 B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {b}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.39 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.03

method	result	size
default	\(B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+A \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\)	\(109\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 439, normalized size of antiderivative = 4.14 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=\left [\frac {9 \, B a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 24 \, A a^{\frac {3}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{48 \, b}, -\frac {9 \, B a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 12 \, A a^{\frac {3}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{24 \, b}, \frac {48 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 9 \, B a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{48 \, b}, -\frac {9 \, B a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 24 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{24 \, b}\right ] \]

[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]

Sympy [A] (verification not implemented)

Time = 4.26 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.58 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=- A a^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {A a^{2}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A a \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + A b \left (\begin {cases} \frac {a \sqrt {a + b x^{2}}}{3 b} + \frac {x^{2} \sqrt {a + b x^{2}}}{3} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + B a \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) + B b \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a x \sqrt {a + b x^{2}}}{8 b} + \frac {x^{3} \sqrt {a + b x^{2}}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \]

[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((a*sqrt(a + b*x**2)/(3*b) + x**2*sqrt(a + b*x**2)/3, Ne(b, 0)), (sqrt(a)*x**2/2, True)
) + B*a*Piecewise((a*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x
**2), True))/2 + x*sqrt(a + b*x**2)/2, Ne(b, 0)), (sqrt(a)*x, True)) + B*b*Piecewise((-a**2*Piecewise((log(2*s
qrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(8*b) + a*x*sqrt(a + b*x**
2)/(8*b) + x**3*sqrt(a + b*x**2)/4, Ne(b, 0)), (sqrt(a)*x**3/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=\frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B x + \frac {3}{8} \, \sqrt {b x^{2} + a} B a x + \frac {3 \, B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} - A a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A + \sqrt {b x^{2} + a} A a \]

[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

Giac [F(-2)]

Exception generated. \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 6.14 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=\frac {A\,{\left (b\,x^2+a\right )}^{3/2}}{3}-A\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )+A\,a\,\sqrt {b\,x^2+a}+\frac {B\,x\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \]

[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)