3.1.14 \(\int \frac {(A+B x) (a+b x^2)^{3/2}}{x^3} \, dx\) [14]

Optimal. Leaf size=111 \[ -\frac {3 (a B-A b x) \sqrt {a+b x^2}}{2 x}-\frac {(A-B x) \left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {3}{2} a \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {3}{2} \sqrt {a} A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 111, 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 = {827, 858, 223, 212, 272, 65, 214} \begin {gather*} -\frac {\left (a+b x^2\right )^{3/2} (A-B x)}{2 x^2}-\frac {3 \sqrt {a+b x^2} (a B-A b x)}{2 x}-\frac {3}{2} \sqrt {a} A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {3}{2} a \sqrt {b} 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^3,x]

[Out]

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

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 827

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 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*} \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^3} \, dx &=-\frac {(A-B x) \left (a+b x^2\right )^{3/2}}{2 x^2}-\frac {3}{8} \int \frac {(-4 a B-4 A b x) \sqrt {a+b x^2}}{x^2} \, dx\\ &=-\frac {3 (a B-A b x) \sqrt {a+b x^2}}{2 x}-\frac {(A-B x) \left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {3}{16} \int \frac {8 a A b+8 a b B x}{x \sqrt {a+b x^2}} \, dx\\ &=-\frac {3 (a B-A b x) \sqrt {a+b x^2}}{2 x}-\frac {(A-B x) \left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {1}{2} (3 a A b) \int \frac {1}{x \sqrt {a+b x^2}} \, dx+\frac {1}{2} (3 a b B) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=-\frac {3 (a B-A b x) \sqrt {a+b x^2}}{2 x}-\frac {(A-B x) \left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {1}{4} (3 a A b) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )+\frac {1}{2} (3 a b B) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=-\frac {3 (a B-A b x) \sqrt {a+b x^2}}{2 x}-\frac {(A-B x) \left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {3}{2} a \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\frac {1}{2} (3 a A) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )\\ &=-\frac {3 (a B-A b x) \sqrt {a+b x^2}}{2 x}-\frac {(A-B x) \left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {3}{2} a \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {3}{2} \sqrt {a} A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.36, size = 109, normalized size = 0.98 \begin {gather*} \frac {1}{2} \left (\frac {\sqrt {a+b x^2} \left (b x^2 (2 A+B x)-a (A+2 B x)\right )}{x^2}+6 \sqrt {a} A b \tanh ^{-1}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )-3 a \sqrt {b} B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 157, normalized size = 1.41

method	result	size
risch	\(-\frac {a \sqrt {b \,x^{2}+a}\, \left (2 B x +A \right )}{2 x^{2}}+\frac {b B x \sqrt {b \,x^{2}+a}}{2}+\frac {3 \sqrt {b}\, B a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2}+b A \sqrt {b \,x^{2}+a}-\frac {3 b A \sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2}\)	\(102\)
default	\(A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )+B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{a x}+\frac {4 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{a}\right )\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-1/2/a/x^2*(b*x^2+a)^(5/2)+3/2*b/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))))+B*(-1/a/x*(b*x^2+a)^(5/2)+4*b/a*(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)))))

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Maxima [A]
time = 0.27, size = 112, normalized size = 1.01 \begin {gather*} \frac {3}{2} \, \sqrt {b x^{2} + a} B b x + \frac {3}{2} \, B a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {3}{2} \, A \sqrt {a} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {b x^{2} + a} A b + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{2 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{x} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{2 \, a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.15, size = 425, normalized size = 3.83 \begin {gather*} \left [\frac {3 \, B a \sqrt {b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 3 \, A \sqrt {a} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (B b x^{3} + 2 \, A b x^{2} - 2 \, B a x - A a\right )} \sqrt {b x^{2} + a}}{4 \, x^{2}}, -\frac {6 \, B a \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 3 \, A \sqrt {a} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (B b x^{3} + 2 \, A b x^{2} - 2 \, B a x - A a\right )} \sqrt {b x^{2} + a}}{4 \, x^{2}}, \frac {6 \, A \sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 3 \, B a \sqrt {b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (B b x^{3} + 2 \, A b x^{2} - 2 \, B a x - A a\right )} \sqrt {b x^{2} + a}}{4 \, x^{2}}, -\frac {3 \, B a \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 3 \, A \sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (B b x^{3} + 2 \, A b x^{2} - 2 \, B a x - A a\right )} \sqrt {b x^{2} + a}}{2 \, x^{2}}\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^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 3.78, size = 182, normalized size = 1.64 \begin {gather*} - \frac {3 A \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2} - \frac {A a \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} + \frac {A a \sqrt {b}}{x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{\frac {3}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} - \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+A)*(b*x**2+a)**(3/2)/x**3,x)

[Out]

-3*A*sqrt(a)*b*asinh(sqrt(a)/(sqrt(b)*x))/2 - A*a*sqrt(b)*sqrt(a/(b*x**2) + 1)/(2*x) + A*a*sqrt(b)/(x*sqrt(a/(
b*x**2) + 1)) + A*b**(3/2)*x/sqrt(a/(b*x**2) + 1) - 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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (88) = 176\).
time = 1.58, size = 191, normalized size = 1.72 \begin {gather*} \frac {3 \, A a b \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3}{2} \, B a \sqrt {b} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \frac {1}{2} \, {\left (B b x + 2 \, A b\right )} \sqrt {b x^{2} + a} + \frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} A a b + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{2} \sqrt {b} + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a^{2} b - 2 \, B a^{3} \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.12, size = 91, normalized size = 0.82 \begin {gather*} A\,b\,\sqrt {b\,x^2+a}-\frac {A\,a\,\sqrt {b\,x^2+a}}{2\,x^2}-\frac {3\,A\,\sqrt {a}\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2}-\frac {B\,{\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^3,x)

[Out]

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