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

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

________________________________________________________________________________________

Rubi [A]  time = 0.08, 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 = {813, 844, 217, 206, 266, 63, 208} \begin {gather*} -\frac {\left (a+c x^2\right )^{3/2} (A-B x)}{2 x^2}-\frac {3 \sqrt {a+c x^2} (a B-A c x)}{2 x}-\frac {3}{2} \sqrt {a} A c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )+\frac {3}{2} a B \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [C]  time = 0.04, size = 90, normalized size = 0.81 \begin {gather*} \frac {A c \left (a+c x^2\right )^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {c x^2}{a}+1\right )}{5 a^2}-\frac {a B \sqrt {a+c x^2} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {c x^2}{a}\right )}{x \sqrt {\frac {c x^2}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*B*Sqrt[a + c*x^2]*Hypergeometric2F1[-3/2, -1/2, 1/2, -((c*x^2)/a)])/(x*Sqrt[1 + (c*x^2)/a])) + (A*c*(a +
c*x^2)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 + (c*x^2)/a])/(5*a^2)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.47, size = 115, normalized size = 1.04 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-a A-2 a B x+2 A c x^2+B c x^3\right )}{2 x^2}+3 \sqrt {a} A c \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )-\frac {3}{2} a B \sqrt {c} \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 425, normalized size = 3.83 \begin {gather*} \left [\frac {3 \, B a \sqrt {c} x^{2} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 3 \, A \sqrt {a} c x^{2} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (B c x^{3} + 2 \, A c x^{2} - 2 \, B a x - A a\right )} \sqrt {c x^{2} + a}}{4 \, x^{2}}, -\frac {6 \, B a \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 3 \, A \sqrt {a} c x^{2} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (B c x^{3} + 2 \, A c x^{2} - 2 \, B a x - A a\right )} \sqrt {c x^{2} + a}}{4 \, x^{2}}, \frac {6 \, A \sqrt {-a} c x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + 3 \, B a \sqrt {c} x^{2} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (B c x^{3} + 2 \, A c x^{2} - 2 \, B a x - A a\right )} \sqrt {c x^{2} + a}}{4 \, x^{2}}, -\frac {3 \, B a \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 3 \, A \sqrt {-a} c x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (B c x^{3} + 2 \, A c x^{2} - 2 \, B a x - A a\right )} \sqrt {c x^{2} + a}}{2 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.25, size = 191, normalized size = 1.72 \begin {gather*} \frac {3 \, A a c \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3}{2} \, B a \sqrt {c} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right ) + \frac {1}{2} \, {\left (B c x + 2 \, A c\right )} \sqrt {c x^{2} + a} + \frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} A a c + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} B a^{2} \sqrt {c} + {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} A a^{2} c - 2 \, B a^{3} \sqrt {c}}{{\left ({\left (\sqrt {c} x - \sqrt {c 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)*(c*x^2+a)^(3/2)/x^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 150, normalized size = 1.35 \begin {gather*} -\frac {3 A \sqrt {a}\, c \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2}+\frac {3 B a \sqrt {c}\, \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2}+\frac {3 \sqrt {c \,x^{2}+a}\, B c x}{2}+\frac {3 \sqrt {c \,x^{2}+a}\, A c}{2}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} B c x}{a}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A c}{2 a}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B}{a x}-\frac {\left (c \,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]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.56, size = 112, normalized size = 1.01 \begin {gather*} \frac {3}{2} \, \sqrt {c x^{2} + a} B c x + \frac {3}{2} \, B a \sqrt {c} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) - \frac {3}{2} \, A \sqrt {a} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {c x^{2} + a} A c + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A c}{2 \, a} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B}{x} - \frac {{\left (c 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)*(c*x^2+a)^(3/2)/x^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 2.20, size = 91, normalized size = 0.82 \begin {gather*} A\,c\,\sqrt {c\,x^2+a}-\frac {A\,a\,\sqrt {c\,x^2+a}}{2\,x^2}-\frac {3\,A\,\sqrt {a}\,c\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{2}-\frac {B\,{\left (c\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {c\,x^2}{a}\right )}{x\,{\left (\frac {c\,x^2}{a}+1\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 8.19, size = 182, normalized size = 1.64 \begin {gather*} - \frac {3 A \sqrt {a} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{2} - \frac {A a \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{2 x} + \frac {A a \sqrt {c}}{x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {A c^{\frac {3}{2}} x}{\sqrt {\frac {a}{c x^{2}} + 1}} - \frac {B a^{\frac {3}{2}}}{x \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {B \sqrt {a} c x \sqrt {1 + \frac {c x^{2}}{a}}}{2} - \frac {B \sqrt {a} c x}{\sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 B a \sqrt {c} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*A*sqrt(a)*c*asinh(sqrt(a)/(sqrt(c)*x))/2 - A*a*sqrt(c)*sqrt(a/(c*x**2) + 1)/(2*x) + A*a*sqrt(c)/(x*sqrt(a/(
c*x**2) + 1)) + A*c**(3/2)*x/sqrt(a/(c*x**2) + 1) - B*a**(3/2)/(x*sqrt(1 + c*x**2/a)) + B*sqrt(a)*c*x*sqrt(1 +
c*x**2/a)/2 - B*sqrt(a)*c*x/sqrt(1 + c*x**2/a) + 3*B*a*sqrt(c)*asinh(sqrt(c)*x/sqrt(a))/2

________________________________________________________________________________________