3.4.34 \(\int \frac {(A+B x) (a+c x^2)^{3/2}}{x^4} \, dx\) [334]

Optimal. Leaf size=109 \[ -\frac {c (2 A-3 B x) \sqrt {a+c x^2}}{2 x}-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}+A c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {3}{2} \sqrt {a} B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right ) \]

[Out]

-1/6*(3*B*x+2*A)*(c*x^2+a)^(3/2)/x^3+A*c^(3/2)*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))-3/2*B*c*arctanh((c*x^2+a)^(1
/2)/a^(1/2))*a^(1/2)-1/2*c*(-3*B*x+2*A)*(c*x^2+a)^(1/2)/x

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Rubi [A]
time = 0.06, antiderivative size = 109, 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 = {825, 827, 858, 223, 212, 272, 65, 214} \begin {gather*} -\frac {c \sqrt {a+c x^2} (2 A-3 B x)}{2 x}-\frac {\left (a+c x^2\right )^{3/2} (2 A+3 B x)}{6 x^3}+A c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {3}{2} \sqrt {a} B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(3/2))/x^4,x]

[Out]

-1/2*(c*(2*A - 3*B*x)*Sqrt[a + c*x^2])/x - ((2*A + 3*B*x)*(a + c*x^2)^(3/2))/(6*x^3) + A*c^(3/2)*ArcTanh[(Sqrt
[c]*x)/Sqrt[a + c*x^2]] - (3*Sqrt[a]*B*c*ArcTanh[Sqrt[a + c*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 825

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^
(m + 1))*((a + c*x^2)^p/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)))*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*
p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e^2) + 2*c*d*p*(e*f - d*g))*x), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^
2 + a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p +
1) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e
^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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+c x^2\right )^{3/2}}{x^4} \, dx &=-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}-\frac {\int \frac {(-4 a A c-6 a B c x) \sqrt {a+c x^2}}{x^2} \, dx}{4 a}\\ &=-\frac {c (2 A-3 B x) \sqrt {a+c x^2}}{2 x}-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}+\frac {\int \frac {12 a^2 B c+8 a A c^2 x}{x \sqrt {a+c x^2}} \, dx}{8 a}\\ &=-\frac {c (2 A-3 B x) \sqrt {a+c x^2}}{2 x}-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}+\frac {1}{2} (3 a B c) \int \frac {1}{x \sqrt {a+c x^2}} \, dx+\left (A c^2\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx\\ &=-\frac {c (2 A-3 B x) \sqrt {a+c x^2}}{2 x}-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}+\frac {1}{4} (3 a B c) \text {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )+\left (A c^2\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )\\ &=-\frac {c (2 A-3 B x) \sqrt {a+c x^2}}{2 x}-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}+A c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )+\frac {1}{2} (3 a B) \text {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )\\ &=-\frac {c (2 A-3 B x) \sqrt {a+c x^2}}{2 x}-\frac {(2 A+3 B x) \left (a+c x^2\right )^{3/2}}{6 x^3}+A c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {3}{2} \sqrt {a} B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 109, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {a+c x^2} \left (2 a A+3 a B x+8 A c x^2-6 B c x^3\right )}{6 x^3}+3 \sqrt {a} B c \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )-A c^{3/2} \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(3/2))/x^4,x]

[Out]

-1/6*(Sqrt[a + c*x^2]*(2*a*A + 3*a*B*x + 8*A*c*x^2 - 6*B*c*x^3))/x^3 + 3*Sqrt[a]*B*c*ArcTanh[(Sqrt[c]*x - Sqrt
[a + c*x^2])/Sqrt[a]] - A*c^(3/2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(180\) vs. \(2(87)=174\).
time = 0.56, size = 181, normalized size = 1.66

method	result	size
risch	\(-\frac {\sqrt {c \,x^{2}+a}\, \left (8 A c \,x^{2}+3 B a x +2 A a \right )}{6 x^{3}}+A \,c^{\frac {3}{2}} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )-\frac {3 B \sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right ) c}{2}+B \sqrt {c \,x^{2}+a}\, c\)	\(96\)
default	\(B \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 c \left (\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {c \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )+A \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}}}{3 a \,x^{3}}+\frac {2 c \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}}}{a x}+\frac {4 c \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{a}\right )}{3 a}\right )\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(3/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 136, normalized size = 1.25 \begin {gather*} \frac {\sqrt {c x^{2} + a} A c^{2} x}{a} + A c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) - \frac {3}{2} \, B \sqrt {a} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {c x^{2} + a} B c + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B c}{2 \, a} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A c}{3 \, a x} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B}{2 \, a x^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A}{3 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 4.36, size = 426, normalized size = 3.91 \begin {gather*} \left [\frac {6 \, A c^{\frac {3}{2}} x^{3} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 9 \, B \sqrt {a} c x^{3} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (6 \, B c x^{3} - 8 \, A c x^{2} - 3 \, B a x - 2 \, A a\right )} \sqrt {c x^{2} + a}}{12 \, x^{3}}, -\frac {12 \, A \sqrt {-c} c x^{3} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 9 \, B \sqrt {a} c x^{3} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (6 \, B c x^{3} - 8 \, A c x^{2} - 3 \, B a x - 2 \, A a\right )} \sqrt {c x^{2} + a}}{12 \, x^{3}}, \frac {9 \, B \sqrt {-a} c x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + 3 \, A c^{\frac {3}{2}} x^{3} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + {\left (6 \, B c x^{3} - 8 \, A c x^{2} - 3 \, B a x - 2 \, A a\right )} \sqrt {c x^{2} + a}}{6 \, x^{3}}, -\frac {6 \, A \sqrt {-c} c x^{3} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 9 \, B \sqrt {-a} c x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (6 \, B c x^{3} - 8 \, A c x^{2} - 3 \, B a x - 2 \, A a\right )} \sqrt {c x^{2} + a}}{6 \, x^{3}}\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^4,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (100) = 200\).
time = 5.06, size = 202, normalized size = 1.85 \begin {gather*} - \frac {A \sqrt {a} c}{x \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {A a \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{3 x^{2}} - \frac {A c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{3} + A c^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )} - \frac {A c^{2} x}{\sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {3 B \sqrt {a} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{2} - \frac {B a \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{2 x} + \frac {B a \sqrt {c}}{x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {B c^{\frac {3}{2}} x}{\sqrt {\frac {a}{c x^{2}} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(3/2)/x**4,x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (87) = 174\).
time = 1.33, size = 211, normalized size = 1.94 \begin {gather*} \frac {3 \, B a c \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - A c^{\frac {3}{2}} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right ) + \sqrt {c x^{2} + a} B c + \frac {3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} B a c + 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a c^{\frac {3}{2}} - 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a^{2} c^{\frac {3}{2}} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{3} c + 8 \, A a^{3} c^{\frac {3}{2}}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^4,x, algorithm="giac")

[Out]

3*B*a*c*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) - A*c^(3/2)*log(abs(-sqrt(c)*x + sqrt(c*x^2 +
a))) + sqrt(c*x^2 + a)*B*c + 1/3*(3*(sqrt(c)*x - sqrt(c*x^2 + a))^5*B*a*c + 12*(sqrt(c)*x - sqrt(c*x^2 + a))^
4*A*a*c^(3/2) - 12*(sqrt(c)*x - sqrt(c*x^2 + a))^2*A*a^2*c^(3/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^3*c + 8
*A*a^3*c^(3/2))/((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^3

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^{3/2}\,\left (A+B\,x\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(3/2)*(A + B*x))/x^4,x)

[Out]

int(((a + c*x^2)^(3/2)*(A + B*x))/x^4, x)

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