\(\int \frac {(a+b x^2)^2}{x \sqrt {c+d x^2}} \, dx\) [641]

Optimal result

Integrand size = 24, antiderivative size = 75 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=-\frac {b (b c-2 a d) \sqrt {c+d x^2}}{d^2}+\frac {b^2 \left (c+d x^2\right )^{3/2}}{3 d^2}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \]

[Out]

1/3*b^2*(d*x^2+c)^(3/2)/d^2-a^2*arctanh((d*x^2+c)^(1/2)/c^(1/2))/c^(1/2)-b*(-2*a*d+b*c)*(d*x^2+c)^(1/2)/d^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {457, 90, 65, 214} \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {b \sqrt {c+d x^2} (b c-2 a d)}{d^2}+\frac {b^2 \left (c+d x^2\right )^{3/2}}{3 d^2} \]

[In]

Int[(a + b*x^2)^2/(x*Sqrt[c + d*x^2]),x]

[Out]

-((b*(b*c - 2*a*d)*Sqrt[c + d*x^2])/d^2) + (b^2*(c + d*x^2)^(3/2))/(3*d^2) - (a^2*ArcTanh[Sqrt[c + d*x^2]/Sqrt
[c]])/Sqrt[c]

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 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^2}{x \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {b (b c-2 a d)}{d \sqrt {c+d x}}+\frac {a^2}{x \sqrt {c+d x}}+\frac {b^2 \sqrt {c+d x}}{d}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b (b c-2 a d) \sqrt {c+d x^2}}{d^2}+\frac {b^2 \left (c+d x^2\right )^{3/2}}{3 d^2}+\frac {1}{2} a^2 \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = -\frac {b (b c-2 a d) \sqrt {c+d x^2}}{d^2}+\frac {b^2 \left (c+d x^2\right )^{3/2}}{3 d^2}+\frac {a^2 \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{d} \\ & = -\frac {b (b c-2 a d) \sqrt {c+d x^2}}{d^2}+\frac {b^2 \left (c+d x^2\right )^{3/2}}{3 d^2}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\frac {b \sqrt {c+d x^2} \left (-2 b c+6 a d+b d x^2\right )}{3 d^2}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \]

[In]

Integrate[(a + b*x^2)^2/(x*Sqrt[c + d*x^2]),x]

[Out]

(b*Sqrt[c + d*x^2]*(-2*b*c + 6*a*d + b*d*x^2))/(3*d^2) - (a^2*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/Sqrt[c]

Maple [A] (verified)

Time = 2.86 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(-\frac {a^{2} d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )-2 \sqrt {d \,x^{2}+c}\, b \left (-\frac {c^{\frac {3}{2}} b}{3}+d \sqrt {c}\, \left (\frac {b \,x^{2}}{6}+a \right )\right )}{\sqrt {c}\, d^{2}}\)	\(63\)
default	\(b^{2} \left (\frac {x^{2} \sqrt {d \,x^{2}+c}}{3 d}-\frac {2 c \sqrt {d \,x^{2}+c}}{3 d^{2}}\right )-\frac {a^{2} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{\sqrt {c}}+\frac {2 a b \sqrt {d \,x^{2}+c}}{d}\)	\(86\)

[In]

int((b*x^2+a)^2/x/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-(a^2*d^2*arctanh((d*x^2+c)^(1/2)/c^(1/2))-2*(d*x^2+c)^(1/2)*b*(-1/3*c^(3/2)*b+d*c^(1/2)*(1/6*b*x^2+a)))/c^(1/
2)/d^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.09 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\left [\frac {3 \, a^{2} \sqrt {c} d^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (b^{2} c d x^{2} - 2 \, b^{2} c^{2} + 6 \, a b c d\right )} \sqrt {d x^{2} + c}}{6 \, c d^{2}}, \frac {3 \, a^{2} \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (b^{2} c d x^{2} - 2 \, b^{2} c^{2} + 6 \, a b c d\right )} \sqrt {d x^{2} + c}}{3 \, c d^{2}}\right ] \]

[In]

integrate((b*x^2+a)^2/x/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/6*(3*a^2*sqrt(c)*d^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(b^2*c*d*x^2 - 2*b^2*c^2 + 6*a
*b*c*d)*sqrt(d*x^2 + c))/(c*d^2), 1/3*(3*a^2*sqrt(-c)*d^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (b^2*c*d*x^2 - 2*
b^2*c^2 + 6*a*b*c*d)*sqrt(d*x^2 + c))/(c*d^2)]

Sympy [A] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\frac {\begin {cases} \frac {2 a^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + \frac {2 b^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}{3 d^{2}} + \frac {2 \sqrt {c + d x^{2}} \cdot \left (2 a b d - b^{2} c\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {a^{2} \log {\left (x^{2} \right )} + 2 a b x^{2} + \frac {b^{2} x^{4}}{2}}{\sqrt {c}} & \text {otherwise} \end {cases}}{2} \]

[In]

integrate((b*x**2+a)**2/x/(d*x**2+c)**(1/2),x)

[Out]

Piecewise((2*a**2*atan(sqrt(c + d*x**2)/sqrt(-c))/sqrt(-c) + 2*b**2*(c + d*x**2)**(3/2)/(3*d**2) + 2*sqrt(c +
d*x**2)*(2*a*b*d - b**2*c)/d**2, Ne(d, 0)), ((a**2*log(x**2) + 2*a*b*x**2 + b**2*x**4/2)/sqrt(c), True))/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{2} x^{2}}{3 \, d} - \frac {a^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{\sqrt {c}} - \frac {2 \, \sqrt {d x^{2} + c} b^{2} c}{3 \, d^{2}} + \frac {2 \, \sqrt {d x^{2} + c} a b}{d} \]

[In]

integrate((b*x^2+a)^2/x/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

1/3*sqrt(d*x^2 + c)*b^2*x^2/d - a^2*arcsinh(c/(sqrt(c*d)*abs(x)))/sqrt(c) - 2/3*sqrt(d*x^2 + c)*b^2*c/d^2 + 2*
sqrt(d*x^2 + c)*a*b/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\frac {a^{2} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d^{4} - 3 \, \sqrt {d x^{2} + c} b^{2} c d^{4} + 6 \, \sqrt {d x^{2} + c} a b d^{5}}{3 \, d^{6}} \]

[In]

integrate((b*x^2+a)^2/x/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

a^2*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) + 1/3*((d*x^2 + c)^(3/2)*b^2*d^4 - 3*sqrt(d*x^2 + c)*b^2*c*d^4 +
6*sqrt(d*x^2 + c)*a*b*d^5)/d^6

Mupad [B] (verification not implemented)

Time = 5.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\frac {b^2\,{\left (d\,x^2+c\right )}^{3/2}}{3\,d^2}-\frac {a^2\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )}{\sqrt {c}}-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )\,\sqrt {d\,x^2+c} \]

[In]

int((a + b*x^2)^2/(x*(c + d*x^2)^(1/2)),x)

[Out]

(b^2*(c + d*x^2)^(3/2))/(3*d^2) - (a^2*atanh((c + d*x^2)^(1/2)/c^(1/2)))/c^(1/2) - ((2*b^2*c - 2*a*b*d)/d^2 -
(b^2*c)/d^2)*(c + d*x^2)^(1/2)