3.6.84 \(\int \frac {(a+b x^2)^2 \sqrt {c+d x^2}}{x^3} \, dx\)

Optimal. Leaf size=109 \[ -\frac {a^2 \left (c+d x^2\right )^{3/2}}{2 c x^2}+\frac {a \sqrt {c+d x^2} (a d+4 b c)}{2 c}-\frac {a (a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 \sqrt {c}}+\frac {b^2 \left (c+d x^2\right )^{3/2}}{3 d} \]

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Rubi [A]  time = 0.09, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 89, 80, 50, 63, 208} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{3/2}}{2 c x^2}+\frac {a \sqrt {c+d x^2} (a d+4 b c)}{2 c}-\frac {a (a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 \sqrt {c}}+\frac {b^2 \left (c+d x^2\right )^{3/2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*b*c + a*d)*Sqrt[c + d*x^2])/(2*c) + (b^2*(c + d*x^2)^(3/2))/(3*d) - (a^2*(c + d*x^2)^(3/2))/(2*c*x^2) -
(a*(4*b*c + a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*Sqrt[c])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 446

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

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Mathematica [A]  time = 0.11, size = 87, normalized size = 0.80 \begin {gather*} \frac {1}{6} \left (\frac {\sqrt {c+d x^2} \left (-3 a^2 d+12 a b d x^2+2 b^2 x^2 \left (c+d x^2\right )\right )}{d x^2}-\frac {3 a (a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[c + d*x^2]*(-3*a^2*d + 12*a*b*d*x^2 + 2*b^2*x^2*(c + d*x^2)))/(d*x^2) - (3*a*(4*b*c + a*d)*ArcTanh[Sqrt
[c + d*x^2]/Sqrt[c]])/Sqrt[c])/6

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IntegrateAlgebraic [A]  time = 0.13, size = 94, normalized size = 0.86 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-3 a^2 d+12 a b d x^2+2 b^2 c x^2+2 b^2 d x^4\right )}{6 d x^2}+\frac {\left (a^2 (-d)-4 a b c\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-3*a^2*d + 2*b^2*c*x^2 + 12*a*b*d*x^2 + 2*b^2*d*x^4))/(6*d*x^2) + ((-4*a*b*c - a^2*d)*ArcTan
h[Sqrt[c + d*x^2]/Sqrt[c]])/(2*Sqrt[c])

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fricas [A]  time = 1.60, size = 211, normalized size = 1.94 \begin {gather*} \left [\frac {3 \, {\left (4 \, a b c d + a^{2} d^{2}\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (2 \, b^{2} c d x^{4} - 3 \, a^{2} c d + 2 \, {\left (b^{2} c^{2} + 6 \, a b c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, c d x^{2}}, \frac {3 \, {\left (4 \, a b c d + a^{2} d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (2 \, b^{2} c d x^{4} - 3 \, a^{2} c d + 2 \, {\left (b^{2} c^{2} + 6 \, a b c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, c d x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^3,x, algorithm="fricas")

[Out]

[1/12*(3*(4*a*b*c*d + a^2*d^2)*sqrt(c)*x^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(2*b^2*c*d*
x^4 - 3*a^2*c*d + 2*(b^2*c^2 + 6*a*b*c*d)*x^2)*sqrt(d*x^2 + c))/(c*d*x^2), 1/6*(3*(4*a*b*c*d + a^2*d^2)*sqrt(-
c)*x^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (2*b^2*c*d*x^4 - 3*a^2*c*d + 2*(b^2*c^2 + 6*a*b*c*d)*x^2)*sqrt(d*x^2
+ c))/(c*d*x^2)]

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giac [A]  time = 0.38, size = 89, normalized size = 0.82 \begin {gather*} \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} + 12 \, \sqrt {d x^{2} + c} a b d - \frac {3 \, \sqrt {d x^{2} + c} a^{2} d}{x^{2}} + \frac {3 \, {\left (4 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^3,x, algorithm="giac")

[Out]

1/6*(2*(d*x^2 + c)^(3/2)*b^2 + 12*sqrt(d*x^2 + c)*a*b*d - 3*sqrt(d*x^2 + c)*a^2*d/x^2 + 3*(4*a*b*c*d + a^2*d^2
)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c))/d

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maple [A]  time = 0.01, size = 132, normalized size = 1.21 \begin {gather*} -\frac {a^{2} d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{2 \sqrt {c}}-2 a b \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )+\frac {\sqrt {d \,x^{2}+c}\, a^{2} d}{2 c}+2 \sqrt {d \,x^{2}+c}\, a b +\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2}}{3 d}-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2}}{2 c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^3,x)

[Out]

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

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maxima [A]  time = 1.06, size = 109, normalized size = 1.00 \begin {gather*} -2 \, a b \sqrt {c} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {a^{2} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{2 \, \sqrt {c}} + 2 \, \sqrt {d x^{2} + c} a b + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2}}{3 \, d} + \frac {\sqrt {d x^{2} + c} a^{2} d}{2 \, c} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{2 \, c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.06, size = 103, normalized size = 0.94 \begin {gather*} \frac {b^2\,{\left (d\,x^2+c\right )}^{3/2}}{3\,d}-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d}-\frac {2\,b^2\,c}{d}\right )\,\sqrt {d\,x^2+c}-\frac {a^2\,\sqrt {d\,x^2+c}}{2\,x^2}+\frac {a\,\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,\left (a\,d+4\,b\,c\right )\,1{}\mathrm {i}}{2\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^3,x)

[Out]

(b^2*(c + d*x^2)^(3/2))/(3*d) - ((2*b^2*c - 2*a*b*d)/d - (2*b^2*c)/d)*(c + d*x^2)^(1/2) - (a^2*(c + d*x^2)^(1/
2))/(2*x^2) + (a*atan(((c + d*x^2)^(1/2)*1i)/c^(1/2))*(a*d + 4*b*c)*1i)/(2*c^(1/2))

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sympy [A]  time = 73.98, size = 148, normalized size = 1.36 \begin {gather*} - \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} - \frac {a^{2} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{2 \sqrt {c}} - 2 a b \sqrt {c} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )} + \frac {2 a b c}{\sqrt {d} x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a b \sqrt {d} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + b^{2} \left (\begin {cases} \frac {\sqrt {c} x^{2}}{2} & \text {for}\: d = 0 \\\frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/x**3,x)

[Out]

-a**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(2*x) - a**2*d*asinh(sqrt(c)/(sqrt(d)*x))/(2*sqrt(c)) - 2*a*b*sqrt(c)*asinh
(sqrt(c)/(sqrt(d)*x)) + 2*a*b*c/(sqrt(d)*x*sqrt(c/(d*x**2) + 1)) + 2*a*b*sqrt(d)*x/sqrt(c/(d*x**2) + 1) + b**2
*Piecewise((sqrt(c)*x**2/2, Eq(d, 0)), ((c + d*x**2)**(3/2)/(3*d), True))

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