3.4.47 \(\int \frac {\sqrt {-c+d x} \sqrt {c+d x} (a+b x^2)}{x^4} \, dx\) [347]

Optimal. Leaf size=84 \[ -\frac {b \sqrt {-c+d x} \sqrt {c+d x}}{x}+\frac {a (-c+d x)^{3/2} (c+d x)^{3/2}}{3 c^2 x^3}+2 b d \tanh ^{-1}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {465, 99, 12, 65, 223, 212} \begin {gather*} \frac {a (d x-c)^{3/2} (c+d x)^{3/2}}{3 c^2 x^3}-\frac {b \sqrt {d x-c} \sqrt {c+d x}}{x}+2 b d \tanh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 99

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

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 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 465

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)
*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*((a2 + b2*x^(n/2))^(p + 1)/(a1*a2*e*(
m + 1))), x] + Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(a1*a2*e^n*(m + 1)), Int[(e*x)^(m + n)*(a1
+ b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, p}, x] && EqQ[non2, n/2] && Eq
Q[a2*b1 + a1*b2, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1
])) &&  !ILtQ[p, -1]

Rubi steps

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

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Mathematica [A]
time = 0.16, size = 81, normalized size = 0.96 \begin {gather*} -\frac {\sqrt {-c+d x} \sqrt {c+d x} \left (3 b c^2 x^2+a \left (c^2-d^2 x^2\right )\right )}{3 c^2 x^3}+2 b d \tanh ^{-1}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Sqrt[-c + d*x]*Sqrt[c + d*x]*(3*b*c^2*x^2 + a*(c^2 - d^2*x^2)))/(c^2*x^3) + 2*b*d*ArcTanh[Sqrt[-c + d*x]
/Sqrt[c + d*x]]

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.30, size = 153, normalized size = 1.82

method	result	size
risch	\(\frac {\sqrt {d x +c}\, \left (-d x +c \right ) \left (-a \,d^{2} x^{2}+3 b \,c^{2} x^{2}+c^{2} a \right )}{3 x^{3} c^{2} \sqrt {d x -c}}+\frac {b \,d^{2} \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-c^{2}}\right ) \sqrt {\left (d x -c \right ) \left (d x +c \right )}}{\sqrt {d^{2}}\, \sqrt {d x -c}\, \sqrt {d x +c}}\)	\(124\)
default	\(-\frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (-3 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\left (d \right )+d x \right ) \mathrm {csgn}\left (d \right )\right ) b \,c^{2} d \,x^{3}-\mathrm {csgn}\left (d \right ) a \,d^{2} x^{2} \sqrt {d^{2} x^{2}-c^{2}}+3 \,\mathrm {csgn}\left (d \right ) b \,c^{2} x^{2} \sqrt {d^{2} x^{2}-c^{2}}+\mathrm {csgn}\left (d \right ) a \,c^{2} \sqrt {d^{2} x^{2}-c^{2}}\right ) \mathrm {csgn}\left (d \right )}{3 \sqrt {d^{2} x^{2}-c^{2}}\, c^{2} x^{3}}\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(d*x-c)^(1/2)*(d*x+c)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.56, size = 75, normalized size = 0.89 \begin {gather*} b d \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right ) - \frac {\sqrt {d^{2} x^{2} - c^{2}} b}{x} + \frac {{\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} a}{3 \, c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x-c)^(1/2)*(d*x+c)^(1/2)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.35, size = 100, normalized size = 1.19 \begin {gather*} -\frac {3 \, b c^{2} d x^{3} \log \left (-d x + \sqrt {d x + c} \sqrt {d x - c}\right ) + {\left (3 \, b c^{2} d - a d^{3}\right )} x^{3} + {\left (a c^{2} + {\left (3 \, b c^{2} - a d^{2}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{3 \, c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x-c)^(1/2)*(d*x+c)^(1/2)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: MellinTransformStripError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(d*x-c)**(1/2)*(d*x+c)**(1/2)/x**4,x)

[Out]

Exception raised: MellinTransformStripError >> Pole inside critical strip?

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (70) = 140\).
time = 0.63, size = 171, normalized size = 2.04 \begin {gather*} -\frac {3 \, b d^{2} \log \left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4}\right ) + \frac {16 \, {\left (3 \, b c^{2} d^{2} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{8} - 3 \, a d^{4} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{8} + 24 \, b c^{4} d^{2} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 48 \, b c^{6} d^{2} - 16 \, a c^{4} d^{4}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 4 \, c^{2}\right )}^{3}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(d*x-c)^(1/2)*(d*x+c)^(1/2)/x^4,x, algorithm="giac")

[Out]

-1/6*(3*b*d^2*log((sqrt(d*x + c) - sqrt(d*x - c))^4) + 16*(3*b*c^2*d^2*(sqrt(d*x + c) - sqrt(d*x - c))^8 - 3*a
*d^4*(sqrt(d*x + c) - sqrt(d*x - c))^8 + 24*b*c^4*d^2*(sqrt(d*x + c) - sqrt(d*x - c))^4 + 48*b*c^6*d^2 - 16*a*
c^4*d^4)/((sqrt(d*x + c) - sqrt(d*x - c))^4 + 4*c^2)^3)/d

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Mupad [B]
time = 3.44, size = 236, normalized size = 2.81 \begin {gather*} \frac {b\,d+\frac {5\,b\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^2}}{\frac {4\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{\sqrt {-c}-\sqrt {d\,x-c}}+\frac {4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^3}}-4\,b\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}-\sqrt {c}}{\sqrt {-c}-\sqrt {d\,x-c}}\right )-\frac {\left (\frac {a\,\sqrt {c+d\,x}}{3}-\frac {a\,d^2\,x^2\,\sqrt {c+d\,x}}{3\,c^2}\right )\,\sqrt {d\,x-c}}{x^3}+\frac {b\,d\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{4\,\left (\sqrt {-c}-\sqrt {d\,x-c}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)*(c + d*x)^(1/2)*(d*x - c)^(1/2))/x^4,x)

[Out]

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

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