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

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

[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]]

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Rubi [A]  time = 0.0803436, antiderivative size = 84, normalized size of antiderivative = 1., 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 = {454, 97, 12, 63, 217, 206} \[ \frac{a (d x-c)^{3/2} (c+d x)^{3/2}}{3 c^2 x^3}-\frac{b \sqrt{d x-c} \sqrt{c+d x}}{x}+2 b d \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right ) \]

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 454

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]

Rule 97

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 12

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

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) \operatorname{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) \operatorname{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*}

Mathematica [A]  time = 0.075709, size = 105, normalized size = 1.25 \[ -\frac{\sqrt{d x-c} \sqrt{c+d x} \left (\sqrt{1-\frac{d^2 x^2}{c^2}} \left (a \left (c^2-d^2 x^2\right )+3 b c^2 x^2\right )+3 b c d x^3 \sin ^{-1}\left (\frac{d x}{c}\right )\right )}{3 c^2 x^3 \sqrt{1-\frac{d^2 x^2}{c^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.016, size = 153, normalized size = 1.8 \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{3\,{c}^{2}{x}^{3}}\sqrt{dx-c}\sqrt{dx+c} \left ( 3\,\ln \left ( \left ( \sqrt{{d}^{2}{x}^{2}-{c}^{2}}{\it csgn} \left ( d \right ) +dx \right ){\it csgn} \left ( d \right ) \right ){x}^{3}b{c}^{2}d+{\it csgn} \left ( d \right ){x}^{2}a{d}^{2}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}-3\,{\it csgn} \left ( d \right ){x}^{2}b{c}^{2}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}-{\it csgn} \left ( d \right ) a{c}^{2}\sqrt{{d}^{2}{x}^{2}-{c}^{2}} \right ){\frac{1}{\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}}} \end{align*}

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)

[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))*x^3*b*c^2*d+csgn(d)*x^2*a*d^2
*(d^2*x^2-c^2)^(1/2)-3*csgn(d)*x^2*b*c^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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.53373, size = 216, normalized size = 2.57 \begin{align*} -\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{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: MellinTransformStripError} \end{align*}

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

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Giac [B]  time = 1.32079, size = 231, normalized size = 2.75 \begin{align*} -\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{align*}

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