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

Optimal. Leaf size=104 \[ \frac{1}{2} x \sqrt{d x-c} \sqrt{c+d x} \left (b-\frac{2 a d^2}{c^2}\right )-\frac{\left (b c^2-2 a d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{d}+\frac{a (d x-c)^{3/2} (c+d x)^{3/2}}{c^2 x} \]

[Out]

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

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Rubi [A]  time = 0.0869517, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {454, 38, 63, 217, 206} \[ \frac{1}{2} x \sqrt{d x-c} \sqrt{c+d x} \left (b-\frac{2 a d^2}{c^2}\right )-\frac{\left (b c^2-2 a d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{d}+\frac{a (d x-c)^{3/2} (c+d x)^{3/2}}{c^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

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

Mathematica [A]  time = 0.0802886, size = 101, normalized size = 0.97 \[ \frac{\sqrt{d x-c} \sqrt{c+d x} \left (c d \left (b x^2-2 a\right ) \sqrt{1-\frac{d^2 x^2}{c^2}}+x \left (b c^2-2 a d^2\right ) \sin ^{-1}\left (\frac{d x}{c}\right )\right )}{2 c d x \sqrt{1-\frac{d^2 x^2}{c^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.014, size = 153, normalized size = 1.5 \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{2\,dx}\sqrt{dx-c}\sqrt{dx+c} \left ({\it csgn} \left ( d \right ){x}^{2}bd\sqrt{{d}^{2}{x}^{2}-{c}^{2}}+2\,\ln \left ( \left ( \sqrt{{d}^{2}{x}^{2}-{c}^{2}}{\it csgn} \left ( d \right ) +dx \right ){\it csgn} \left ( d \right ) \right ) xa{d}^{2}-\ln \left ( \left ( \sqrt{{d}^{2}{x}^{2}-{c}^{2}}{\it csgn} \left ( d \right ) +dx \right ){\it csgn} \left ( d \right ) \right ) xb{c}^{2}-2\,{\it csgn} \left ( d \right ) d\sqrt{{d}^{2}{x}^{2}-{c}^{2}}a \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^2,x)

[Out]

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

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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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57855, size = 181, normalized size = 1.74 \begin{align*} -\frac{2 \, a d^{2} x -{\left (b c^{2} - 2 \, a d^{2}\right )} x \log \left (-d x + \sqrt{d x + c} \sqrt{d x - c}\right ) -{\left (b d x^{2} - 2 \, a d\right )} \sqrt{d x + c} \sqrt{d x - c}}{2 \, d x} \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^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right ) \sqrt{- c + d x} \sqrt{c + d x}}{x^{2}}\, dx \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**2,x)

[Out]

Integral((a + b*x**2)*sqrt(-c + d*x)*sqrt(c + d*x)/x**2, x)

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Giac [A]  time = 1.24287, size = 149, normalized size = 1.43 \begin{align*} -\frac{\frac{6144 \, a c^{2} d^{2}}{{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{4} + 4 \, c^{2}} - 2 \,{\left ({\left (d x + c\right )} b - b c\right )} \sqrt{d x + c} \sqrt{d x - c} -{\left (b c^{2} - 2 \, a d^{2}\right )} \log \left ({\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{4}\right )}{768 \, 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^2,x, algorithm="giac")

[Out]

-1/768*(6144*a*c^2*d^2/((sqrt(d*x + c) - sqrt(d*x - c))^4 + 4*c^2) - 2*((d*x + c)*b - b*c)*sqrt(d*x + c)*sqrt(
d*x - c) - (b*c^2 - 2*a*d^2)*log((sqrt(d*x + c) - sqrt(d*x - c))^4))/d