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

Optimal. Leaf size=96 \[ -\frac{\left (2 b c^2-a d^2\right ) \tan ^{-1}\left (\frac{\sqrt{d x-c} \sqrt{c+d x}}{c}\right )}{2 c}-\frac{a \sqrt{d x-c} \sqrt{c+d x}}{2 x^2}+b \sqrt{d x-c} \sqrt{c+d x} \]

[Out]

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

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Rubi [A]  time = 0.0840813, antiderivative size = 114, normalized size of antiderivative = 1.19, 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, 101, 12, 92, 205} \[ \frac{1}{2} \sqrt{d x-c} \sqrt{c+d x} \left (2 b-\frac{a d^2}{c^2}\right )-\frac{\left (2 b c^2-a d^2\right ) \tan ^{-1}\left (\frac{\sqrt{d x-c} \sqrt{c+d x}}{c}\right )}{2 c}+\frac{a (d x-c)^{3/2} (c+d x)^{3/2}}{2 c^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[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 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0490826, size = 114, normalized size = 1.19 \[ -\frac{\sqrt{d x-c} \sqrt{c+d x} \left (c \left (a-2 b x^2\right ) \sqrt{d^2 x^2-c^2}+x^2 \left (2 b c^2-a d^2\right ) \tan ^{-1}\left (\frac{\sqrt{d^2 x^2-c^2}}{c}\right )\right )}{2 c x^2 \sqrt{d^2 x^2-c^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.014, size = 182, normalized size = 1.9 \begin{align*} -{\frac{1}{2\,{x}^{2}}\sqrt{dx-c}\sqrt{dx+c} \left ( \ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{2}a{d}^{2}-2\,\ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{2}b{c}^{2}-2\,{x}^{2}b\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}+\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}a \right ){\frac{1}{\sqrt{-{c}^{2}}}}{\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^3,x)

[Out]

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

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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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54694, size = 182, normalized size = 1.9 \begin{align*} -\frac{2 \,{\left (2 \, b c^{2} - a d^{2}\right )} x^{2} \arctan \left (-\frac{d x - \sqrt{d x + c} \sqrt{d x - c}}{c}\right ) -{\left (2 \, b c x^{2} - a c\right )} \sqrt{d x + c} \sqrt{d x - c}}{2 \, c x^{2}} \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^3,x, algorithm="fricas")

[Out]

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

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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^{3}}\, 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**3,x)

[Out]

Integral((a + b*x**2)*sqrt(-c + d*x)*sqrt(c + d*x)/x**3, x)

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Giac [A]  time = 1.18163, size = 212, normalized size = 2.21 \begin{align*} \frac{\sqrt{d x + c} \sqrt{d x - c} b d + \frac{{\left (2 \, b c^{2} d - a d^{3}\right )} \arctan \left (\frac{{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2}}{2 \, c}\right )}{c} + \frac{2 \,{\left (a d^{3}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{6} - 4 \, a c^{2} d^{3}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2}\right )}}{{\left ({\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{4} + 4 \, c^{2}\right )}^{2}}}{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^3,x, algorithm="giac")

[Out]

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