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

Optimal. Leaf size=80 \[ a \sqrt{d x-c} \sqrt{c+d x}-a c \tan ^{-1}\left (\frac{\sqrt{d x-c} \sqrt{c+d x}}{c}\right )+\frac{b (d x-c)^{3/2} (c+d x)^{3/2}}{3 d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0782181, antiderivative size = 80, 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 = {460, 101, 12, 92, 205} \[ a \sqrt{d x-c} \sqrt{c+d x}-a c \tan ^{-1}\left (\frac{\sqrt{d x-c} \sqrt{c+d x}}{c}\right )+\frac{b (d x-c)^{3/2} (c+d x)^{3/2}}{3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 460

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

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

Mathematica [A]  time = 0.161454, size = 85, normalized size = 1.06 \[ \frac{1}{3} \sqrt{d x-c} \sqrt{c+d x} \left (-\frac{3 a c \tan ^{-1}\left (\frac{\sqrt{d^2 x^2-c^2}}{c}\right )}{\sqrt{d^2 x^2-c^2}}+3 a+b \left (x^2-\frac{c^2}{d^2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.036, size = 174, normalized size = 2.2 \begin{align*}{\frac{1}{3\,{d}^{2}}\sqrt{dx-c}\sqrt{dx+c} \left ({x}^{2}b{d}^{2}\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}+3\,\ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ) a{c}^{2}{d}^{2}+3\,\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}a{d}^{2}-b{c}^{2}\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}} \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,x)

[Out]

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

________________________________________________________________________________________

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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.55539, size = 174, normalized size = 2.17 \begin{align*} -\frac{6 \, a c d^{2} \arctan \left (-\frac{d x - \sqrt{d x + c} \sqrt{d x - c}}{c}\right ) -{\left (b d^{2} x^{2} - b c^{2} + 3 \, a d^{2}\right )} \sqrt{d x + c} \sqrt{d x - c}}{3 \, d^{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,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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}\, 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,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.69464, size = 109, normalized size = 1.36 \begin{align*} 2 \, a c \arctan \left (\frac{{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2}}{2 \, c}\right ) + \frac{1}{1920} \,{\left (3 \, a d^{6} +{\left ({\left (d x + c\right )} b d^{4} - 2 \, b c d^{4}\right )}{\left (d x + c\right )}\right )} \sqrt{d x + c} \sqrt{d x - c} \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,x, algorithm="giac")

[Out]

2*a*c*arctan(1/2*(sqrt(d*x + c) - sqrt(d*x - c))^2/c) + 1/1920*(3*a*d^6 + ((d*x + c)*b*d^4 - 2*b*c*d^4)*(d*x +
c))*sqrt(d*x + c)*sqrt(d*x - c)