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]

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

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Rubi [A]  time = 0.08, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 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 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 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]

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]

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*}

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Mathematica [A]  time = 0.24, 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

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fricas [A]  time = 0.94, size = 80, normalized size = 1.00 \[ -\frac {6 \, a c d^{2} \arctan \left (-\frac {d x - \sqrt {d x + c} \sqrt {d x - c}}{c}\right ) - {\left (b d^{2} x^{2} - b c^{2} + 3 \, a d^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{3 \, d^{2}} \]

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

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giac [A]  time = 0.32, size = 78, normalized size = 0.98 \[ 2 \, a c \arctan \left (\frac {{\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}}{2 \, c}\right ) + \frac {1}{3} \, \sqrt {d x + c} \sqrt {d x - c} {\left ({\left (d x + c\right )} {\left (\frac {{\left (d x + c\right )} b}{d^{2}} - \frac {2 \, b c}{d^{2}}\right )} + 3 \, a\right )} \]

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

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maple [B]  time = 0.10, size = 174, normalized size = 2.18 \[ \frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (3 a \,c^{2} d^{2} \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right )+\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, b \,d^{2} x^{2}+3 \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, a \,d^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, b \,c^{2}\right )}{3 \sqrt {d^{2} x^{2}-c^{2}}\, \sqrt {-c^{2}}\, d^{2}} \]

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)

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maxima [A]  time = 1.30, size = 52, normalized size = 0.65 \[ a c \arcsin \left (\frac {c}{d {\left | x \right |}}\right ) + \sqrt {d^{2} x^{2} - c^{2}} a + \frac {{\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} b}{3 \, d^{2}} \]

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]

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

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mupad [B]  time = 3.60, size = 248, normalized size = 3.10 \[ a\,\sqrt {-c}\,\sqrt {c}\,\ln \left (\frac {{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^2}+1\right )-a\,\sqrt {-c}\,\sqrt {c}\,\ln \left (\frac {\sqrt {c+d\,x}-\sqrt {c}}{\sqrt {-c}-\sqrt {d\,x-c}}\right )-\frac {b\,\left (c^2-d^2\,x^2\right )\,\sqrt {c+d\,x}\,\sqrt {d\,x-c}}{3\,d^2}-\frac {8\,a\,\sqrt {-c}\,\sqrt {c}\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^2\,\left (\frac {{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^4}-\frac {2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^2}+1\right )} \]

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,x)

[Out]

a*(-c)^(1/2)*c^(1/2)*log(((c + d*x)^(1/2) - c^(1/2))^2/((-c)^(1/2) - (d*x - c)^(1/2))^2 + 1) - a*(-c)^(1/2)*c^
(1/2)*log(((c + d*x)^(1/2) - c^(1/2))/((-c)^(1/2) - (d*x - c)^(1/2))) - (b*(c^2 - d^2*x^2)*(c + d*x)^(1/2)*(d*
x - c)^(1/2))/(3*d^2) - (8*a*(-c)^(1/2)*c^(1/2)*((c + d*x)^(1/2) - c^(1/2))^2)/(((-c)^(1/2) - (d*x - c)^(1/2))
^2*(((c + d*x)^(1/2) - c^(1/2))^4/((-c)^(1/2) - (d*x - c)^(1/2))^4 - (2*((c + d*x)^(1/2) - c^(1/2))^2)/((-c)^(
1/2) - (d*x - c)^(1/2))^2 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right ) \sqrt {- c + d x} \sqrt {c + d x}}{x}\, dx \]

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)

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