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

Optimal. Leaf size=72 \[ \frac {\sqrt {d x-c} \sqrt {c+d x} \left (3 a d^2+2 b c^2\right )}{3 d^4}+\frac {b x^2 \sqrt {d x-c} \sqrt {c+d x}}{3 d^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {460, 74} \[ \frac {\sqrt {d x-c} \sqrt {c+d x} \left (3 a d^2+2 b c^2\right )}{3 d^4}+\frac {b x^2 \sqrt {d x-c} \sqrt {c+d x}}{3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*c^2 + 3*a*d^2)*Sqrt[-c + d*x]*Sqrt[c + d*x])/(3*d^4) + (b*x^2*Sqrt[-c + d*x]*Sqrt[c + d*x])/(3*d^2)

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

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 {x \left (a+b x^2\right )}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx &=\frac {b x^2 \sqrt {-c+d x} \sqrt {c+d x}}{3 d^2}-\frac {1}{3} \left (-3 a-\frac {2 b c^2}{d^2}\right ) \int \frac {x}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx\\ &=\frac {\left (2 b c^2+3 a d^2\right ) \sqrt {-c+d x} \sqrt {c+d x}}{3 d^4}+\frac {b x^2 \sqrt {-c+d x} \sqrt {c+d x}}{3 d^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 61, normalized size = 0.85 \[ \frac {\left (d^2 x^2-c^2\right ) \left (3 a d^2+2 b c^2+b d^2 x^2\right )}{3 d^4 \sqrt {d x-c} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-c^2 + d^2*x^2)*(2*b*c^2 + 3*a*d^2 + b*d^2*x^2))/(3*d^4*Sqrt[-c + d*x]*Sqrt[c + d*x])

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fricas [A]  time = 1.14, size = 42, normalized size = 0.58 \[ \frac {{\left (b d^{2} x^{2} + 2 \, b c^{2} + 3 \, a d^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{3 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 65, normalized size = 0.90 \[ \frac {\sqrt {d x + c} \sqrt {d x - c} {\left ({\left (d x + c\right )} {\left (\frac {{\left (d x + c\right )} b}{d^{3}} - \frac {2 \, b c}{d^{3}}\right )} + \frac {3 \, {\left (b c^{2} d^{9} + a d^{11}\right )}}{d^{12}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

1/3*sqrt(d*x + c)*sqrt(d*x - c)*((d*x + c)*((d*x + c)*b/d^3 - 2*b*c/d^3) + 3*(b*c^2*d^9 + a*d^11)/d^12)/d

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maple [A]  time = 0.04, size = 43, normalized size = 0.60 \[ \frac {\sqrt {d x +c}\, \left (b \,d^{2} x^{2}+3 a \,d^{2}+2 b \,c^{2}\right ) \sqrt {d x -c}}{3 d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.64, size = 69, normalized size = 0.96 \[ \frac {\sqrt {d^{2} x^{2} - c^{2}} b x^{2}}{3 \, d^{2}} + \frac {2 \, \sqrt {d^{2} x^{2} - c^{2}} b c^{2}}{3 \, d^{4}} + \frac {\sqrt {d^{2} x^{2} - c^{2}} a}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.66, size = 76, normalized size = 1.06 \[ \frac {\sqrt {d\,x-c}\,\left (\frac {2\,b\,c^3+3\,a\,c\,d^2}{3\,d^4}+\frac {b\,x^3}{3\,d}+\frac {x\,\left (2\,b\,c^2\,d+3\,a\,d^3\right )}{3\,d^4}+\frac {b\,c\,x^2}{3\,d^2}\right )}{\sqrt {c+d\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 44.70, size = 223, normalized size = 3.10 \[ \frac {a c {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} + \frac {i a c {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} + \frac {b c^{3} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {5}{4}, - \frac {3}{4} & -1, -1, - \frac {1}{2}, 1 \\- \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{4}} + \frac {i b c^{3} {G_{6, 6}^{2, 6}\left (\begin {matrix} -2, - \frac {7}{4}, - \frac {3}{2}, - \frac {5}{4}, -1, 1 & \\- \frac {7}{4}, - \frac {5}{4} & -2, - \frac {3}{2}, - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x**2+a)/(d*x-c)**(1/2)/(d*x+c)**(1/2),x)

[Out]

a*c*meijerg(((-1/4, 1/4), (0, 0, 1/2, 1)), ((-1/2, -1/4, 0, 1/4, 1/2, 0), ()), c**2/(d**2*x**2))/(4*pi**(3/2)*
d**2) + I*a*c*meijerg(((-1, -3/4, -1/2, -1/4, 0, 1), ()), ((-3/4, -1/4), (-1, -1/2, -1/2, 0)), c**2*exp_polar(
2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**2) + b*c**3*meijerg(((-5/4, -3/4), (-1, -1, -1/2, 1)), ((-3/2, -5/4, -1,
-3/4, -1/2, 0), ()), c**2/(d**2*x**2))/(4*pi**(3/2)*d**4) + I*b*c**3*meijerg(((-2, -7/4, -3/2, -5/4, -1, 1), (
)), ((-7/4, -5/4), (-2, -3/2, -3/2, 0)), c**2*exp_polar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**4)

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