3.366 \(\int \frac {a+b x^2}{x^4 \sqrt {-c+d x} \sqrt {c+d x}} \, dx\)

Optimal. Leaf size=75 \[ \frac {\sqrt {d x-c} \sqrt {c+d x} \left (2 a d^2+3 b c^2\right )}{3 c^4 x}+\frac {a \sqrt {d x-c} \sqrt {c+d x}}{3 c^2 x^3} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {454, 95} \[ \frac {\sqrt {d x-c} \sqrt {c+d x} \left (2 a d^2+3 b c^2\right )}{3 c^4 x}+\frac {a \sqrt {d x-c} \sqrt {c+d x}}{3 c^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 95

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] /; FreeQ[{a, b, c, d,
e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && EqQ[a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1), 0
] && NeQ[m, -1]

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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.88 \[ -\frac {\left (c^2-d^2 x^2\right ) \left (a \left (c^2+2 d^2 x^2\right )+3 b c^2 x^2\right )}{3 c^4 x^3 \sqrt {d x-c} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 67, normalized size = 0.89 \[ \frac {{\left (3 \, b c^{2} d + 2 \, a d^{3}\right )} x^{3} + {\left (a c^{2} + {\left (3 \, b c^{2} + 2 \, a d^{2}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{3 \, c^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^4/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.24, size = 137, normalized size = 1.83 \[ \frac {8 \, {\left (3 \, b d^{2} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{8} + 24 \, b c^{2} d^{2} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 24 \, a d^{4} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 48 \, b c^{4} d^{2} + 32 \, a c^{2} d^{4}\right )}}{3 \, {\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 4 \, c^{2}\right )}^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^4/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

8/3*(3*b*d^2*(sqrt(d*x + c) - sqrt(d*x - c))^8 + 24*b*c^2*d^2*(sqrt(d*x + c) - sqrt(d*x - c))^4 + 24*a*d^4*(sq
rt(d*x + c) - sqrt(d*x - c))^4 + 48*b*c^4*d^2 + 32*a*c^2*d^4)/(((sqrt(d*x + c) - sqrt(d*x - c))^4 + 4*c^2)^3*d
)

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maple [A]  time = 0.04, size = 49, normalized size = 0.65 \[ \frac {\sqrt {d x +c}\, \left (2 a \,d^{2} x^{2}+3 b \,c^{2} x^{2}+a \,c^{2}\right ) \sqrt {d x -c}}{3 c^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.32, size = 75, normalized size = 1.00 \[ \frac {\sqrt {d^{2} x^{2} - c^{2}} b}{c^{2} x} + \frac {2 \, \sqrt {d^{2} x^{2} - c^{2}} a d^{2}}{3 \, c^{4} x} + \frac {\sqrt {d^{2} x^{2} - c^{2}} a}{3 \, c^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^4/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.77, size = 79, normalized size = 1.05 \[ \frac {\sqrt {d\,x-c}\,\left (\frac {a}{3\,c}+\frac {x^2\,\left (3\,b\,c^3+2\,a\,c\,d^2\right )}{3\,c^4}+\frac {x^3\,\left (3\,b\,c^2\,d+2\,a\,d^3\right )}{3\,c^4}+\frac {a\,d\,x}{3\,c^2}\right )}{x^3\,\sqrt {c+d\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 70.80, size = 170, normalized size = 2.27 \[ - \frac {a d^{3} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {9}{4}, \frac {11}{4}, 1 & \frac {5}{2}, \frac {5}{2}, 3 \\2, \frac {9}{4}, \frac {5}{2}, \frac {11}{4}, 3 & 0 \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{4}} - \frac {i a d^{3} {G_{6, 6}^{2, 6}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4}, 2, \frac {9}{4}, \frac {5}{2}, 1 & \\\frac {7}{4}, \frac {9}{4} & \frac {3}{2}, 2, 2, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{4}} - \frac {b d {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4}, 1 & \frac {3}{2}, \frac {3}{2}, 2 \\1, \frac {5}{4}, \frac {3}{2}, \frac {7}{4}, 2 & 0 \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{2}} - \frac {i b d {G_{6, 6}^{2, 6}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2}, 1 & \\\frac {3}{4}, \frac {5}{4} & \frac {1}{2}, 1, 1, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)/x**4/(d*x-c)**(1/2)/(d*x+c)**(1/2),x)

[Out]

-a*d**3*meijerg(((9/4, 11/4, 1), (5/2, 5/2, 3)), ((2, 9/4, 5/2, 11/4, 3), (0,)), c**2/(d**2*x**2))/(4*pi**(3/2
)*c**4) - I*a*d**3*meijerg(((3/2, 7/4, 2, 9/4, 5/2, 1), ()), ((7/4, 9/4), (3/2, 2, 2, 0)), c**2*exp_polar(2*I*
pi)/(d**2*x**2))/(4*pi**(3/2)*c**4) - b*d*meijerg(((5/4, 7/4, 1), (3/2, 3/2, 2)), ((1, 5/4, 3/2, 7/4, 2), (0,)
), c**2/(d**2*x**2))/(4*pi**(3/2)*c**2) - I*b*d*meijerg(((1/2, 3/4, 1, 5/4, 3/2, 1), ()), ((3/4, 5/4), (1/2, 1
, 1, 0)), c**2*exp_polar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*c**2)

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