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

Optimal. Leaf size=65 \[ -\frac {\frac {a}{c^2}+\frac {b}{d^2}}{\sqrt {d x-c} \sqrt {c+d x}}-\frac {a \tan ^{-1}\left (\frac {\sqrt {d x-c} \sqrt {c+d x}}{c}\right )}{c^3} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {458, 92, 205} \[ -\frac {\frac {a}{c^2}+\frac {b}{d^2}}{\sqrt {d x-c} \sqrt {c+d x}}-\frac {a \tan ^{-1}\left (\frac {\sqrt {d x-c} \sqrt {c+d x}}{c}\right )}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 458

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)
*(x_)^(n_)), x_Symbol] :> -Simp[((b1*b2*c - a1*a2*d)*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*(a2 + b2*x^(n/2))
^(p + 1))/(a1*a2*b1*b2*e*n*(p + 1)), x] - Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(a1*a2*b1*b2*n*
(p + 1)), Int[(e*x)^m*(a1 + b1*x^(n/2))^(p + 1)*(a2 + b2*x^(n/2))^(p + 1), x], x] /; FreeQ[{a1, b1, a2, b2, c,
d, e, m, n}, x] && EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -
5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1, m, -(n*(p + 1))]))

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 84, normalized size = 1.29 \[ -\frac {a d^2 \sqrt {d^2 x^2-c^2} \tan ^{-1}\left (\frac {\sqrt {d^2 x^2-c^2}}{c}\right )+a c d^2+b c^3}{c^3 d^2 \sqrt {d x-c} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.10, size = 101, normalized size = 1.55 \[ -\frac {{\left (b c^{3} + a c d^{2}\right )} \sqrt {d x + c} \sqrt {d x - c} + 2 \, {\left (a d^{4} x^{2} - a c^{2} d^{2}\right )} \arctan \left (-\frac {d x - \sqrt {d x + c} \sqrt {d x - c}}{c}\right )}{c^{3} d^{4} x^{2} - c^{5} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-((b*c^3 + a*c*d^2)*sqrt(d*x + c)*sqrt(d*x - c) + 2*(a*d^4*x^2 - a*c^2*d^2)*arctan(-(d*x - sqrt(d*x + c)*sqrt(
d*x - c))/c))/(c^3*d^4*x^2 - c^5*d^2)

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giac [B]  time = 0.36, size = 115, normalized size = 1.77 \[ \frac {2 \, a \arctan \left (\frac {{\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}}{2 \, c}\right )}{c^{3}} - \frac {{\left (b c^{2} + a d^{2}\right )} \sqrt {d x + c}}{2 \, \sqrt {d x - c} c^{3} d^{2}} + \frac {2 \, {\left (b c^{2} + a d^{2}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2} + 2 \, c\right )} c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.08, size = 188, normalized size = 2.89 \[ \frac {a \,d^{4} x^{2} \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right )-a \,c^{2} d^{2} \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right )-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, a \,d^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, b \,c^{2}}{\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, \sqrt {d x +c}\, \sqrt {d x -c}\, c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)/x/(d*x-c)^(3/2)/(d*x+c)^(3/2),x)

[Out]

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

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maxima [A]  time = 1.46, size = 58, normalized size = 0.89 \[ \frac {a \arcsin \left (\frac {c}{d {\left | x \right |}}\right )}{c^{3}} - \frac {a}{\sqrt {d^{2} x^{2} - c^{2}} c^{2}} - \frac {b}{\sqrt {d^{2} x^{2} - c^{2}} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {b\,x^2+a}{x\,{\left (c+d\,x\right )}^{3/2}\,{\left (d\,x-c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)/(x*(c + d*x)^(3/2)*(d*x - c)^(3/2)),x)

[Out]

int((a + b*x^2)/(x*(c + d*x)^(3/2)*(d*x - c)^(3/2)), x)

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sympy [C]  time = 136.44, size = 172, normalized size = 2.65 \[ a \left (- \frac {{G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4}, 1 & 1, 2, \frac {5}{2} \\\frac {5}{4}, \frac {3}{2}, \frac {7}{4}, 2, \frac {5}{2} & 0 \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} c^{3}} - \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, 1 & \\\frac {3}{4}, \frac {5}{4} & 0, \frac {1}{2}, \frac {3}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} c^{3}}\right ) + b \left (- \frac {{G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4}, 1 & 0, 1, \frac {3}{2} \\\frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} c d^{2}} - \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, 1 & \\- \frac {1}{4}, \frac {1}{4} & -1, - \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} c d^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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