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/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

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Rubi [A]  time = 0.0805375, antiderivative size = 65, normalized size of antiderivative = 1., 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 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))]))

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{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*}

Mathematica [A]  time = 0.0388531, 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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Maple [B]  time = 0.021, size = 188, normalized size = 2.9 \begin{align*}{\frac{1}{{c}^{2}{d}^{2}} \left ( \ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{2}a{d}^{4}-\ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ) a{c}^{2}{d}^{2}-\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}}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{dx-c}}}} \end{align*}

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53798, size = 201, normalized size = 3.09 \begin{align*} -\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}} \end{align*}

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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Sympy [C]  time = 51.5601, size = 172, normalized size = 2.65 \begin{align*} 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 ) \end{align*}

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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Giac [B]  time = 1.31751, size = 155, normalized size = 2.38 \begin{align*} \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}} \end{align*}

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)