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

Optimal. Leaf size=76 \[ \frac{\left (a d^2+2 b c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d x-c} \sqrt{c+d x}}{c}\right )}{2 c^3}+\frac{a \sqrt{d x-c} \sqrt{c+d x}}{2 c^2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.072997, antiderivative size = 76, 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 = {454, 92, 205} \[ \frac{\left (a d^2+2 b c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d x-c} \sqrt{c+d x}}{c}\right )}{2 c^3}+\frac{a \sqrt{d x-c} \sqrt{c+d x}}{2 c^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

Mathematica [A]  time = 0.0572556, size = 102, normalized size = 1.34 \[ \frac{x^2 \sqrt{d^2 x^2-c^2} \left (a d^2+2 b c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d^2 x^2-c^2}}{c}\right )+a \left (c d^2 x^2-c^3\right )}{2 c^3 x^2 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.022, size = 158, normalized size = 2.1 \begin{align*} -{\frac{1}{2\,{c}^{2}{x}^{2}}\sqrt{dx-c}\sqrt{dx+c} \left ( \ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{2}a{d}^{2}+2\,\ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{2}b{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}a \right ){\frac{1}{\sqrt{-{c}^{2}}}}{\frac{1}{\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d*x-c)^(1/2)*(d*x+c)^(1/2)/c^2*(ln(-2*(c^2-(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2))/x)*x^2*a*d^2+2*ln(-2*(c^2-(
-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2))/x)*x^2*b*c^2-(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2)*a)/(d^2*x^2-c^2)^(1/2)/(-c^2)^(
1/2)/x^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^3/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51693, size = 165, normalized size = 2.17 \begin{align*} \frac{2 \,{\left (2 \, b c^{2} + a d^{2}\right )} x^{2} \arctan \left (-\frac{d x - \sqrt{d x + c} \sqrt{d x - c}}{c}\right ) + \sqrt{d x + c} \sqrt{d x - c} a c}{2 \, c^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^3/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 28.438, size = 162, normalized size = 2.13 \begin{align*} - \frac{a d^{2}{G_{6, 6}^{5, 3}\left (\begin{matrix} \frac{7}{4}, \frac{9}{4}, 1 & 2, 2, \frac{5}{2} \\\frac{3}{2}, \frac{7}{4}, 2, \frac{9}{4}, \frac{5}{2} & 0 \end{matrix} \middle |{\frac{c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} c^{3}} + \frac{i a d^{2}{G_{6, 6}^{2, 6}\left (\begin{matrix} 1, \frac{5}{4}, \frac{3}{2}, \frac{7}{4}, 2, 1 & \\\frac{5}{4}, \frac{7}{4} & 1, \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}} c^{3}} - \frac{b{G_{6, 6}^{5, 3}\left (\begin{matrix} \frac{3}{4}, \frac{5}{4}, 1 & 1, 1, \frac{3}{2} \\\frac{1}{2}, \frac{3}{4}, 1, \frac{5}{4}, \frac{3}{2} & 0 \end{matrix} \middle |{\frac{c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} c} + \frac{i b{G_{6, 6}^{2, 6}\left (\begin{matrix} 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, 1 & \\\frac{1}{4}, \frac{3}{4} & 0, \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}} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)/x**3/(d*x-c)**(1/2)/(d*x+c)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.24414, size = 190, normalized size = 2.5 \begin{align*} -\frac{\frac{{\left (2 \, b c^{2} d + a d^{3}\right )} \arctan \left (\frac{{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2}}{2 \, c}\right )}{c^{3}} + \frac{2 \,{\left (a d^{3}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{6} - 4 \, a c^{2} d^{3}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2}\right )}}{{\left ({\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{4} + 4 \, c^{2}\right )}^{2} c^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^3/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

-((2*b*c^2*d + a*d^3)*arctan(1/2*(sqrt(d*x + c) - sqrt(d*x - c))^2/c)/c^3 + 2*(a*d^3*(sqrt(d*x + c) - sqrt(d*x
- c))^6 - 4*a*c^2*d^3*(sqrt(d*x + c) - sqrt(d*x - c))^2)/(((sqrt(d*x + c) - sqrt(d*x - c))^4 + 4*c^2)^2*c^2))
/d