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

Optimal. Leaf size=117 \[ -\frac{3 a d^2+2 b c^2}{2 c^4 \sqrt{d x-c} \sqrt{c+d x}}-\frac{\left (3 a d^2+2 b c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d x-c} \sqrt{c+d x}}{c}\right )}{2 c^5}+\frac{a}{2 c^2 x^2 \sqrt{d x-c} \sqrt{c+d x}} \]

[Out]

-(2*b*c^2 + 3*a*d^2)/(2*c^4*Sqrt[-c + d*x]*Sqrt[c + d*x]) + a/(2*c^2*x^2*Sqrt[-c + d*x]*Sqrt[c + d*x]) - ((2*b
*c^2 + 3*a*d^2)*ArcTan[(Sqrt[-c + d*x]*Sqrt[c + d*x])/c])/(2*c^5)

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Rubi [A]  time = 0.0975361, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {454, 104, 21, 92, 205} \[ -\frac{3 a d^2+2 b c^2}{2 c^4 \sqrt{d x-c} \sqrt{c+d x}}-\frac{\left (3 a d^2+2 b c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d x-c} \sqrt{c+d x}}{c}\right )}{2 c^5}+\frac{a}{2 c^2 x^2 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*b*c^2 + 3*a*d^2)/(2*c^4*Sqrt[-c + d*x]*Sqrt[c + d*x]) + a/(2*c^2*x^2*Sqrt[-c + d*x]*Sqrt[c + d*x]) - ((2*b
*c^2 + 3*a*d^2)*ArcTan[(Sqrt[-c + d*x]*Sqrt[c + d*x])/c])/(2*c^5)

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 104

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] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegersQ[2*m, 2*n, 2*p]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx &=\frac{a}{2 c^2 x^2 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{1}{2} \left (2 b+\frac{3 a d^2}{c^2}\right ) \int \frac{1}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\\ &=-\frac{2 b c^2+3 a d^2}{2 c^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{a}{2 c^2 x^2 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{\left (-2 b-\frac{3 a d^2}{c^2}\right ) \int \frac{c d+d^2 x}{x \sqrt{-c+d x} (c+d x)^{3/2}} \, dx}{2 c^2 d}\\ &=-\frac{2 b c^2+3 a d^2}{2 c^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{a}{2 c^2 x^2 \sqrt{-c+d x} \sqrt{c+d x}}-\frac{\left (2 b c^2+3 a d^2\right ) \int \frac{1}{x \sqrt{-c+d x} \sqrt{c+d x}} \, dx}{2 c^4}\\ &=-\frac{2 b c^2+3 a d^2}{2 c^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{a}{2 c^2 x^2 \sqrt{-c+d x} \sqrt{c+d x}}-\frac{\left (d \left (2 b c^2+3 a d^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 )}{2 c^4}\\ &=-\frac{2 b c^2+3 a d^2}{2 c^4 \sqrt{-c+d x} \sqrt{c+d x}}+\frac{a}{2 c^2 x^2 \sqrt{-c+d x} \sqrt{c+d x}}-\frac{\left (2 b c^2+3 a d^2\right ) \tan ^{-1}\left (\frac{\sqrt{-c+d x} \sqrt{c+d x}}{c}\right )}{2 c^5}\\ \end{align*}

Mathematica [C]  time = 0.0282738, size = 75, normalized size = 0.64 \[ \frac{a c^2-x^2 \left (3 a d^2+2 b c^2\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};1-\frac{d^2 x^2}{c^2}\right )}{2 c^4 x^2 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^2 - (2*b*c^2 + 3*a*d^2)*x^2*Hypergeometric2F1[-1/2, 1, 1/2, 1 - (d^2*x^2)/c^2])/(2*c^4*x^2*Sqrt[-c + d*x]
*Sqrt[c + d*x])

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Maple [B]  time = 0.024, size = 315, normalized size = 2.7 \begin{align*}{\frac{1}{2\,{c}^{4}{x}^{2}} \left ( 3\,\ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{4}a{d}^{4}+2\,\ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{4}b{c}^{2}{d}^{2}-3\,\ln \left ( -2\,{\frac{{c}^{2}-\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}{x}} \right ){x}^{2}a{c}^{2}{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}^{4}-3\,\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}{x}^{2}a{d}^{2}-2\,\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}{x}^{2}b{c}^{2}+\sqrt{-{c}^{2}}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}a{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^3/(d*x-c)^(3/2)/(d*x+c)^(3/2),x)

[Out]

1/2/c^4*(3*ln(-2*(c^2-(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2))/x)*x^4*a*d^4+2*ln(-2*(c^2-(-c^2)^(1/2)*(d^2*x^2-c^2)^(
1/2))/x)*x^4*b*c^2*d^2-3*ln(-2*(c^2-(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2))/x)*x^2*a*c^2*d^2-2*ln(-2*(c^2-(-c^2)^(1/
2)*(d^2*x^2-c^2)^(1/2))/x)*x^2*b*c^4-3*(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2)*x^2*a*d^2-2*(-c^2)^(1/2)*(d^2*x^2-c^2)
^(1/2)*x^2*b*c^2+(-c^2)^(1/2)*(d^2*x^2-c^2)^(1/2)*a*c^2)/x^2/(-c^2)^(1/2)/(d^2*x^2-c^2)^(1/2)/(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^3/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59939, size = 278, normalized size = 2.38 \begin{align*} \frac{{\left (a c^{3} -{\left (2 \, b c^{3} + 3 \, a c d^{2}\right )} x^{2}\right )} \sqrt{d x + c} \sqrt{d x - c} - 2 \,{\left ({\left (2 \, b c^{2} d^{2} + 3 \, a d^{4}\right )} x^{4} -{\left (2 \, b c^{4} + 3 \, a c^{2} d^{2}\right )} x^{2}\right )} \arctan \left (-\frac{d x - \sqrt{d x + c} \sqrt{d x - c}}{c}\right )}{2 \,{\left (c^{5} d^{2} x^{4} - c^{7} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*c^3 - (2*b*c^3 + 3*a*c*d^2)*x^2)*sqrt(d*x + c)*sqrt(d*x - c) - 2*((2*b*c^2*d^2 + 3*a*d^4)*x^4 - (2*b*c
^4 + 3*a*c^2*d^2)*x^2)*arctan(-(d*x - sqrt(d*x + c)*sqrt(d*x - c))/c))/(c^5*d^2*x^4 - c^7*x^2)

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Sympy [C]  time = 79.3979, size = 165, normalized size = 1.41 \begin{align*} a \left (- \frac{d^{2}{G_{6, 6}^{5, 3}\left (\begin{matrix} \frac{9}{4}, \frac{11}{4}, 1 & 2, 3, \frac{7}{2} \\\frac{9}{4}, \frac{5}{2}, \frac{11}{4}, 3, \frac{7}{2} & 0 \end{matrix} \middle |{\frac{c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac{3}{2}} c^{5}} - \frac{i d^{2}{G_{6, 6}^{2, 6}\left (\begin{matrix} 1, \frac{3}{2}, \frac{7}{4}, 2, \frac{9}{4}, 1 & \\\frac{7}{4}, \frac{9}{4} & 1, \frac{3}{2}, \frac{5}{2}, 0 \end{matrix} \middle |{\frac{c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac{3}{2}} c^{5}}\right ) + b \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.38762, size = 285, normalized size = 2.44 \begin{align*} \frac{{\left (2 \, b c^{2} + 3 \, a d^{2}\right )} \arctan \left (\frac{{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{2}}{2 \, c}\right )}{c^{5}} - \frac{{\left (b c^{2} + a d^{2}\right )} \sqrt{d x + c}}{2 \, \sqrt{d x - c} c^{5}} + \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^{4}} + \frac{2 \,{\left (a d^{2}{\left (\sqrt{d x + c} - \sqrt{d x - c}\right )}^{6} - 4 \, a c^{2} d^{2}{\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^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*c^2 + 3*a*d^2)*arctan(1/2*(sqrt(d*x + c) - sqrt(d*x - c))^2/c)/c^5 - 1/2*(b*c^2 + a*d^2)*sqrt(d*x + c)/(s
qrt(d*x - c)*c^5) + 2*(b*c^2 + a*d^2)/(((sqrt(d*x + c) - sqrt(d*x - c))^2 + 2*c)*c^4) + 2*(a*d^2*(sqrt(d*x + c
) - sqrt(d*x - c))^6 - 4*a*c^2*d^2*(sqrt(d*x + c) - sqrt(d*x - c))^2)/(((sqrt(d*x + c) - sqrt(d*x - c))^4 + 4*
c^2)^2*c^4)