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

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 117, normalized size of antiderivative = 1.00, 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 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 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 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 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^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*}

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Mathematica [C]  time = 0.03, 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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fricas [A]  time = 1.07, size = 138, normalized size = 1.18 \[ \frac {{\left (a c^{3} - {\left (2 \, b c^{3} + 3 \, a c d^{2}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c} - 2 \, {\left ({\left (2 \, b c^{2} d^{2} + 3 \, a d^{4}\right )} x^{4} - {\left (2 \, b c^{4} + 3 \, a c^{2} d^{2}\right )} x^{2}\right )} \arctan \left (-\frac {d x - \sqrt {d x + c} \sqrt {d x - c}}{c}\right )}{2 \, {\left (c^{5} d^{2} x^{4} - c^{7} x^{2}\right )}} \]

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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giac [B]  time = 0.54, size = 211, normalized size = 1.80 \[ \frac {{\left (2 \, b c^{2} + 3 \, a d^{2}\right )} \arctan \left (\frac {{\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}}{2 \, c}\right )}{c^{5}} - \frac {{\left (b c^{2} + a d^{2}\right )} \sqrt {d x + c}}{2 \, \sqrt {d x - c} c^{5}} + \frac {2 \, {\left (b c^{2} + a d^{2}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2} + 2 \, c\right )} c^{4}} + \frac {2 \, {\left (a d^{2} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{6} - 4 \, a c^{2} d^{2} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 4 \, c^{2}\right )}^{2} c^{4}} \]

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)

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maple [B]  time = 0.09, size = 315, normalized size = 2.69 \[ \frac {3 a \,d^{4} x^{4} \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right )+2 b \,c^{2} d^{2} x^{4} \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right )-3 a \,c^{2} d^{2} x^{2} \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right )-2 b \,c^{4} x^{2} \ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right )-3 \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, a \,d^{2} x^{2}-2 \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, b \,c^{2} x^{2}+\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, a \,c^{2}}{2 \sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\, \sqrt {d x +c}\, \sqrt {d x -c}\, c^{4} x^{2}} \]

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

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maxima [A]  time = 1.26, size = 104, normalized size = 0.89 \[ \frac {b \arcsin \left (\frac {c}{d {\left | x \right |}}\right )}{c^{3}} + \frac {3 \, a d^{2} \arcsin \left (\frac {c}{d {\left | x \right |}}\right )}{2 \, c^{5}} - \frac {b}{\sqrt {d^{2} x^{2} - c^{2}} c^{2}} - \frac {3 \, a d^{2}}{2 \, \sqrt {d^{2} x^{2} - c^{2}} c^{4}} + \frac {a}{2 \, \sqrt {d^{2} x^{2} - c^{2}} c^{2} x^{2}} \]

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]

b*arcsin(c/(d*abs(x)))/c^3 + 3/2*a*d^2*arcsin(c/(d*abs(x)))/c^5 - b/(sqrt(d^2*x^2 - c^2)*c^2) - 3/2*a*d^2/(sqr
t(d^2*x^2 - c^2)*c^4) + 1/2*a/(sqrt(d^2*x^2 - c^2)*c^2*x^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {b\,x^2+a}{x^3\,{\left (c+d\,x\right )}^{3/2}\,{\left (d\,x-c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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