3.376 \(\int \frac {a+b x^2}{x^4 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=119 \[ -\frac {2 d^2 x \left (4 a d^2+3 b c^2\right )}{3 c^6 \sqrt {d x-c} \sqrt {c+d x}}+\frac {4 a d^2+3 b c^2}{3 c^4 x \sqrt {d x-c} \sqrt {c+d x}}+\frac {a}{3 c^2 x^3 \sqrt {d x-c} \sqrt {c+d x}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {454, 103, 12, 39} \[ -\frac {2 d^2 x \left (4 a d^2+3 b c^2\right )}{3 c^6 \sqrt {d x-c} \sqrt {c+d x}}+\frac {4 a d^2+3 b c^2}{3 c^4 x \sqrt {d x-c} \sqrt {c+d x}}+\frac {a}{3 c^2 x^3 \sqrt {d x-c} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a/(3*c^2*x^3*Sqrt[-c + d*x]*Sqrt[c + d*x]) + (3*b*c^2 + 4*a*d^2)/(3*c^4*x*Sqrt[-c + d*x]*Sqrt[c + d*x]) - (2*d
^2*(3*b*c^2 + 4*a*d^2)*x)/(3*c^6*Sqrt[-c + d*x]*Sqrt[c + d*x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rule 103

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] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

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^4 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx &=\frac {a}{3 c^2 x^3 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {1}{3} \left (3 b+\frac {4 a d^2}{c^2}\right ) \int \frac {1}{x^2 (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\\ &=\frac {a}{3 c^2 x^3 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {3 b c^2+4 a d^2}{3 c^4 x \sqrt {-c+d x} \sqrt {c+d x}}+\frac {\left (3 b+\frac {4 a d^2}{c^2}\right ) \int \frac {2 d^2}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx}{3 c^2}\\ &=\frac {a}{3 c^2 x^3 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {3 b c^2+4 a d^2}{3 c^4 x \sqrt {-c+d x} \sqrt {c+d x}}+\frac {\left (2 d^2 \left (3 b+\frac {4 a d^2}{c^2}\right )\right ) \int \frac {1}{(-c+d x)^{3/2} (c+d x)^{3/2}} \, dx}{3 c^2}\\ &=\frac {a}{3 c^2 x^3 \sqrt {-c+d x} \sqrt {c+d x}}+\frac {3 b c^2+4 a d^2}{3 c^4 x \sqrt {-c+d x} \sqrt {c+d x}}-\frac {2 d^2 \left (3 b c^2+4 a d^2\right ) x}{3 c^6 \sqrt {-c+d x} \sqrt {c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 77, normalized size = 0.65 \[ \frac {a \left (c^4+4 c^2 d^2 x^2-8 d^4 x^4\right )+3 b c^2 x^2 \left (c^2-2 d^2 x^2\right )}{3 c^6 x^3 \sqrt {d x-c} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*c^2*x^2*(c^2 - 2*d^2*x^2) + a*(c^4 + 4*c^2*d^2*x^2 - 8*d^4*x^4))/(3*c^6*x^3*Sqrt[-c + d*x]*Sqrt[c + d*x])

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fricas [A]  time = 0.88, size = 132, normalized size = 1.11 \[ -\frac {2 \, {\left (3 \, b c^{2} d^{3} + 4 \, a d^{5}\right )} x^{5} - 2 \, {\left (3 \, b c^{4} d + 4 \, a c^{2} d^{3}\right )} x^{3} - {\left (a c^{4} - 2 \, {\left (3 \, b c^{2} d^{2} + 4 \, a d^{4}\right )} x^{4} + {\left (3 \, b c^{4} + 4 \, a c^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{3 \, {\left (c^{6} d^{2} x^{5} - c^{8} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^4/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-1/3*(2*(3*b*c^2*d^3 + 4*a*d^5)*x^5 - 2*(3*b*c^4*d + 4*a*c^2*d^3)*x^3 - (a*c^4 - 2*(3*b*c^2*d^2 + 4*a*d^4)*x^4
+ (3*b*c^4 + 4*a*c^2*d^2)*x^2)*sqrt(d*x + c)*sqrt(d*x - c))/(c^6*d^2*x^5 - c^8*x^3)

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giac [B]  time = 0.73, size = 242, normalized size = 2.03 \[ -\frac {{\left (b c^{2} d + a d^{3}\right )} \sqrt {d x + c}}{2 \, \sqrt {d x - c} c^{6}} - \frac {2 \, {\left (b c^{2} d + a d^{3}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2} + 2 \, c\right )} c^{5}} - \frac {8 \, {\left (3 \, b c^{2} d {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{8} + 3 \, a d^{3} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{8} + 24 \, b c^{4} d {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 48 \, a c^{2} d^{3} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 48 \, b c^{6} d + 80 \, a c^{4} d^{3}\right )}}{3 \, {\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 4 \, c^{2}\right )}^{3} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/x^4/(d*x-c)^(3/2)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

-1/2*(b*c^2*d + a*d^3)*sqrt(d*x + c)/(sqrt(d*x - c)*c^6) - 2*(b*c^2*d + a*d^3)/(((sqrt(d*x + c) - sqrt(d*x - c
))^2 + 2*c)*c^5) - 8/3*(3*b*c^2*d*(sqrt(d*x + c) - sqrt(d*x - c))^8 + 3*a*d^3*(sqrt(d*x + c) - sqrt(d*x - c))^
8 + 24*b*c^4*d*(sqrt(d*x + c) - sqrt(d*x - c))^4 + 48*a*c^2*d^3*(sqrt(d*x + c) - sqrt(d*x - c))^4 + 48*b*c^6*d
+ 80*a*c^4*d^3)/(((sqrt(d*x + c) - sqrt(d*x - c))^4 + 4*c^2)^3*c^4)

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maple [A]  time = 0.05, size = 73, normalized size = 0.61 \[ \frac {-8 a \,d^{4} x^{4}-6 b \,c^{2} d^{2} x^{4}+4 a \,c^{2} d^{2} x^{2}+3 b \,c^{4} x^{2}+a \,c^{4}}{3 \sqrt {d x +c}\, \sqrt {d x -c}\, c^{6} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)/x^4/(d*x-c)^(3/2)/(d*x+c)^(3/2),x)

[Out]

1/3*(-8*a*d^4*x^4-6*b*c^2*d^2*x^4+4*a*c^2*d^2*x^2+3*b*c^4*x^2+a*c^4)/(d*x+c)^(1/2)/x^3/c^6/(d*x-c)^(1/2)

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maxima [A]  time = 1.34, size = 125, normalized size = 1.05 \[ -\frac {2 \, b d^{2} x}{\sqrt {d^{2} x^{2} - c^{2}} c^{4}} - \frac {8 \, a d^{4} x}{3 \, \sqrt {d^{2} x^{2} - c^{2}} c^{6}} + \frac {b}{\sqrt {d^{2} x^{2} - c^{2}} c^{2} x} + \frac {4 \, a d^{2}}{3 \, \sqrt {d^{2} x^{2} - c^{2}} c^{4} x} + \frac {a}{3 \, \sqrt {d^{2} x^{2} - c^{2}} c^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b*d^2*x/(sqrt(d^2*x^2 - c^2)*c^4) - 8/3*a*d^4*x/(sqrt(d^2*x^2 - c^2)*c^6) + b/(sqrt(d^2*x^2 - c^2)*c^2*x) +
4/3*a*d^2/(sqrt(d^2*x^2 - c^2)*c^4*x) + 1/3*a/(sqrt(d^2*x^2 - c^2)*c^2*x^3)

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mupad [B]  time = 2.90, size = 104, normalized size = 0.87 \[ \frac {\sqrt {d\,x-c}\,\left (\frac {a}{3\,c^2\,d}+\frac {x^2\,\left (3\,b\,c^4+4\,a\,c^2\,d^2\right )}{3\,c^6\,d}-\frac {x^4\,\left (6\,b\,c^2\,d^2+8\,a\,d^4\right )}{3\,c^6\,d}\right )}{x^4\,\sqrt {c+d\,x}-\frac {c\,x^3\,\sqrt {c+d\,x}}{d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)/(x^4*(c + d*x)^(3/2)*(d*x - c)^(3/2)),x)

[Out]

((d*x - c)^(1/2)*(a/(3*c^2*d) + (x^2*(3*b*c^4 + 4*a*c^2*d^2))/(3*c^6*d) - (x^4*(8*a*d^4 + 6*b*c^2*d^2))/(3*c^6
*d)))/(x^4*(c + d*x)^(1/2) - (c*x^3*(c + d*x)^(1/2))/d)

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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**4/(d*x-c)**(3/2)/(d*x+c)**(3/2),x)

[Out]

Timed out

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