3.4.73 \(\int \frac {a+b x^2}{x (-c+d x)^{3/2} (c+d x)^{3/2}} \, dx\) [373]

Optimal. Leaf size=65 \[ -\frac {\frac {a}{c^2}+\frac {b}{d^2}}{\sqrt {-c+d x} \sqrt {c+d x}}-\frac {a \tan ^{-1}\left (\frac {\sqrt {-c+d x} \sqrt {c+d x}}{c}\right )}{c^3} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 65, normalized size of antiderivative = 1.00, 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 = {469, 94, 211} \begin {gather*} -\frac {a \text {ArcTan}\left (\frac {\sqrt {d x-c} \sqrt {c+d x}}{c}\right )}{c^3}-\frac {\frac {a}{c^2}+\frac {b}{d^2}}{\sqrt {d x-c} \sqrt {c+d x}} \end {gather*}

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 94

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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 469

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)]))

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) \text {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*}

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Mathematica [A]
time = 0.15, size = 70, normalized size = 1.08 \begin {gather*} \frac {-\frac {2 \left (b c^3+a c d^2\right )}{d^2 \sqrt {-c+d x} \sqrt {c+d x}}+4 a \tan ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {-c+d x}}\right )}{2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(193\) vs. \(2(58)=116\).
time = 0.30, size = 194, normalized size = 2.98

method	result	size
default	\(\frac {\sqrt {d x -c}\, \left (-\ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right ) a \,d^{4} x^{2}+\ln \left (-\frac {2 \left (c^{2}-\sqrt {-c^{2}}\, \sqrt {d^{2} x^{2}-c^{2}}\right )}{x}\right ) a \,c^{2} d^{2}+\sqrt {d^{2} x^{2}-c^{2}}\, \sqrt {-c^{2}}\, a \,d^{2}+\sqrt {d^{2} x^{2}-c^{2}}\, \sqrt {-c^{2}}\, b \,c^{2}\right )}{c^{2} \sqrt {-c^{2}}\, \left (-d x +c \right ) \sqrt {d^{2} x^{2}-c^{2}}\, d^{2} \sqrt {d x +c}}\)	\(194\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.53, size = 58, normalized size = 0.89 \begin {gather*} \frac {a \arcsin \left (\frac {c}{d {\left | x \right |}}\right )}{c^{3}} - \frac {a}{\sqrt {d^{2} x^{2} - c^{2}} c^{2}} - \frac {b}{\sqrt {d^{2} x^{2} - c^{2}} d^{2}} \end {gather*}

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]

a*arcsin(c/(d*abs(x)))/c^3 - a/(sqrt(d^2*x^2 - c^2)*c^2) - b/(sqrt(d^2*x^2 - c^2)*d^2)

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Fricas [A]
time = 3.13, size = 101, normalized size = 1.55 \begin {gather*} -\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 {gather*}

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

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]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (57) = 114\).
time = 0.59, size = 115, normalized size = 1.77 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {b\,x^2+a}{x\,{\left (c+d\,x\right )}^{3/2}\,{\left (d\,x-c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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