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

Optimal. Leaf size=63 \[ \frac {2 b \tanh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )}{d^3}-\frac {x \left (\frac {a}{c^2}+\frac {b}{d^2}\right )}{\sqrt {d x-c} \sqrt {c+d x}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {386, 63, 217, 206} \[ \frac {2 b \tanh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )}{d^3}-\frac {x \left (\frac {a}{c^2}+\frac {b}{d^2}\right )}{\sqrt {d x-c} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 386

Int[((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symb
ol] :> -Simp[((b1*b2*c - a1*a2*d)*x*(a1 + b1*x^(n/2))^(p + 1)*(a2 + b2*x^(n/2))^(p + 1))/(a1*a2*b1*b2*n*(p + 1
)), x] - Dist[(a1*a2*d - b1*b2*c*(n*(p + 1) + 1))/(a1*a2*b1*b2*n*(p + 1)), Int[(a1 + b1*x^(n/2))^(p + 1)*(a2 +
b2*x^(n/2))^(p + 1), x], x] /; FreeQ[{a1, b1, a2, b2, c, d, n}, x] && EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0]
&& (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 86, normalized size = 1.37 \[ \frac {2 b c^{5/2} \sqrt {\frac {d x}{c}+1} \sinh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {2} \sqrt {c}}\right )-\frac {d x \left (a d^2+b c^2\right )}{\sqrt {d x-c}}}{c^2 d^3 \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((d*(b*c^2 + a*d^2)*x)/Sqrt[-c + d*x]) + 2*b*c^(5/2)*Sqrt[1 + (d*x)/c]*ArcSinh[Sqrt[-c + d*x]/(Sqrt[2]*Sqrt[
c])])/(c^2*d^3*Sqrt[c + d*x])

________________________________________________________________________________________

fricas [B]  time = 1.41, size = 129, normalized size = 2.05 \[ \frac {b c^{4} + a c^{2} d^{2} - {\left (b c^{2} d + a d^{3}\right )} \sqrt {d x + c} \sqrt {d x - c} x - {\left (b c^{2} d^{2} + a d^{4}\right )} x^{2} - {\left (b c^{2} d^{2} x^{2} - b c^{4}\right )} \log \left (-d x + \sqrt {d x + c} \sqrt {d x - c}\right )}{c^{2} d^{5} x^{2} - c^{4} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.28, size = 113, normalized size = 1.79 \[ -\frac {b \log \left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2}\right )}{d^{3}} - \frac {2 \, {\left (b c^{2} + a d^{2}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2} + 2 \, c\right )} c d^{3}} - \frac {{\left (b c^{2} d^{3} + a d^{5}\right )} \sqrt {d x + c}}{2 \, \sqrt {d x - c} c^{2} d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.07, size = 160, normalized size = 2.54 \[ \frac {\left (b \,c^{2} d^{2} x^{2} \ln \left (\left (d x +\sqrt {\left (d x -c \right ) \left (d x +c \right )}\, \mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )\right )-\sqrt {d^{2} x^{2}-c^{2}}\, a \,d^{3} x \,\mathrm {csgn}\relax (d )-b \,c^{4} \ln \left (\left (d x +\sqrt {\left (d x -c \right ) \left (d x +c \right )}\, \mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )\right )-\sqrt {d^{2} x^{2}-c^{2}}\, b \,c^{2} d x \,\mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )}{\sqrt {d^{2} x^{2}-c^{2}}\, \sqrt {d x +c}\, \sqrt {d x -c}\, c^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.55, size = 76, normalized size = 1.21 \[ -\frac {a x}{\sqrt {d^{2} x^{2} - c^{2}} c^{2}} - \frac {b x}{\sqrt {d^{2} x^{2} - c^{2}} d^{2}} + \frac {b \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 112.36, size = 182, normalized size = 2.89 \[ a \left (- \frac {{G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & \frac {1}{2}, \frac {3}{2}, 2 \\\frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2}, 2 & 0 \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} c^{2} d} + \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1 & \\\frac {1}{4}, \frac {3}{4} & - \frac {1}{2}, 0, 1, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} c^{2} d}\right ) + b \left (\frac {{G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, \frac {1}{2}, 1, 1 \\- \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1, 0 & \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} d^{3}} + \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {3}{2}, -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 1 & \\- \frac {3}{4}, - \frac {1}{4} & - \frac {3}{2}, -1, 0, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} d^{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________