3.362 \(\int \frac {a+b x^2}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx\)

Optimal. Leaf size=68 \[ \frac {\left (2 a d^2+b c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )}{d^3}+\frac {b x \sqrt {d x-c} \sqrt {c+d x}}{2 d^2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 68, 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 = {389, 63, 217, 206} \[ \frac {\left (2 a d^2+b c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )}{d^3}+\frac {b x \sqrt {d x-c} \sqrt {c+d x}}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x*Sqrt[-c + d*x]*Sqrt[c + d*x])/(2*d^2) + ((b*c^2 + 2*a*d^2)*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 389

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

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 119, normalized size = 1.75 \[ \frac {4 \left (a d^2+b c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )-\frac {2 b c^{5/2} \sqrt {\frac {d x}{c}+1} \sinh ^{-1}\left (\frac {\sqrt {d x-c}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {c+d x}}+b d x \sqrt {d x-c} \sqrt {c+d x}}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x*Sqrt[-c + d*x]*Sqrt[c + d*x] - (2*b*c^(5/2)*Sqrt[1 + (d*x)/c]*ArcSinh[Sqrt[-c + d*x]/(Sqrt[2]*Sqrt[c])]
)/Sqrt[c + d*x] + 4*(b*c^2 + a*d^2)*ArcTanh[Sqrt[-c + d*x]/Sqrt[c + d*x]])/(2*d^3)

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fricas [A]  time = 1.17, size = 63, normalized size = 0.93 \[ \frac {\sqrt {d x + c} \sqrt {d x - c} b d x - {\left (b c^{2} + 2 \, a d^{2}\right )} \log \left (-d x + \sqrt {d x + c} \sqrt {d x - c}\right )}{2 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.26, size = 79, normalized size = 1.16 \[ \frac {\sqrt {d x + c} \sqrt {d x - c} {\left (\frac {{\left (d x + c\right )} b}{d^{2}} - \frac {b c}{d^{2}}\right )} - \frac {2 \, {\left (b c^{2} + 2 \, a d^{2}\right )} \log \left ({\left | -\sqrt {d x + c} + \sqrt {d x - c} \right |}\right )}{d^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

1/2*(sqrt(d*x + c)*sqrt(d*x - c)*((d*x + c)*b/d^2 - b*c/d^2) - 2*(b*c^2 + 2*a*d^2)*log(abs(-sqrt(d*x + c) + sq
rt(d*x - c)))/d^2)/d

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maple [C]  time = 0.07, size = 124, normalized size = 1.82 \[ \frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (2 a \,d^{2} \ln \left (\left (d x +\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )\right )+b \,c^{2} \ln \left (\left (d x +\sqrt {d^{2} x^{2}-c^{2}}\, \mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )\right )+\sqrt {d^{2} x^{2}-c^{2}}\, b d x \,\mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )}{2 \sqrt {d^{2} x^{2}-c^{2}}\, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x)

[Out]

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

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maxima [A]  time = 0.46, size = 89, normalized size = 1.31 \[ \frac {b c^{2} \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{2 \, d^{3}} + \frac {a \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{d} + \frac {\sqrt {d^{2} x^{2} - c^{2}} b x}{2 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 10.80, size = 417, normalized size = 6.13 \[ \frac {\frac {2\,b\,c^2\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{\sqrt {-c}-\sqrt {d\,x-c}}+\frac {14\,b\,c^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^3}+\frac {14\,b\,c^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^5}+\frac {2\,b\,c^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^7}}{d^3-\frac {4\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^2}+\frac {6\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^4}-\frac {4\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^6}+\frac {d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^8}}+\frac {4\,a\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}{\sqrt {-d^2}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )}{\sqrt {-d^2}}-\frac {2\,b\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}-\sqrt {c}}{\sqrt {-c}-\sqrt {d\,x-c}}\right )}{d^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)/((c + d*x)^(1/2)*(d*x - c)^(1/2)),x)

[Out]

((2*b*c^2*((c + d*x)^(1/2) - c^(1/2)))/((-c)^(1/2) - (d*x - c)^(1/2)) + (14*b*c^2*((c + d*x)^(1/2) - c^(1/2))^
3)/((-c)^(1/2) - (d*x - c)^(1/2))^3 + (14*b*c^2*((c + d*x)^(1/2) - c^(1/2))^5)/((-c)^(1/2) - (d*x - c)^(1/2))^
5 + (2*b*c^2*((c + d*x)^(1/2) - c^(1/2))^7)/((-c)^(1/2) - (d*x - c)^(1/2))^7)/(d^3 - (4*d^3*((c + d*x)^(1/2) -
c^(1/2))^2)/((-c)^(1/2) - (d*x - c)^(1/2))^2 + (6*d^3*((c + d*x)^(1/2) - c^(1/2))^4)/((-c)^(1/2) - (d*x - c)^
(1/2))^4 - (4*d^3*((c + d*x)^(1/2) - c^(1/2))^6)/((-c)^(1/2) - (d*x - c)^(1/2))^6 + (d^3*((c + d*x)^(1/2) - c^
(1/2))^8)/((-c)^(1/2) - (d*x - c)^(1/2))^8) + (4*a*atan((d*((-c)^(1/2) - (d*x - c)^(1/2)))/((-d^2)^(1/2)*((c +
d*x)^(1/2) - c^(1/2)))))/(-d^2)^(1/2) - (2*b*c^2*atanh(((c + d*x)^(1/2) - c^(1/2))/((-c)^(1/2) - (d*x - c)^(1
/2))))/d^3

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sympy [C]  time = 41.78, size = 199, normalized size = 2.93 \[ \frac {a {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} - \frac {i a {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d} + \frac {b c^{2} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{4} & - \frac {1}{2}, - \frac {1}{2}, 0, 1 \\-1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 0 & \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{3}} - \frac {i b c^{2} {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 1 & \\- \frac {5}{4}, - \frac {3}{4} & - \frac {3}{2}, -1, -1, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)/(d*x-c)**(1/2)/(d*x+c)**(1/2),x)

[Out]

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

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