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]

(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

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Rubi [A]  time = 0.0323963, antiderivative size = 68, normalized size of antiderivative = 1., 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.261142, 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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Maple [C]  time = 0.016, size = 124, normalized size = 1.8 \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{2\,{d}^{3}}\sqrt{dx-c}\sqrt{dx+c} \left ({\it csgn} \left ( d \right ) d\sqrt{{d}^{2}{x}^{2}-{c}^{2}}xb+2\,\ln \left ( \left ( \sqrt{{d}^{2}{x}^{2}-{c}^{2}}{\it csgn} \left ( d \right ) +dx \right ){\it csgn} \left ( d \right ) \right ) a{d}^{2}+\ln \left ( \left ( \sqrt{{d}^{2}{x}^{2}-{c}^{2}}{\it csgn} \left ( d \right ) +dx \right ){\it csgn} \left ( d \right ) \right ) b{c}^{2} \right ){\frac{1}{\sqrt{{d}^{2}{x}^{2}-{c}^{2}}}}} \end{align*}

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

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Maxima [A]  time = 0.976806, size = 140, normalized size = 2.06 \begin{align*} \frac{a \log \left (2 \, d^{2} x + 2 \, \sqrt{d^{2} x^{2} - c^{2}} \sqrt{d^{2}}\right )}{\sqrt{d^{2}}} + \frac{b c^{2} \log \left (2 \, d^{2} x + 2 \, \sqrt{d^{2} x^{2} - c^{2}} \sqrt{d^{2}}\right )}{2 \, \sqrt{d^{2}} d^{2}} + \frac{\sqrt{d^{2} x^{2} - c^{2}} b x}{2 \, d^{2}} \end{align*}

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]

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

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Fricas [A]  time = 1.52222, size = 142, normalized size = 2.09 \begin{align*} \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}} \end{align*}

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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Sympy [C]  time = 16.4835, size = 199, normalized size = 2.93 \begin{align*} \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}} \end{align*}

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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Giac [A]  time = 1.20247, size = 107, normalized size = 1.57 \begin{align*} \frac{{\left ({\left (d x + c\right )} b d^{4} - b c d^{4}\right )} \sqrt{d x + c} \sqrt{d x - c} - 2 \,{\left (b c^{2} d^{4} + 2 \, a d^{6}\right )} \log \left ({\left | -\sqrt{d x + c} + \sqrt{d x - c} \right |}\right )}{384 \, d} \end{align*}

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]

1/384*(((d*x + c)*b*d^4 - b*c*d^4)*sqrt(d*x + c)*sqrt(d*x - c) - 2*(b*c^2*d^4 + 2*a*d^6)*log(abs(-sqrt(d*x + c
) + sqrt(d*x - c))))/d