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

Optimal. Leaf size=114 \[ \frac{x \sqrt{d x-c} \sqrt{c+d x} \left (4 a d^2+b c^2\right )}{8 d^2}-\frac{c^2 \left (4 a d^2+b c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{4 d^3}+\frac{b x (d x-c)^{3/2} (c+d x)^{3/2}}{4 d^2} \]

[Out]

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

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Rubi [A]  time = 0.0465434, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {389, 38, 63, 217, 206} \[ \frac{x \sqrt{d x-c} \sqrt{c+d x} \left (4 a d^2+b c^2\right )}{8 d^2}-\frac{c^2 \left (4 a d^2+b c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{c+d x}}\right )}{4 d^3}+\frac{b x (d x-c)^{3/2} (c+d x)^{3/2}}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c^2 + 4*a*d^2)*x*Sqrt[-c + d*x]*Sqrt[c + d*x])/(8*d^2) + (b*x*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(4*d^2) -
(c^2*(b*c^2 + 4*a*d^2)*ArcTanh[Sqrt[-c + d*x]/Sqrt[c + d*x]])/(4*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 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 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 \sqrt{-c+d x} \sqrt{c+d x} \left (a+b x^2\right ) \, dx &=\frac{b x (-c+d x)^{3/2} (c+d x)^{3/2}}{4 d^2}-\frac{\left (-b c^2-4 a d^2\right ) \int \sqrt{-c+d x} \sqrt{c+d x} \, dx}{4 d^2}\\ &=\frac{\left (b c^2+4 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{8 d^2}+\frac{b x (-c+d x)^{3/2} (c+d x)^{3/2}}{4 d^2}+\frac{\left (c^2 \left (-b c^2-4 a d^2\right )\right ) \int \frac{1}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx}{8 d^2}\\ &=\frac{\left (b c^2+4 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{8 d^2}+\frac{b x (-c+d x)^{3/2} (c+d x)^{3/2}}{4 d^2}-\frac{\left (c^2 \left (b c^2+4 a d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 c+x^2}} \, dx,x,\sqrt{-c+d x}\right )}{4 d^3}\\ &=\frac{\left (b c^2+4 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{8 d^2}+\frac{b x (-c+d x)^{3/2} (c+d x)^{3/2}}{4 d^2}-\frac{\left (c^2 \left (b c^2+4 a d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt{-c+d x}}{\sqrt{c+d x}}\right )}{4 d^3}\\ &=\frac{\left (b c^2+4 a d^2\right ) x \sqrt{-c+d x} \sqrt{c+d x}}{8 d^2}+\frac{b x (-c+d x)^{3/2} (c+d x)^{3/2}}{4 d^2}-\frac{c^2 \left (b c^2+4 a d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{-c+d x}}{\sqrt{c+d x}}\right )}{4 d^3}\\ \end{align*}

Mathematica [A]  time = 0.234537, size = 129, normalized size = 1.13 \[ \frac{d x \left (c^2-d^2 x^2\right ) \left (b \left (c^2-2 d^2 x^2\right )-4 a d^2\right )-2 c^{5/2} \sqrt{d x-c} \sqrt{\frac{d x}{c}+1} \left (4 a d^2+b c^2\right ) \sinh ^{-1}\left (\frac{\sqrt{d x-c}}{\sqrt{2} \sqrt{c}}\right )}{8 d^3 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x*(c^2 - d^2*x^2)*(-4*a*d^2 + b*(c^2 - 2*d^2*x^2)) - 2*c^(5/2)*(b*c^2 + 4*a*d^2)*Sqrt[-c + d*x]*Sqrt[1 + (d
*x)/c]*ArcSinh[Sqrt[-c + d*x]/(Sqrt[2]*Sqrt[c])])/(8*d^3*Sqrt[-c + d*x]*Sqrt[c + d*x])

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Maple [C]  time = 0.01, size = 180, normalized size = 1.6 \begin{align*} -{\frac{{\it csgn} \left ( d \right ) }{8\,{d}^{3}}\sqrt{dx-c}\sqrt{dx+c} \left ( -2\,{\it csgn} \left ( d \right ){x}^{3}b{d}^{3}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}-4\,{\it csgn} \left ( d \right ){d}^{3}\sqrt{{d}^{2}{x}^{2}-{c}^{2}}xa+{\it csgn} \left ( d \right ) d\sqrt{{d}^{2}{x}^{2}-{c}^{2}}xb{c}^{2}+4\,\ln \left ( \left ( \sqrt{{d}^{2}{x}^{2}-{c}^{2}}{\it csgn} \left ( d \right ) +dx \right ){\it csgn} \left ( d \right ) \right ) a{c}^{2}{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}^{4} \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]

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

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Maxima [A]  time = 0.942171, size = 205, normalized size = 1.8 \begin{align*} -\frac{a c^{2} \log \left (2 \, d^{2} x + 2 \, \sqrt{d^{2} x^{2} - c^{2}} \sqrt{d^{2}}\right )}{2 \, \sqrt{d^{2}}} - \frac{b c^{4} \log \left (2 \, d^{2} x + 2 \, \sqrt{d^{2} x^{2} - c^{2}} \sqrt{d^{2}}\right )}{8 \, \sqrt{d^{2}} d^{2}} + \frac{1}{2} \, \sqrt{d^{2} x^{2} - c^{2}} a x + \frac{\sqrt{d^{2} x^{2} - c^{2}} b c^{2} x}{8 \, d^{2}} + \frac{{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} b x}{4 \, 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]

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

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Fricas [A]  time = 1.58721, size = 190, normalized size = 1.67 \begin{align*} \frac{{\left (2 \, b d^{3} x^{3} -{\left (b c^{2} d - 4 \, a d^{3}\right )} x\right )} \sqrt{d x + c} \sqrt{d x - c} +{\left (b c^{4} + 4 \, a c^{2} d^{2}\right )} \log \left (-d x + \sqrt{d x + c} \sqrt{d x - c}\right )}{8 \, 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]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b x^{2}\right ) \sqrt{- c + d x} \sqrt{c + d x}\, dx \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]

Integral((a + b*x**2)*sqrt(-c + d*x)*sqrt(c + d*x), x)

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Giac [A]  time = 1.2615, size = 204, normalized size = 1.79 \begin{align*} \frac{4 \,{\left (\sqrt{d x + c} \sqrt{d x - c} d x + 2 \, c^{2} \log \left ({\left | -\sqrt{d x + c} + \sqrt{d x - c} \right |}\right )\right )} a +{\left (\frac{2 \, c^{4} \log \left ({\left | -\sqrt{d x + c} + \sqrt{d x - c} \right |}\right )}{d^{2}} +{\left ({\left (d x + c\right )}{\left (2 \,{\left (d x + c\right )}{\left (\frac{d x + c}{d^{2}} - \frac{3 \, c}{d^{2}}\right )} + \frac{5 \, c^{2}}{d^{2}}\right )} - \frac{c^{3}}{d^{2}}\right )} \sqrt{d x + c} \sqrt{d x - c}\right )} b}{8 \, 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]

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