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

Optimal. Leaf size=67 \[ \frac{(d x-c)^{3/2} (c+d x)^{3/2} \left (a d^2+b c^2\right )}{3 d^4}+\frac{b (d x-c)^{5/2} (c+d x)^{5/2}}{5 d^4} \]

[Out]

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

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Rubi [A]  time = 0.0443008, antiderivative size = 72, normalized size of antiderivative = 1.07, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {460, 74} \[ \frac{(d x-c)^{3/2} (c+d x)^{3/2} \left (5 a d^2+2 b c^2\right )}{15 d^4}+\frac{b x^2 (d x-c)^{3/2} (c+d x)^{3/2}}{5 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*c^2 + 5*a*d^2)*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(15*d^4) + (b*x^2*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(5*
d^2)

Rule 460

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

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rubi steps

\begin{align*} \int x \sqrt{-c+d x} \sqrt{c+d x} \left (a+b x^2\right ) \, dx &=\frac{b x^2 (-c+d x)^{3/2} (c+d x)^{3/2}}{5 d^2}-\frac{1}{5} \left (-5 a-\frac{2 b c^2}{d^2}\right ) \int x \sqrt{-c+d x} \sqrt{c+d x} \, dx\\ &=\frac{\left (2 b c^2+5 a d^2\right ) (-c+d x)^{3/2} (c+d x)^{3/2}}{15 d^4}+\frac{b x^2 (-c+d x)^{3/2} (c+d x)^{3/2}}{5 d^2}\\ \end{align*}

Mathematica [A]  time = 0.03065, size = 62, normalized size = 0.93 \[ \frac{\sqrt{d x-c} \sqrt{c+d x} \left (d^2 x^2-c^2\right ) \left (5 a d^2+2 b c^2+3 b d^2 x^2\right )}{15 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[-c + d*x]*Sqrt[c + d*x]*(-c^2 + d^2*x^2)*(2*b*c^2 + 5*a*d^2 + 3*b*d^2*x^2))/(15*d^4)

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Maple [A]  time = 0.002, size = 44, normalized size = 0.7 \begin{align*}{\frac{3\,b{d}^{2}{x}^{2}+5\,a{d}^{2}+2\,b{c}^{2}}{15\,{d}^{4}} \left ( dx+c \right ) ^{{\frac{3}{2}}} \left ( dx-c \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(d*x+c)^(3/2)*(3*b*d^2*x^2+5*a*d^2+2*b*c^2)*(d*x-c)^(3/2)/d^4

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Maxima [A]  time = 0.96054, size = 95, normalized size = 1.42 \begin{align*} \frac{{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} b x^{2}}{5 \, d^{2}} + \frac{2 \,{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} b c^{2}}{15 \, d^{4}} + \frac{{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} a}{3 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(d^2*x^2 - c^2)^(3/2)*b*x^2/d^2 + 2/15*(d^2*x^2 - c^2)^(3/2)*b*c^2/d^4 + 1/3*(d^2*x^2 - c^2)^(3/2)*a/d^2

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Fricas [A]  time = 1.5493, size = 140, normalized size = 2.09 \begin{align*} \frac{{\left (3 \, b d^{4} x^{4} - 2 \, b c^{4} - 5 \, a c^{2} d^{2} -{\left (b c^{2} d^{2} - 5 \, a d^{4}\right )} x^{2}\right )} \sqrt{d x + c} \sqrt{d x - c}}{15 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b*d^4*x^4 - 2*b*c^4 - 5*a*c^2*d^2 - (b*c^2*d^2 - 5*a*d^4)*x^2)*sqrt(d*x + c)*sqrt(d*x - c)/d^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \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(x*(b*x**2+a)*(d*x-c)**(1/2)*(d*x+c)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.51827, size = 126, normalized size = 1.88 \begin{align*} \frac{{\left ({\left (d x + c\right )}{\left (3 \,{\left (d x + c\right )}{\left (\frac{d x + c}{d^{3}} - \frac{4 \, c}{d^{3}}\right )} + \frac{17 \, c^{2}}{d^{3}}\right )} - \frac{10 \, c^{3}}{d^{3}}\right )}{\left (d x + c\right )}^{\frac{3}{2}} \sqrt{d x - c} b + \frac{5 \,{\left (d x + c\right )}^{\frac{3}{2}}{\left (d x - c\right )}^{\frac{3}{2}} a}{d}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(((d*x + c)*(3*(d*x + c)*((d*x + c)/d^3 - 4*c/d^3) + 17*c^2/d^3) - 10*c^3/d^3)*(d*x + c)^(3/2)*sqrt(d*x -
c)*b + 5*(d*x + c)^(3/2)*(d*x - c)^(3/2)*a/d)/d