3.338 \(\int x^3 \sqrt{-c+d x} \sqrt{c+d x} (a+b x^2) \, dx\)

Optimal. Leaf size=109 \[ \frac{(d x-c)^{5/2} (c+d x)^{5/2} \left (a d^2+2 b c^2\right )}{5 d^6}+\frac{c^2 (d x-c)^{3/2} (c+d x)^{3/2} \left (a d^2+b c^2\right )}{3 d^6}+\frac{b (d x-c)^{7/2} (c+d x)^{7/2}}{7 d^6} \]

[Out]

(c^2*(b*c^2 + a*d^2)*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(3*d^6) + ((2*b*c^2 + a*d^2)*(-c + d*x)^(5/2)*(c + d*x)
^(5/2))/(5*d^6) + (b*(-c + d*x)^(7/2)*(c + d*x)^(7/2))/(7*d^6)

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Rubi [A]  time = 0.0867472, antiderivative size = 118, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {460, 100, 12, 74} \[ \frac{x^2 (d x-c)^{3/2} (c+d x)^{3/2} \left (7 a d^2+4 b c^2\right )}{35 d^4}+\frac{2 c^2 (d x-c)^{3/2} (c+d x)^{3/2} \left (7 a d^2+4 b c^2\right )}{105 d^6}+\frac{b x^4 (d x-c)^{3/2} (c+d x)^{3/2}}{7 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*(4*b*c^2 + 7*a*d^2)*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(105*d^6) + ((4*b*c^2 + 7*a*d^2)*x^2*(-c + d*x)^(
3/2)*(c + d*x)^(3/2))/(35*d^4) + (b*x^4*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(7*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 100

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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^3 \sqrt{-c+d x} \sqrt{c+d x} \left (a+b x^2\right ) \, dx &=\frac{b x^4 (-c+d x)^{3/2} (c+d x)^{3/2}}{7 d^2}-\frac{1}{7} \left (-7 a-\frac{4 b c^2}{d^2}\right ) \int x^3 \sqrt{-c+d x} \sqrt{c+d x} \, dx\\ &=\frac{\left (4 b c^2+7 a d^2\right ) x^2 (-c+d x)^{3/2} (c+d x)^{3/2}}{35 d^4}+\frac{b x^4 (-c+d x)^{3/2} (c+d x)^{3/2}}{7 d^2}+\frac{\left (4 b c^2+7 a d^2\right ) \int 2 c^2 x \sqrt{-c+d x} \sqrt{c+d x} \, dx}{35 d^4}\\ &=\frac{\left (4 b c^2+7 a d^2\right ) x^2 (-c+d x)^{3/2} (c+d x)^{3/2}}{35 d^4}+\frac{b x^4 (-c+d x)^{3/2} (c+d x)^{3/2}}{7 d^2}+\frac{\left (2 c^2 \left (4 b c^2+7 a d^2\right )\right ) \int x \sqrt{-c+d x} \sqrt{c+d x} \, dx}{35 d^4}\\ &=\frac{2 c^2 \left (4 b c^2+7 a d^2\right ) (-c+d x)^{3/2} (c+d x)^{3/2}}{105 d^6}+\frac{\left (4 b c^2+7 a d^2\right ) x^2 (-c+d x)^{3/2} (c+d x)^{3/2}}{35 d^4}+\frac{b x^4 (-c+d x)^{3/2} (c+d x)^{3/2}}{7 d^2}\\ \end{align*}

Mathematica [A]  time = 0.0528235, size = 88, normalized size = 0.81 \[ \frac{\sqrt{d x-c} \sqrt{c+d x} \left (d^2 x^2-c^2\right ) \left (7 a d^2 \left (2 c^2+3 d^2 x^2\right )+b \left (12 c^2 d^2 x^2+8 c^4+15 d^4 x^4\right )\right )}{105 d^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*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)*(7*a*d^2*(2*c^2 + 3*d^2*x^2) + b*(8*c^4 + 12*c^2*d^2*x^2 + 15*d
^4*x^4)))/(105*d^6)

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Maple [A]  time = 0.006, size = 68, normalized size = 0.6 \begin{align*}{\frac{15\,b{d}^{4}{x}^{4}+21\,a{d}^{4}{x}^{2}+12\,b{c}^{2}{d}^{2}{x}^{2}+14\,a{c}^{2}{d}^{2}+8\,b{c}^{4}}{105\,{d}^{6}} \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^3*(b*x^2+a)*(d*x-c)^(1/2)*(d*x+c)^(1/2),x)

[Out]

1/105*(d*x+c)^(3/2)*(15*b*d^4*x^4+21*a*d^4*x^2+12*b*c^2*d^2*x^2+14*a*c^2*d^2+8*b*c^4)*(d*x-c)^(3/2)/d^6

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Maxima [A]  time = 0.984139, size = 167, normalized size = 1.53 \begin{align*} \frac{{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} b x^{4}}{7 \, d^{2}} + \frac{4 \,{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} b c^{2} x^{2}}{35 \, d^{4}} + \frac{{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} a x^{2}}{5 \, d^{2}} + \frac{8 \,{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} b c^{4}}{105 \, d^{6}} + \frac{2 \,{\left (d^{2} x^{2} - c^{2}\right )}^{\frac{3}{2}} a c^{2}}{15 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)*(d*x-c)^(1/2)*(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/7*(d^2*x^2 - c^2)^(3/2)*b*x^4/d^2 + 4/35*(d^2*x^2 - c^2)^(3/2)*b*c^2*x^2/d^4 + 1/5*(d^2*x^2 - c^2)^(3/2)*a*x
^2/d^2 + 8/105*(d^2*x^2 - c^2)^(3/2)*b*c^4/d^6 + 2/15*(d^2*x^2 - c^2)^(3/2)*a*c^2/d^4

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Fricas [A]  time = 1.67887, size = 193, normalized size = 1.77 \begin{align*} \frac{{\left (15 \, b d^{6} x^{6} - 8 \, b c^{6} - 14 \, a c^{4} d^{2} - 3 \,{\left (b c^{2} d^{4} - 7 \, a d^{6}\right )} x^{4} -{\left (4 \, b c^{4} d^{2} + 7 \, a c^{2} d^{4}\right )} x^{2}\right )} \sqrt{d x + c} \sqrt{d x - c}}{105 \, d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)*(d*x-c)^(1/2)*(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

1/105*(15*b*d^6*x^6 - 8*b*c^6 - 14*a*c^4*d^2 - 3*(b*c^2*d^4 - 7*a*d^6)*x^4 - (4*b*c^4*d^2 + 7*a*c^2*d^4)*x^2)*
sqrt(d*x + c)*sqrt(d*x - c)/d^6

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \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**3*(b*x**2+a)*(d*x-c)**(1/2)*(d*x+c)**(1/2),x)

[Out]

Integral(x**3*(a + b*x**2)*sqrt(-c + d*x)*sqrt(c + d*x), x)

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Giac [A]  time = 1.19113, size = 227, normalized size = 2.08 \begin{align*} \frac{7 \,{\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} a +{\left ({\left (3 \,{\left ({\left (d x + c\right )}{\left (5 \,{\left (d x + c\right )}{\left (\frac{d x + c}{d^{5}} - \frac{6 \, c}{d^{5}}\right )} + \frac{74 \, c^{2}}{d^{5}}\right )} - \frac{96 \, c^{3}}{d^{5}}\right )}{\left (d x + c\right )} + \frac{203 \, c^{4}}{d^{5}}\right )}{\left (d x + c\right )} - \frac{70 \, c^{5}}{d^{5}}\right )}{\left (d x + c\right )}^{\frac{3}{2}} \sqrt{d x - c} b}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)*(d*x-c)^(1/2)*(d*x+c)^(1/2),x, algorithm="giac")

[Out]

1/105*(7*((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)*a + ((3*((d*x + c)*(5*(d*x + c)*((d*x + c)/d^5 - 6*c/d^5) + 74*c^2/d^5) - 96*c^3/d^5)*(d*x + c) + 203*c
^4/d^5)*(d*x + c) - 70*c^5/d^5)*(d*x + c)^(3/2)*sqrt(d*x - c)*b)/d