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

Optimal. Leaf size=118 \[ \frac{x^2 \sqrt{d x-c} \sqrt{c+d x} \left (5 a d^2+4 b c^2\right )}{15 d^4}+\frac{2 c^2 \sqrt{d x-c} \sqrt{c+d x} \left (5 a d^2+4 b c^2\right )}{15 d^6}+\frac{b x^4 \sqrt{d x-c} \sqrt{c+d x}}{5 d^2} \]

[Out]

(2*c^2*(4*b*c^2 + 5*a*d^2)*Sqrt[-c + d*x]*Sqrt[c + d*x])/(15*d^6) + ((4*b*c^2 + 5*a*d^2)*x^2*Sqrt[-c + d*x]*Sq
rt[c + d*x])/(15*d^4) + (b*x^4*Sqrt[-c + d*x]*Sqrt[c + d*x])/(5*d^2)

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Rubi [A]  time = 0.0864362, antiderivative size = 118, normalized size of antiderivative = 1., 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} \[ \frac{x^2 \sqrt{d x-c} \sqrt{c+d x} \left (5 a d^2+4 b c^2\right )}{15 d^4}+\frac{2 c^2 \sqrt{d x-c} \sqrt{c+d x} \left (5 a d^2+4 b c^2\right )}{15 d^6}+\frac{b x^4 \sqrt{d x-c} \sqrt{c+d x}}{5 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2*(4*b*c^2 + 5*a*d^2)*Sqrt[-c + d*x]*Sqrt[c + d*x])/(15*d^6) + ((4*b*c^2 + 5*a*d^2)*x^2*Sqrt[-c + d*x]*Sq
rt[c + d*x])/(15*d^4) + (b*x^4*Sqrt[-c + d*x]*Sqrt[c + d*x])/(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 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 \frac{x^3 \left (a+b x^2\right )}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx &=\frac{b x^4 \sqrt{-c+d x} \sqrt{c+d x}}{5 d^2}-\frac{1}{5} \left (-5 a-\frac{4 b c^2}{d^2}\right ) \int \frac{x^3}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx\\ &=\frac{\left (4 b c^2+5 a d^2\right ) x^2 \sqrt{-c+d x} \sqrt{c+d x}}{15 d^4}+\frac{b x^4 \sqrt{-c+d x} \sqrt{c+d x}}{5 d^2}+\frac{\left (4 b c^2+5 a d^2\right ) \int \frac{2 c^2 x}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx}{15 d^4}\\ &=\frac{\left (4 b c^2+5 a d^2\right ) x^2 \sqrt{-c+d x} \sqrt{c+d x}}{15 d^4}+\frac{b x^4 \sqrt{-c+d x} \sqrt{c+d x}}{5 d^2}+\frac{\left (2 c^2 \left (4 b c^2+5 a d^2\right )\right ) \int \frac{x}{\sqrt{-c+d x} \sqrt{c+d x}} \, dx}{15 d^4}\\ &=\frac{2 c^2 \left (4 b c^2+5 a d^2\right ) \sqrt{-c+d x} \sqrt{c+d x}}{15 d^6}+\frac{\left (4 b c^2+5 a d^2\right ) x^2 \sqrt{-c+d x} \sqrt{c+d x}}{15 d^4}+\frac{b x^4 \sqrt{-c+d x} \sqrt{c+d x}}{5 d^2}\\ \end{align*}

Mathematica [A]  time = 0.0544678, size = 87, normalized size = 0.74 \[ \frac{\left (d^2 x^2-c^2\right ) \left (5 a d^2 \left (2 c^2+d^2 x^2\right )+b \left (4 c^2 d^2 x^2+8 c^4+3 d^4 x^4\right )\right )}{15 d^6 \sqrt{d x-c} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-c^2 + d^2*x^2)*(5*a*d^2*(2*c^2 + d^2*x^2) + b*(8*c^4 + 4*c^2*d^2*x^2 + 3*d^4*x^4)))/(15*d^6*Sqrt[-c + d*x]*
Sqrt[c + d*x])

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Maple [A]  time = 0.006, size = 68, normalized size = 0.6 \begin{align*}{\frac{3\,b{d}^{4}{x}^{4}+5\,a{d}^{4}{x}^{2}+4\,b{c}^{2}{d}^{2}{x}^{2}+10\,a{c}^{2}{d}^{2}+8\,b{c}^{4}}{15\,{d}^{6}}\sqrt{dx+c}\sqrt{dx-c}} \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]

1/15*(d*x+c)^(1/2)*(3*b*d^4*x^4+5*a*d^4*x^2+4*b*c^2*d^2*x^2+10*a*c^2*d^2+8*b*c^4)/d^6*(d*x-c)^(1/2)

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Maxima [A]  time = 0.966855, size = 167, normalized size = 1.42 \begin{align*} \frac{\sqrt{d^{2} x^{2} - c^{2}} b x^{4}}{5 \, d^{2}} + \frac{4 \, \sqrt{d^{2} x^{2} - c^{2}} b c^{2} x^{2}}{15 \, d^{4}} + \frac{\sqrt{d^{2} x^{2} - c^{2}} a x^{2}}{3 \, d^{2}} + \frac{8 \, \sqrt{d^{2} x^{2} - c^{2}} b c^{4}}{15 \, d^{6}} + \frac{2 \, \sqrt{d^{2} x^{2} - c^{2}} a c^{2}}{3 \, 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]

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

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Fricas [A]  time = 1.57225, size = 144, normalized size = 1.22 \begin{align*} \frac{{\left (3 \, b d^{4} x^{4} + 8 \, b c^{4} + 10 \, a c^{2} d^{2} +{\left (4 \, b c^{2} d^{2} + 5 \, a d^{4}\right )} x^{2}\right )} \sqrt{d x + c} \sqrt{d x - c}}{15 \, 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]

1/15*(3*b*d^4*x^4 + 8*b*c^4 + 10*a*c^2*d^2 + (4*b*c^2*d^2 + 5*a*d^4)*x^2)*sqrt(d*x + c)*sqrt(d*x - c)/d^6

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Sympy [C]  time = 60.904, size = 240, normalized size = 2.03 \begin{align*} \frac{a c^{3}{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{5}{4}, - \frac{3}{4} & -1, -1, - \frac{1}{2}, 1 \\- \frac{3}{2}, - \frac{5}{4}, -1, - \frac{3}{4}, - \frac{1}{2}, 0 & \end{matrix} \middle |{\frac{c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d^{4}} + \frac{i a c^{3}{G_{6, 6}^{2, 6}\left (\begin{matrix} -2, - \frac{7}{4}, - \frac{3}{2}, - \frac{5}{4}, -1, 1 & \\- \frac{7}{4}, - \frac{5}{4} & -2, - \frac{3}{2}, - \frac{3}{2}, 0 \end{matrix} \middle |{\frac{c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d^{4}} + \frac{b c^{5}{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{9}{4}, - \frac{7}{4} & -2, -2, - \frac{3}{2}, 1 \\- \frac{5}{2}, - \frac{9}{4}, -2, - \frac{7}{4}, - \frac{3}{2}, 0 & \end{matrix} \middle |{\frac{c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} d^{6}} + \frac{i b c^{5}{G_{6, 6}^{2, 6}\left (\begin{matrix} -3, - \frac{11}{4}, - \frac{5}{2}, - \frac{9}{4}, -2, 1 & \\- \frac{11}{4}, - \frac{9}{4} & -3, - \frac{5}{2}, - \frac{5}{2}, 0 \end{matrix} \middle |{\frac{c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac{3}{2}} 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)

[Out]

a*c**3*meijerg(((-5/4, -3/4), (-1, -1, -1/2, 1)), ((-3/2, -5/4, -1, -3/4, -1/2, 0), ()), c**2/(d**2*x**2))/(4*
pi**(3/2)*d**4) + I*a*c**3*meijerg(((-2, -7/4, -3/2, -5/4, -1, 1), ()), ((-7/4, -5/4), (-2, -3/2, -3/2, 0)), c
**2*exp_polar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**4) + b*c**5*meijerg(((-9/4, -7/4), (-2, -2, -3/2, 1)), ((-5
/2, -9/4, -2, -7/4, -3/2, 0), ()), c**2/(d**2*x**2))/(4*pi**(3/2)*d**6) + I*b*c**5*meijerg(((-3, -11/4, -5/2,
-9/4, -2, 1), ()), ((-11/4, -9/4), (-3, -5/2, -5/2, 0)), c**2*exp_polar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**6
)

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Giac [A]  time = 1.21164, size = 151, normalized size = 1.28 \begin{align*} \frac{{\left (15 \, b c^{4} d^{25} + 15 \, a c^{2} d^{27} -{\left (20 \, b c^{3} d^{25} + 10 \, a c d^{27} -{\left (22 \, b c^{2} d^{25} + 5 \, a d^{27} + 3 \,{\left ({\left (d x + c\right )} b d^{25} - 4 \, b c d^{25}\right )}{\left (d x + c\right )}\right )}{\left (d x + c\right )}\right )}{\left (d x + c\right )}\right )} \sqrt{d x + c} \sqrt{d x - c}}{276480 \, 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]

1/276480*(15*b*c^4*d^25 + 15*a*c^2*d^27 - (20*b*c^3*d^25 + 10*a*c*d^27 - (22*b*c^2*d^25 + 5*a*d^27 + 3*((d*x +
c)*b*d^25 - 4*b*c*d^25)*(d*x + c))*(d*x + c))*(d*x + c))*sqrt(d*x + c)*sqrt(d*x - c)/d