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

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

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 118, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.06, size = 87, normalized size = 0.74 \[ \frac {\left (d^2 x^2-c^2\right ) \left (5 a d^2 \left (2 c^2+d^2 x^2\right )+b \left (8 c^4+4 c^2 d^2 x^2+3 d^4 x^4\right )\right )}{15 d^6 \sqrt {d x-c} \sqrt {c+d x}} \]

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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fricas [A]  time = 1.10, size = 66, normalized size = 0.56 \[ \frac {{\left (3 \, b d^{4} x^{4} + 8 \, b c^{4} + 10 \, a c^{2} d^{2} + {\left (4 \, b c^{2} d^{2} + 5 \, a d^{4}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{15 \, d^{6}} \]

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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giac [A]  time = 0.25, size = 124, normalized size = 1.05 \[ \frac {{\left ({\left ({\left (d x + c\right )} {\left (3 \, {\left (d x + c\right )} {\left (\frac {{\left (d x + c\right )} b}{d^{5}} - \frac {4 \, b c}{d^{5}}\right )} + \frac {22 \, b c^{2} d^{25} + 5 \, a d^{27}}{d^{30}}\right )} - \frac {10 \, {\left (2 \, b c^{3} d^{25} + a c d^{27}\right )}}{d^{30}}\right )} {\left (d x + c\right )} + \frac {15 \, {\left (b c^{4} d^{25} + a c^{2} d^{27}\right )}}{d^{30}}\right )} \sqrt {d x + c} \sqrt {d x - c}}{15 \, d} \]

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]

1/15*(((d*x + c)*(3*(d*x + c)*((d*x + c)*b/d^5 - 4*b*c/d^5) + (22*b*c^2*d^25 + 5*a*d^27)/d^30) - 10*(2*b*c^3*d
^25 + a*c*d^27)/d^30)*(d*x + c) + 15*(b*c^4*d^25 + a*c^2*d^27)/d^30)*sqrt(d*x + c)*sqrt(d*x - c)/d

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maple [A]  time = 0.05, size = 68, normalized size = 0.58 \[ \frac {\sqrt {d x +c}\, \left (3 b \,d^{4} x^{4}+5 a \,d^{4} x^{2}+4 b \,c^{2} d^{2} x^{2}+10 a \,c^{2} d^{2}+8 b \,c^{4}\right ) \sqrt {d x -c}}{15 d^{6}} \]

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 = 0.55, size = 124, normalized size = 1.05 \[ \frac {\sqrt {d^{2} x^{2} - c^{2}} b x^{4}}{5 \, d^{2}} + \frac {4 \, \sqrt {d^{2} x^{2} - c^{2}} b c^{2} x^{2}}{15 \, d^{4}} + \frac {\sqrt {d^{2} x^{2} - c^{2}} a x^{2}}{3 \, d^{2}} + \frac {8 \, \sqrt {d^{2} x^{2} - c^{2}} b c^{4}}{15 \, d^{6}} + \frac {2 \, \sqrt {d^{2} x^{2} - c^{2}} a c^{2}}{3 \, d^{4}} \]

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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mupad [B]  time = 2.70, size = 130, normalized size = 1.10 \[ \frac {\sqrt {d\,x-c}\,\left (\frac {8\,b\,c^5+10\,a\,c^3\,d^2}{15\,d^6}+\frac {x^3\,\left (4\,b\,c^2\,d^3+5\,a\,d^5\right )}{15\,d^6}+\frac {x\,\left (8\,b\,c^4\,d+10\,a\,c^2\,d^3\right )}{15\,d^6}+\frac {b\,x^5}{5\,d}+\frac {x^2\,\left (4\,b\,c^3\,d^2+5\,a\,c\,d^4\right )}{15\,d^6}+\frac {b\,c\,x^4}{5\,d^2}\right )}{\sqrt {c+d\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*x^2))/((c + d*x)^(1/2)*(d*x - c)^(1/2)),x)

[Out]

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

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sympy [C]  time = 70.84, size = 240, normalized size = 2.03 \[ \frac {a c^{3} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {5}{4}, - \frac {3}{4} & -1, -1, - \frac {1}{2}, 1 \\- \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{4}} + \frac {i a c^{3} {G_{6, 6}^{2, 6}\left (\begin {matrix} -2, - \frac {7}{4}, - \frac {3}{2}, - \frac {5}{4}, -1, 1 & \\- \frac {7}{4}, - \frac {5}{4} & -2, - \frac {3}{2}, - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{4}} + \frac {b c^{5} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {9}{4}, - \frac {7}{4} & -2, -2, - \frac {3}{2}, 1 \\- \frac {5}{2}, - \frac {9}{4}, -2, - \frac {7}{4}, - \frac {3}{2}, 0 & \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{6}} + \frac {i b c^{5} {G_{6, 6}^{2, 6}\left (\begin {matrix} -3, - \frac {11}{4}, - \frac {5}{2}, - \frac {9}{4}, -2, 1 & \\- \frac {11}{4}, - \frac {9}{4} & -3, - \frac {5}{2}, - \frac {5}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{6}} \]

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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