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

Optimal. Leaf size=115 \[ \frac {2 \sqrt {d x-c} \sqrt {c+d x} \left (3 a d^2+4 b c^2\right )}{3 d^6}-\frac {x^2 \left (3 a d^2+4 b c^2\right )}{3 d^4 \sqrt {d x-c} \sqrt {c+d x}}+\frac {b x^4}{3 d^2 \sqrt {d x-c} \sqrt {c+d x}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 115, 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, 98, 21, 74} \[ -\frac {x^2 \left (3 a d^2+4 b c^2\right )}{3 d^4 \sqrt {d x-c} \sqrt {c+d x}}+\frac {2 \sqrt {d x-c} \sqrt {c+d x} \left (3 a d^2+4 b c^2\right )}{3 d^6}+\frac {b x^4}{3 d^2 \sqrt {d x-c} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + 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 98

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

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

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Mathematica [A]  time = 0.05, size = 72, normalized size = 0.63 \[ \frac {-6 a c^2 d^2+3 a d^4 x^2-8 b c^4+4 b c^2 d^2 x^2+b d^4 x^4}{3 d^6 \sqrt {d x-c} \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.18, size = 80, normalized size = 0.70 \[ \frac {{\left (b d^{4} x^{4} - 8 \, b c^{4} - 6 \, a c^{2} d^{2} + {\left (4 \, b c^{2} d^{2} + 3 \, a d^{4}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{3 \, {\left (d^{8} x^{2} - c^{2} d^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.43, size = 200, normalized size = 1.74 \[ \frac {{\left (2 \, {\left (d x + c\right )} {\left ({\left (d x + c\right )} {\left (\frac {{\left (d x + c\right )} b}{d^{6}} - \frac {4 \, b c}{d^{6}}\right )} + \frac {10 \, b c^{2} d^{24} + 3 \, a d^{26}}{d^{30}}\right )} - \frac {3 \, {\left (9 \, b c^{3} d^{24} + 5 \, a c d^{26}\right )}}{d^{30}}\right )} \sqrt {d x + c}}{6 \, \sqrt {d x - c}} + \frac {2 \, {\left (b^{2} c^{8} + 2 \, a b c^{6} d^{2} + a^{2} c^{4} d^{4}\right )}}{{\left (b c^{4} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2} + a c^{2} d^{2} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{2} + 2 \, b c^{5} + 2 \, a c^{3} d^{2}\right )} d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(d*x + c)*((d*x + c)*((d*x + c)*b/d^6 - 4*b*c/d^6) + (10*b*c^2*d^24 + 3*a*d^26)/d^30) - 3*(9*b*c^3*d^24
+ 5*a*c*d^26)/d^30)*sqrt(d*x + c)/sqrt(d*x - c) + 2*(b^2*c^8 + 2*a*b*c^6*d^2 + a^2*c^4*d^4)/((b*c^4*(sqrt(d*x
+ c) - sqrt(d*x - c))^2 + a*c^2*d^2*(sqrt(d*x + c) - sqrt(d*x - c))^2 + 2*b*c^5 + 2*a*c^3*d^2)*d^6)

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maple [A]  time = 0.05, size = 68, normalized size = 0.59 \[ -\frac {-b \,d^{4} x^{4}-3 a \,d^{4} x^{2}-4 b \,c^{2} d^{2} x^{2}+6 a \,c^{2} d^{2}+8 b \,c^{4}}{3 \sqrt {d x +c}\, \sqrt {d x -c}\, d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 123, normalized size = 1.07 \[ \frac {b x^{4}}{3 \, \sqrt {d^{2} x^{2} - c^{2}} d^{2}} + \frac {4 \, b c^{2} x^{2}}{3 \, \sqrt {d^{2} x^{2} - c^{2}} d^{4}} + \frac {a x^{2}}{\sqrt {d^{2} x^{2} - c^{2}} d^{2}} - \frac {8 \, b c^{4}}{3 \, \sqrt {d^{2} x^{2} - c^{2}} d^{6}} - \frac {2 \, a c^{2}}{\sqrt {d^{2} x^{2} - c^{2}} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.80, size = 90, normalized size = 0.78 \[ \frac {\sqrt {d\,x-c}\,\left (\frac {x^2\,\left (4\,b\,c^2\,d^2+3\,a\,d^4\right )}{3\,d^7}-\frac {8\,b\,c^4+6\,a\,c^2\,d^2}{3\,d^7}+\frac {b\,x^4}{3\,d^3}\right )}{x\,\sqrt {c+d\,x}-\frac {c\,\sqrt {c+d\,x}}{d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 177.26, size = 226, normalized size = 1.97 \[ a \left (\frac {c {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{4} & -1, 0, \frac {1}{2}, 1 \\- \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} d^{4}} - \frac {i c {G_{6, 6}^{2, 6}\left (\begin {matrix} -2, - \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, 1 & \\- \frac {5}{4}, - \frac {3}{4} & -2, - \frac {3}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} d^{4}}\right ) + b \left (\frac {c^{3} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {7}{4}, - \frac {5}{4} & -2, -1, - \frac {1}{2}, 1 \\- \frac {7}{4}, - \frac {3}{2}, - \frac {5}{4}, -1, - \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} d^{6}} - \frac {i c^{3} {G_{6, 6}^{2, 6}\left (\begin {matrix} -3, - \frac {5}{2}, - \frac {9}{4}, -2, - \frac {7}{4}, 1 & \\- \frac {9}{4}, - \frac {7}{4} & -3, - \frac {5}{2}, - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}} d^{6}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x**2+a)/(d*x-c)**(3/2)/(d*x+c)**(3/2),x)

[Out]

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

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