3.787 \(\int \frac {x^2}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=151 \[ \frac {2 \sqrt {a+b x} \left (-3 a^2 d^2-6 a b c d+b^2 c^2\right )}{3 b d \sqrt {c+d x} (b c-a d)^3}-\frac {2 \sqrt {a+b x} \left (3 a^2 d^2+b^2 c^2\right )}{3 b^2 d (c+d x)^{3/2} (b c-a d)^2}-\frac {2 a^2}{b^2 \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {89, 78, 37} \[ \frac {2 \sqrt {a+b x} \left (-3 a^2 d^2-6 a b c d+b^2 c^2\right )}{3 b d \sqrt {c+d x} (b c-a d)^3}-\frac {2 \sqrt {a+b x} \left (3 a^2 d^2+b^2 c^2\right )}{3 b^2 d (c+d x)^{3/2} (b c-a d)^2}-\frac {2 a^2}{b^2 \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x]

[Out]

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

Rule 37

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

Rule 78

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rubi steps

\begin {align*} \int \frac {x^2}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx &=-\frac {2 a^2}{b^2 (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}+\frac {2 \int \frac {-\frac {1}{2} a (b c+3 a d)+\frac {1}{2} b (b c-a d) x}{\sqrt {a+b x} (c+d x)^{5/2}} \, dx}{b^2 (b c-a d)}\\ &=-\frac {2 a^2}{b^2 (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {2 \left (b^2 c^2+3 a^2 d^2\right ) \sqrt {a+b x}}{3 b^2 d (b c-a d)^2 (c+d x)^{3/2}}+\frac {\left (b^2 c^2-6 a b c d-3 a^2 d^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx}{3 b d (b c-a d)^2}\\ &=-\frac {2 a^2}{b^2 (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {2 \left (b^2 c^2+3 a^2 d^2\right ) \sqrt {a+b x}}{3 b^2 d (b c-a d)^2 (c+d x)^{3/2}}+\frac {2 \left (b^2 c^2-6 a b c d-3 a^2 d^2\right ) \sqrt {a+b x}}{3 b d (b c-a d)^3 \sqrt {c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 82, normalized size = 0.54 \[ \frac {-2 a^2 \left (8 c^2+12 c d x+3 d^2 x^2\right )-4 a b c x (2 c+3 d x)+2 b^2 c^2 x^2}{3 \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x]

[Out]

(2*b^2*c^2*x^2 - 4*a*b*c*x*(2*c + 3*d*x) - 2*a^2*(8*c^2 + 12*c*d*x + 3*d^2*x^2))/(3*(b*c - a*d)^3*Sqrt[a + b*x
]*(c + d*x)^(3/2))

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fricas [B]  time = 2.25, size = 275, normalized size = 1.82 \[ -\frac {2 \, {\left (8 \, a^{2} c^{2} - {\left (b^{2} c^{2} - 6 \, a b c d - 3 \, a^{2} d^{2}\right )} x^{2} + 4 \, {\left (a b c^{2} + 3 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 2.36, size = 395, normalized size = 2.62 \[ -\frac {4 \, \sqrt {b d} a^{2} b}{{\left (b^{2} c^{2} {\left | b \right |} - 2 \, a b c d {\left | b \right |} + a^{2} d^{2} {\left | b \right |}\right )} {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}} + \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (b^{6} c^{4} d {\left | b \right |} - 8 \, a b^{5} c^{3} d^{2} {\left | b \right |} + 13 \, a^{2} b^{4} c^{2} d^{3} {\left | b \right |} - 6 \, a^{3} b^{3} c d^{4} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{7} c^{5} d - 5 \, a b^{6} c^{4} d^{2} + 10 \, a^{2} b^{5} c^{3} d^{3} - 10 \, a^{3} b^{4} c^{2} d^{4} + 5 \, a^{4} b^{3} c d^{5} - a^{5} b^{2} d^{6}} - \frac {6 \, {\left (a b^{6} c^{4} d {\left | b \right |} - 3 \, a^{2} b^{5} c^{3} d^{2} {\left | b \right |} + 3 \, a^{3} b^{4} c^{2} d^{3} {\left | b \right |} - a^{4} b^{3} c d^{4} {\left | b \right |}\right )}}{b^{7} c^{5} d - 5 \, a b^{6} c^{4} d^{2} + 10 \, a^{2} b^{5} c^{3} d^{3} - 10 \, a^{3} b^{4} c^{2} d^{4} + 5 \, a^{4} b^{3} c d^{5} - a^{5} b^{2} d^{6}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

-4*sqrt(b*d)*a^2*b/((b^2*c^2*abs(b) - 2*a*b*c*d*abs(b) + a^2*d^2*abs(b))*(b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x
+ a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)) + 2/3*sqrt(b*x + a)*((b^6*c^4*d*abs(b) - 8*a*b^5*c^3*d^2*abs(b
) + 13*a^2*b^4*c^2*d^3*abs(b) - 6*a^3*b^3*c*d^4*abs(b))*(b*x + a)/(b^7*c^5*d - 5*a*b^6*c^4*d^2 + 10*a^2*b^5*c^
3*d^3 - 10*a^3*b^4*c^2*d^4 + 5*a^4*b^3*c*d^5 - a^5*b^2*d^6) - 6*(a*b^6*c^4*d*abs(b) - 3*a^2*b^5*c^3*d^2*abs(b)
+ 3*a^3*b^4*c^2*d^3*abs(b) - a^4*b^3*c*d^4*abs(b))/(b^7*c^5*d - 5*a*b^6*c^4*d^2 + 10*a^2*b^5*c^3*d^3 - 10*a^3
*b^4*c^2*d^4 + 5*a^4*b^3*c*d^5 - a^5*b^2*d^6))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2)

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maple [A]  time = 0.01, size = 111, normalized size = 0.74 \[ \frac {2 a^{2} d^{2} x^{2}+4 a b c d \,x^{2}-\frac {2}{3} b^{2} c^{2} x^{2}+8 a^{2} c d x +\frac {8}{3} a b \,c^{2} x +\frac {16}{3} a^{2} c^{2}}{\sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{2}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x+a)^(3/2)/(d*x+c)^(5/2),x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [B]  time = 1.63, size = 139, normalized size = 0.92 \[ \frac {\sqrt {c+d\,x}\,\left (\frac {x^2\,\left (6\,a^2\,d^2+12\,a\,b\,c\,d-2\,b^2\,c^2\right )}{3\,d^2\,{\left (a\,d-b\,c\right )}^3}+\frac {16\,a^2\,c^2}{3\,d^2\,{\left (a\,d-b\,c\right )}^3}+\frac {8\,a\,c\,x\,\left (3\,a\,d+b\,c\right )}{3\,d^2\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,\sqrt {a+b\,x}+\frac {c^2\,\sqrt {a+b\,x}}{d^2}+\frac {2\,c\,x\,\sqrt {a+b\,x}}{d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x)

[Out]

((c + d*x)^(1/2)*((x^2*(6*a^2*d^2 - 2*b^2*c^2 + 12*a*b*c*d))/(3*d^2*(a*d - b*c)^3) + (16*a^2*c^2)/(3*d^2*(a*d
- b*c)^3) + (8*a*c*x*(3*a*d + b*c))/(3*d^2*(a*d - b*c)^3)))/(x^2*(a + b*x)^(1/2) + (c^2*(a + b*x)^(1/2))/d^2 +
(2*c*x*(a + b*x)^(1/2))/d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b*x+a)**(3/2)/(d*x+c)**(5/2),x)

[Out]

Integral(x**2/((a + b*x)**(3/2)*(c + d*x)**(5/2)), x)

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