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

Optimal. Leaf size=158 \[ -\frac {2 c}{3 d (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}-\frac {16 d \sqrt {a+b x} (a d+b c)}{3 \sqrt {c+d x} (b c-a d)^4}-\frac {8 (a d+b c)}{3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^3}+\frac {2 (a d+b c)}{3 d (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^2} \]

[Out]

-2/3*c/d/(-a*d+b*c)/(b*x+a)^(3/2)/(d*x+c)^(3/2)+2/3*(a*d+b*c)/d/(-a*d+b*c)^2/(b*x+a)^(3/2)/(d*x+c)^(1/2)-8/3*(
a*d+b*c)/(-a*d+b*c)^3/(b*x+a)^(1/2)/(d*x+c)^(1/2)-16/3*d*(a*d+b*c)*(b*x+a)^(1/2)/(-a*d+b*c)^4/(d*x+c)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {78, 45, 37} \[ -\frac {2 c}{3 d (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}-\frac {16 d \sqrt {a+b x} (a d+b c)}{3 \sqrt {c+d x} (b c-a d)^4}-\frac {8 (a d+b c)}{3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^3}+\frac {2 (a d+b c)}{3 d (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c)/(3*d*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2)) + (2*(b*c + a*d))/(3*d*(b*c - a*d)^2*(a + b*x)^(3/2)*
Sqrt[c + d*x]) - (8*(b*c + a*d))/(3*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x]) - (16*d*(b*c + a*d)*Sqrt[a + b*
x])/(3*(b*c - a*d)^4*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 45

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] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 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]))))

Rubi steps

\begin {align*} \int \frac {x}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx &=-\frac {2 c}{3 d (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}-\frac {(b c+a d) \int \frac {1}{(a+b x)^{5/2} (c+d x)^{3/2}} \, dx}{d (b c-a d)}\\ &=-\frac {2 c}{3 d (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}+\frac {2 (b c+a d)}{3 d (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}+\frac {(4 (b c+a d)) \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx}{3 (b c-a d)^2}\\ &=-\frac {2 c}{3 d (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}+\frac {2 (b c+a d)}{3 d (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}-\frac {8 (b c+a d)}{3 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}-\frac {(8 d (b c+a d)) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx}{3 (b c-a d)^3}\\ &=-\frac {2 c}{3 d (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}+\frac {2 (b c+a d)}{3 d (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}-\frac {8 (b c+a d)}{3 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}-\frac {16 d (b c+a d) \sqrt {a+b x}}{3 (b c-a d)^4 \sqrt {c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 134, normalized size = 0.85 \[ -\frac {2 \left (a^3 d^2 (2 c+3 d x)+3 a^2 b d \left (4 c^2+7 c d x+4 d^2 x^2\right )+a b^2 \left (2 c^3+21 c^2 d x+24 c d^2 x^2+8 d^3 x^3\right )+b^3 c x \left (3 c^2+12 c d x+8 d^2 x^2\right )\right )}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a^3*d^2*(2*c + 3*d*x) + 3*a^2*b*d*(4*c^2 + 7*c*d*x + 4*d^2*x^2) + b^3*c*x*(3*c^2 + 12*c*d*x + 8*d^2*x^2)
+ a*b^2*(2*c^3 + 21*c^2*d*x + 24*c*d^2*x^2 + 8*d^3*x^3)))/(3*(b*c - a*d)^4*(a + b*x)^(3/2)*(c + d*x)^(3/2))

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fricas [B]  time = 2.79, size = 468, normalized size = 2.96 \[ -\frac {2 \, {\left (2 \, a b^{2} c^{3} + 12 \, a^{2} b c^{2} d + 2 \, a^{3} c d^{2} + 8 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{3} + 12 \, {\left (b^{3} c^{2} d + 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} + 3 \, {\left (b^{3} c^{3} + 7 \, a b^{2} c^{2} d + 7 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a^{2} b^{4} c^{6} - 4 \, a^{3} b^{3} c^{5} d + 6 \, a^{4} b^{2} c^{4} d^{2} - 4 \, a^{5} b c^{3} d^{3} + a^{6} c^{2} d^{4} + {\left (b^{6} c^{4} d^{2} - 4 \, a b^{5} c^{3} d^{3} + 6 \, a^{2} b^{4} c^{2} d^{4} - 4 \, a^{3} b^{3} c d^{5} + a^{4} b^{2} d^{6}\right )} x^{4} + 2 \, {\left (b^{6} c^{5} d - 3 \, a b^{5} c^{4} d^{2} + 2 \, a^{2} b^{4} c^{3} d^{3} + 2 \, a^{3} b^{3} c^{2} d^{4} - 3 \, a^{4} b^{2} c d^{5} + a^{5} b d^{6}\right )} x^{3} + {\left (b^{6} c^{6} - 9 \, a^{2} b^{4} c^{4} d^{2} + 16 \, a^{3} b^{3} c^{3} d^{3} - 9 \, a^{4} b^{2} c^{2} d^{4} + a^{6} d^{6}\right )} x^{2} + 2 \, {\left (a b^{5} c^{6} - 3 \, a^{2} b^{4} c^{5} d + 2 \, a^{3} b^{3} c^{4} d^{2} + 2 \, a^{4} b^{2} c^{3} d^{3} - 3 \, a^{5} b c^{2} d^{4} + a^{6} c d^{5}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(2*a*b^2*c^3 + 12*a^2*b*c^2*d + 2*a^3*c*d^2 + 8*(b^3*c*d^2 + a*b^2*d^3)*x^3 + 12*(b^3*c^2*d + 2*a*b^2*c*d
^2 + a^2*b*d^3)*x^2 + 3*(b^3*c^3 + 7*a*b^2*c^2*d + 7*a^2*b*c*d^2 + a^3*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^
2*b^4*c^6 - 4*a^3*b^3*c^5*d + 6*a^4*b^2*c^4*d^2 - 4*a^5*b*c^3*d^3 + a^6*c^2*d^4 + (b^6*c^4*d^2 - 4*a*b^5*c^3*d
^3 + 6*a^2*b^4*c^2*d^4 - 4*a^3*b^3*c*d^5 + a^4*b^2*d^6)*x^4 + 2*(b^6*c^5*d - 3*a*b^5*c^4*d^2 + 2*a^2*b^4*c^3*d
^3 + 2*a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*d^5 + a^5*b*d^6)*x^3 + (b^6*c^6 - 9*a^2*b^4*c^4*d^2 + 16*a^3*b^3*c^3*d^3
- 9*a^4*b^2*c^2*d^4 + a^6*d^6)*x^2 + 2*(a*b^5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3*b^3*c^4*d^2 + 2*a^4*b^2*c^3*d^3 -
3*a^5*b*c^2*d^4 + a^6*c*d^5)*x)

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giac [B]  time = 3.79, size = 829, normalized size = 5.25 \[ -\frac {2 \, {\left (\frac {\sqrt {b x + a} {\left (\frac {{\left (5 \, b^{8} c^{4} d^{3} {\left | b \right |} - 12 \, a b^{7} c^{3} d^{4} {\left | b \right |} + 6 \, a^{2} b^{6} c^{2} d^{5} {\left | b \right |} + 4 \, a^{3} b^{5} c d^{6} {\left | b \right |} - 3 \, a^{4} b^{4} d^{7} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{9} c^{7} d - 7 \, a b^{8} c^{6} d^{2} + 21 \, a^{2} b^{7} c^{5} d^{3} - 35 \, a^{3} b^{6} c^{4} d^{4} + 35 \, a^{4} b^{5} c^{3} d^{5} - 21 \, a^{5} b^{4} c^{2} d^{6} + 7 \, a^{6} b^{3} c d^{7} - a^{7} b^{2} d^{8}} + \frac {3 \, {\left (2 \, b^{9} c^{5} d^{2} {\left | b \right |} - 7 \, a b^{8} c^{4} d^{3} {\left | b \right |} + 8 \, a^{2} b^{7} c^{3} d^{4} {\left | b \right |} - 2 \, a^{3} b^{6} c^{2} d^{5} {\left | b \right |} - 2 \, a^{4} b^{5} c d^{6} {\left | b \right |} + a^{5} b^{4} d^{7} {\left | b \right |}\right )}}{b^{9} c^{7} d - 7 \, a b^{8} c^{6} d^{2} + 21 \, a^{2} b^{7} c^{5} d^{3} - 35 \, a^{3} b^{6} c^{4} d^{4} + 35 \, a^{4} b^{5} c^{3} d^{5} - 21 \, a^{5} b^{4} c^{2} d^{6} + 7 \, a^{6} b^{3} c d^{7} - a^{7} b^{2} d^{8}}\right )}}{{\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, \sqrt {b d} b^{8} c^{3} - \sqrt {b d} a b^{7} c^{2} d - 7 \, \sqrt {b d} a^{2} b^{6} c d^{2} + 5 \, \sqrt {b d} a^{3} b^{5} d^{3} - 6 \, \sqrt {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} b^{6} c^{2} - 6 \, \sqrt {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} a b^{5} c d + 12 \, \sqrt {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} a^{2} b^{4} d^{2} + 3 \, \sqrt {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 )}^{4} b^{4} c + 3 \, \sqrt {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 )}^{4} a b^{3} d\right )}}{{\left (b^{3} c^{3} {\left | b \right |} - 3 \, a b^{2} c^{2} d {\left | b \right |} + 3 \, a^{2} b c d^{2} {\left | b \right |} - a^{3} d^{3} {\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 )}^{3}}\right )}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(sqrt(b*x + a)*((5*b^8*c^4*d^3*abs(b) - 12*a*b^7*c^3*d^4*abs(b) + 6*a^2*b^6*c^2*d^5*abs(b) + 4*a^3*b^5*c*
d^6*abs(b) - 3*a^4*b^4*d^7*abs(b))*(b*x + a)/(b^9*c^7*d - 7*a*b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^3 - 35*a^3*b^6*c^
4*d^4 + 35*a^4*b^5*c^3*d^5 - 21*a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^7*b^2*d^8) + 3*(2*b^9*c^5*d^2*abs(b) - 7
*a*b^8*c^4*d^3*abs(b) + 8*a^2*b^7*c^3*d^4*abs(b) - 2*a^3*b^6*c^2*d^5*abs(b) - 2*a^4*b^5*c*d^6*abs(b) + a^5*b^4
*d^7*abs(b))/(b^9*c^7*d - 7*a*b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^3 - 35*a^3*b^6*c^4*d^4 + 35*a^4*b^5*c^3*d^5 - 21*
a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^7*b^2*d^8))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) + 2*(3*sqrt(b*d)*b^8*c
^3 - sqrt(b*d)*a*b^7*c^2*d - 7*sqrt(b*d)*a^2*b^6*c*d^2 + 5*sqrt(b*d)*a^3*b^5*d^3 - 6*sqrt(b*d)*(sqrt(b*d)*sqrt
(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^6*c^2 - 6*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c
+ (b*x + a)*b*d - a*b*d))^2*a*b^5*c*d + 12*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d -
a*b*d))^2*a^2*b^4*d^2 + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^4*c +
3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^3*d)/((b^3*c^3*abs(b) - 3*a*
b^2*c^2*d*abs(b) + 3*a^2*b*c*d^2*abs(b) - a^3*d^3*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)^3))/b

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maple [A]  time = 0.01, size = 198, normalized size = 1.25 \[ -\frac {2 \left (8 a \,b^{2} d^{3} x^{3}+8 b^{3} c \,d^{2} x^{3}+12 a^{2} b \,d^{3} x^{2}+24 a \,b^{2} c \,d^{2} x^{2}+12 b^{3} c^{2} d \,x^{2}+3 a^{3} d^{3} x +21 a^{2} b c \,d^{2} x +21 a \,b^{2} c^{2} d x +3 b^{3} c^{3} x +2 a^{3} c \,d^{2}+12 a^{2} b \,c^{2} d +2 a \,b^{2} c^{3}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}} \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(8*a*b^2*d^3*x^3+8*b^3*c*d^2*x^3+12*a^2*b*d^3*x^2+24*a*b^2*c*d^2*x^2+12*b^3*c^2*d*x^2+3*a^3*d^3*x+21*a^2*
b*c*d^2*x+21*a*b^2*c^2*d*x+3*b^3*c^3*x+2*a^3*c*d^2+12*a^2*b*c^2*d+2*a*b^2*c^3)/(b*x+a)^(3/2)/(d*x+c)^(3/2)/(a^
4*d^4-4*a^3*b*c*d^3+6*a^2*b^2*c^2*d^2-4*a*b^3*c^3*d+b^4*c^4)

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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/(b*x+a)^(5/2)/(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 = 1.85, size = 238, normalized size = 1.51 \[ -\frac {\sqrt {c+d\,x}\,\left (\frac {8\,x^2\,{\left (a\,d+b\,c\right )}^2}{d\,{\left (a\,d-b\,c\right )}^4}+\frac {16\,b\,x^3\,\left (a\,d+b\,c\right )}{3\,{\left (a\,d-b\,c\right )}^4}+\frac {x\,\left (6\,a^3\,d^3+42\,a^2\,b\,c\,d^2+42\,a\,b^2\,c^2\,d+6\,b^3\,c^3\right )}{3\,b\,d^2\,{\left (a\,d-b\,c\right )}^4}+\frac {4\,a\,c\,\left (a^2\,d^2+6\,a\,b\,c\,d+b^2\,c^2\right )}{3\,b\,d^2\,{\left (a\,d-b\,c\right )}^4}\right )}{x^3\,\sqrt {a+b\,x}+\frac {a\,c^2\,\sqrt {a+b\,x}}{b\,d^2}+\frac {x^2\,\left (a\,d+2\,b\,c\right )\,\sqrt {a+b\,x}}{b\,d}+\frac {c\,x\,\left (2\,a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b\,d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c + d*x)^(1/2)*((8*x^2*(a*d + b*c)^2)/(d*(a*d - b*c)^4) + (16*b*x^3*(a*d + b*c))/(3*(a*d - b*c)^4) + (x*(6*
a^3*d^3 + 6*b^3*c^3 + 42*a*b^2*c^2*d + 42*a^2*b*c*d^2))/(3*b*d^2*(a*d - b*c)^4) + (4*a*c*(a^2*d^2 + b^2*c^2 +
6*a*b*c*d))/(3*b*d^2*(a*d - b*c)^4)))/(x^3*(a + b*x)^(1/2) + (a*c^2*(a + b*x)^(1/2))/(b*d^2) + (x^2*(a*d + 2*b
*c)*(a + b*x)^(1/2))/(b*d) + (c*x*(2*a*d + b*c)*(a + b*x)^(1/2))/(b*d^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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