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*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])

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Rubi [A]  time = 0.0550696, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, 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 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 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 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]

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*}

Mathematica [A]  time = 0.0407968, size = 134, normalized size = 0.85 \[ -\frac{2 \left (3 a^2 b d \left (4 c^2+7 c d x+4 d^2 x^2\right )+a^3 d^2 (2 c+3 d x)+a b^2 \left (21 c^2 d x+2 c^3+24 c d^2 x^2+8 d^3 x^3\right )+b^3 c x \left (3 c^2+12 c d x+8 d^2 x^2\right )\right )}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)^4} \]

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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Maple [A]  time = 0.006, size = 198, normalized size = 1.3 \begin{align*} -{\frac{16\,a{b}^{2}{d}^{3}{x}^{3}+16\,{b}^{3}c{d}^{2}{x}^{3}+24\,{a}^{2}b{d}^{3}{x}^{2}+48\,a{b}^{2}c{d}^{2}{x}^{2}+24\,{b}^{3}{c}^{2}d{x}^{2}+6\,{a}^{3}{d}^{3}x+42\,{a}^{2}bc{d}^{2}x+42\,a{b}^{2}{c}^{2}dx+6\,{b}^{3}{c}^{3}x+4\,{a}^{3}c{d}^{2}+24\,{a}^{2}b{c}^{2}d+4\,a{b}^{2}{c}^{3}}{3\,{a}^{4}{d}^{4}-12\,{a}^{3}bc{d}^{3}+18\,{a}^{2}{b}^{2}{c}^{2}{d}^{2}-12\,a{b}^{3}{c}^{3}d+3\,{b}^{4}{c}^{4}} \left ( bx+a \right ) ^{-{\frac{3}{2}}} \left ( dx+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 9.48107, size = 936, normalized size = 5.92 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (a + b x\right )^{\frac{5}{2}} \left (c + d x\right )^{\frac{5}{2}}}\, dx \end{align*}

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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Giac [B]  time = 2.23239, size = 933, normalized size = 5.91 \begin{align*} \frac{\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^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}} + \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^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}}\right )}}{{\left (b^{2} c +{\left (b x + a\right )} b d - a b d\right )}^{\frac{3}{2}}} - \frac{16 \,{\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}}}{12 \, b} \end{align*}

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]

1/12*(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^8*c^2*d^4 - 2*a*b^7*c*d^5 + a^2*b^6*d^6) + 3*(2*b^9*c^5*d^2*ab
s(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^8*c^2*d^4 - 2*a*b^7*c*d^5 + a^2*b^6*d^6))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 16*(
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