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

Optimal. Leaf size=189 \[ \frac {4 c \sqrt {a+b x} \left (-3 a^2 d^2-6 a b c d+b^2 c^2\right )}{3 b d \sqrt {c+d x} (b c-a d)^4}-\frac {4 c \sqrt {a+b x} \left (3 a^2 d^2+b^2 c^2\right )}{3 b^2 d (c+d x)^{3/2} (b c-a d)^3}-\frac {4 a^2 c}{b^2 \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)^2}-\frac {2 x^3}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)} \]

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Rubi [A]  time = 0.15, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {94, 89, 78, 37} \begin {gather*} \frac {4 c \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)^4}-\frac {4 c \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)^3}-\frac {4 a^2 c}{b^2 \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)^2}-\frac {2 x^3}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 94

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

Rubi steps

\begin {align*} \int \frac {x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx &=-\frac {2 x^3}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}+\frac {(2 c) \int \frac {x^2}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx}{b c-a d}\\ &=-\frac {2 x^3}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}-\frac {4 a^2 c}{b^2 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/2}}+\frac {(4 c) \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)^2}\\ &=-\frac {2 x^3}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}-\frac {4 a^2 c}{b^2 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/2}}-\frac {4 c \left (b^2 c^2+3 a^2 d^2\right ) \sqrt {a+b x}}{3 b^2 d (b c-a d)^3 (c+d x)^{3/2}}+\frac {\left (2 c \left (b^2 c^2-6 a b c d-3 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx}{3 b d (b c-a d)^3}\\ &=-\frac {2 x^3}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}-\frac {4 a^2 c}{b^2 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/2}}-\frac {4 c \left (b^2 c^2+3 a^2 d^2\right ) \sqrt {a+b x}}{3 b^2 d (b c-a d)^3 (c+d x)^{3/2}}+\frac {4 c \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)^4 \sqrt {c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 125, normalized size = 0.66 \begin {gather*} -\frac {2 \left (a^3 \left (16 c^3+24 c^2 d x+6 c d^2 x^2-d^3 x^3\right )+3 a^2 b c x \left (8 c^2+12 c d x+3 d^2 x^2\right )+3 a b^2 c^2 x^2 (2 c+3 d x)-b^3 c^3 x^3\right )}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-(b^3*c^3*x^3) + 3*a*b^2*c^2*x^2*(2*c + 3*d*x) + 3*a^2*b*c*x*(8*c^2 + 12*c*d*x + 3*d^2*x^2) + a^3*(16*c^3
+ 24*c^2*d*x + 6*c*d^2*x^2 - d^3*x^3)))/(3*(b*c - a*d)^4*(a + b*x)^(3/2)*(c + d*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.14, size = 92, normalized size = 0.49 \begin {gather*} \frac {2 (a+b x)^{3/2} \left (\frac {a^3 (c+d x)^3}{(a+b x)^3}-\frac {9 a^2 c (c+d x)^2}{(a+b x)^2}-\frac {9 a c^2 (c+d x)}{a+b x}+c^3\right )}{3 (c+d x)^{3/2} (b c-a d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(3/2)*(c^3 - (9*a*c^2*(c + d*x))/(a + b*x) - (9*a^2*c*(c + d*x)^2)/(a + b*x)^2 + (a^3*(c + d*x)^3
)/(a + b*x)^3))/(3*(b*c - a*d)^4*(c + d*x)^(3/2))

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fricas [B]  time = 3.55, size = 448, normalized size = 2.37 \begin {gather*} -\frac {2 \, {\left (16 \, a^{3} c^{3} - {\left (b^{3} c^{3} - 9 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{3} + 6 \, {\left (a b^{2} c^{3} + 6 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} x^{2} + 24 \, {\left (a^{2} b c^{3} + a^{3} c^{2} d\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(16*a^3*c^3 - (b^3*c^3 - 9*a*b^2*c^2*d - 9*a^2*b*c*d^2 + a^3*d^3)*x^3 + 6*(a*b^2*c^3 + 6*a^2*b*c^2*d + a^
3*c*d^2)*x^2 + 24*(a^2*b*c^3 + a^3*c^2*d)*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 = 3.97, size = 771, normalized size = 4.08 \begin {gather*} \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (b^{7} c^{6} d {\left | b \right |} - 12 \, a b^{6} c^{5} d^{2} {\left | b \right |} + 30 \, a^{2} b^{5} c^{4} d^{3} {\left | b \right |} - 28 \, a^{3} b^{4} c^{3} d^{4} {\left | b \right |} + 9 \, a^{4} b^{3} c^{2} d^{5} {\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 {9 \, {\left (a b^{7} c^{6} d {\left | b \right |} - 4 \, a^{2} b^{6} c^{5} d^{2} {\left | b \right |} + 6 \, a^{3} b^{5} c^{4} d^{3} {\left | b \right |} - 4 \, a^{4} b^{4} c^{3} d^{4} {\left | b \right |} + a^{5} b^{3} c^{2} d^{5} {\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 )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} - \frac {4 \, {\left (9 \, \sqrt {b d} a^{2} b^{5} c^{3} - 19 \, \sqrt {b d} a^{3} b^{4} c^{2} d + 11 \, \sqrt {b d} a^{4} b^{3} c d^{2} - \sqrt {b d} a^{5} b^{2} d^{3} - 18 \, \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^{3} c^{2} + 18 \, \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^{3} b^{2} c d + 9 \, \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^{2} b 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^{3} d\right )}}{3 \, {\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(b*x + a)*((b^7*c^6*d*abs(b) - 12*a*b^6*c^5*d^2*abs(b) + 30*a^2*b^5*c^4*d^3*abs(b) - 28*a^3*b^4*c^3*d^
4*abs(b) + 9*a^4*b^3*c^2*d^5*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) - 9*(a*b^7*c^6*d*abs(b) - 4
*a^2*b^6*c^5*d^2*abs(b) + 6*a^3*b^5*c^4*d^3*abs(b) - 4*a^4*b^4*c^3*d^4*abs(b) + a^5*b^3*c^2*d^5*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) - 4/3*(9*sqrt(b*d)*a^2*b^5*c^3 - 19*sqrt(
b*d)*a^3*b^4*c^2*d + 11*sqrt(b*d)*a^4*b^3*c*d^2 - sqrt(b*d)*a^5*b^2*d^3 - 18*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^3*c^2 + 18*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^2*a^3*b^2*c*d + 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*
b*d))^4*a^2*b*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^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)

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maple [A]  time = 0.01, size = 182, normalized size = 0.96 \begin {gather*} -\frac {2 \left (-a^{3} d^{3} x^{3}+9 a^{2} b c \,d^{2} x^{3}+9 a \,b^{2} c^{2} d \,x^{3}-b^{3} c^{3} x^{3}+6 a^{3} c \,d^{2} x^{2}+36 a^{2} b \,c^{2} d \,x^{2}+6 a \,b^{2} c^{3} x^{2}+24 a^{3} c^{2} d x +24 a^{2} b \,c^{3} x +16 a^{3} 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(-a^3*d^3*x^3+9*a^2*b*c*d^2*x^3+9*a*b^2*c^2*d*x^3-b^3*c^3*x^3+6*a^3*c*d^2*x^2+36*a^2*b*c^2*d*x^2+6*a*b^2*
c^3*x^2+24*a^3*c^2*d*x+24*a^2*b*c^3*x+16*a^3*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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(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 [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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