3.2259 \(\int \frac {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=186 \[ -\frac {2 (B d-A e)}{3 e (a+b x)^{3/2} (d+e x)^{3/2} (b d-a e)}-\frac {16 e \sqrt {a+b x} (a B e-2 A b e+b B d)}{3 \sqrt {d+e x} (b d-a e)^4}-\frac {8 (a B e-2 A b e+b B d)}{3 \sqrt {a+b x} \sqrt {d+e x} (b d-a e)^3}+\frac {2 (a B e-2 A b e+b B d)}{3 e (a+b x)^{3/2} \sqrt {d+e x} (b d-a e)^2} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {78, 45, 37} \[ -\frac {2 (B d-A e)}{3 e (a+b x)^{3/2} (d+e x)^{3/2} (b d-a e)}-\frac {16 e \sqrt {a+b x} (a B e-2 A b e+b B d)}{3 \sqrt {d+e x} (b d-a e)^4}-\frac {8 (a B e-2 A b e+b B d)}{3 \sqrt {a+b x} \sqrt {d+e x} (b d-a e)^3}+\frac {2 (a B e-2 A b e+b B d)}{3 e (a+b x)^{3/2} \sqrt {d+e x} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(B*d - A*e))/(3*e*(b*d - a*e)*(a + b*x)^(3/2)*(d + e*x)^(3/2)) + (2*(b*B*d - 2*A*b*e + a*B*e))/(3*e*(b*d -
a*e)^2*(a + b*x)^(3/2)*Sqrt[d + e*x]) - (8*(b*B*d - 2*A*b*e + a*B*e))/(3*(b*d - a*e)^3*Sqrt[a + b*x]*Sqrt[d +
e*x]) - (16*e*(b*B*d - 2*A*b*e + a*B*e)*Sqrt[a + b*x])/(3*(b*d - a*e)^4*Sqrt[d + e*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 {A+B x}{(a+b x)^{5/2} (d+e x)^{5/2}} \, dx &=-\frac {2 (B d-A e)}{3 e (b d-a e) (a+b x)^{3/2} (d+e x)^{3/2}}-\frac {(b B d-2 A b e+a B e) \int \frac {1}{(a+b x)^{5/2} (d+e x)^{3/2}} \, dx}{e (b d-a e)}\\ &=-\frac {2 (B d-A e)}{3 e (b d-a e) (a+b x)^{3/2} (d+e x)^{3/2}}+\frac {2 (b B d-2 A b e+a B e)}{3 e (b d-a e)^2 (a+b x)^{3/2} \sqrt {d+e x}}+\frac {(4 (b B d-2 A b e+a B e)) \int \frac {1}{(a+b x)^{3/2} (d+e x)^{3/2}} \, dx}{3 (b d-a e)^2}\\ &=-\frac {2 (B d-A e)}{3 e (b d-a e) (a+b x)^{3/2} (d+e x)^{3/2}}+\frac {2 (b B d-2 A b e+a B e)}{3 e (b d-a e)^2 (a+b x)^{3/2} \sqrt {d+e x}}-\frac {8 (b B d-2 A b e+a B e)}{3 (b d-a e)^3 \sqrt {a+b x} \sqrt {d+e x}}-\frac {(8 e (b B d-2 A b e+a B e)) \int \frac {1}{\sqrt {a+b x} (d+e x)^{3/2}} \, dx}{3 (b d-a e)^3}\\ &=-\frac {2 (B d-A e)}{3 e (b d-a e) (a+b x)^{3/2} (d+e x)^{3/2}}+\frac {2 (b B d-2 A b e+a B e)}{3 e (b d-a e)^2 (a+b x)^{3/2} \sqrt {d+e x}}-\frac {8 (b B d-2 A b e+a B e)}{3 (b d-a e)^3 \sqrt {a+b x} \sqrt {d+e x}}-\frac {16 e (b B d-2 A b e+a B e) \sqrt {a+b x}}{3 (b d-a e)^4 \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 107, normalized size = 0.58 \[ \frac {2 \left (\frac {(-d-e x) \left ((b d-a e)^2-4 e (a+b x) (a e+b (d+2 e x))\right ) (a B e-2 A b e+b B d)}{(b d-a e)^3}-A e+B d\right )}{3 e (a+b x)^{3/2} (d+e x)^{3/2} (a e-b d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(B*d - A*e + ((b*B*d - 2*A*b*e + a*B*e)*(-d - e*x)*((b*d - a*e)^2 - 4*e*(a + b*x)*(a*e + b*(d + 2*e*x))))/(
b*d - a*e)^3))/(3*e*(-(b*d) + a*e)*(a + b*x)^(3/2)*(d + e*x)^(3/2))

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fricas [B]  time = 12.16, size = 565, normalized size = 3.04 \[ -\frac {2 \, {\left (A a^{3} e^{3} + {\left (2 \, B a b^{2} + A b^{3}\right )} d^{3} + 3 \, {\left (4 \, B a^{2} b - 3 \, A a b^{2}\right )} d^{2} e + {\left (2 \, B a^{3} - 9 \, A a^{2} b\right )} d e^{2} + 8 \, {\left (B b^{3} d e^{2} + {\left (B a b^{2} - 2 \, A b^{3}\right )} e^{3}\right )} x^{3} + 12 \, {\left (B b^{3} d^{2} e + 2 \, {\left (B a b^{2} - A b^{3}\right )} d e^{2} + {\left (B a^{2} b - 2 \, A a b^{2}\right )} e^{3}\right )} x^{2} + 3 \, {\left (B b^{3} d^{3} + {\left (7 \, B a b^{2} - 2 \, A b^{3}\right )} d^{2} e + {\left (7 \, B a^{2} b - 12 \, A a b^{2}\right )} d e^{2} + {\left (B a^{3} - 2 \, A a^{2} b\right )} e^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{3 \, {\left (a^{2} b^{4} d^{6} - 4 \, a^{3} b^{3} d^{5} e + 6 \, a^{4} b^{2} d^{4} e^{2} - 4 \, a^{5} b d^{3} e^{3} + a^{6} d^{2} e^{4} + {\left (b^{6} d^{4} e^{2} - 4 \, a b^{5} d^{3} e^{3} + 6 \, a^{2} b^{4} d^{2} e^{4} - 4 \, a^{3} b^{3} d e^{5} + a^{4} b^{2} e^{6}\right )} x^{4} + 2 \, {\left (b^{6} d^{5} e - 3 \, a b^{5} d^{4} e^{2} + 2 \, a^{2} b^{4} d^{3} e^{3} + 2 \, a^{3} b^{3} d^{2} e^{4} - 3 \, a^{4} b^{2} d e^{5} + a^{5} b e^{6}\right )} x^{3} + {\left (b^{6} d^{6} - 9 \, a^{2} b^{4} d^{4} e^{2} + 16 \, a^{3} b^{3} d^{3} e^{3} - 9 \, a^{4} b^{2} d^{2} e^{4} + a^{6} e^{6}\right )} x^{2} + 2 \, {\left (a b^{5} d^{6} - 3 \, a^{2} b^{4} d^{5} e + 2 \, a^{3} b^{3} d^{4} e^{2} + 2 \, a^{4} b^{2} d^{3} e^{3} - 3 \, a^{5} b d^{2} e^{4} + a^{6} d e^{5}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 4.26, size = 1102, normalized size = 5.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(b*x + a)*((5*B*b^7*d^4*abs(b)*e^3 - 12*B*a*b^6*d^3*abs(b)*e^4 - 8*A*b^7*d^3*abs(b)*e^4 + 6*B*a^2*b^5
*d^2*abs(b)*e^5 + 24*A*a*b^6*d^2*abs(b)*e^5 + 4*B*a^3*b^4*d*abs(b)*e^6 - 24*A*a^2*b^5*d*abs(b)*e^6 - 3*B*a^4*b
^3*abs(b)*e^7 + 8*A*a^3*b^4*abs(b)*e^7)*(b*x + a)/(b^9*d^7*e - 7*a*b^8*d^6*e^2 + 21*a^2*b^7*d^5*e^3 - 35*a^3*b
^6*d^4*e^4 + 35*a^4*b^5*d^3*e^5 - 21*a^5*b^4*d^2*e^6 + 7*a^6*b^3*d*e^7 - a^7*b^2*e^8) + 3*(2*B*b^8*d^5*abs(b)*
e^2 - 7*B*a*b^7*d^4*abs(b)*e^3 - 3*A*b^8*d^4*abs(b)*e^3 + 8*B*a^2*b^6*d^3*abs(b)*e^4 + 12*A*a*b^7*d^3*abs(b)*e
^4 - 2*B*a^3*b^5*d^2*abs(b)*e^5 - 18*A*a^2*b^6*d^2*abs(b)*e^5 - 2*B*a^4*b^4*d*abs(b)*e^6 + 12*A*a^3*b^5*d*abs(
b)*e^6 + B*a^5*b^3*abs(b)*e^7 - 3*A*a^4*b^4*abs(b)*e^7)/(b^9*d^7*e - 7*a*b^8*d^6*e^2 + 21*a^2*b^7*d^5*e^3 - 35
*a^3*b^6*d^4*e^4 + 35*a^4*b^5*d^3*e^5 - 21*a^5*b^4*d^2*e^6 + 7*a^6*b^3*d*e^7 - a^7*b^2*e^8))/(b^2*d + (b*x + a
)*b*e - a*b*e)^(3/2) - 4/3*(3*B*b^(15/2)*d^3*e^(1/2) - B*a*b^(13/2)*d^2*e^(3/2) - 8*A*b^(15/2)*d^2*e^(3/2) - 6
*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*B*b^(11/2)*d^2*e^(1/2) - 7*B*a^2*b^(1
1/2)*d*e^(5/2) + 16*A*a*b^(13/2)*d*e^(5/2) - 6*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a
*b*e))^2*B*a*b^(9/2)*d*e^(3/2) + 18*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2*A*
b^(11/2)*d*e^(3/2) + 3*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^4*B*b^(7/2)*d*e^(
1/2) + 5*B*a^3*b^(9/2)*e^(7/2) - 8*A*a^2*b^(11/2)*e^(7/2) + 12*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (
b*x + a)*b*e - a*b*e))^2*B*a^2*b^(7/2)*e^(5/2) - 18*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*
e - a*b*e))^2*A*a*b^(9/2)*e^(5/2) + 3*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^4*
B*a*b^(5/2)*e^(3/2) - 6*(sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^4*A*b^(7/2)*e^(3
/2))/((b^3*d^3*abs(b) - 3*a*b^2*d^2*abs(b)*e + 3*a^2*b*d*abs(b)*e^2 - a^3*abs(b)*e^3)*(b^2*d - a*b*e - (sqrt(b
*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2)^3)

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maple [A]  time = 0.01, size = 320, normalized size = 1.72 \[ -\frac {2 \left (-16 A \,b^{3} e^{3} x^{3}+8 B a \,b^{2} e^{3} x^{3}+8 B \,b^{3} d \,e^{2} x^{3}-24 A a \,b^{2} e^{3} x^{2}-24 A \,b^{3} d \,e^{2} x^{2}+12 B \,a^{2} b \,e^{3} x^{2}+24 B a \,b^{2} d \,e^{2} x^{2}+12 B \,b^{3} d^{2} e \,x^{2}-6 A \,a^{2} b \,e^{3} x -36 A a \,b^{2} d \,e^{2} x -6 A \,b^{3} d^{2} e x +3 B \,a^{3} e^{3} x +21 B \,a^{2} b d \,e^{2} x +21 B a \,b^{2} d^{2} e x +3 B \,b^{3} d^{3} x +A \,a^{3} e^{3}-9 A \,a^{2} b d \,e^{2}-9 A a \,b^{2} d^{2} e +A \,b^{3} d^{3}+2 B \,a^{3} d \,e^{2}+12 B \,a^{2} b \,d^{2} e +2 B a \,b^{2} d^{3}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} \left (e x +d \right )^{\frac {3}{2}} \left (e^{4} a^{4}-4 b \,e^{3} d \,a^{3}+6 b^{2} e^{2} d^{2} a^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(5/2),x)

[Out]

-2/3*(-16*A*b^3*e^3*x^3+8*B*a*b^2*e^3*x^3+8*B*b^3*d*e^2*x^3-24*A*a*b^2*e^3*x^2-24*A*b^3*d*e^2*x^2+12*B*a^2*b*e
^3*x^2+24*B*a*b^2*d*e^2*x^2+12*B*b^3*d^2*e*x^2-6*A*a^2*b*e^3*x-36*A*a*b^2*d*e^2*x-6*A*b^3*d^2*e*x+3*B*a^3*e^3*
x+21*B*a^2*b*d*e^2*x+21*B*a*b^2*d^2*e*x+3*B*b^3*d^3*x+A*a^3*e^3-9*A*a^2*b*d*e^2-9*A*a*b^2*d^2*e+A*b^3*d^3+2*B*
a^3*d*e^2+12*B*a^2*b*d^2*e+2*B*a*b^2*d^3)/(b*x+a)^(3/2)/(e*x+d)^(3/2)/(a^4*e^4-4*a^3*b*d*e^3+6*a^2*b^2*d^2*e^2
-4*a*b^3*d^3*e+b^4*d^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((B*x+A)/(b*x+a)^(5/2)/(e*x+d)^(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*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [B]  time = 2.50, size = 304, normalized size = 1.63 \[ -\frac {\sqrt {d+e\,x}\,\left (\frac {16\,b\,x^3\,\left (B\,a\,e-2\,A\,b\,e+B\,b\,d\right )}{3\,{\left (a\,e-b\,d\right )}^4}+\frac {4\,B\,a^3\,d\,e^2+2\,A\,a^3\,e^3+24\,B\,a^2\,b\,d^2\,e-18\,A\,a^2\,b\,d\,e^2+4\,B\,a\,b^2\,d^3-18\,A\,a\,b^2\,d^2\,e+2\,A\,b^3\,d^3}{3\,b\,e^2\,{\left (a\,e-b\,d\right )}^4}+\frac {8\,x^2\,\left (a\,e+b\,d\right )\,\left (B\,a\,e-2\,A\,b\,e+B\,b\,d\right )}{e\,{\left (a\,e-b\,d\right )}^4}+\frac {2\,x\,\left (a^2\,e^2+6\,a\,b\,d\,e+b^2\,d^2\right )\,\left (B\,a\,e-2\,A\,b\,e+B\,b\,d\right )}{b\,e^2\,{\left (a\,e-b\,d\right )}^4}\right )}{x^3\,\sqrt {a+b\,x}+\frac {a\,d^2\,\sqrt {a+b\,x}}{b\,e^2}+\frac {x^2\,\left (a\,e+2\,b\,d\right )\,\sqrt {a+b\,x}}{b\,e}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {a+b\,x}}{b\,e^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((a + b*x)^(5/2)*(d + e*x)^(5/2)),x)

[Out]

-((d + e*x)^(1/2)*((16*b*x^3*(B*a*e - 2*A*b*e + B*b*d))/(3*(a*e - b*d)^4) + (2*A*a^3*e^3 + 2*A*b^3*d^3 + 4*B*a
*b^2*d^3 + 4*B*a^3*d*e^2 - 18*A*a*b^2*d^2*e - 18*A*a^2*b*d*e^2 + 24*B*a^2*b*d^2*e)/(3*b*e^2*(a*e - b*d)^4) + (
8*x^2*(a*e + b*d)*(B*a*e - 2*A*b*e + B*b*d))/(e*(a*e - b*d)^4) + (2*x*(a^2*e^2 + b^2*d^2 + 6*a*b*d*e)*(B*a*e -
2*A*b*e + B*b*d))/(b*e^2*(a*e - b*d)^4)))/(x^3*(a + b*x)^(1/2) + (a*d^2*(a + b*x)^(1/2))/(b*e^2) + (x^2*(a*e
+ 2*b*d)*(a + b*x)^(1/2))/(b*e) + (d*x*(2*a*e + b*d)*(a + b*x)^(1/2))/(b*e^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)**(5/2)/(e*x+d)**(5/2),x)

[Out]

Timed out

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