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*(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])

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Rubi [A]  time = 0.113714, antiderivative size = 186, normalized size of antiderivative = 1., 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 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{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*}

Mathematica [A]  time = 0.254602, 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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Maple [A]  time = 0.008, size = 320, normalized size = 1.7 \begin{align*} -{\frac{-32\,A{b}^{3}{e}^{3}{x}^{3}+16\,Ba{b}^{2}{e}^{3}{x}^{3}+16\,B{b}^{3}d{e}^{2}{x}^{3}-48\,Aa{b}^{2}{e}^{3}{x}^{2}-48\,A{b}^{3}d{e}^{2}{x}^{2}+24\,B{a}^{2}b{e}^{3}{x}^{2}+48\,Ba{b}^{2}d{e}^{2}{x}^{2}+24\,B{b}^{3}{d}^{2}e{x}^{2}-12\,A{a}^{2}b{e}^{3}x-72\,Aa{b}^{2}d{e}^{2}x-12\,A{b}^{3}{d}^{2}ex+6\,B{a}^{3}{e}^{3}x+42\,B{a}^{2}bd{e}^{2}x+42\,Ba{b}^{2}{d}^{2}ex+6\,B{b}^{3}{d}^{3}x+2\,A{a}^{3}{e}^{3}-18\,A{a}^{2}bd{e}^{2}-18\,Aa{b}^{2}{d}^{2}e+2\,A{b}^{3}{d}^{3}+4\,B{a}^{3}d{e}^{2}+24\,B{a}^{2}b{d}^{2}e+4\,Ba{b}^{2}{d}^{3}}{3\,{e}^{4}{a}^{4}-12\,b{e}^{3}d{a}^{3}+18\,{b}^{2}{e}^{2}{d}^{2}{a}^{2}-12\,a{b}^{3}{d}^{3}e+3\,{b}^{4}{d}^{4}} \left ( bx+a \right ) ^{-{\frac{3}{2}}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}} \end{align*}

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 45.9045, size = 1141, normalized size = 6.13 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [B]  time = 5.82896, size = 1310, normalized size = 7.04 \begin{align*} \text{result too large to display} \end{align*}

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]

1/48*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^8*d^2*e^4 - 2*a*b^7*d*e^5 + a^2*b^6*e^6) + 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^8*d^2*e^4 - 2*a*b^7*d*e^5 + a^2*b^6*e^6))/(
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^(11/2)*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)